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AP 10th Class Maths Model Paper 2023 with Solutions

February 3, 2024January 31, 2024 by AP Board

Regularly solving AP 10th Class Maths Model Papers and AP 10th Class Maths Board Model Paper 2023 contributes to the development of problem-solving skills.

AP SSC Maths Model Paper 2023 with Solutions

Time : 3.15 hours
Max. Marks : 100

Instructions:

In the duration of 3 hours 15 minutes, 15 minutes of time is allotted to read the question paper.
All answers shall be written in the answer booklet only.
Question paper consists of 4 Sections which includes 33 questions.
Internal choice is available in section – IV only.
Answers shall be written neatly and legibly.

Section – I
(12 × 1 = 12 M)

Note:

Answer all the questions in one word or phrase.
Each question carries 1 mark.

Question 1.
“The cost of 2 pens and 5 pencils is Rs. 20”. Express this data as a linear equation.
Solution:
Let the pens = X and pencils = y
The cost of 2 pens and 5 pencils = ₹ 20.
∴ 2x + 5y = 20

Question 2.
Choose the correct answer satisfying the following statements.
Statement (A) : Number of maximum zeroes of a cubic polynomial is 3.
Statement (B) : Zero of a linear polynomial x – 2 is -2.
i) Both (A) and (B) are TRUE.
ii) (A) is TRUE, (B) is FALSE
iii) (A) is FALSE, (B) is TRUE,
iv) Both (A) and (B) are FALSE.
Solution:
(ii) [A is TRUE, B is FALSE]

Question 3.
Find the value of Log₆36.
Solution:
Let log₆36 = x
6^x = 36
6^x = 6²
x = 2
∴ log₆36 = 2

Question 4.
Find the discriminant of quadratic equation x² – 5x + 6 = 0.
Solution:
x² – 5x + 6 = 0
where a = 1, b = -5, c = 6.
discriminant = b² – 4ac
= (-5)² – 4 × 1 × 6
= 25 – 24
= 1

Question 5.
Find the volume of a cylinder whose base radius is 3 cm and height is 7 cm.
Solution:
Base radius of a cylinder = r = 3 cm
height = h = 7 cm
Volume of a cylinder = πr²h
= $\frac{22}{7}$ × 3² × 7
= 22 × 9
= 198 cm³

Question 6.
If n(A) = 10, n(B) = 6 and n(A∪B) = 12, then n(A∩B) = ………….
Solution:
n(A) = 10, n(B) = 6, and n(A∪B) = 12, n(A∩B) = ?
n(A∪B) = n(A) + n(B) – n(A∩B)
12 = 10 + 6 – n(A∩B)
n(A∩B) = 16 – 12
n(A∩B) = 4

Question 7.
Given two examples of pair of similar figures.
Solution:
i) A pair of circles are similar
ii) A pair of square are similar

Question 8.
Match the following:
AP 10th Class Maths Model Paper 2023 2
a) A → (i), B → (ii), C → (iii)
b) A → (ii), B → (iii), C → (i)
c) A → (iii), B → (i), C → (ii)
d) A → (ii), B → (i), C → (iii)
Solution:
c) A→(iii), B→(i), C→(ii)]

Question 9.
Midpoint of a line joining the two points (0, 0) and (4, 6) is ………..
Solution:
The midpoint of a line joining the two points (0, 0) and (4, 6) is
= $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
= $\left(\frac{0+4}{2}, \frac{0+6}{2}\right)$ = $\left(\frac{4}{2}, \frac{6}{2}\right)$ = (2, 3)

Question 10.
Draw a rough diagram to the given situation.
“A person observed a top of a tree 10 m away from its foot at the angle of elevation is 45°.”
Solution:
Height of the tree = AB
The distance of a person from foot of the tree = BC = 10 m
Angle of elevation = ∠BCA = 45°
AP 10th Class Maths Model Paper 2023 3

Question 11.
Assertion : Mode of sin 0°, cos 0°, sin 90° and tan 45″ is 0
Reason : $\bar{x}$ = $\frac{\sum f_i x_i}{\sum f}$
Choose the correct answer
A) Both Assertion and Reason are true. Reason is supporting the Assertion.
B) Both Assertion and Reason are true. But Reason is not supporting the Assertion.
C) Assertion is true but the Reason is false.
D) Assertion is false but the Reason is true.
Answer:
D) Assertion is false and Reason is true

Question 12.
The number of common tangents can be drawn to two concentric circles is ……..
Answer:
0

Section – II
(8 × 2 = 16 M)

Note:

Answer all the questions.
Each question carries 2 marks.

Question 13.
If x + 2y = 7 and 4x – ay = 10 has no solution, then find the value of ‘a’.
Solution:
x + 2y = 7 and 4x – ay = 10
a₁ = 1; b₁ = 2; c₁ = 7
a₂ = 4; b₂ = -a; c₂ = 10
Given pair of linear equations has no solution.
So they are parallel lines
∴ $\frac{a_1}{a_2}$ = $\frac{b_1}{b_2}$ ≠ $\frac{c_1}{c_2}$
$\frac{1}{4}$ = $\frac{2}{-a}$ ≠ $\frac{7}{10}$
$\frac{1}{4}$ = $\frac{2}{-a}$ = -a = 2 × 4
-a = 8
a = -8

Question 14.
In a construction of kitchen shed at school, a truck unloaded the sand which was formed in the shape of a cone. The base radius of the cone is 2.7 m. and its height is 7 m. Find the volume of sand unloaded there.
Solution:
The unloaded sand was formed in the shape of a cone.
The base radius of cone = r = 2.7 m
height = h = 7 m
AP 10th Class Maths Model Paper 2023 4
∴ The volume of sand = volume of cone

Question 15.
Express the following sets in set-builder form :
i) A = {1, 8, 27, 64}
ii) B = {-3, -2, -1, 0, 1, 2}
Solution:
i) A = {1, 8, 27,64}
= {1³, 2³, 3³, 4³}
= [x : x = n³, n ∈ N and n ≤ 4}

ii) B = {-3, -2, -1, 0, 1, 2}
= {x : x ∈ Z and -3 < x < 2}

Question 16.
Find the HCF of 220 and 40 by using Euclid Division Lemma.
Solution:
When 220 is divided by 40
220 = 40 × 5 + 20
40 = 20 × 2 + 0
∴ The H.C.F. of (220, 40) is 20.

Question 17.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other is a secant to the circle.
Solution:
AP 10th Class Maths Model Paper 2023 6
In the figure l is a given line and l || m || n such that m is a tangent and n is secant of the circle.

Question 18.
If tanθ = $\frac{7}{24}$, then find the value of sec θ.
Solution:
tan θ = $\frac{7}{24}$
sec²θ – tan²θ = 1
sec²θ = 1 + tan²θ
= 1 + $\left(\frac{7}{24}\right)^2$
= 1 + $\frac{49}{576}$
= $\frac{576+49}{576}$ = $\frac{625}{576}$
secθ = $\sqrt{\frac{625}{576}}$ = $\frac{25}{24}$

Question 19.
Write the formula to find mean of grouped data in direct method and explain the terms in it.
Solution:
Mean of grouped data by direct method
$(\bar{x})$ = $\frac{\sum f_i x_i}{\sum f_i}$
f_i = Frequency of corresponding class
x_i = Mid value (class mark)
Σf_i = Total frequency

Question 20.
Show that the centroid of a triangle formed by the vertices (o, o), (2, 0) and (1, 3) is (1, 1).
Solution:
Centroid of a triangle formed by the vertices (0, 0), (2, 0), and (1,3) is
AP 10th Class Maths Model Paper 2023 7

Section – III
(8 × 4 = 32)

Note:

Answer all the questions.
Each question carries 4 marks.

Question 21.
‘Find the zeroes of the polynomial x² + 7x + 10 and verify the relation between the co-efficients of the polynomial and the zeroes of the polynomial
Solution:
We have
x² + 7x + 10 = (x + 2) (x + 5)
So, the value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0,
i.e., when x = -2 or x = -5.
Therefore, the zeroes of x² + 7x + 10 are -2 and -5.
Now, sum of the zeroes= -2 + (-5) = – (7)
= $\frac{-(7)}{1}$ = $\frac{-(\text { coefficient of } x)}{\text { coefficient of } x^2}$
Product of the zeroes = -2 × (-5) = 10
= $\frac{10}{1}$ = $=\frac{\text { Constant term }}{\text { Coefficient of } x^2}$

Question 22.
Kavitha went to a bank to withdraw Rs. 8,000. She asked the cashier to give the cash in Rs. 100 and Rs. 500 notes only. She got 32 notes in all. Can you tell how many notes each of Rs. 100 and Rs. 500 she received?
Solution:
Let the number of ₹ 100 notes be x
Let the number of ₹ 500 notes be y
She got 32 notes in all
∴ x + y = 32 → 1
The amount of money withdraw = ₹ 8000
100 x + 500 y = 8000 → (2)
From eq(1) y = 32 – x
y = 32 – x is subsituting in eq(2)
100x + 500(32 – x) = 8000
100x + 16000 – 500x = 8000
-400x = 8000 – 16000
-400x = -8000
x = $\frac{8000}{400}$ = 20
x = 20
x = 20 is subsituting in y = 32 – x
y = 32 – 20
y = 12
∴ Number of ₹ 100 notes = 20
Number of ₹ 500 notes = 12

Question 23.
If A = {x: x is a factor of 6}, B = {x: x is a positive even number less than 10}, then find
(i) A∪B,
(ii) A∩B,
(iii) A – B by Venn diagram.
Solution:
A = {x : x is a factor of 6} = {1, 2, 3, 6}
B = {x : x is a positive even number less than 10}
= {2, 4, 6, 8}
AP 10th Class Maths Model Paper 2023 8
i) A∪B
A ∪ B = {1, 2, 3, 6} ∪ {2, 4, 6, 8}
A ∪ B = {1, 2, 3, 4, 6, 8}

ii) A ∩ B
A ∩ B = {1, 2, 3, 6} ∩ {2, 4, 6, 8}
= {2, 6}

iii) A – B
A – B = {1, 2, 3, 6} – {2, 4, 6, 8}
= {1, 3}

Question 24.
If the Geometric Progressions 162, 54, 18, ……. and $\frac{2}{81}$, $\frac{2}{27}$, $\frac{2}{9}$,…….. have their n^th terms equal, then find the value of ‘n’.
Solution:
Given geometric progression = 162, 54, 18…
AP 10th Class Maths Model Paper 2023 9
Given n^th term of these geometric progressions are equal.

Question 25.
Find the area of shaded D region in figure, if AB CD is a square of side 10.5 cm and APD, BPC are semicircles
AP 10th Class Maths Model Paper 2023 1
(Use π = $\frac{22}{7}$).
Solution:
Side of a square = 10.5 cm
Area of a square
= 10.5 × 10.5
= 110.25 cm²

Radius of each semi circle = $\frac{10.5}{2}$
= 5.25 cm

Similarly area of semicircle BPC = 43.3125 cm²
∴ Sum of areas of two semicircles = 2 × 43.3125
= 86.625 cm²
∴ Area of shaded region
= Area of square ABCD – sum of areas of two semicircles
= 110.25 – 86.625
= 23.625 cm²

Question 26.
Express the following in terms of trigonometric ratios of angles between 0° and 45°.
i) sin 81° + tan 75°
ii) cos 65° + cot 75°
Solution:
i) sin 81° + tan 75°
= sin(90 – 9)° + tan(90 – 15)°
= cos 9° + cot 15°
[∵ sin(90 – θ) = cosθ
tan(90 – θ) = cotθ]

ii) cos 65° + cot 75°
= cos(90-25)° + cot(90-15)°
= sin 25° + tan 15°
[∵ cos(90 – θ) = sin θ
cot (90 – θ) = tan θ]

Question 27.
When a die is thrown once, find the probability of getting on its face :
i) 8
ii) a number less than 6
iii) a prime number
iv) a composite number
Solution:
When a die is thrown once then possible outcomes = {1, 2, 3, 4, 5, 6}
Total possible outcomes = 6.

i) 8
Let ‘E’ be the event of getting on its face is 8
No. of favourable outcomes = 0.
Probability of an event P (E)
$=\frac{\text { No.of favourable outcomes }}{\text { Total possible outcomes }}$ = $\frac{0}{6}$ = 0

ii) A number less than 6.
Let ‘F’ be the event of getting on its face is number less than 6.
No. of favourable outcomes = 5
Probability of an event = P(F) = $\frac{5}{6}$

iii) A prime number
Let ‘G’ be the event of getting on its face is prime number.
No. of favourable outcomes = 3 i.e {2, 3, 5}
Probability of an event P(G) = $\frac{3}{6}$ = $\frac{1}{2}$

iv) a composite number
Let ‘H’ be the event of getting on its face is a composite number.
No. of favourable outcomes 2
Probability of an event = P(H) = $\frac{2}{6}$ = $\frac{1}{3}$

Question 28.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point ‘O’.
Show that $\frac{\mathrm{AO}}{\mathrm{BO}}$ = $\frac{\mathrm{CO}}{\mathrm{DO}}$.
Solution:
Given : ABCD is trapezium in which AB || DC.
Its diagonals intersect each other at the point ‘O’
To Prove: $\frac{\mathrm{AO}}{\mathrm{BO}}$ = $\frac{\mathrm{CO}}{\mathrm{DO}}$
AP 10th Class Maths Model Paper 2023 13
Construction: Draw a line OE parallel to AB or CD through ‘O’ and intersecting AD at E
Proof: In ∆ABD we have OE || AB
∴ By Basic proportionality theorem,
$\frac{\mathrm{AE}}{\mathrm{ED}}$ = $\frac{\mathrm{BO}}{\mathrm{OD}}$ ……. (1)
In ∆ADC, we have EO || DC
⇒ $\frac{\mathrm{AE}}{\mathrm{ED}}$ = $\frac{\mathrm{AO}}{\mathrm{OC}}$ ……. (2)
From (1) and (2) we have
$\frac{B O}{O D}$ = $\frac{A O}{O C}$ ⇒ $\frac{A O}{B O}$ = $\frac{C O}{D O}$
Hence proved.

Section – IV
(5 × 8 = 40 M)

Note:

Answer all the questions.
Each question carries 8 marks.
Each question has internal choice.

Question 29.
a) Prove that $\sqrt{2}$ + $\sqrt{5}$ is an irrational.
OR
b) Check whether – 150 a term of the AP : 11, 8, 5, 2, ……
Solution:
a) Let us assume that $\sqrt{2}$ + $\sqrt{5}$ is a rational.
$\sqrt{2}$ + $\sqrt{5}$ = $\frac{a}{b}$ [where a, b are integers and b ≠ 0]
squaring on both sides
AP 10th Class Maths Model Paper 2023 14
∴ Since a, b are integers $\frac{a^2+3 b^2}{2 a b}$ is rational
So $\sqrt{5}$ is rational.
This contradicts the fact $\sqrt{5}$ is irrational.
∴ Hence $\sqrt{2}$ + $\sqrt{5}$ is irrational.

b) We have,
a₂ – a₁ = 8 – 11 = -3;
a₃ – a₂ = 5 – 8 = -3;
a₄ – a₃ = 2 – 5 = -3;
a_k+1 – a_k is same for k = 1, 2, 3, … . the given list of numbers is an AP.
For this AP; a = 11; d = -3
We choose to begin with the assumption that -150 is a term. Say the n^th term of this AP. We will see if an ‘n’ exists for which
a_n = -150
a + (n – 1)d = -150
11 + (n – 1)(-3) = -150
11 – 3n + 3 = -150
14 – 3n = -150
-3n = -150 – 14
-3n = 164
n = $\frac{-164}{-3}$ = $\frac{164}{3}$
But n should be a positive integer.
So, -150 is not a term in the given AP.

Question 30.
a) Find the roots of the quadratic equation x² + 3x -18 = 0 by the method of completing the square.
OR
b) Two cubes each of volume 64 cm³ are joined end to end together. Find the lateral surface area and total surface area of resulting cuboid.
Solution:
a) x² + 3x – 18 = 0
x² + 3x = 18
x² + $\frac{2}{2}$ × 3 × x = 18
x² + 2 × x × $\frac{3}{2}$ = 18
Adding $\left(\frac{3}{2}\right)^2$ on both sidés
AP 10th Class Maths Model Paper 2023 15

b) Volume of cube (v) = 64 cm³
AP 10th Class Maths Model Paper 2023 17
s³ = 64 cm³ = (4 cm)³
⇒ s = 4 cm
Length of the cuboid = 4 cm + 4cm = 8 cm
Surface area of the cuboid
= 2(lb + lh + bh)
= 2[8 × 4 + 8 × 4 + 4 × 4] cm²
= 2[32 + 32 + 16] cm² = 160 cm²
Lateral surface area of cuboid = 2h[l + b]
= 2 × 4[8 + 4]
= 8 × 12
= 96 cm²

Question 31.
a) Find the area of the quadrilateral whose vertices taken in order are A(1, 1), B(7, -3), C(12, 2), D(7, 21).
OR
b) If the median of 60 observations given below is 28.5, then find the values of x and y.

Class interval	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	5	X	20	15.	y	5

Solution:
a) A(1, 1), B(7, -3), C(12, 2) and D(7, 21) are the vertices of quadrilateral ABCD.
Join BD, ∆ABD and ∆BCD are formed.
AP 10th Class Maths Model Paper 2023 18
Area of ∆BCD
= $\frac{1}{2}$ |7[(2 – 21) + 12(21 – (-3) + 7(-3 – 2)] |
= $\frac{1}{2}$ |7(-19) + 12(24) + 7(-5)|
= $\frac{1}{2}$ |-133 + 288 – 35| = $\frac{1}{2}$|-120| = 60 sq.units
∴ Area of quadrilateral ABCD
= Area of ∆ABD + Area of ∆BCD
= 72 + 60
= 132 sq.units.

b) Given median value 28.5 is in the class 20-30.
So, 20-30 is the median class.
AP 10th Class Maths Model Paper 2023 19
Given, total number of observations,
n = 45 + x + y = 60
⇒ x + y = 15 → 1
Also, $\frac{\mathrm{n}}{2}$ = 30, l = 20, f = 20, cf = 5 + x, h = 10

Substitute x value in eq.(1)
8 + y = -15; y = 15 – 8 = 7
So, required values; x = 8, y = 7

Question 32.
a) A straight highway leads to the foot of a tower. Ramaiah standing at the top of the tower observes a car at an angle of depression 30°. The car is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
OR
b) One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting :
i) A queen of black colour
ii) a face card
iii) a jack of diamond
iv) a club card
Solution:
a) Let the distance travelled by the car in 6 seconds = AB = x meters
Height of the tower CD = h meters
The remaining distance to be travelled by the car BC = d meters
and AC = AB + BC = (x + d) meters
∠PDA = ∠DAC = 30°
(Alternate interior angles)
∠PDB = ∠DB×C = 60°
(Alternate interior angles)
AP 10th Class Maths Model Paper 2023 21
x + d = 3d; x = 2d ; d = $\frac{x}{2}$
Time taken to travel ‘x’ meters = 6 seconds.
Time taken to travel the distance of d meters, i.e., $\frac{x}{2}$ meters = 3 seconds.

b) Total number of cards in a deck = 52
Total possible outcomes = 52

i) A queen of black colour
Let ‘E’ be the event of getting a queen of black colour.
No. of favourable outcomes = 2
Probability of an event E is
P(E) $=\frac{\text { No. favourable outcomes }}{\text { Total possible outcomes }}$ = $\frac{2}{52}$
= $\frac{1}{26}$

ii) A face card
Let ‘F’ be the event of getting a face card.
favourable outcomes = 12
Probability of an event F is P(F) = $\frac{12}{52}$
= $\frac{3}{13}$

iii) A jack of diamond
Let ‘G’ be the event of getting a jack of diamond
Favourable outcomes = 1
probability of an event = P(G) = $\frac{1}{52}$

iv) A club card
Let ‘H’ be the event of getting a club card.
Favourable outcomes = 13
Probability of an event = P(H) = $\frac{13}{52}$ = $\frac{1}{4}$

Question 33.
a) Draw the graph of the polynomial p(x) = x² – 2x – 8 and find its zeroes.
OR
b) Construct an isosceles triangle whose base is 6 cm and altitude is 3 cm. Then draw another triangle whose sides are 1$\frac{1}{3}$ times the corresponding sides of the isosceles triangle.
Solution:
a) p(x) = x² – 2x – 8
Let y = x² – 2x – 8
AP 10th Class Maths Model Paper 2023 22

Result:

p(x) = x² – 2x – 8 is a quadratic equation that forms parabola.
p (x) = x² – 2x – 8 intersects X-axis at (-2, 0) and (4, 0)
Hence zeroes of p(x) are -2 and 4.

Verification:

⇒ x² – 2x -8 = 0
⇒ x² + 2x – 4x – 8 = 0
⇒ x(x + 2) – 4(x – 4) = 0
(x + 2) (x – 4) = 0
x + 2 = 0 and x – 4 = 0
x = -2 and x = 4

b)
AP 10th Class Maths Model Paper 2023 23

AP 10th Class Maths Model Paper 2023 24

Steps of Construction:

Construct an isosceles triangle ABC with given measurements
Draw a line segment AB 6 cm and draw a perpendicular bisector of AB intersects it at ‘P.
Draw an arc with measure of 3 cm at the point P, it intersects perpendicular bisector at ‘C’. Join AC and BC.
Draw a ray AX, making an acute angle with AB on the side opposite to vertex ‘C’.
Locate 4 points A₁, A₂, A₃, A₄ on AX so that AA₁ = AA₂ = A₂A₃ = A₃A₄.
Join A₃B and draw a line through A₄ parallel to A₃B intersecting AB at B’.

AP SSC 10th Class Telugu Solutions Chapter 1 మాతృభావన

January 27, 2024January 25, 2024 by AP Board

AP State Board Syllabus AP SSC 10th Class Telugu Solutions 1st Lesson మాతృభావన Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Telugu Solutions 1st Lesson మాతృభావన

10th Class Telugu 1st Lesson మాతృభావన Textbook Questions and Answers

ఉన్ముఖీకరణ : చదవండి – ఆలోచించి చెప్పండి

తే. సర్వతీర్ధాంబువులకంటె సమధికంబు
పావనంబైన జనయిత్రి పాదజలము
వరతనూజున కఖిలదేవతల కంటె
జనని యెక్కుడు సన్నుతాచారనిరత

ప్రశ్నలు – జవాబులు

ప్రశ్న1.
“పావనంబైన జనయిత్రి పాదజలము” అంటే ఏమిటి?
జవాబు:
జనయిత్రి అంటే తల్లి. జన్మనిచ్చిన తల్లి సర్వదేవతల కంటే ఎక్కువ. అటువంటి తల్లి యొక్క పాదాలు కడిగిన నీరు చాలా పవిత్రమైనది. విష్ణువు పాదాల నుండి జన్మించింది గంగ. అది ఎంతో పవిత్రమైంది. అటువంటి పవిత్రత కలిగిందే తల్లి పాదాలు కడిగిన నీరు.

ప్రశ్న2.
తల్లి పాదజలం దేనికంటే గొప్పదని తెలుసుకొన్నారు? ఎందువల్ల?
జవాబు:
తల్లి పాదజలం అన్ని తీర్థాలలోని (పుణ్యనదులలోని) నీటి కంటే పవిత్రమైనదని తెలుసుకొన్నాం. ఆ నదులలోని నీరు ఆ నదీ తీరాలలోని దైవం లేదా దైవాల పాదాలకు తగలడం వల్ల అవి పవిత్రమై పుణ్యనదులుగా లెక్కింపబడతాయి. కానీ, తల్లి సమస్త దేవతల కంటే ఎక్కువ కనుక తల్లి పాదాలు కడిగిన నీరు పుణ్యనదీ జలం కంటే గొప్పది.

ప్రశ్న3.
కుమారునికి అన్నింటికంటే ఎవరు మిన్న? ఎందుకు?
జవాబు:
కుమారునికి అంటే సంతానమందరికీ అన్నింటికంటే తల్లి యొక్క పాదాలు కడిగిన నీరు పరమ పవిత్రమైనది. ఎందుకంటే తన కడుపులో 9 నెలలు మోసి, కని, పెంచి, పోషిస్తూ, రక్షించే తల్లి దైవం కంటే గొప్పది. దైవం కనబడడు. తల్లి కనబడే దైవం. అటువంటి తల్లి యొక్క పాదాలు కడిగిన నీరు దేవుడికి అభిషేకం చేసిన నీటికంటే పవిత్రమైనది.

ప్రశ్న4.
ఈ పద్యం ద్వారా తల్లికి గల స్థానమేమిటని గ్రహించారు?
జవాబు:
మన సంప్రదాయం, మన సంస్కృతి తల్లికి అత్యున్నత స్థానమిచ్చింది. ఈ పద్యం కూడా తల్లి యొక్క మహోన్నత స్థానం గుర్తుచేసింది. ‘మాతృదేవోభవ, పితృదేవోభవ, ఆచార్యదేవోభవ…….’ అని ఉపనిషత్తులు కూడా తల్లికి మొదటిస్థానం ఇచ్చాయి. దైవం కంటే గొప్పదైన తల్లికి నమస్కరించాలి. ఆమె పాదజలం సంతానానికి శిరోధార్యం అని ఈ పద్యం ద్వారా గ్రహించాము.

ప్రశ్న5.
“ప్రతి స్త్రీమూర్తీ మనకు తల్లితో సమానం” అని ఎందుకంటారు?
జవాబు:
స్త్రీ లేకపోతే సృష్టి లేదు. భగవంతుడు అందరి వద్దా ఉండలేడు కనుక తనకు మారుగా తల్లిని సృష్టించాడు. ప్రతి స్త్రీలోనూ తన తల్లిని చూసుకోగలిగినవాడే మహాత్ముడు. రామకృష్ణ పరమహంస తన భార్య శారదాదేవిలో కూడా తన తల్లిని, జగన్మాతను సందర్శించి పూజించాడు. అందుచేత ప్రతి స్త్రీని తల్లిలాగా చూడాలి. గౌరవించాలి. ఆదరించాలి.

ఆలోచించి చెప్పండి

ప్రశ్న1.
‘విజయగర్వంతో నీవు చేసిన పని సరికాదని’ అనే మాటలనుబట్టి శివాజీ ఎలాంటివాడని భావిస్తున్నారు?
జవాబు:
గర్వం ప్రమాదకరం. విజయగర్వం మరీ ప్రమాదకరం. విజయం వచ్చినపుడు చాలా జాగ్రత్తగా ఉండాలి. లేకపోతే ఆ విజయగర్వంతో చాలా తప్పులు చేసే అవకాశం ఉంది. కనుక శివాజీది ఎదిగే కొద్దీ ఒదిగి ఉండే స్వభావం అని తెలిసింది. విజయం సాధించిన ప్రతిసారీ ఆత్మవిమర్శ చేసుకొనేవాడు. వినయం పెంచుకొనేవాడు. శివాజీ గర్వం లేని వీరుడు.

ప్రశ్న2.
స్త్రీలపట్ల మర్యాదగా ప్రవర్తించడం అంటే ఏమిటి?
జవాబు:
స్త్రీలు శారీరకంగా, మానసికంగా సున్నితంగా ఉంటారు. వారి మనసు బాధపడేలా మాట్లాడకూడదు. కించపరచ కూడదు. వెకిలిగా ప్రవర్తించకూడదు. వారికి చట్టపరంగా సంక్రమించవలసిన హక్కులను పొందేలా చూడడం, సహాయం చేయడం, మన తల్లి, సోదరి పట్ల ఎలా ప్రవర్తిస్తామో ప్రతి స్త్రీ పట్ల అలా ప్రవర్తించడం మర్యాద.

ప్రశ్న3.
శివాజీ కోపానికి కారణమేమిటి ? కోపంలో శివాజీ ఎలా ఉన్నాడు?
జవాబు:
ఓడిపోయిన వీరుని సో దేవుడు బంధించి తెచ్చాడు. అతనితో బాటు అతని రాణివాసాన్ని కూడా బంధించి తెచ్చాడు. రాణివాసాన్ని బంధించి తేవడమే శివాజీ కోపానికి కారణమైంది.

కోపంలో శివాజీకి కళ్లు ఎఱ్ఱబడ్డాయి. పెదవులు అదిరాయి. బొమముడి కదుల్తోంది. హుంకరిస్తున్నాడు. గర్జిస్తున్నాడు. శివాజీని చూడడానికి కూడా రాజసభ జంకింది. అంటే ప్రళయకాల రుద్రుడిలా ఉన్నాడు శివాజీ.

ప్రశ్న4.
“సరభసోత్సాహంబు కన్జప్పె” అంటే మీకేమర్థమైంది?
జవాబు:
సరభస ఉత్సాహము అంటే ఉవ్విళ్ళూరు ఉత్సాహం. అంటే ఒక విజయం సాధించినపుడు చాలా ఉత్సాహం వస్తుంది. కన్దప్పడము అంటే ఆ ఉత్సాహంలో సాధించిన విజయం తప్ప కళ్లకు ఏదీ కనబడదు. అంటే ఇతరుల బాధలు కానీ, తప్పులు కానీ, భయాలు కానీ, ఏవీ కళ్లకు కనబడవు- ఆ విజయం తప్ప.

ఆలోచించి చెప్పండి

ప్రశ్న1.
స్త్రీలను ఎవరితో పోల్చారు? ఎందుకు?
జవాబు:
స్త్రీలను సీత, సావిత్రి, అనసూయ, సుమతి మొదలైన పతివ్రతలతో పోల్చారు. స్త్రీలను దేవతావృక్షాలతో పోల్చారు. పతివ్రతా స్త్రీలు అగ్నిజ్వాలల వంటి వారన్నారు. ఎందుకంటే – రాముడు అగ్నిపరీక్ష చేశాడు. సీతాదేవి ఆ అగ్నిని పూలరాశిగా భావించింది. సీత యొక్క పవిత్రతకు అగ్ని కూడా చల్లబడింది. అంతటి మహాపతివ్రత సీత.

యమధర్మరాజును ప్రార్థించి, పోరాడి, మెప్పించి, తన భర్త సత్యవంతుని ప్రాణాలు తిరిగి తెచ్చింది సావిత్రి. యమధర్మాన్ని కూడా తన పాతివ్రత్య మహిమతో మార్చి తన భర్తను బ్రతికించుకొంది.

బ్రహ్మ, విష్ణు, మహేశ్వరులను పసిపిల్లలుగా మార్చి జోలపాడింది అనసూయ. ఈమె అత్రి మహాముని భార్య.

సూర్యోదయం అయితే భర్త మరణిస్తాడని, భర్తకు మరణం రాకుంటకు సూర్యోదయాన్ని ఆపిన మహా పతివ్రత సుమతి.

దేవతావృక్షాలు కోరిన కోరికలు తీరుస్తాయి. అవి ఉన్నచోట అశాంతి, అనారోగ్యం, ముసలితనం వంటి బాధలు ఉండవు. స్త్రీలు ఉన్న ఇల్లు కళకళలాడుతుంది. అశాంతికి అవకాశం లేదు.

ప్రశ్న2.
స్త్రీల పట్ల సమాజంలో ఎలాంటి అభిప్రాయాలు ఉన్నాయి? దీనిపై మీ అభిప్రాయాలు తెల్పండి.
జవాబు:
స్త్రీల పట్ల సమాజంలో గౌరవ భావమే ఉన్నది. కానీ,
సమాజంలో కొంతమంది స్త్రీలను చులకనగా చూస్తారు. చదువుకోనివారు, వివేకం లేనివారు, గౌరవం లేనివారు మాత్రమే స్త్రీలను తక్కువగా చూసే ప్రయత్నం చేస్తారు. స్త్రీలు బలహీనులనే భావం కూడా కొంతమందికి ఉంది. అది తప్పు.

ప్రశ్న3.
స్త్రీల వల్ల భారత కీర్తి ప్రతిష్ఠలు పెరుగుతున్నాయనడానికి ఉదాహరణలు తెల్పండి.
జవాబు:
స్త్రీల వలన ఏ దేశపు కీర్తి ప్రతిష్ఠలైనా పెరుగుతాయి. మన భారతదేశ స్త్రీలు అన్ని రంగాలలోనూ మగవారితో సమానంగా ఉన్నారు. యుద్ధరంగంలో రుద్రమదేవి, ఝాన్సీ లక్ష్మీబాయి, చాంద్ బీబీ మొదలైనవారు శత్రువులను గడగడలాడించారు.

రాజకీయ రంగంలో ఇందిరాగాంధీ, మీరాకుమార్, షీలాదీక్షిత్ మొదలైనవారు ధ్రువతారలు. రచనారంగంలో మొల్ల, రంగాజమ్మ మొదలైనవారు కావ్యాలు రాశారు.

మాలతీ చందూర్, యద్దనపూడి సులోచనారాణి, కోడూరి కౌసల్యాదేవి మొదలైనవారు నవలా రచయిత్రులుగా ఖ్యాతి గడించారు.

పి.టి. ఉష, అశ్వనీ నాచప్ప, కుంజరాణి, మిథాలీ రాజ్, కరణం మల్లీశ్వరి మొదలైనవారు క్రీడారంగంలో మణిపూసలు.

కస్తూరిబా గాంధీ, సరోజినీనాయుడు, దుర్గాబాయ్ దేశ్ ముఖ్ మొదలైనవారు స్వాతంత్ర్యోద్యమంలో పాల్గొన్నారు.

ప్రశ్న4.
“అనలజ్యోతుల………. సాగునే? ” అనే పద్యం ద్వారా మీకేమర్థమైంది?
జవాబు:
అగ్ని వంటి తేజస్సు కలవారు పతివ్రతలు, అంటే పుణ్యస్త్రీలు. తప్పుడు ఆలోచనలతో వారిని సమీపించడం కూడా తప్పు. అలా చేస్తే ఎంత గొప్పవారికైనా మరణం తప్పదు. నాశనం తప్పదు. వారి వంశం కూడా నిలబడదు.

రావణాసురుడు మహాభక్తుడు. గొప్ప పండితుడు. మహా బలవంతుడు, కానీ, సీతాదేవిని ఎత్తుకొని వచ్చాడు. తనను పెళ్ళి చేసుకోమని బాధించాడు. దాని ఫలితంగా రాముని చేతిలో మరణించాడు. యుద్ధంలో బంధువులు, స్నేహితులు అందరూ మరణించారు.

అంటే ఎంత గొప్పవారైనా స్త్రీని అవమానపరిస్తే నాశనం తప్పదని తెలిసింది.

ఆలోచించి చెప్పండి

ప్రశ్న1.
తల్లిగా గౌరవించడం అంటే ఏమిటి? ఆ ప్రవర్తన ఎలా ఉంటుంది?
జవాబు:
తల్లిని మించిన దైవం లేదు. తల్లి ప్రత్యక్ష దైవం. తొమ్మిది నెలలు కడుపులో పెట్టుకొని మోసి, కని, పెంచిన తల్లిని ఎంతగా గౌరవించినా తక్కువే. తల్లితో సమానంగా ప్రతి స్త్రీని గౌరవించాలి. ప్రతి స్త్రీలోనూ అమ్మను చూడాలి. అమ్మలోని కారుణ్యం చూడాలి. అదే, తల్లిగా గౌరవించడ
మంటే.

ప్రశ్న2.
సన్మార్గంలో నడవడం అంటే ఏమిటి? విద్యార్థులుగా మీరు చేయాల్సిన కొన్ని పనులను తెల్పండి.
జవాబు:
సన్మార్గం అంటే మంచి మార్గం. సన్మార్గంలో నడవడ మంటే చక్కని ప్రవర్తన కలిగి ఉండడం. “సాధించ వలసిన లక్ష్యమే కాదు. దానిని సాధించే మార్గం కూడా మంచిది కావాలి” అన్నాడు గాంధీజీ.. విద్యార్థులు మంచి ప్రవర్తన కలిగి ఉండాలి. సంఘంలో చాలా చెడులు ఉన్నాయి. వాటిని సంస్కరించాలి. ప్రజలను చైతన్యపరచాలి.

చదువురాని వారికి చదవటం, రాయడం నేర్పాలి. సమాజంలో జరిగే అనేక మోసాలను గూర్చి చెప్పాలి. మన చట్టాలపై అవగాహన కల్గించాలి.

వరకట్నం ఇవ్వడం, తీసుకోవడం తప్పని చెప్పాలి. ఆరోగ్యకరమైన అలవాట్లు నేర్పాలి. పరిశుభ్రత నేర్పాలి. మన గ్రామ, రాష్ట్ర, దేశ, అంతర్జాతీయ సమస్యలపై
అవగాహన కల్గించాలి. ఓటుహక్కు వినియోగం చెప్పాలి.

ప్రశ్న3.
“స్త్రీ రత్నముల్ పూజ్య, లేయవమానంబు ఘటింపరాదు,” అంటే ఏమిటి ?
జవాబు:
స్త్రీలు గౌరవింపదగినవారు. పూజింపతగినవారు. వారికి ఏ అవమానం జరగకూడదు. స్త్రీలను గౌరవించడం మన సంస్కారం. అది మన సంస్కృతి. అది మన విధి. వారిని మన మాటలతో గాని, ప్రవర్తనతో గాని బాధ పెట్టకూడదు.

ప్రశ్న4.
“హితసూక్తిన్ బల్కి” అంటే ఏమిటి?
జవాబు:
సు + ఉక్తి – సూక్తి అంటే మంచి మాట. హితసూక్తి అంటే ఇష్టాన్ని కలిగించే మంచి మాట. అంటే మంచి మాట అయినా ఇతరులు బాధ పడేలాగా చెప్పకూడదు. వినేవారికి సంతోషం కలగాలి. శివాజీ స్త్రీని గౌరవించాడు. సత్కరించాడు. తన వారు చేసిన తప్పును క్షమించ మన్నాడు. శత్రువీరుడిని విడిచిపెట్టాడు. అపుడు ‘హితసూక్తి’ చెప్పాడు.

I. అవగాహన – ప్రతిస్పందన

1. కింది అంశాల గురించి చర్చించండి.

అ) “ప్రస్తుతం స్త్రీలపై జరిగే దాడులకు కారణాలు – నివారణోపాయాలు”
జవాబు:
కారణాలు:
ప్రస్తుత సమాజంలో గురువుల పట్ల, పెద్దలపట్ల, తల్లిదండ్రుల పట్ల గౌరవ భావన తగ్గుతోంది. కారణాలు ఏమైనా కావచ్చును. నైతికత కూడా లోపించింది. దైవభక్తి తగ్గింది. ‘పాపం’ అనే భావన, భయం తగ్గింది. స్త్రీల పట్ల, బలహీనుల పట్ల, వృద్ధుల పట్ల బాధ్యత తగింది. దీనికి కారణం ప్రధానంగా సినిమాలు. సినిమాలలో, టి.వీ సీరియళ్ళలో స్త్రీలను అసభ్యకరంగా, కేవలం విలాసవస్తువుగా చూపిస్తున్నారు. ప్రేమికులు విచ్చలవిడిగా ప్రవర్తిస్తున్నారు. చెడు వ్యసనాలు కూడా మితిమీరి పోయాయి. రెచ్చగొట్టే ప్రవర్తనలు కూడా కారణం. మానవుని ఆలోచనా విధానం మారిపోయింది. చట్టాలన్నా కొందరికి భయం లేదు. అందుచేతనే స్త్రీలపై దాడులు పెరుగుతున్నాయి.

నివారణోపాయాలు :
చలనచిత్రాలలో స్త్రీని ఉన్నతంగా చూపించాలి. సాహిత్యంలో కూడా స్త్రీలను అంగాంగ వర్ణన చేయకూడదు. స్త్రీల పట్ల గౌరవం పెరిగే పాఠ్యాంశాలు పెట్టాలి. ఎవరైనా స్త్రీని కించపరుస్తున్నా, అవమానిస్తున్నా చూసీ చూడనట్లు వదలకూడదు. పిల్లలకు చిన్నతనం నుంచీ మంచి మంచి కథలు చెప్పాలి. స్త్రీని మాతృమూర్తిగా చూసే భావన పెంపొందాలి. ప్రేమికులు బహిరంగ ప్రదర్శనలు మానాలి. దుర్వ్యసనాలు నిరోధించాలి. సమాజాన్ని చైతన్యపరచాలి. స్త్రీ విద్యను ప్రోత్సహించాలి. సమాజంలో సంస్కారం, నీతి పెంచాలి. స్త్రీలకు రక్షణ పెంచాలి. చట్టాలు కచ్చితంగా అమలుచేయాలి. విదేశీ విజ్ఞానం ఆర్జించాలి గాని విదేశీ సంస్కృతి, అలంకరణలు కాదు. స్త్రీలు కూడా తగిన జాగ్రత్తలు పాటించాలి. ధైర్యం పెంచుకోవాలి. ఒంటరిగా తిరగకూడదు.

ఆ) “కుటుంబం – సమాజం అభివృద్ధిలో స్త్రీల పాత్ర”
జవాబు:
వ్యక్తులు లేనిదే కుటుంబం లేదు. కుటుంబాలు లేనిదే సమాజం లేదు. వ్యక్తిని బట్టి కుటుంబం అభివృద్ధి చెందుతుంది. కుటుంబాలను బట్టి సమాజం అభివృద్ధి చెందుతుంది.

కుటుంబమైనా, సమాజమైనా ఏర్పడాలన్నా, అభివృద్ధి చెందాలన్నా స్త్రీలది కీలకపాత్ర. “ఒక స్త్రీ విద్యావంతురాలైతే ఆ కుటుంబమంతా విద్యావంతమౌతుంది” అన్నారు జవహర్‌లాల్ నెహ్రూ. కుటుంబంలోని వ్యక్తుల ఆరోగ్యం, ఆలోచనలు, సంస్కారం అన్నీ స్త్రీల చేతిలోనే ఉంటాయి. – స్త్రీ విద్య దేశాభివృద్ధికి దిక్సూచి. దైవభక్తి, నైతికత, తెలివితేటలు, అంకిత భావన స్త్రీలకు ఎక్కువ. స్త్రీ తన కుటుంబం చల్లగా ఉండాలని, కుటుంబమంతా ఆరోగ్యంగా ఉండాలని దైవాన్ని రోజూ కోరుకుంటుంది. స్త్రీ తన ప్రాధాన్యతను కోరుకోకుండానే కుటుంబ అభివృద్ధికి కష్టపడుతుంది.

అటువంటి స్త్రీల వలన కుటుంబం, సమాజం అభివృద్ధి చెందుతుంది. ప్రతి మగవాడి విజయం వెనుక ఒక స్త్రీ ఉంటుంది అని ఒక రచయిత అన్నమాట అక్షర సత్యం. “ముదితల్ నేర్వగ రాని విద్య కలదే ముద్దార నేర్పించినన్” అన్నారు. ఆధునిక కవిగారు.

‘స్త్రీలకు మగవారి కంటె తెలివి, సాహసం ఎక్కువ” అని ఆర్యోక్తి.

అందుచేత స్త్రీ నిరంతర చైతన్యానికి గుర్తు. క్లిష్ట పరిస్థితులలో కూడా తల్లిగా, సోదరిగా, భార్యగా, ……….. అనేక విధాల విశ్వరూపం ధరించి స్త్రీ కుటుంబాన్ని, సమాజాన్ని అభివృద్ధి చేస్తోంది.

2. * గుర్తుగల పద్యాలకు ప్రతిపదార్థాలు రాయండి.

పద్యం -1

శా॥ “ఆ-యేమీ ? ……….. మౌహిత్య మోర్వన్ జుమీ”
ప్రతిపదార్థం :
ఆ – యేమీ = (ఆశ్చర్యం, కోపం కలిపి) ఆ ఏమిటి ?
పుణ్య + ఆవాసమున్ = పుణ్యానికి నిలయమైన
ఒక = ఒక
రాణివాసమును = రాణివాసాన్ని
తెచ్చినావా = బందీగా తీసుకొచ్చావా?
ఏ హైందవుఁడు + ఐననూ = హైందవుడు ఎవడైనా
ఈ గతిన్ – = ఈ విధంగా
అమర్యాదన్ = మర్యాద తప్పి (మర్యాద లేకుండా)
ప్రవర్తించును + ఏ = ప్రవర్తిస్తాడా?
మా + ఆజ్ఞన్ = నా ఆజ్ఞను
గమనింపవు + ఓ = గమనించలేదా? (పట్టించుకోలేదా?)
మద + ఉన్మాదంబునన్ = గర్వం మత్తులో
రేఁగి = అతిశయించి
నీ = నీ యొక్క
ఆయుః + సూత్రములు = ప్రాణాలనే సూత్రాలు (దారాలు)
ఈవ = నీవే
త్రుంచుకొనెదు + ఓ : తెంచుకుంటావా?
ఔద్ధత్యము = ఈ తెగింపును (గర్వమును)
ఓర్వన్ + చుమీ = సహించను సుమా !

పద్యం -4

మ| శివరాజంతట …………….తప్పు సైరింపుమీ !
ప్రతిపదార్థం :
శివరాజు = శివాజీ మహారాజు
అంతటన్ = అప్పుడు
మేల్ముసుంగుఁదెరలోన్; మేల్ముసుంగు = సువాసినీ స్త్రీలు వేసుకొనే మేలు ముసుగు యొక్క (బురఖా)
జయ = (యుద్ధంలో) విజయం పొందిన
తెరలోన్ = తెరలోపల
స్నిగ్దాంబుదచ్ఛాయలోన్, (స్నిగ్ధ + అంబుద + ఛాయలోన్) = దట్టమైన
అంబుద = మేఘము యొక్క
ఛాయలోన్ = నీడలో (మాటున నున్న)
నవసౌదామినిన్ = కొత్త మెరుపు తీగను
పోలు = పోలినట్లు ఉన్న
ఆ, యవన కాంతారత్నమున్ = ఆ రత్నము వంటి యవనకాంతను (మహమ్మదీయ స్త్రీని)
భక్తి గౌరవముల్ = భక్తియునూ, గౌరవమునూ
పాఱగన్ + చూచి = స్ఫురించేటట్లు చూసి
పల్కెన్ = ఈ విధంగా అన్నాడు
వనితారత్నంబులు = రత్నముల వంటి స్త్రీలు (శ్రేష్ఠులైన స్త్రీలు)
ఈ = ఈ
భవ్య హైందవ భూ జంగమ పుణ్యదేవతలు; భవ్య = శుభప్రదమైన
హైందవ భూ = భారత భూమిపై
జంగమ = సంచరించే (తిరుగాడే)
పుణ్యదేవతలు = పుణ్యప్రదమైన దేవతల వంటివారు
మాతా! = అమ్మా
తప్పున్ = మా వారు చేసిన తప్పును
సైరింపుమీ = మన్నింపుము (క్షమింపుము)

పద్యం -6

మ|| అనలజ్యోతుల ………… దుశ్చారిత్రముల్ సాగునే?
ప్రతిపదార్థం :
అనల జ్యోతులన్ = అగ్ని జ్వా లల వంటి,
ఈ పతివ్రతలన్ – ఈ పతివ్రతలను
పాపాచారులై (పాప + ఆచారులు + ఐ) = అపచారం చేసేవారై
డాయు = కలిసే
భూజనులు + ఎల్లన్ = భూమి పైనున్న ప్రజలు అందరునూ
నిజ సంపదల్ = తమ సంపదలను
తొఱగి = వీడి (పోగొట్టుకొని)
అసద్వస్తులై (అసద్వస్తులు + ఐ) = సర్వ నాశనమైనవారై
పోరె = పోకుండా ఉంటారా?
విత్తనమే – విత్తనము (వారి వంశవృక్షం
యొక్క విత్తనం)
నిల్చునె = నిలుస్తుందా? (అనగా వంశం నిలుస్తుందా?)
మున్ను = పూర్వం
పులస్త్య బ్రహ్మ సంతానమున్ = పులస్త్య బ్రహ్మ యొక్క కుమారుడైన రావణుని గూర్చి
ఎఱుంగమై = మనకు తెలియదా?
హైందవ భూమిని – భారత భూమియందు
ఈ పగిది = ఇటువంటి
దుశ్చరిత్రముల్ = చెడు పనులు (దుశ్చర్యలు)
సాగునే = సాగుతాయా? (సాగవు)

పద్యం -8

శా॥ మా సర్దారుఁడు ………….. దాల్ని సారింపుమీ!
ప్రతిపదార్థం :
మా సర్దారుడు = మా సర్దార్ సోన్ దేవుడు బ
తొందరన్ బడి = తొందరపాటుపడి
అసన్మార్గంబునన్ (అసత్ + మార్గంబునన్) = తప్పుడు మార్గంలో
పోయెన్ = వెళ్ళాడు (పొరపాటున నిన్ను బంధించి తెచ్చాడు)
ఈ దోసంబున్ = ఈ దోషాన్ని
కని = చూచి
నొచ్చుకోకు = బాధపడకు
ఇప్పుడే = ఇప్పుడే
నినున్ = నిన్ను
నీ గృహంబున్ = నీ ఇంటిని (నీ ఇంటికి)
చేరున్ = చేరుస్తాను
నా సైన్యంబున్ = సైన్యాన్ని
తోడుగాన్ = నీకు సాయంగా
పనిచెదన్ = పంపిస్తాను
నా తల్లిగాన్ = నా యొక్క తల్లివలెనూ
తోడుగాన్ = నా తోడబుట్టిన సోదరిగానూ
దోసిళ్లన్ = (నా) అరచేతులపై
నడిపింతున్ = నడిపిస్తాను (నిన్ను కాలుక్రింద పెట్టకుండా నా అరచేతులపై సగౌరవంగా నడిపించి మీ ఇంటికి పంపిస్తాను)
నీ కనులయందున్ = నీ కళ్లల్లో
తాల్మిన్ = ఓర్పును
సారింపుమీ = ప్రసరింప చెయ్యి (చూపించుము)

3. పాఠం ఆధారంగా కింది ప్రశ్నలకు జవాబులు రాయండి.

అ) శివాజీ కొలువులోని వారంతా నిశ్చేష్టులవడానికి కారణం ఏమిటి?
జవాబు:
సో దేవుడు విజయోత్సాహంతో ఉన్నాడు. ఓడిపోయిన వీరుని, అతని రాణివాసాన్ని బంధించి తీసుకొని వచ్చాడు. పుణ్యవాసమైన రాణివాసాన్ని బంధించి తెచ్చినందుకు శివాజీకి చాలా కోపం వచ్చింది. ఏ హిందువుడూ ఆ విధంగా ప్రవర్తించడన్నాడు. తన ఆజ్ఞ పట్టించుకోలేదని ఆగ్రహించాడు. సో దేవుడు తన ప్రాణాలు తానే పోగొట్టుకొంటున్నాడని హెచ్చరించాడు. గర్వాన్ని సహించనన్నాడు.

శివాజీ కళ్లు ఎఱ్ఱబారాయి. పెదవులు కోపంతో వణికాయి. కనుబొమ్మలు కదిలాయి. ఆయన హుంకరించాడు. కోపంతో గర్జించాడు. ఈ పరిస్థితికి శివాజీ కొలువులోని వారంతా భయపడ్డారు. నిశ్చేష్టులయ్యారు.

ఆ) సోన్ దేవుడు శివాజీని ఎలా శాంతపరిచాడు?
జవాబు:
సోన్ దేవుడు ఛత్రపతి శివాజీ ఆజ్ఞననుసరించి రాణివాసపు బంధనాలు తొలగించాడు. వారిని తీసుకొని వచ్చినందుకు తనను క్షమించమని కోరాడు. ఓడిపోయిన వీరుడిని తెచ్చే విజయోత్సాహం కళ్లకు క్రమ్మేసిందని అన్నాడు. చెడు ఆలోచన లేదన్నాడు. చక్రవర్తి పాదాల సాక్షిగా చక్రవర్తి ఆజ్ఞను ధిక్కరించే గర్వం లేదన్నాడు. ఈ మాటలు విన్న శివాజీ కొద్దిగా శాంతించాడు.

ఇ) భారతదేశ భాగ్య కల్పలతలని శివాజీ ఎవరిని, ఎలా కీర్తించాడు?
(లేదా)
భారతదేశ భాగ్య కల్పలతలుగా ఎవరెవరిని ఏ విధంగా శివాజీ ప్రస్తుతించాడో రాయండి.
జవాబు:
స్త్రీలను భారతదేశపు దేవతావృక్షాలని శివాజీ కీర్తించాడు. హరిహరబ్రహ్మలను చంటి పిల్లలుగా చేసిన అనసూయను కీర్తించాడు. యమధర్మరాజు పాశాన్ని తెంచి పతిప్రాణాలు కాపాడిన సావిత్రిని పావన చరిత్రగా నుతించాడు. అగ్నిని పూలరాశిగా భావించిన సీతామాతను సాధ్వీమతల్లిగా సన్నుతించాడు. భర్త ప్రాణాల కోసం సూర్యోదయాన్ని ఆపుచేసిన సుమతిని పుణ్యాలపంటగా ప్రశంసించాడు. పుట్టినింటికి, మెట్టినింటికి కీర్తి ప్రతిష్టలు పెంచే పుణ్యసతులను స్తుతించాడు.

ఈ) శివాజీ యవన కాంత పట్ల చూపిన ఆదరాభిమానాలు ఎటువంటివి?
జవాబు:
ఛత్రపతి శివాజీ మేలిముసుగులోని యవన కాంతను చూశాడు. భక్తి, గౌరవాలతో ఆమెతో మాట్లాడాడు. స్త్రీలు హిందూదేశ వాసులకు దేవతలు అన్నాడు. తల్లీ! తప్పు క్షమించు అని వేడుకొన్నాడు.

హరిహరబ్రహ్మలను పురిటి బిడ్డలుగా చేసిన అనసూయ మా భారతదేశపు గృహిణి అన్నాడు. యమధర్మరాజును ఎదిరించి పతి ప్రాణాలు సంపాదించిన సావిత్రి పావన చరిత్ర కలది అన్నాడు. అగ్నిని పూలరాశిగా భావించి నడయాడిన సీత మా సాధ్వీమతల్లి అన్నాడు. పతికోసం సూర్యోదయాన్ని ఆపిన సుమతి పుణ్యాలపంట అన్నాడు. పుట్టినింటికి, అత్తవారింటికి పేరు తెచ్చే స్త్రీలు దేవతావృక్షాల వంటివారన్నాడు.

స్త్రీలను బాధిస్తే మరణం, నాశనం తప్పదన్నాడు. రావణాసురుని ఉదాహరించాడు. నీవు నన్ను కనని తల్లినన్నాడు. ఇప్పుడే పుట్టింటి మర్యాదతో నీ ఇంటికి చేరుస్తానన్నాడు. బంధించబడిన ఆమె భర్తను కూడా విడిచిపెట్టాడు. ఇద్దరినీ సాదరంగా వారి ఇంటికి సాగనంపాడు.

II. వ్యక్తికరణ-సృజనాత్మకత

1. క్రింది ప్రశ్నలకు ఐదేసి వాక్యాల్లో సమాధానాలు రాయండి.

అ) సో దేవుని మనస్తత్వాన్ని గురించి పాఠం ఆధారంగా సొంతమాటల్లో రాయండి.
జవాబు:
రాణివాసాన్ని సో దేవుడు బంధించి తెచ్చినందుకు శివాజీ ఆగ్రహించాడు. వెంటనే వారిని విడిపించి తీసుకొని రమ్మని శివాజీ ఆజ్ఞాపించాడు. శివాజీ ఆజ్ఞానుసారం సో దేవుడు రాణివాసాన్ని వెంటనే బంధనాలు తొలగించి తీసుకొని వచ్చాడు.

దీనిని బట్టి శివాజీ ఆజ్ఞను వెంటనే అమలు జరిపే నమ్మినబంటు సో దేవుడని తెలుస్తోంది. ముందు వెనుకలు ఆలోచించకుండా రాజభక్తితో రాజాజ్ఞను అమలు జరిపే మనస్తత్వం కలవాడు సో దేవుడు. సో దేవునకు స్వామిభక్తి ఎక్కువ.

“దేవా! నన్ను మన్నించు. ఈ వీరుడిని బంధించిన విజయం నా కళ్లకు కప్పింది. చెడు ఆలోచన లేదు. తమ ఆజ్ఞను ఉల్లంఘించే గర్వం లేదు. మీ పాదాల సాక్షిగా కావాలని తప్పుచేయలేదు” అన్నాడు సో దేవుడు శివాజీతో.

పై మాటలను బట్టి తను తప్పుచేస్తే సో దేవుడు క్షమార్పణ కోరతాడు. ఆత్మ విమర్శ చేసుకొని తన తప్పునకు కారణం తెలుసుకొంటాడు. సిగ్గుపడకుండా దానిని చెబుతాడు. అహంకారం లేదు. గర్వం లేదు. నిజాయితీ కలవాడు. నిర్భయంగా నిజం చెబుతాడు. మంచి స్వభావం గల సైన్యాధికారి. కొంచెం తొందరపాటు గలవాడు. తనను తాను సరిచేసుకుంటాడు.

ఆ) శివాజీ రాజై ఉండీ తన వద్దకు బందీగా తెచ్చిన యవన కాంతతో “మాతా! తప్పు సైరింపుమీ!” అన్నాడు. దీనిమీద మీ అభిప్రాయాలేమిటి?
జవాబు:
శివాజీకి స్త్రీలంటే గౌరవం ఎక్కువ. స్త్రీలకు అవమానం జరిగితే సహించలేడు. దీనికి కారణం శివాజీ చిన్నతనం నుండి వినిన మంచి కథలు కావచ్చును. వాళ్ల అమ్మగారు పురాణ కథలు చెప్పి ఉండవచ్చును. మన భారతీయ సాహిత్యం చదివి ఉండవచ్చును. అందుచేతనే ఆ యవన కాంతను ‘అమ్మా!’ అని సంబోధించాడు. తను చదివిన ఉత్తమమైన సాహిత్యం అతనికి ఆ సంస్కారం నేర్పింది. అందుకే తను రాజునని కూడా మరచిపోయాడు. అహంకారం ప్రదర్శించలేదు. తన వలన తప్పు జరిగిందని తెలుసుకొన్నాడు. అందుకే క్షమార్పణ కోరాడు. అది శివాజీ ఉత్తమ సంస్కారానికి నిదర్శనం.

ఇ) మీ తోటి బాలికలను మీరెలా గౌరవిస్తారు?
జవాబు:
మా తోటి బాలికలను మాతో సమానంగా గౌరవిస్తాం. కలసి ఆడుకొంటాం. చదువుకొంటాం. అల్లరి చేస్తాం. పాఠాలు వింటాం. ఆడపిల్లలను అగౌరవించం. సహాయం చేస్తాం. మా అక్కచెల్లెళ్లలా భావిస్తాం. ఏ అమ్మాయిలోనైనా మా అక్కనో, చెల్లినో చూస్తాం. ఎవరైనా అమ్మాయిల్ని అగౌరవపరిస్తే సహించం. కించపరిస్తే ఊరుకోం. ఆకతాయిలెవరైనా అల్లరి పెడితే, అందరం కలిసి బుద్ధి చెబుతాం. అమ్మాయిలు ధైర్యంగా ఉండేలాగా చేస్తాం. వారికి అన్నదమ్ములు లాగా తోడు నీడ ఔతాం.

2. క్రింది ప్రశ్నలకు పదేసి వాక్యాల్లో సమాధానాలు రాయండి.

అ) మీ పాఠం ఆధారంగా శివాజీ వ్యక్తిత్వాన్ని విశ్లేషించండి.
(లేదా)
మీ పాఠంలో శివాజీ ప్రవర్తనను బట్టి ఆయన వ్యక్తిత్వాన్ని విశ్లేషించండి.
(లేదా)
‘పరస్త్రీలను కన్నతల్లిలాగా చూడాలి’ అని సర్దారులను ఆదేశించిందెవరు? ఆ మహావీరుని యొక్క వ్యక్తిత్వాన్ని విశ్లేషించండి.
(లేదా)
పరస్త్రీని తల్లిగా భావించడమనేది మన సంప్రదాయం . ఆ సంప్రదాయాన్ని చక్రవర్తియైన శివాజీ కొనసాగించాడు కదా ! “మాతృభావన” పాఠం ఆధారంగా ఆయన వ్యక్తిత్వాన్ని విశ్లేషించండి.
జవాబు:
శివాజీ వ్యక్తిత్వము : వ్యక్తిత్వం అంటే, మాటలకూ చేతలకూ తేడా లేనితనం.
1) ధర్మమూర్తి :
శివాజీ ధర్మప్రభువు. ఇతడు శత్రు దుర్గాలపై దండయాత్రకు పోయినప్పుడు, అక్కడ స్త్రీలకు హాని చేయవద్దని తన సర్దారులను ఆజ్ఞాపించేవాడు.

2) తప్పు చేస్తే శిక్ష :
సో దేవుడు కళ్యాణి దుర్గాన్ని జయించినా, రాణివాస స్త్రీని బంధించాడని, అతడిపై కోపించి ప్రాణం తీస్తానని శివాజీ హెచ్చరించాడు.

3) పశ్చాత్తాపం కలవాడు :
యవనకాంతను విడిపించి, తన సర్దారు తప్పు చేశాడనీ, అందుకు తన్ను మన్నించమనీ కోరి, ఆమెను పూజించి మర్యాదగా ఆమెను ఇంటికి పంపాడు.

4) క్షమామూర్తి :
సో దేవుడు తాను కావాలని తప్పు చేయలేదనీ, కోటను జయించిన ఉత్సాహంతో తాను తప్పు చేశాననీ, తన్ను మన్నించమని కోరగా, శివాజీ అతడిని క్షమించి విడిచాడు.

5) స్త్రీలపై గౌరవం :
పతివ్రతలు భూలోకంలో తిరిగే పుణ్య దేవతలని శివాజీ భావన. పతివ్రతలు భారత భాగ్య కల్పలతలని శివాజీ మెచ్చుకున్నాడు. స్త్రీలు అగ్నిజ్వాలలవంటి వారని, అపచారం చేస్తే వారు నశిస్తారనీ శివాజీ నమ్మకం.

6) తప్పును సరిదిద్దడం :
ధర్మ ప్రభువైన శివాజీ, యవనకాంతను విడిపించి, ఆమెను గౌరవించి, తన సర్దారు చేసిన తప్పును సరిదిద్దాడు. శివాజీ ఈ విధంగా గొప్ప వ్యక్తిత్వం కలవాడు.

ఆ) “స్త్రీ రత్నములు పూజ్యలు” అన్న శివాజీ మాటలను మీ సొంత అనుభవాల ఆధారంగా సమర్థించండి.
జవాబు:
స్త్రీ రత్నములు అంటే ఉత్తమ స్త్రీలు. వారు పూజింపదగినవారు అని శివాజీ చెప్పాడు. ఆ మాట సత్యమైనది.
నా సొంత అనుభవాలు :
1) ఒకసారి గోదావరిలో స్నానం చేస్తున్నాను. నా పక్కన కళాశాల ఆడపిల్లలు కూడా స్నానాలు చేస్తున్నారు. ఆడపిల్లలను ఆ తడి బట్టలలో చూసి, కొందరు ఆకతాయిలు వారిని ఆటపట్టిస్తున్నారు. నేను వెంటనే వారితో తగవు పెట్టుకున్నాను. గట్టున ఉన్న పోలీసును పిలిచాను. అల్లరి పిల్లలు వెంటనే పారిపోయారు. కాలేజీ బాలికలు నన్ను గౌరవంగా చూశారు.

2) మా గ్రామంలో ఒక వితంతువు ఉంది. ఆమె చాలా మంచిది. ఆమెను గ్రామంలో కొందరు దుషులు మాటలతో వేధిస్తున్నారు. ఆమె తన గోడును మా అమ్మగారి దగ్గర చెప్పుకొని ఏడ్చేది. నేనూ మా అమ్మగారూ, ఆ విషయాన్ని మా నాన్నగార్కి చెప్పాం. మా నాన్నగారు ఆ గ్రామ సర్పంచి. విషయము మా నాన్నగారి దృష్టికి రాగానే, ఆయన అల్లరిచేస్తున్న వారిని గట్టిగా హెచ్చరించారు.

స్త్రీ రత్నాలు పూజ్యలన్న శివాజీ అభిప్రాయాన్ని మగవారు 70దరూ గ్రహించి నడచుకోవాలి.

3. కింది అంశాల గురించి సృజనాత్మకంగా ప్రశంసిస్తూ రాయండి.

అ) పాఠ్యాంశాన్ని “ఏకాంకిక” లేదా శివాజీ ఏకపాత్ర రూపంలో రాసి ప్రదర్శించండి.
జవాబు:
(స్త్రీ మూర్తి (ఏకాంకిక)
పాత్రలు – శివాజీ, సో దేవుడు, భటులు, శత్రువీరుడు, అతని భార్య.
దృశ్యం -సభ. (శివాజీ ఒంటరిగా కూర్చొని ఉంటాడు.)

శివాజీ : (తనలో) ఆహా! ఈ ప్రకృతి ఎంత బాగుంది? ఈ పైరగాలి అమ్మ పాడే జోలపాటలా హాయిగా ఉంది. ఈ రోజెందుకో చాలా ఆనందంగా ఉంది.

భటుడు : (ప్రవేశిస్తూ) రాజాధిరాజ! రాజమార్తాండ! మహారాజా! సార్వభౌమా! ఛత్రపతి గారికి జయము! జయము!’

శివాజీ : ఏమది?

భటుడు : ఆ ప్రభూ!

శివాజీ : ఊ…..

భటుడు : తమ ఆజ్ఞానుసారం కళ్యాణి దుర్గం జయించారు. శ్రీ సో దేవుడు గారు తమ దర్శనానికి వేచి ఉన్నారు.

శివాజీ : (నవ్వుతూ) చాలా మంచి మాట చెప్పావు. వెంటనే ప్రవేశపెట్టు.

సోన్ దేవుడు : జయము ! జయము ! మహారాజా!

శివాజీ : మన పౌరుషం రుచి చూపించారు. యుద్ధ విశేషాలు చెప్పండి. దుర్గం లొంగదీసుకోవడం కష్టమైందా? తొందరగా చెప్పండి.

సోన్ దేవుడు : మన బలగాలను చూసేసరికి ఆ సర్దారు ఠారెత్తిపోయాడు. అయినా గట్టిగా ప్రతిఘటించాడు.

శివాజీ : చివరకు మరణించాడా? లొంగిపోయాడా?

సోన్ దేవుడు : లొంగిపోయాడు.

శివాజీ : (పకపక నవ్వుతూ) శభాష్, ఇది నా కల. (మీసాలు మెలివేస్తూ) ఇక మనకు ఎదురు లేదు. ఇదిగో ఈ వజ్రాలహారం స్వీకరించండి.

సోన్ దేవుడు : మహా ప్రసాదం. మహారాజా! బందీలను ప్రవేశపెట్టమంటారా?

శివాజీ : బందీలా? అంటే సైన్యాన్ని కూడా బంధించారా?

సోన్ దేవుడు : ఆ సర్దారను, రాణివాసాన్ని కూడా బంధించి తెచ్చాం మహారాజా!

శివాజీ : (కోపంగా) ఆ … ఏమిటీ పుణ్యావాసమైన రాణివాసాన్ని బంధించి తెచ్చావా? ఏ భారతీయుడైనా ఇలా చేస్తాడా? మా ఆజ్ఞ లెక్కలేదా? నీ ప్రాణాలు నీవే పోగొట్టుకొంటావా? గర్వాన్ని సహించను.

సోన్ దేవుడు : అదికాదు ప్రభూ! నేను చెప్పేది వినండి దేవా!

శివాజీ : (చాలా కోపంతో) చేసినది చాలు. ఇప్పటికైనా వాళ్లను బంధ విముక్తులను చేసి, ప్రవేశ పెట్టండి.

సోన్ దేవుడు : (రాణిని ప్రవేశపెట్టి) ప్రభూ! నన్ను క్షమించండి. విజయోత్సాహంతో తప్పు చేశాను. ‘నాకు చెడు ఆలోచన లేదు. తమ ఆజ్ఞను ఉల్లంఘించే గర్వం లేదు. మీ పాదాల సాక్షిగా తప్పు చేయలేదు.

శివాజీ : (శాంతించి, రాణి వైపు తిరిగి) : అమ్మా! మాకు స్త్రీలు ఈ భూమిపై తిరిగే దేవతలు. తల్లీ! మా తప్పును మన్నించు.

రాణి : మీ తప్పు లేదు. స్త్రీగా పుట్టడం నేను చేసిన తప్పు.

శివాజీ : అలా అనకమ్మా! హరిహరబ్రహ్మలను పురిటిబిడ్డలను చేసిన అనసూయ మహా పతివ్రత. యమధర్మరాజును ఎదిరించి తన భర్త ప్రాణాలు తెచ్చిన సావిత్రి పావన చరిత్ర కలది. అగ్నిరాశిని పూలరాశిగా భావించిన సీత మహాసాధ్వి. భర్త జీవించడం కోసం సూర్యోదయం ఆపిన సుమతి పుణ్యాల పంట.

రాణి : అది పురాణ కాలం.

శివాజీ : అలాంటి వారు ఎంతోమంది భరతమాత బిడ్డలు ఇప్పటికీ ఉన్నారు. ఇటువంటి పుణ్యసతులు ఎంతోమంది పుట్టినింటికి, మెట్టినింటికి పేరు తెస్తున్నారు.

రాణి : ఎంత పేరు తెచ్చినా మాకు అవమానాలు తప్పడంలేదు.

శివాజీ : లేదమ్మా! స్త్రీలను అవమానించిన వారెవరికీ వంశం నిలబడదు. నాశనం తప్పదు. రావణాసురుడు నాశనం కాలేదా? నీవు నా తల్లివమ్మా! నిన్నూ, నీ భర్తనూ సగౌరవంగా పంపుతాను.

సర్దారు : మీరు మంచివారని విన్నాం. కానీ, ఇంతమంచి వారనుకోలేదు.

శివాజీ : పుణ్యస్త్రీల ఆశీస్సులే మా అభివృద్ధికి కారణం.

ఆ) ఈ పాఠం ఆధారంగా స్త్రీల పట్ల మనం ఎలా ప్రవర్తించాలో, మన బాధ్యతలు ఏమిటో తెలిపేలా నినాదాలు /సూక్తులు రాయండి.
జవాబు:

నినాదాలు :	సూక్తులు :
1) స్త్రీలకు రక్షణ కావాలి. స్త్రీలను బాధించే వారికి శిక్షలు పెరగాలి.	1) తల్లిని మించిన దైవం లేదు.
2) మీ అమ్మ కూడా స్త్రీయే. ప్రతి స్త్రీ మీ అమ్మవంటిదే!	2) తల్లి మొదటి గురువు.
3) అమ్మ లేకుంటే సృష్టిలేదు. అమ్మతనం లేకుంటే మనుగడ లేదు.	3) స్త్రీ ఓర్పులో భూమాత వంటిది.
4) స్త్రీలను గౌరవించు, గౌరవంగా జీవించు.	4) స్త్రీలకు జాలి ఎక్కువ.
5) స్త్రీల సంతోషం సంపదలకు స్వాగతం.	5) స్త్రీ విద్య ప్రగతికి సోపానం.

భాషా కార్యకలాపాలు ప్రాజెక్టు పని

* స్త్రీల అభ్యున్నతికి కృషి చేసిన సంస్కర్తల వివరాలు సేకరించి ప్రదర్శించండి.
జవాబు:
స్త్రీల అభ్యున్నతికి కృషి చేసిన సంస్కర్తల వివరాలు :
1) రాజారామమోహన్ రాయ్ :

ఇతడు భారతదేశంలో బెంగాలు రాష్ట్రంలో జన్మించాడు. ‘సతీసహగమనము’ అనే దురాచార నిర్మూలనకు కృషిచేసి, విలియం బెంటింక్ ద్వారా నిషేధ చట్టాన్ని చేయించాడు.

2) వీరేశలింగం పంతులు :

విధవా పునర్వివాహములను ప్రోత్సహించాడు. స్త్రీలకు పాఠశాలలు ఏర్పాటు చేశాడు. స్త్రీలకు విద్యాభివృద్ధికై ‘సతీహితబోధిని’ పత్రిక స్థాపించాడు.

3) జ్యోతిరావుఫూలే :

ఈయన పునా(పూణె)లో జన్మించాడు. స్త్రీ చదువుకుంటేనే కుటుంబం బాగుపడుతుందని భార్య సావిత్రికి చదువు చెప్పి, ఆమెను మొదటి పంతులమ్మను చేశాడు. తన సొంత డబ్బుతో ఆడపిల్లల కోసం బడి పెట్టాడు.

4) గురజాడ వెంకట అప్పారావు :

ఈయన ఆంధ్రదేశంలో విజయనగరం జిల్లావాడు. సమాజంలో ఉన్న ‘కన్యాశుల్కం’ అనే దురాచారాన్ని పోగొట్టడానికి “కన్యాశుల్కం” అనే నాటకాన్ని రచించాడు.

5) కనుపర్తి వరలక్ష్మమ్మ :

ఈమె భర్త ప్రోత్సాహంతో “స్త్రీ హితైషిణీ మండలి”ని స్థాపించి, స్త్రీ విద్యను ప్రోత్సహించింది. స్త్రీలకు ఓటుహక్కు కోసం ప్రయత్నించింది.

6) దుర్గాబాయి దేశ్ ముఖ్ :

ఈమె మద్రాసు, హైదరాబాదు నగరాలలో ఆంధ్ర మహిళాసభ ద్వారా స్త్రీలకు పాఠశాలలు, కళాశాలలు స్థాపించింది. స్త్రీలకు నర్సింగ్, కుట్టుపని వంటి వాటిలో శిక్షణ ఇప్పించింది.

(లేదా)

వివిధ రంగాలలో ప్రసిద్ధిగాంచిన స్త్రీల వివరాలను సేకరించి ప్రదర్శించండి.
జవాబు:
1) ఝాన్సీ లక్ష్మీబాయి : స్వాతంత్ర్య ఉద్యమంలో కత్తిపట్టి బ్రిటిష్ వారితో పోరాడి ప్రాణాలు కోల్పోయింది.

2) ఇందిరాగాంధీ : సుమారు 17 సంవత్సరాలు భారతదేశ ప్రధానమంత్రిగా పనిచేసింది.

3) సునీతా విలియమ్స్ : భారత సంతతికి చెందిన అమెరికా అంతరిక్ష పరిశోధకురాలు.

4) మార్గరెట్ థాచర్ : బ్రిటన్ ప్రధానమంత్రి.

5) శ్రీమతి భండారునాయకే : శ్రీలంక అధ్యక్షురాలు.

6) – సరోజినీ నాయుడు : స్వరాజ్య సమరంలో పాల్గొంది.

7) కల్పనా చావ్లా : అంతరిక్షంలో ఎగిరిన మహిళ

8) దుర్గాబాయి దేశ్ ముఖ్ : మహిళాభివృద్ధికి కృషి చేసింది.

9) సానియా మీర్జా గొప్ప టెన్నిస్ క్రీడాకారిణి.

10) సైనానెహ్వాల్ : బ్యాడ్మింటన్ క్రీడాకారిణి.

11) సావిత్రీబాయి ఫూలే : స్త్రీలకు విద్య నేర్పడం – సమాజ సేవ.

12) కరణం మల్లేశ్వరి . : సుప్రసిద్ధ వెయిట్ లిఫ్టర్ (ఒలింపిక్ పతక గ్రహీత)

III. భాషాంశాలు

పదజాలం

1. కింది పర్యాయపదాలకు సంబంధించిన పదాన్ని పాఠంలో గుర్తించి గడిలో రాయండి.

అ) …………… – ఆదేశము, ఆన, ఉత్తరువు, నిర్దేశము.
ఆ) …………… – అక్షి, చక్షువు, నేత్రము, నయనము.
ఇ) …………… – అగ్ని, వహ్ని, జ్వలనుడు.
ఈ) …………… – మగువ, కొమ్మ, ఇంతి, పడతి
జవాబు:
అ) ఆజ్ఞ
ఆ) కన్ను
ఇ) అనలము
ఈ) సతి

2. కింది ఆధారాలను బట్టి గళ్ళను పూరించండి.

అడ్డం :	నిలువు :
1. సీతకు అగ్నిగుండం కూడా ఇలా ఉంటుంది (4)	2. సోన్ దేవుడు దీన్ని బంధించాడనే శివాజీ కోపించింది (4)
4. ‘అంబుదం’ దీన్నే ఇలా కూడా అంటారు (2)	6. రావణుని తాత (4)
3. శివాజీ గౌరవించిన కాంత వంశం (3)	7. యవన కాంత స్వస్థలం (4)
5. సావిత్రి చరిత్ర విశేషణం (3)	8. సోన్ దేవుని మదోన్మాదానికి కారణం (2)
6. పాపం కాదు పుణ్యానికి నిలయం (4)	11. శివాజీని సో దేవుడు పిలిచినట్లు మీరూ పిలవండి (2)
9. కుడివైపు నుండి సీతకు మరో పేరు (3)	13. శీర్షాసనం వేసిన త్వరితం, వేగం (2)
10. కుడివైపు నుండి శివాజీ కోపించిన సేనాని (4)
12. ఈ పాఠం కవి ఇంటి పేరు (4)
14. పాఠంలో శివాజీ తొలిపలుకు (1)

3. కింది ప్రకృతులకు సరైన వికృతులను జతపరచండి.
వికృతి

ప్రకృతి	వికృతి
అ) రాజ్జి	1) ఆన
ఆ) ఆజ్ఞ	2) రతనము
ఇ) ఛాయ	3) బత్తి
ఈ) రత్నము	4) రాణి
ఉ) భక్తి	5) చాయ

జవాబు:

ప్రకృతి	వికృతి
అ) రాజ్జి	4) రాణి
ఆ) ఆజ్ఞ	1) ఆన
ఇ) ఛాయ	5) చాయ
ఈ) రత్నము	2) రతనము
ఉ) భక్తి	3) బత్తి

4. ఈ కింది పదాలకు వ్యుత్పత్యర్థాలు రాయండి.
శివుడు : సాధువుల హృదయాన శయనించి ఉండువాడు, మంగళప్రదుడు (ఈశ్వరుడు)
పతివ్రత : పతిని సేవించుటయే వ్రతంగా కలిగినది (సాధ్వి)
పురంధి : గృహమును ధరించునది (గృహిణి)
అంగన : చక్కని అవయవముల అమరిక కలది (అందగత్తె)

5. ఈ కింది పదాలకు నానార్థాలు రాయండి.
వాసము : ఇల్లు, వస్త్రం
సూత్రము : నూలిపోగు, తీగె, త్రాడు
చరణము : పాదము, కిరణము, పద్యపాదము
హరి : యముడు, సింహము, ఇంద్రుడు
రత్నము : మణి, స్త్రీ, ముంత

6. కింది పదాల్లోని ప్రకృతి – వికృతి పదాలను వేరుచేసి రాయండి.

ప్రకృతి	వికృతి
గౌరవము	గారవము
పుణ్యము	పున్నెం
రాశి	రాసి
అంబ	అమ్మ
దోషము	దోసము
బ్రహ్మ	బమ్మ
జ్యోతి	జోతి
గృహము	గీము
భాగ్యము	బాగ్గెము

వ్యాకరణాంశాలు

1. కింది పదాలు పరిశీలించండి. వాటిలో సవర్ణదీర్ఘ గుణ, వృద్ధి సంధులున్నాయి. గుర్తించి, విడదీసి సూత్రాలు రాయండి.
అ) పుణ్యావాసము
ఆ) మదోన్మాదము
ఇ) స్నిగ్గాంబుద
ఈ) సరభసోత్సాహం
ఉ) గుణోద్ధత్యం
ఊ) రసైకస్థితి

అ) సవర్ణదీర్ఘ సంధి
సూత్రము ‘అ, ఇ, ఉ, ఋ లకు అవే (సవర్ణ) అచ్చులు పరములయినచో వానికి దీర్ఘములు వచ్చును.
అ) పుణ్యవాసము = పుణ్య + ఆవాసము – (అ + ఆ = ఆ)
ఇ) స్నిగ్లాంబుద = స్నిగ + అంబుద . (అ + అ = ఆ)

ఆ) గుణ సంధి –
సూత్రము ‘అకారమునకు ఇ, ఉ, ఋ లు పరమైన వానికి క్రమముగా ఏ, ఓ, అర్లు ఆదేశమగును.
ఆ) మదోన్మాదము – మద + ఉన్మాదము – (అ + ఉ = ఓ)
ఈ) సరభసోత్సాహం = సరభస + ఉత్సాహం – (అ + ఉ = ఓ)

ఇ) వృద్ధి సంధి
సూత్రము అకారమునకు ఏ, ఐ లు పరమైన ‘ఐ’ కారం, ఓ, ఔ లు పరమైన ‘జై’ కారం ఆదేశమగును.
ఉ) గుణోద్ధత్యం – గుణ + ఔద్దత్యం – (అ + ఔ – ఔ)
ఊ) రసైకస్థితి : రస + ఏకసితి – (అ + ఏ = ఐ)

2. కింది పదాల్లో ఉత్వ, త్రిక, రుగాగమ, లులనల సంధులున్నాయి. పదాలు విడదీసి, సంధి జరిగిన తీరును చర్చించండి.
అ బంధమూడ్చి
ఆ) అవ్వారల
ఇ) భక్తురాలు
ఈ) బాలెంతరాలు
ఉ) గుణవంతురాలు
ఊ) దేశాల
ఋ) పుస్తకాలు
ఋా) సమయాన

ఉత్వ సంధి
సూత్రము :
ఉత్తునకచ్చు పరమగునపుడు సంధియగు.
అ) బంధమూడ్చి = బంధము + ఊడ్చి – (ఉ + ఊ – ఊ)

త్రిక సంధి
సూత్రము :

ఆ, ఈ, ఏ లు త్రికమనబడును – (ఆ + వారల)
త్రికము మీది అసంయుక్త హల్లునకు ద్విత్వంబు బహుళంబుగానగు – (ఆ + వ్వారల)
ద్విరుక్తంబగు హల్లు పరమగునపుడు ఆచ్చికమగు దీర్ఘమునకు హ్రస్వంబగు – (అవ్వారల)

ఆ) అవ్వారల = ఆ + వారల – త్రిక సంధి

రుగాగమ సంధి
సూత్రము :కర్మధారయము నందు తత్సమంబులకు ‘ఆలు’ ‘శబ్దం పరమగునపుడు అత్వంబునకు ఉత్వమును, – రుగాగమంబును అగును.
ఇ) భక్తురాలు : భక్త + ఆలు – భక్తురు(క్) + ఆలు
ఉ) గుణవంతురాలు – గుణవంత + ఆలు – గుణవంతురు(క్) + ఆలు

సూత్రము :పేదాది శబ్దములకు ‘ఆలు’ శబ్దం పరమగునపుడు రుగాగమంబగు.
ఈ) బాలెంతరాలు : బాలెంత + ఆలు – రుగాగమ సంధి

లు ల న ల సంధి
సూత్రము : లు ల న లు పరంబగునపుడు ఒకానొకచోట ముగాగమంబునకు లోపంబును, దాని పూర్వస్వరమునకు దీర్ఘమును విభాషనగు.
ఊ) దేశాల = దేశము + ల – (‘ము’ లోపం – దాని పూర్వపు ‘శ’ కు దీరం వచ్చింది.)
ఋ) పుస్తకాలు : పుస్తకము + లు – (‘ము’ లోపం – దాని పూర్వపు ‘క’ కు దీర్ఘం వచ్చింది.)
ఋా) సమయాన = సమయము + న – (‘ము’ లోపం – దాని పూర్వపు ‘య’ కు దీర్ఘం వచ్చింది.)

3. కింది పద్యపాదాల్లోని అలంకారాన్ని గుర్తించండి. లక్షణాలను సరిచూసుకోండి. అ) అనుచున్ జేవుఱుమీజు కన్నుఁగవతో నాస్పందితోష్ఠంబుతో ఘన హుంకారముతో నటద్ర్భుకుటితో గర్జిల్లు నా భోలే శునిఁ జూడన్ ………
జవాబు:
ఈ పద్యపాదాలలో స్వభావోక్తి అలంకారం ఉంది. భానసలేశుని కోపాన్ని ఉన్నదున్నట్లుగా వర్ణించారు కనుక ఇది స్వభావోక్తి అలంకారం.

4. కింది పద్యపాదాలకు గురులఘువులను గుర్తించి, గణవిభజనచేసి, అవి ఏ పద్యాలకు సంబంధించినవో నిర్ణయించండి. లక్షణాలను చర్చించండి.

అ) ఆ – యేమీ యొక రాణివాసమును బుణ్యవాసమున్ దెచ్చినా
జవాబు:

లక్షణాలు :

ఈ పద్యపాదం ‘శార్దూలం’ వృత్తానికి చెందింది.
యతి 13వ అక్షరం – ‘ఆ’ కు 13వ అక్షరమైన ‘జ్యా’ లో ‘య’ తో యతి.
ప్రాస నియమం కలదు.
4 పాదాలుంటాయి.

ఆ) అనలజ్యోతుల నీ పతివ్రతలఁ బాపాచారులై డాయు భూ
జవాబు:

లక్షణాలు :

ఈ పద్యపాదం ‘మత్తేభం’ వృత్తానికి చెందింది.
యతి 14వ అక్షరం – ‘అ’ కు 14వ అక్షరమైన ‘పాప + ఆచారులు’ లోని పరపదమైన ‘ఆచారులు’ లోని ‘ఆ’ తో యతి చెల్లినది.
ప్రాస నియమం కలదు.
4 పాదాలుంటాయి.

5. కింది పదాలను విడదీయండి.
అ) వాజ్మయం = వాక్ + మయం – ‘క్’ స్థానంలో ‘ఙ’ వచ్చింది.
ఆ) రాణ్మహేంద్రవరం = రాట్ + మహేంద్రవరం – ‘ట్’ కు బదులుగా ‘ణ’ వచ్చింది.
ఇ) జగన్నాథుడు = జగత్ + నాథుడు – ‘త్’ కు బదులుగా ‘న’ వచ్చింది.

అంటే మొదటి పదంలోని క కారం పోయి క వర్గ అనునాసికమైన ఙ (క, ఖ, గ, ఘ, ఙ), ట కారం పోయి ట వర్గ అనునాసికమైన ‘ణ’ (ట, ఠ, డ, ఢ, ణ), ‘త’ కారం పోయి త వర్గ అనునాసికమైన ‘న’ (త, థ, ద, ధ, న) వచ్చాయి కదా! అలాగే మొదటి పదం చివర ‘చ’ కారం ఉంటే చ వర్గ అనునాసికమైన ‘ఞ’ (చ, ఛ, జ, ఝ, ఞ), ‘ప’ కారం ఉంటే పవర్గ అనునాసికమైన ‘మ’ (ప, ఫ, బ, భ, మ) వస్తాయి.

దీనిని సూత్రీకరిస్తే : క, చ, ట, త, ప వరాక్షరాలకు న, మ లు పరమైతే వాని వాని అనునాసికాక్షరాలు వికల్పంగా వస్తాయి. దీనినే ‘అనునాసిక సంధి’ అంటారు.

కింది పదాలను విడదీసి, అనునాసిక సంధి సూత్రంతో అన్వయించి చూడండి.
అ) తన్మయము
ఆ) రాణ్మణి
ఇ) మరున్నందనుడు
జవాబు:
అ) తన్మయము = తత్ + మయము . ‘త్’ కు బదులుగా ‘మ’ వచ్చింది.
ఆ) రాణ్మణి = రాట్ + మణి – ‘ట్’ కు బదులుగా ‘ణ’ వచ్చింది.
ఇ) మరున్నందనుడు = మరుత్ + నందనుడు – ‘త్’ కు బదులుగా ‘న’ వచ్చింది.
అంటే క, చ, ట, త, ప వర్గాక్షరాలకు న, మ లు పరమైతే వాని అనునాసికాక్షరాలు వికల్పంగా వచ్చును.

6. ఉపజాతి పద్యాల్లో తేటగీతి, ఆటవెలది పద్యాల లక్షణాలను తెలుసుకున్నారు కదా ! ఇప్పుడు సీసపద్య లక్షణాలను పరిశీలిద్దాం.
తేటగీతి:

ఇది ఉపజాతి పద్యం .
దీనిలో 4 పాదాలు ఉంటాయి.
ప్రతి పాదంలోను వరుసగా ఒక సూర్యగణం, 2 ఇంద్రగణాలు, 2 సూర్యగణాలు ఉంటాయి.
4వ గణం మొదటి అక్షరం యతి. ప్రాసయతి అయినా వేయవచ్చును.
ప్రాస నియమం లేదు.

ఉదా :

ఆటవెలది :

ఇది ఉపజాతి పద్యం .
దీనిలో 4 పాదాలు ఉంటాయి.
1వ పాదంలో వరుసగా 3 సూర్యగణాలు, 2 ఇంద్రగణాలు ఉంటాయి.
3వ పాదంలో కూడా ఇలానే ఉంటాయి.
2వ పాదంలోను, 4వ పాదంలోను వరుసగా 5 సూర్యగణాలు ఉంటాయి.
ప్రతి పాదంలోను యతి 4వ గణం మొదటి అక్షరం.
ప్రాసయతిని అయినా వేయవచ్చును.
ప్రాస నియమం లేదు.

ఉదా :

సీసపద్యం :

సీసపద్యంలో ప్రతిపాదం రెండు భాగాలుగా ఉంటుంది. ప్రతి భాగంలో నాల్గేసి గణాల చొప్పున ఒక్కొక్క పాదంలో ఎనిమిది గణాలుంటాయి. ఈ 8 గణాల్లో మొదటి ఆరు ఇంద్రగణాలు. చివరి రెండు సూర్యగణాలు. (పాదం మొదటి భాగంలో 4 ఇంద్రగణాలు, 2వ భాగంలో వరుసగా రెండు ఇంద్రగణాలు, రెండు సూర్యగణాలుంటాయి.)
ఉదా :

లక్షణాలు :

4 పాదాలుంటాయి.
ప్రతి పాదం 2 భాగాలుగా ఉంటుంది.
మొదటి భాగంలో 4 గణాలుంటాయి. 2వ భాగంలో 4 గణాలుంటాయి.
రెండు భాగాలలోను 3వ గణం మొదటి అక్షరం యతి. లేక ప్రాసయతి చెల్లుతుంది.
మొదటి భాగంలో 4 ఇంద్రగణాలుంటాయి.
2వ భాగంలో 2 ఇంద్ర, 2 సూర్య గణాలుంటాయి.
ప్రాస నియమం లేదు.
4 పాదాల (8 పాదభాగాలు) తర్వాత తేటగీతి గాని, ఆటవెలది గాని తప్పనిసరిగా ఉండాలి.

ఈ కింది పద్య పాదాన్ని గణ విభజన చేసి లక్షణ సమన్వయం చేయండి.

ధగధగ ద్దహనమధ్యము పూలరాసిగా
విహరించియున్న సాధ్వీమతల్లి.

మీ పాఠంలోని 5వ పద్యం సీసం. ఆ పద్యం లక్షణాలు సరిచూడండి.

అదనపు సమాచారము

సంస్కృత సంధులు

1. సవర్ణదీర్ఘ సంధి:
1) భారతావని భారత + అవని – సవర్ణదీర్ఘ సంధి
2) దుశ్చరితాలోచన దుశ్చరిత + ఆలోచన – సవర్ణదీర్ఘ సంధి
3) పాపాచారులు = పాప + ఆచారులు – సవర్ణదీర్ఘ సంధి
4) భరతాంబ = భరత + అంబ – సవర్ణదీర్ఘ సంధి
5) మదీయాదర్శము = మదీయ + ఆదర్శము – సవర్ణదీర్ఘ సంధి
6) సూక్తి = సు + ఉక్తి – సవర్ణదీర్ఘ సంధి

2. గుణ సంధి :
7) బోన్ సలేశుడు = బోన్ సల + ఈశుడు – గుణసంధి
8) అజోల్లంఘన = ఆజ్ఞ + ఉల్లంఘన – గుణసంధి
9) ఉల్లంఘనోద్వృత్తి = ఉల్లంఘన + ఉద్వృతి – గుణసంధి

3. జశ్వ సంధి:
10) నటద్ర్భుకుటి = నటత్ + భ్రుకుటి – జత్త్వసంధి
11) భవదాజ్ఞ = భవత్ + ఆజ్ఞ – జత్త్వసంధి

4. అనునాసిక సంధి :
12) అసన్మార్గంబు = అసత్ + మార్గంబు – అనునాసిక సంధి

5. శ్చుత్వ సంధి:
13) దుశ్చరితము = దుస్ +చరితము – శ్చుత్వసంధి
14) దుశ్చరిత్రము = దుస్ + చారిత్రము – శ్చుత్వసంధి
15) అస్మచ్ఛబ్దము = అస్మత్ + శబ్దము – శ్చుత్వసంధి

తెలుగు సంధులు

1. అత్వ సంధి:
1) పుట్టినిల్లు = పుట్టిన + ఇల్లు – అత్వసంధి
2) మెట్టినిల్లు = మెట్టిన + ఇల్లు – అత్వసంధి

2. ఉత్వ సంధి:
3) తోడంపు = తోడు + అంపు – ఉత్వసంధి
4) పుయిలోడు = పుయిలు + ఓడు – ఉత్వసంధి

3. గసడదవాదేశ సంధి :
5) భాగ్యములు వోసి = భాగ్యములు + పోసి – గసడదవాదేశ సంధి
6) భిక్షగొన్న = భిక్ష + కొన్న – గసడదవాదేశ సంధి

4. నుగాగమ సంధి :
7) భగవానునుదయము= భగవాను + ఉదయము – నుగాగమ సంధి
8) కన్నుఁగవ = కన్ను + కవ (కన్ను + న్ + కవ) – నుగాగమ సంధి
9) ముసుంగుఁదెర = ముసుంగు + తెర (ముసుంగు + న్ + తెర) – నుగాగమ సంధి

5. యడాగమ సంధి:
10) మాయాజ్ఞ = మా + ఆజ్ఞ – యడాగమ సంధి
11) ఈ యాజ్ఞ = ఈ + ఆజ్ఞ – యడాగమ సంధి

సమాసాలు

ప్రకృతి – వికృతి

జ్యోతి – జోతి
మర్యాద – మరియాద
రాట్టు – ఱేడు
ఈర్ష్య – ఈసు
రాశి – రాసి
బంధము – బందము
సూక్ష్మత – సుంత
బిక్ష – బిచ్చము, బికిరము
భక్తి – బత్తి
మణి – మిన్
భాగ్యము – బాగైం
రూపము – రూపు
ఛాయ – చాయ
భూమి – బూమి
పుత్రుడు – బొట్టె
రాజ్ఞి – రాణి
బ్రహ్మ – బమ్మ, బొమ్మ
దోషము – దోసము, దొసగు
పుణ్యము – పున్నెము
గృహము – గీము
భయము – పుయిలు
సూక్తి – సుద్ది
ద్వంద్వము – దొందము
ముఖము – మొగము
గౌరవము – గారవము
స్త్రీ – ఇంతి
రత్నము – రతనము
ఆజ్ఞ – ఆన
ఓష్ఠము – ఔడు

నానార్థాలు

1. బలము : సత్తువ, సేన, వాసన
2. తోడు : సహాయము, నీరువంటి వాటిని పైకి లాగడం, తోడబుట్టినవాడు
3. పాశము : తాడు, గుంపు, బాణము, ఆయుధము
4. పుణ్యము : ధర్మము, పవిత్రత, నీరు
5. సూత్రము : నూలిపోగు, తీగె, త్రాడు
6. బంధము : కట్ట, దారము, సంకెల, దేహము
7. రూపము : ఆకృతి, సౌందర్యము
8. చరణము : పాదము, కిరణము, పద్యపాదము
9. సంపద : ఐశ్వర్యము, సౌఖ్యము, లాభము, ధనము
10. ఛాయ : నీడ, పార్వతి, పోలిక
11. భిక్షము : బిచ్చము, కూలి, కొలువు
12. గౌరవము : బరువు, మన్నన, గొప్పతనము
18. సంతానము : బిడ్డ, కులము, వరుస
14. హరి : విష్ణువు, ఇంద్రుడు, సూర్యుడు, గుఱ్ఱము, కోతి
15. దోసము : పాపము, తప్పు, లోపము
16. మర్యా ద : కట్టుబాటు, పొలిమేర, నడత, నిష్ఠ

పర్యాయపదాలు

1. తల్లి : జనయిత్రి, మాత, అమ్మ, జనని
2. ఆజ్ఞ : ఆదేశము, ఆన, ఉత్తరువు, ఆనతి, ఆజ్ఞప్తి
3. కన్ను : చక్షువు, నేత్రము, నయనము, అక్షి
4. పతివ్రత : సాధ్వి, పురంధి, పతిదేవత, సతి
5. దోషము : దోసము, దొసగు, తప్పు, అపరాధము
6. దేవతలు : అమరులు, వేల్పులు, విబుధులు, నిర్జరులు
7. అంబుధి : ఉదధి, పారావారము, కడలి, సముద్రము
8. హరి : విష్ణువు, చక్రి, నారాయణుడు, వైకుంఠుడు
9. బ్రహ్మ : పద్మభవుడు, చతుర్ముఖుడు, నలువ
10. కాంత : స్త్రీ, వనిత, చెలువ, మహిళ, ఇంతి, ఆడుది, యువతి
11. బిడ్డ : కొడుకు, శిశువు, బాలుడు
12. అంబుదము : మేఘము, మొగులు, అంభోదము, జలదము, ఘనము
13. అనలము : అగ్ని, దహనము, శుచి, వహ్ని
14. ముఖము : మొగము, ఆననము, వదనము, మోము
15. భూమి : ధరణి, అవని, ధర, పృథివి

వ్యుత్పత్త్యర్థాలు

1. అంబుదము : నీటినిచ్చునది (మేఘము)
2. పురంధి : గృహమును ధరించునది (ఇల్లాలు)
3. పతివ్రత : పతిని సేవించుటయే వ్రతముగా గలది (సాధ్వి)
4. జనని : సంతానమును ఉత్పత్తి చేయునది (తల్లి)
5. దహనము : కాల్చుటకు సాధనమైనది (అగ్ని)

కవి పరిచయం

పేరు : డా|| గడియారం వేంకటశేష శాస్త్రి

తల్లితండ్రి : తల్లి నరసమాంబ, తండ్రి రామయ్య, కడప జిల్లా, జమ్మలమడుగు తాలుకా
నెమళ్ళ దిన్నె గ్రామంలో 1894లో జన్మించారు. కడప మండలం ప్రొద్దుటూరు మునిసిపల్ హైస్కూల్లో తెలుగు పండితులుగా
పనిచేశారు. వీరు శతావధాని.

రచనలు : రాజశేఖర శతావధాని గారితో కలిసి కొన్ని కావ్యాలు, నాటకాలు రచించారు. ‘శ్రీ శివభారతం’ వీరికి చాలా పేరు తెచ్చిన కావ్యం. పారతంత్ర్యాన్ని నిరసించి స్వాతంత్ర్యకాంక్షను అణువణువునా రగుల్కొల్పిన మహాకావ్యం ఇది. మురారి, పుష్పబాణ విలాసము, వాస్తు జంత్రి (అముద్రిత వచన రచన), మల్లికామారుతము, శ్రీనాథ కవితా సామ్రాజ్యము (విమర్శ), రఘునాథీయము అనే కావ్యాలు రచించారు.

బిరుదులు :
కవితావతంస, కవిసింహ, అవధాన పంచానన అనేవి వారి బిరుదులు.

పద్యాలు – ప్రతిపదార్థాలు – భావాలు

అవగాహన – ప్రతిస్పందన

పద్యం – 1 : కంఠస్థ పద్యం

శా॥ “ఆ యేమీ ? యొక రాణివాసమును బుణ్యవాసమున్ దెచ్చినా
వా? యే హైందవుఁడైన నీ గతి నమర్యాదన్ బ్రవర్తించునే?
మా యాజ్ఞన్ గమనింపవో ? జయ మదోన్మాదంబునన్ రేఁగి, నీ
యాయుస్సూత్రము లీవ క్రుంచుకొనేదో ? యౌధ్ధత్య మోర్వన్ జుమీ”
ప్రతిపదార్థం :
ఆ – యేమీ? = ఆ, ఏమిటీ? (ఆశ్చర్యం, కోపంతో)
పుణ్యవాసముల్ (పుణ్య + ఆవాసమున్) = పుణ్యానికి నిలయమైన
ఒక = ఒక
రాణివాసమును = అంతఃపురమును (మహారాణిని)
తెచ్చినావా? = బందీగా తీసుకొని వచ్చావా?
ఏ, హైందవుడు + ఐనన్ = ఏ హిందువైనా (భారతీయుడెవరైనా)
ఈ గతిన్ = ఈ విధంగా
అమర్యాదన్ ప్రవర్తించునే = గౌరవం లేకుండా
ప్రవర్తించును + ఏ = ప్రవర్తిస్తాడా? (ప్రవర్తించడు)
మా + ఆజ్ఞన్ = మా ఆజ్ఞను (రాజాజ్ఞను)
గమనింపవు + ఓ = పట్టించుకోవా?
జయ = జయం వలన
మద = గర్వంతో
ఉన్నాదంబునన్ = మితిమీరిన పిచ్చితనముతో
రేఁగి = విజృంభించి
నీ = నీ యొక్క
ఆయుస్సూత్రములు = ఆయుర్దాయపు నూలిపోగులు (ప్రాణాలు)
ఈవ త్రుంచుకొనెదు + ఓ = త్రెంచుకొంటావా?
ఔద్ధత్యము = గర్వంతో చేసే పనులను
ఓర్వన్ = సహించను
చుమీ = సుమా!

భావం :
“ఆ-ఏమిటీ? పుణ్యానికి నిలయమైన ఒక రాణి వాసాన్ని బంధించి తీసుకొనివచ్చావా? ఏ భారతీయుడైనా ఈ విధంగా గౌరవం లేకుండా ప్రవర్తిస్తాడా? రాజాజ్ఞను కూడా పట్టించుకోవా? జయం వలన గర్వంతో, మితిమీరిన పిచ్చితనంతో విజృంభిస్తావా? నీ ప్రాణాలు నీవే తెంచుకొంటావా? గర్వాన్ని సహించను సుమా !” అని శివాజీ, సో దేవునితో ఆగ్రహంగా అన్నాడు.

పద్యం – 2

మ|| | అనుచున్ జేవుజు మీ జు కన్నుఁగవతో నాస్పందితోష్ణంబుతో
ఘన హుంకారముతో నటద్భుకుటితో గర్జిల్లు నా భోసలే
శునిఁ జూదన్ బుయిలోడెఁ గొల్వు శివుఁడీసున్ గుత్తుకన్ మ్రింగి, బో
రన నవ్వారల బంధ మూడ్చి గొని తేరన్ బంచె సోన్ దేవునిన్
ప్రతిపదార్థం :
అనుచున్ = శివాజీ అలా హెచ్చరిస్తూ (ఆ విధంగా చెపుతూ)
జేవుఱుమీటు = జేగురు రంగును (ఎరుపు రంగును) అతిశయించే (జేగురు రంగు కంటే ఎఱ్ఱగా నున్న)
కన్నుఁగవతోన్ = కనుల జంటతో
ఆస్పందదోష్ఠంబుతోన్; ఆస్పందత్ = కొలదిగా కదులుతున్న
ఓష్ఠంబుతోన్ = పెదవితో
ఘనహుంకారముతోన్ = గొప్ప హుంకార ధ్వనితో
నటద్ర్భుకుటితోన్; నటత్ = నాట్యము చేయుచున్న (బాగా కదలి ఆడుచున్న)
భ్రుకుటీతోన్ = కనుబొమల ముడితో
గర్జీల్లు = గర్జిస్తున్న
ఆ ఫోన్సలేశునిన్ (ఆ ఫోన్సల + ఈశునిన్) = ఆభోంసల వంశ ప్రభువైన శివాజీని
చూడన్ = చూడ్డానికి
కొల్వు = రాజసభ
పుయిలోడెన్ = జంకింది (భయపడింది.) (నిశ్చేష్టులయ్యారు)
శివుడు = శివాజీ
ఈసున్ = (తన) కోపాన్ని
కుత్తుకన్ = గొంతుకలో
మ్రింగి = అణచుకొని
బోరనన్ = శీఘ్రముగా (ఇది ‘బోరునన్’) అని ఉండాలి.)
అవ్వారల = వారి యొక్క (కళ్యాణి సర్దారు యొక్క ఆతని అంతఃపురకాంత యొక్క
బంధమూడ్చి (బంధము + ఊడ్చి) – సంకెలలు తొలగించి,
కొనితేరన్ = తీసికొనిరావడానికి (సభలోకి తీసుకురావడానికి)
సోన్ దేవునిన్ = (తన సైన్యాధిపతియైన, వారిని బంధించి తెచ్చిన) సోన్ దేవుడిని
పంచెన్ = ఆజ్ఞాపించెను.

భావం:
అంటూ ఎర్రబడిన కన్నులతో, అదిరిపడే పై పెదవితో, గొప్ప హుంకారముతో, కదలియాడే కనుబొమ్మల ముడితో, గర్జిస్తున్న ఆ ఫోన్సలేశుడైన శివాజీని చూడ్డానికి సభలోనివారు భయపడ్డారు. తరువాత శివాజీ తన కోపాన్ని గొంతుకలో అణచుకొని, వెంటనే వారి సంకెళ్లను తొలగించి, తీసుకొని రమ్మని, సో దేవుడిని ఆజ్ఞాపించాడు.

పద్యం – 3

మ|| | త్వరితుండై యతఁ డట్టులే నలిపి “దేవా! నన్ను మన్నింపు; మీ
సరదారున్ గొని తెచ్చుచో సరభసోత్సాహంబు కగ్గప్పె; దు
శృరితాలోచన లేదు, లేదు భవదాజా లంఘనోద్వృత్తి; మీ
చరణద్వంద్వమునాన” యంచు వినిపించన్, సుంత శాంతించుచున్
ప్రతిపదార్థం :
త్వరితుండు + ఐ = తొందర కలవాడై
అతడు = ఆసోన్ దేవుడు
అట్టులే = ఆ విధంగానే (శివాజీ చెప్పినట్లుగానే)
సలిపి = చెసి
దేవా = దేవా (శివాజీని దైవమా ! అని సంబోధించి)
నన్ను = నన్ను (సోన్ దేవుని)
మన్నింపుము = అపరాధమును క్షమింపుము
ఈ సరదారున్ = (ఓడిపోయిన) ఈ వీరుడిని
కొని తెచ్చుచో = తీసుకొని వచ్చేటపుడు
సరభస + ఉత్సాహంబు = ఉవ్విళ్ళూరు ఉత్సాహము
కన్దప్పె = కళ్లకు కమ్మేసింది
దుస్+చరిత + ఆలోచన = చెడు చేయాలనే తలంపు
లేదు = లేదు
మీ = తమ యొక్క
చరణద్వంద్వంబులు = పాదాలు
ఆన = సాక్షి (ఒట్టు)గా
భవత్ = తమ యొక్క
ఆజ్ఞ = ఆజ్ఞను
ఉల్లంఘన = అతిక్రమించాలనే
ఉద్వృత్తి = గర్వము
లేదు = లేదు
అంచు = అనుచు
వినిపించన్ = నివేదించగా
సుంత = కొద్దిగా
శాంతించుచున్ – శాంతిని పొందినవాడై (కోపం తగ్గినవాడై)

భావం :
శివాజీ ఆజ్ఞాపించిన పనిని సోదేవుడు తొందరగా చేశాడు. “దేవా! నన్ను మన్నించండి. ఓడిపోయిన ఈ వీరుడిని బంధించి తెచ్చేటప్పుడు ఉవ్విళ్ళూరు ఉత్సాహం కళ్లకు కమ్మేసింది. మీ పాదాల సాక్షిగా నాకు చెడు చేయాలనే ఆలోచన లేదు. తమ ఆజ్ఞను అతిక్రమించాలనే గర్వంలేదు.” అని నివేదించగా శివాజీ కొద్దిగా శాంతించాడు.

పద్యం – 4 : కంఠస్థ పద్యం

*మ|| శివరాజంతట మేల్ముసుంగుఁ దెరలో – స్నిగ్జాంబుదద్ఛాయలో
నవసౌదామినిఁ బోలు నా యవనకాంతారత్నమున్ భక్తి గా
రవముల్ వాజఁగఁ జూచి వల్కె “వనితారత్నంబు లీ భవ్యహైం
దవభూజంగమ పుణ్యదేవతలు; మాతా! తప్పు సైరింపుమీ !”
ప్రతిపదార్థం :
శివరాజు = శివాజీ మహారాజు
అంతటన = అప్పుడు
మేల్ముసుంగుఁదెరలోన్; మేల్ముసుంగు = సువాసినీ స్త్రీలు వేసుకొనే మేలు ముసుగు యొక్క (బురఖా)
తెరలోన్ = తెరలోపల
స్నిగ్దాంబుదచ్ఛాయలోస్, (స్నిగ్ధ+ అంబుద + ఛాయలోన్) స్నిగ్ధ = దట్టమైన
అంబుద = మేఘము యొక్క
ఛాయలోన్ = నీడలో (మాటున నున్న)
నవసౌదామినిన్ = కొత్త మెరుపు తీగను
పోలు = పోలినట్లు ఉన్న
ఆ, యవన కాంతారత్నమున్ = ఆ రత్నము వంటి యవనకాంతను (మహమ్మదీయ స్త్రీని)
భక్తి గౌరవముల్ = భక్తియునూ, గౌరవమునూ
పాఱగన్ + చూచి = స్ఫురించేటట్లు చూసి
పల్కెన్ = ఈ విధంగా అన్నాడు
వనితారత్నంబులు = రత్నముల వంటి స్త్రీలు (శ్రేష్ఠులైన స్త్రీలు)
ఈ = ఈ
భవ్య హైందవ భూ జంగమ పుణ్యదేవతలు; భవ్య = శుభప్రదమైన
హైందవ భూ = భారత భూమిపై
జంగమ = సంచరించే (తిరుగాడే)
పుణ్యదేవతలు = పుణ్యప్రదమైన దేవతల వంటివారు
మాతా! = అమ్మా
తప్పున్ = మా వారు చేసిన తప్పును
సైరింపుమీ = మన్నింపుము (క్షమింపుము)

భావం :
శివాజీ మహారాజు అప్పుడు మేలు ముసుగు తెరలో దట్టమైన నీలి మేఘం వెనుక ఉన్న మెరుపు తీగవంటి యవన కాంతను భక్తి గౌరవాలతో చూస్తూ ఇలా అన్నాడు. “స్త్రీలు శుభప్రదమైన ఈ హైందవ భూమిపై సంచరించే పుణ్యదేవతలు. అమ్మా ! మా తప్పును మన్నింపుము.”

చారిత్రక విశేషం :
అబ్బాజీసో దేవుడు అనే శివాజీ యొక్క సైన్యాధిపతి ‘కళ్యాణి’ కోటను పట్టుకొన్నాడు. అక్కడ అతడు ఒక అందమైన అమ్మాయిని బందీగా పట్టుకొన్నాడు. ఆ అమ్మాయి కళ్యాణి కోటకు గవర్నరు (సర్దారు) అయిన మౌలానా అహమ్మదుకు కోడలు. ఆ అమ్మాయిని సో దేవుడు శివాజీకి బహుమతిగా ఇచ్చాడు. అప్పుడు శివాజీ ఆ అమ్మాయితో “అమ్మా! నా తల్లి నీ అంత అందగత్తె అయి ఉన్నట్లయితే, నేను కూడా నీ అంత అందంగా కనబడేవాడిని” అని అన్నాడు. శివాజీ ఆ యవన కాంతను తన కూతురుగా ఆదరించాడు. ఆమెకు వస్త్రాలు ఇచ్చి, ఆమెను ఆమె ఇంటికి – బీజాపూరుకు పంపాడు. (ఇది చరిత్రలలో చెప్పబడింది)

పద్యం – 5

సీ॥ హరి హర బ్రహ్మలం బురిటిబిడ్డలం జేసి
జోలంబాడిన పురంద్రీలలామ,
యమధర్మరాజు పాశముం ద్రుంచి యదలించి
పతిభిక్ష గొన్న పావనచరిత్ర,
ధగధగ దహనమధ్యము పూలరాసిగా
విహరించియున్న సాధ్వీమతల్లి,
పతి నిమిత్తము సూర్యభగవానును దయంబు
నరికట్టి నిలుపు పుణ్యములవంట,
తే|| అట్టి యెందతో భరతాంబ యాఁదుబిద్ద
లమల పతిదేవతాత్వ భాగ్యములు వోసి
పుట్టినిలు మెట్టినిలుఁ బెంచు పుణ్యసతులు
గలరు, భారతావని భాగ్యకల్పలతలు
ప్రతిపదార్థం :
హరి హర బ్రహ్మలన్ = విష్ణువును, శివుని, బ్రహ్మను
పురిటి బిడ్డలన్ + చేసి = పసిపిల్లలుగా చేసి
జోలన్ = జోలపాటను
పాడిన = పాడినటువంటి
పురంధీలలామ = శ్రేష్ఠురాలైన గృహిణి (అనసూయ)
యమధర్మరాజు = మృత్యుదేవత యొక్క
పాశమున్ = త్రాడును
త్రుంచి = తెంచి
అదలించి = గద్దించి
పతిభిక్షన్ = భర్తను భిక్షగా
కొన్న = సంపాదించిన
పావన చరిత్ర = పవిత్రమైన చరిత్ర గలది; (సావిత్రి)
ధగధగత్ = ధగధగ మండుచున్న
దహన మధ్యము = చితి మధ్యభాగము
పూలరాసిగా = పూలకుప్ప వలె
విహరించియున్న = సంచరించి ఉన్నటువంటి
సాధ్వీమ తల్లి = శ్రేష్ఠురాలైన స్త్రీ (సీత)
పతి నిమిత్తము = పతి కొరకు
సూర్యభగవానుని = సూర్యదేవుని యొక్క
ఉదయంబును = ఉదయమును
అరికట్టి = నిరోధించి
నిలుపు = నిలిపిన
రతాంబ
పుణ్యముల పంట = తల్లిదండ్రుల పుణ్యఫలము (సుమతి)
అట్టి = అటువంటి
ఎందఱో = ఎంతోమంది
భరతాంబ = భరతమాత యొక్క
ఆఁడుబిడ్డలు = స్త్రీ సంతానం
అమల = స్వచ్చమైన
తిదేవతాత్వ = పతివ్రతా ధర్మమనెడు
భాగ్యములు + పోసి = సంపదలను ఇచ్చి
అట్టిన + ఇలున్ = పుట్టినింటిని
పెట్టిన + ఇలున్ = అత్తవారింటిని
పెంచు = అభివృద్ధి చేయు
భరత + అవని = భారతదేశము యొక్క
భాగ్య కల్పలతలు = సంపద అనెడు దేవతావృక్షాల వంటి
అణ్యసతులు = పుణ్యాత్ములైన స్త్రీలు
కలరు = ఉన్నారు

భావం :
బ్రహ్మ, విష్ణు, మహేశ్వరులను పసిపిల్లలుగా చేసి ద్రపుచ్చినది అనసూయ అను పేరు గల ఒక గృహిణి. దుమధర్మరాజు పాశమును కూడా ట్రెంచి, గద్దించి, పతి పాణాలు సాధించిన పవిత్రమైన చరిత్ర కలది సావిత్రి. నిప్పుల రాశి మధ్యను పూలరాశిగా సంచరించిన శ్రేష్ఠురాలైన స్త్రీ సీత. -తిప్రాణాలు కాపాడడానికి సూర్యోదయాన్ని నిలిపిన అణ్యాత్మురాలు సుమతి. అటువంటి భరతమాత సంతానమైన స్త్రీలు స్వచ్ఛమైన పతివ్రతలు. వారి పాతివ్రత్య మహిమతో అట్టింటిని, అత్తవారింటిని అభివృద్ధి చేస్తున్నారు. వారు ఈ కారతదేశపు సంపదలనెడు దేవతావృక్షాలు. అటువంటి అణ్యస్త్రీలు ఉన్నారు.

ఇవి తెలుసుకోండి

1. అనసూయ :
అత్రి మహాముని భార్య. ఈమెను పరీక్షించ డానికి బ్రహ్మ, విష్ణు, మహేశ్వరులు ప్రయత్నించారు. వారిని ముగ్గురినీ పసిపిల్లలుగా మార్చింది. వారు కోరినట్లే లాలించింది. ఆమె పాతివ్రత్యానికి దేవతలు సంతోషించారు.

2. సావిత్రి :
సత్యవంతుని భార్య, ‘సత్యవంతుడు మరణిస్తాడు. యమధర్మరాజుని ప్రార్థించి, మెప్పించి, వరాలు పొంది, తన భర్త ప్రాణాలు తిరిగి తెచ్చి, భర్తను బ్రతికించిన మహా పతివ్రత సావిత్రి.

3. సుమతి :
కౌశికుడనే బ్రాహ్మణుని భార్య. అతడు కుష్టురోగి. అతని కోరికపై ఒకచోటుకు తీసుకొని వెడుతోంది. తట్టలో కూర్చోబెట్టుకొని, తలపై పెట్టుకొని, మోసుకొని వెడుతోంది. చీకటిలో అతని కాలు మాండవ్య మహామునికి తగిలింది. సూర్యోదయానికి మరణించాలని శపించాడు. సూర్యోదయం కాకూడదని ఆమె అంది. సూర్యోదయం ఆగిపోయింది.

4. దేవతావృక్షాలు :
కోరిన వస్తువులిచ్చెడు దేవతామ్మకాలు అయిదు. అవి :
1. మందారము,
2. పారిజాతము,
3.సంతానము,
4. కల్పవృక్షము,
5.హరిచందనము.

పద్యం – 6 : కంఠస్థ పద్యం

*మ | అనలజ్యోతుల నీ పతివ్రతలఁ బాపాచారులై దాయు భూ
జనులెల్లన్ నిజసంపదల్ దొలుంగి యస్తద్వసులై పోరి? వి
శనమే నిల్చునా ? మున్నెఱుంగమె పులస్త బ్రహ్మసంతాన? మో
జననీ! హైందవ భూమి నీ పగిది దుశ్చరిత్రముల్ సాగునే?
ప్రతిపదార్థం :
అనల జ్యోతులన్ – అగ్ని జ్వా లల వంటి
ఈ పతివ్రతలన్ = ఈ పతివ్రతలను
పాపాచారులై (పాప + ఆచారులు + ఐ) = అపచారం చేసేవారై
డాయు = కలిసే
భూజనులు + ఎల్లన్ = భూమిపైనున్న ప్రజలు అందరునూ
నిజ సంపదల్ = తమ సంపదలను
తొఱగి = వీడి (పోగొట్టుకొని)
అసద్వస్తులై (అసద్వస్తులు + ఐ) = సర్వ నాశనమైనవారై
పోరె = పోకుండా ఉంటారా?
విత్తనమే = విత్తనము (వారి వంశవృక్షం యొక్క విత్తనం)
నిల్చునె = నిలుస్తుందా ? (అనగా వంశం నిలుస్తుందా?)
మున్ను = పూర్వం
పులస్త్రబ్రహ్మ సంతానమున్ = పులస్త్య బ్రహ్మ యొక్క కుమారుడైన రావణుని గూర్చి
ఎఱుంగమై = మనకు తెలియదా?
హైందవ భూమిని = భారత భూమియందు
ఈ పగిది = ఇటువంటి
దుశ్చారిత్రముల్ = చెడు పనులు (దుశ్చర్యలు)
సాగునే = సాగుతాయా? (సాగవు)

భావం :
ఓ తల్లీ ! అగ్ని జ్వా లల వంటి పతివ్రతల పట్ల అపచారం చేసేవారు, తమ సంపదలు పోగొట్టుకొని, సర్వ నాశనం కారా? అసలు వారి వంశం నిలుస్తుందా? (విత్తనంతో సైతంగా నశించదా?) పులస్తబ్రహ్న సంతానమైన రావణాసురుని పతనం గురించి మనకు తెలియదా? భారతభూమిపై ఇటువంటి దుశ్చర్యలు సాగుతాయా? (సాగవు)

పద్యం -7

తే|| యవన పుణ్యాంగనామణి వగుదుగాక
హైందవులపూజ తల్లియట్లందరాదె?
నీదురూపము నాయందు లేద యైనం
గనని తల్లివిగా నిన్ను గారవింతు
ప్రతిపదార్థం :
యవన = యవన జాతికి చెందిన
పుణ్య + అంగనా మణివి = శ్రేష్ఠమైన పుణ్యస్త్రీవి
అగుదుగాక = అయిన దానివి
తల్లి + అట్లు = మా యొక్క తల్లివలె
హైందవుల = హిందూదేశ వాసుల యొక్క
పూజ = పూజను
అందరాదె = స్వీకరించరాదా ! (స్వీకరించు)
నీదు రూపము = నీ పోలిక
నా + అందు = నాలో
లేదు + ఆ = లేదు
ఐనన్ = ఐనప్పటికీ
కనని = నాకు జన్మనీయని
తల్లివిగా = నా తల్లిగా
నిన్ను = నిన్ను
గారవింతు = గౌరవిస్తాను

భావం:
యవన జాతికి చెందిన పుణ్యస్త్రీవి. అయినా హిందువుల పూజలను మా తల్లివలె స్వీకరించు. నీ పోలిక నాలో లేదు. అయినా నాకు జన్మనివ్వని తల్లిగా నిన్ను గౌరవిస్తాను.

పద్యం – 8: కంఠస్థ పద్యం

*శా॥ మా సర్దారుడు తొందరన్ బడి యసన్మార్గంబునన్ బోయి, నీ
దోసంబున్ గని నొచ్చుకోకు, నినుఁ జేరున్ నీ గృహం బిప్పుడే,
నా సైన్యంబును దోడుగాఁ బనిచెదన్, నాతల్లిగాఁ దోడుగా
దోసిళ్లన్ నడిపింతు; నీ కనులయందున్ దాల్ని సారింపుమీ!
ప్రతిపదార్థం :
మా సర్దారుడు = మా సర్దార్ సో దేవుడు
తొందరన్ బడి = తొందరపాటుపడి
అసన్మార్గంబునన్ = తప్పుడు మార్గంలో
(అసత్ + మార్గంబునన్) పోయెన్ = వెళ్ళాడు. (పొరపాటున నిన్ను బంధించి తెచ్చాడు)
ఈ దోసంబున్ = ఈ దోషాన్ని
కని = చూచి
నొచ్చుకోకు = బాధపడకు
ఇప్పుడే = ఇప్పుడే
నినున్ = నిన్ను
నీ గృహంబున్ = నీ ఇంటిని (నీ ఇంటికి)
చేరున్ = చేరుస్తాను
నా సైన్యంబున్ = నా సైన్యాన్ని
తోడుగాస్ = నీకు సాయంగా
పనిచెదన్ = పంపిస్తాను
నా తల్లిగాన్ = నా యొక్క తల్లివలెనూ
తోడుగాన్ = నా తోడబుట్టిన సోదరిగానూ
దోసిళ్లన్ = (నా) అరచేతులపై
నడిపింతున్ = నడిపిస్తాను (నిన్ను కాలుక్రింద పెట్టకుండా నా అరచేతులపై సగౌరవంగా నడిపించి మీ ఇంటికి పంపిస్తాను)
నీ కనులయందున్ = నీ కళ్లల్లో
తాల్మిన్ = ఓర్పును
సారింపుమా = ప్రసరింప చేయుము. (చూపించుము)

భావం :
మా సర్దారు తొందరపడి తప్పు మార్గంలో నడిచాడు. ఈ దోషాన్ని చూచి బాధపడకు. నిన్ను నీ ఇంటికి ఇప్పుడే చేరుస్తాను. నా సైన్యాన్ని నీకు తోడుగా పంపిస్తాను. నిన్ను నా కన్నులలో ఓరిమిని చూపు. నన్ను సహించి క్షమించు.

పద్యం – 9

మ|| అని కొందాడి, పతివ్రతా హిత సపర్యాధుర్యుందాతండు యా
వన కాంతామణి కరసత్కృతు లొనర్పన్ వేసి, చేసేతఁ జి
క్కిన సర్దారుని గారవించి హితసూక్తిన్ బల్కి బీజాపురం
బునకున్ బోవిదే – వారితోఁ దనబలంబుల్ కొన్ని వాదంపుచున్.
ప్రతిపదార్ధం :
అని = పై విధంగా పలికి
కొండాడి = స్తుతించి
పతివ్రతా = పతివ్రతల యొక్క
హిత = ఇష్టమునకు
సపర్యా = పూజ అనెడు
ధుర్యుడు = భారము వహించువాడు
ఆతండు = ఆ శివాజీ
యావన = యవన సంబంధమైన
కాంతామణికి = శ్రేష్ఠురాలైన ఆ స్త్రీకి
అర్హ = తగినటువంటి
సత్కృతులు = గౌరవాదరాలు
ఒనర్పన్ = అతిశయించునట్లు
చేసి = చేసి
చేత + చేత = చేతులారా
చిక్కిన = తనకు బందీ అయిన
సర్దారుని గారవించి = గౌరవించి
హిత = మంచిని కల్గించే
సు + ఉక్తిన్ – మంచి మాటను
పల్కి = చెప్పి
తన బలంబుల్ = తన సైన్యము
కొన్ని = కొంత
వారితో = ఆ యవన దంపతులతో
తోడు + అంపుచున్ – సహాయంగా పంపుతూ
బీజాపురంబునకున్ = బీజాపూర్‌కు
పోన్ + విడా : పోవుటకు విడిచిపెట్టెను.

భావం :
శివాజీ పై విధంగా ఆ యవనకాంతను స్తుతించాడు. పతివ్రతల ఇష్టానికి తగినట్లు పూజించాడు. ఆ యవనకాంతకు తగిన గౌరవ మర్యాదలు చేశాడు. తనకు చిక్కిన వీరుడైన ఆమె భర్తను గౌరవించాడు. మంచి మాటలు చెప్పాడు. వారికి సహాయంగా తన సైన్యం కొంత పంపాడు. వారిని బీజాపూర్ వెళ్ళడానికి విడిచి పెట్టాడు.

పద్యం – 10

శివరా అంతట సోనదేవుమొగమై సీరత్నముల్ పూజ్య, లే
యవమానంబు ఘటింపరా, దిది మదీయాదర్శ మస్మచ్చమూ
ధవు లీయాజ్ఞ నవశ్య మోమవలె; నీతాత్పర్యమున్ జూచి, లో
కువ చేకూరమి నెంచి, నీయెద దొసంగు బ్లేమి భావించితిన్”
(అని వాక్రుచ్చెను.)
ప్రతిపదార్ధం :
అంతట = అంతలో
శివరాజు = ఛత్రపతి శివాజీ
సోనదేవు మొగమై = సో దేవును వైపు తిరిగి
స్త్రీ రత్నముల్ = శ్రేష్ఠులైన స్త్రీలు
పూజ్యులు = పూజింప తగినవారు
ఏ అవమానంబు = ఏ విధమైన అవమానమును
ఘటింపరాదు = జరుగరాదు
ఇది = ఈ పద్దతి
మదీయ = నా యొక్క
ఆదర్శము = ఆశయము
అస్మ త్ = నా యొక్క
చమూధవులు = సైన్యాధికారులు
ఈ + ఆజ్ఞను = ఈ ఉత్తర్వును
అవశ్యము = తప్పనిసరిగా
ఓమవలె = రక్షించాలి
నీ తాత్పర్యమున్ = నీ భావమును
చూచి = పరిశీలించి
లోకువ = తక్కువ
చేకూరమిన్ = కలుగపోవుటను
ఎంచి = పరిశీలించి
నీ + ఎడ = నీ పట్ల
దొసంగుల్ + లేమి = తప్పులు లేకపోవుటను
భావించితిన్ = గ్రహించితిని

భావం :
అపుడు ఛత్రపతి శివాజీ సో దేవుని వైపు తిరిగి, “స్త్రీలు పూజ్యనీయులు. వారికి ఏ అవమానం జరగకూడదు. ఇది నా ఆశయం. మన సైన్యాధికారులందరూ ఈ ఆజ్ఞను రక్షించాలి. నీ భావం గ్రహించాను. మమ్ము తక్కువ చేయక పోవుటను తెలుసుకొన్నాను. నీ తప్పు లేదని గ్రహించాను” అన్నాడు.

AP State Board 8th Class Maths Notes

November 15, 2023November 14, 2023 by AP Board

Students can go through Andhra Pradesh SCERT AP State Board Syllabus 8th Class Maths Notes Pdf in English Medium and Telugu Medium to understand and remember the concepts easily. Besides, with our AP State 8th Class Maths Notes students can have a complete revision of the subject effectively while focusing on the important chapters and topics. Students can also read AP Board 8th Class Maths Solutions for exam preparation.

AP State Board Syllabus 8th Class Maths Notes

These AP State Board Syllabus 8th Class Maths Notes provide an extra edge and help students to boost their self-confidence before appearing for their final examinations.

AP SSC 10th Class Maths Notes Chapter 1 Real Numbers

November 15, 2023November 14, 2023 by AP Board

Students can go through AP SSC 10th Class Maths Notes Chapter 1 Real Numbers to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 1 Real Numbers

→ “God made the integers. All else is the work of man” …… Leopold Kronecker

→ Euclid’s division lemma: Given positive integers a, b there exists unique pair of integers q and r satisfying
a = bq + r; 0 ≤ r < b
This result was first published / recorded in book VII of Euclid’s “The Elements”.

→ Euclid’s division algorithm is a technique to compute the Highest Common Factor (H.C.F) of two given numbers.
E.g: HCF of 80 and 130
130 = 80 × 1 + 50 80 = 50 × 1 + 30
50 = 30 × 1 + 20 30 = 20 × 1 + 10
20 = 10 × 2 + 0 and H.C.F = 10

→ Euclid’s division algorithm can also be extended to all integers.

→ Numbers which can be expressed in the form p/q, where q ≠ 0 and ‘p and q’ are integers are called rational numbers; represented by Q.
Q = { $\frac{p}{q}$ ; q ≠ 0; p, q ∈ Z} .

→ Every rational number can be expressed either as a terminating decimal or as a non-terminating recurring decimal.

→ Numbers which can’t be expressed in p/q form are called irrational numbers represented by S. You may notice that the first letter of surds is ‘S’.
Eg: √2, √3, √5, …….. etc,

→ The combined set of rationals and irrationals is called the set of Real numbers; represented by R.
R = Q ∪ S.

→ Diagramatic representation of the number system:
AP SSC 10th Class Maths Notes Chapter 1 Real Numbers 1
where N = the set of natural numbers; W = the set of whole numbers;
Z = the set of integers; Q = the set of rational numbers;
S = the set of irrational numbers ; R = the set of real numbers.

→ Fundamental Theorem of Arithmetic: Every composite number can be expressed as a product of primes uniquely, (i.e.,) if x is a composite number, then
x = $p_{1}^{l} \cdot p_{2}^{m} \cdot p_{3}^{n}$ …… where p₁, p₂, p₃, ….. are prime numbers and l, m, n, …… are natural numbers.
Eg: 420 = 2 × 210 = 2 × 2 × 105 = 2 × 2 × 3 × 35 = 2 × 2 × 3 × 5 × 7
i. e. 420 = 2² × 3¹ × 5¹ × 7¹ and the factorisation on the R.H.S is unique.
Note: R.H.S is called exponential form of 420.

→ To find the H.C.F. of two or more numbers:
Step (i): Express given numbers in their exponential form.
Step (ii): Take the common bases.
Step (iii): Assign the respective smallest exponent from their exponential forms.
Step (iv): Take the product of the above.
Eg: H.C.E of 60 and 75 is
Step (i) 60 = 2² × 3 × 5 ; 75 = 3 × 5²
Step (ii) 3^O × 5^O [taking common bases]
Step (iii) 3¹ × 5¹ [∵ smallest exponent among 3¹ and 3¹ is 1]
Step (iv) 3 × 5 = 15 [smallest exponent among 5¹ and 5² is 1]
∴ H.C.F = 15
(i.e.) The highest common factor of the given set of numbers is the product of the com¬mon bases with the respective least exponents.

→ To find the L.C.M. of two or more numbers:
Step – 1: Express the given numbers in their exponential forms.
Step – 2: Take every base.
Step – 3: Assign the respective greatest exponent to each base.
Step – 4: Take the product of the above.
Eg: L.C.M. of 60 and 75 is
Step – 1: 60 = 2² × 3 × 5 ; 75 = 3 × 5²
Step – 2: 2^O × 3^O × 5^O
Step – 3: 2² × 3¹ × 5²
Step – 4: 4 × 3 × 25 = 300
L.C.M = 300

→ We may notice that the product of any two numbers N₁ and N₂ is equal to the product of their L.C.M. (L) and H.C.F. (H).
i.e., N₁ . N₂ = L.H

→ Let x = p/q be a rational number. If the numerator p is divided by the denominator q, we get the decimal form of x. The decimal form of x may or may not be terminating, i.e., every rational number can be expressed either as a terminating decimal or a non-terminating decimal. This gives us the following theorems.
Theorem – 1: Let ‘x’ be a rational number when expressed in decimal form, terminates, then x can be expressed in the form p/q where p, q are co-primes and the prime factorization of q is of the form 2ⁿ × 5^m, where n and m are non-negative integers.
Theorem – 2: Let x = p/q be a rational number, where q is of the form 2ⁿ × 5^m then x has a decimal expansion that terminates.
Theorem – 3: Let x = p/q be a rational number, where p, q are co-primes and the prime factorization of q is not of the form 2ⁿ . 5^m (n, m ∈ Z⁺) then x has a decimal expansion which is non-terminating recurring decimal.
Theorem – 4: Let ‘p’ be a prime number. If p divides a² then p divides a, where ‘a’ is a positive integer.

→ If a is a non-zero rational number and b is any irrational number, then (a + b), (a – b), a/b and ab are all irrational numbers.

→ Properties of Real Numbers:
If a, b and c are any three real numbers we may notice that

a + b is also a real number – closure property w.r.t. addition
a.b is also a real number – closure property w.r.t. multiplication
a + b = b + a – commutative property w.r.t. addition
a . b = b . a – commutative property w.r.t. multiplication
(a + b) + c = a + (b + c) – associative law w.r.t.
addition (a.b).c = a.(b.c) – associative law w.r.t. multiplication
a + 0 = 0 + a = a, where ‘0’ is the additive identity,
a × 1 = 1 × a = a, where 1 is the multiplicative identity,
a + (-a) = (-a) + a = 0 where (a) and (-a) are additive inverse of each other.
a × $\frac{1}{a}$ = $\frac{1}{a}$ × a = 1 where a and $\frac{1}{a}$ are multiplicative inverse of each other.

→ If aⁿ = x, where a and x are positive integers and a ≠ 1, then we define log_ax = n read as logarithm of x to the base a is equal to n.
Eg.: 2⁴ = 16 ⇒ log₂16 = 4

→ log_ax + log_ay = log_axy

→ log_aa = 1

→ log_ax – log_ay = log_a$\frac{x}{y}$

→ log_a1 = 0

→ log_ax^m = m log_ax

→ In general, the bases in the logarithms are 10 (or) e, where e’ is approximated to 2.718.

→ If p is a prime number and p divides a² then p divides ‘a’ also.

AP State Board 9th Class Maths Notes

November 15, 2023November 14, 2023 by AP Board

Students can go through Andhra Pradesh SCERT AP State Board Syllabus 9th Class Maths Notes Pdf in English Medium and Telugu Medium to understand and remember the concepts easily. Besides, with our AP State 9th Class Maths Notes students can have a complete revision of the subject effectively while focusing on the important chapters and topics. Students can also read AP Board 9th Class Maths Solutions for exam preparation.

AP State Board Syllabus 9th Class Maths Notes

These AP State Board Syllabus 9th Class Maths Notes provide an extra edge and help students to boost their self-confidence before appearing for their final examinations.

AP State Board SSC 10th Class Maths Notes

November 15, 2023November 14, 2023 by AP Board

Students can go through Andhra Pradesh SCERT AP State Board Syllabus SSC 10th Class Maths Notes Pdf in English Medium and Telugu Medium to understand and remember the concepts easily. Besides, with our AP State 10th Class Maths Notes students can have a complete revision of the subject effectively while focusing on the important chapters and topics. Students can also read AP SSC 10th Class Maths Solutions for board exams.

AP State Board Syllabus SSC 10th Class Maths Notes

These AP State Board Syllabus SSC 10th Class Maths Notes provide an extra edge and help students to boost their self-confidence before appearing for their final examinations.

AP Board 9th Class Physical Science Solutions Chapter 7 Reflection of Light at Curved Surfaces

November 6, 2023November 5, 2023 by AP Board

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 7 Reflection of Light at Curved Surfaces Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 7th Lesson Reflection of Light at Curved Surfaces

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces Textbook Questions and Answers

Improve Your Learning

Question 1.
“Laws of reflection of light are not applicable to curved surfaces” Is it correct?
Answer:
No, laws of reflection of light are applicable to curved surfaces also.

Question 2.
How do you find the focal length of a concave mirror?
(OR)
What is the name given to distance between pole and focal point? How would you able to find that distance by using an activity? That distance is called focal length.
Answer:

To find the focal length of a concave mirror –

Hold a concave mirror perpendicular to the direction of sunlight.
Take a small paper and slowly move it in front of the mirror and find out the point where you get smallest and brightest spot, which is image of the sun.
The rays coming from sun parallel to the concave mirror are converging at a point.
This point is called Focus or focal point (F) of the concave mirror.
Measure the distance of the spot from the vertex (point P) of the mirror.
This distance gives the focal length of the concave mirror.

Question 3.
Where will the image form when we place an object, on the principal axis of a concave mirror at a point between focus and centre of curvature?
Answer:

When we place an object on the principal axis of a concave mirror at a point between focus and centre of curvature then the image will form at a point beyond the centre of curvature.
The image is enlarged, inverted and real.

Question 4.
Find the distance of the image, when an object is placed on the principal axis at a distance of 10 cm in front of a concave mirror whose radius of curvature is 8 cm.
(OR)
If the object is placed on the principal axis at a distance of 10 cm in front of a concave mirror with curvature is 8 cm, what is the distance of the image?
Answer:
Object distance u = – 10 cm
Radius of curvature (R) = – 8 cm
∴ Focal length f = $\frac{R}{2}=\frac{-8}{2}$ = – 4 cm
Image distance = v = ?

∴ The image distance = 6.67 cm.
i.e., Real image is formed at same side of the mirror.

Question 5.
State the differences between convex and concave mirrors.
(OR)
Distinguish between convex and concave mirrors.
(OR)
Ramu observed a mirror which is used by drivers to see the vehicles behind. Srinu observed a mirror which is used by dentist to see teeth inside the mouth. What are those two mirrors and distinguish them?
Answer:
Those mirrors are (1) Convex, (2) Concave

Convex mirror	Concave mirror
1. After reflection from the mirror the light rays diverge.	1. After reflection from the mirror the light rays converge.
2. The reflecting surface curved outward.	2. The reflecting surface curved inward.
3. Radius of curvature and focal length are taken as positive, in sign conversion.	3. Radius of curvature and focal length are taken as negative, in sign conversion.
4. It’s magnification is positive only.	4. It’s magnification may be both positive and negative.
5. Magnification of convex mirror is in between zero and one.	5. Magnification value of concave mirror having all values.
6. The image formed due to convex mirror is always virtual.	6. The image formed due to concave mirror is generally real. But when an object placed between vertex and focus it is virtual.
7. The image formed due to convex mirror is always diminished.	7. The image formed due to concave mirror is diminished or magnified or same size based on position of object.
8. The image due to convex mirror is always erect.	8. Generally the image is inverted except when the object is placed between P and F.

Question 6.
Distinguish between real and virtual images.
(OR)
Varun sees his image in a concave mirror. When he moves away he was unable to see his image. Write the differences between those two images.
Answer:

Real image	Virtual image
1. Real image is formed due to converging of light rays.	1. Virtual image is formed due to diverging of light rays.
2. This image can be formed on screen.	2. This image cannot be formed on screen.
3. Generally this image is inverted.	3. Generally this image is erect.
4. It cannot be seen in the mirror.	4. It can be seen in the mirror.

Question 7.
How do you get a virtual image using a concave mirror?
(OR)
Ramu moved an object towards a concave mirror. When he reached certain point he was able to see his image. Where does he had placed the object inorder to see the object in the mirror?
Answer:

Place the object at a distance less than the focal length of the mirror.
Draw ray diagram as shown in the figure.
When the rays are extended, they seem to be coming out from a point on the other side of the mirror. (G)
We cannot find the image on the screen, but we can see the image in the mirror. (IG)
This is the virtual, erect and magnified image.

Question 8.
What do you know about the terms given below related to the spherical mirrors?
a) Pole
b) Centre of curvature
c) Focus
d) Radius of curvature
e) Focal length
f) Principal axis
g) Object distance
h) Image distance
i) Magnification
Answer:
a) Pole :
The centre of the spherical mirror is called pole of the mirror.

b) Centre of curvature :
The normal from a concave mirror converge at a point, that point is called centre of curvature.

c) Focus :
A point on the principal axis where a beam of light parallel to the principal axis either converges into or appears to diverge from after reflection from the mirror is called focus (or) focal point (F).

d) Radius of curvature :
The distance between pole and centre of curvature is called radius of curvature. It is denoted by ‘R’.
R = 2f

e) Focal length :
The distance of the focus from the pole is called the focal length (f) of the mirror.

f) Principal axis :
The horizontal line which passes through the centre of curvature and pole.

g) Object distance :
The distance between object and mirror is called object distance (u).

h) Image distance :
The distance between image and mirror is called image distance (v).

i) Magnification :
The ratio of size of image to the size of object is known as magnification.

Question 9.
Write the rules for sign convention.
(OR)
Vinay wants to solve a problem related to mirrors. What are rules to be followed by him in order to solve the problem?
Answer:

Sign conventions used in mirror equation :

All distances should be measured from the pole.
All distances measured in the direction of incident ray are to be taken as positive.
The distances measured in the direction opposite to incident ray are to be taken as negative.
Height of object (h₀) and height of image (h_i) are positive, if measured above principal axis and negative if measured below principal axis.

Question 10.
The magnification produced by a convex mirror is – 1. Do you agree it?
Answer:
The magnification produced by a convex mirror is – 1.1 agree with this statement.

Question 11.
Imagine that spherical mirrors were not known to human beings. Guess the consequences.
(OR)
What are the consequences occur when the spherical mirrors were not known to human beings?
Answer:

Spherical mirrors are used by human beings in their daily life.
Convex mirrors are used in hospitals, hotels, stores, apartments, roads, ATMs, computer monitors, rare-view mirrors in vehicles, and also in sun glasses,
Whereas concave mirrors are used by ENT specialists, dentists to see inner parts of mouth,
Car head lights are used to get parallel beam of light,
Solar cookers are also made from concave mirrors, So everywhere we are using these spherical mirrors,
So without these spherical mirrors the life of human beings is miserable,

Question 12.
By observing steel vessels and different images in them, Surya, a third class student, asked his elder sister Vidya some questions. What may be those questions?
(OR)
By observing the image on steel vessels, Surya asked his sister some questions. What may be those questions?
Answer:

Why do we see our image in some vessels which bulged outwards?
Why are we not able to see our image in some vessels which bulged inwards?
Which type of vessels are useful to reflect light on wall and form our image on wall?
Can we see the images in all vessels?
Is these images real or virtual?
The vessels which are bulged inside behaves like which type of mirrors?

Question 13.
How do you form a diminished image by a concave mirror on a screen?
Answer:
If we place an object beyond centre of curvature of a concave mirror, diminished image will be formed between focus and centre of curvature,

Question 14.
How do you find the focal length of a concave mirror in the lab?
Answer:
Aim :
To find out focal length of a concave mirror,

Material required :
A candle, paper, concave mirror, V-stand, measuring tape, or meter scale.

Procedure:
1) Place the concave mirror on V-stand, a candle, and meter scale as shown in figure.

2) Keep the candle at different distances from the mirror (10 cm to 80 cm) along the axis and by moving the paper (screen) find the position where you get the sharp image on paper.

3) Note the distance of candle from the mirror (u) and distance of the image from the mirror (v) in the given table.
Here u = object distance, v = image distance, f = focal length.

4) Find the average of T. This value of T is the focal length of the given concave mirror.

Question 15.
What do you infer from the experiment which you did with concave mirrors and measured the distance of object and distance of image?
Answer:
I observed the following points from the experiment with concave mirrors, to form images at different places, they are :

As the object distance increases, the image distance decreases.
As the object distance increases, the size of image decreases.

Question 16.
Draw a ray diagram to show the process of converging of the parallel beam by a concave mirror by taking four parallel incident rays.
Answer:

Question 17.
Draw suitable rays by which we can guess the position of the image formed by a concave mirror.
Answer:
Suitable rays by which we can guess the position of the image formed by a concave mirror.

Rule 1 :
All the rays, that are parallel to the axis, get reflected such that they pass through the focal point of the mirror.

Rule 2 :
A ray that passes through the focal point of the mirror will, travel parallel to the axis after reflection.

Rule 3 :
A ray, passing through the centre of curvature of the mirror and incidents on the mirror, after reflection will travel along the same line, but in opposition direction.

Question 18.
Show the formation of image with a ray diagram, when an object is placed on the •principal axis of a concave mirror away from the centre of curvature.
Answer:

Question 19.
Make a solar heater / cooker and explain the process of making.
(OR)
Which cooker uses solar energy to cook food and explain its making?
(OR)
Explain the construction and working of a solar cooker.
Answer:

Construction:

Make a wooden / iron frame in dish shape. It is solar cooker.
Cut the acrylic mirror sheets into 8 or 12 pieces in the shape of isosceles triangles with a height equal to the radius of a dish antenna.
The bases of 8 or 12 triangles together make the circumference of the dish.
Stick the triangle mirrors to the dish as shown in figure.
Thus solar heater/cooker is ready.

Working:

Face it towards the sun.
Find its focal point and place a vessel at that point.
It will get heated. Everyone cook rice in that vessel.

Question 20.
To form the image on the object itself, how should we place the object in front of a concave mirror? Explain with a ray diagram.
(OR)
Draw ray diagram to explain the image formation on the concave mirror if the object is in front of it.
Answer:

The object should be placed on centre of curvature of concave mirror.
When the object placed at centre of curva¬ture after reflection from the mirror the light rays converge at centre of curvature.
So a real and inverted image is formed on the object itself. Ray diagram

Question 21.
How do you appreciate the role of spherical mirrors in our daily life?
(OR)
Write the usage of spherical mirror in daily life situations.
Answer:

Spherical mirrors are useful in our daily life in many ways.
Convex mirrors are used as rear view mirrors in cars, scooters, buses, etc. This helps us to see the traffic behind the vehicle, which avoids accidents while taking turns.
Big convex mirrors are used as shop security mirrors.
Concave mirrors are used by dentists, opthamologists, to see the smaller parts of teeth, eyes, and ears.
Concave mirrors are also used in solar heating devices.
Concave mirrors are used as shaving mirrors to see a large image of the chin (or) face.
Concave mirrors are used as doctor’s head mirrors to focus light coming from a lamp on to the body parts of the patient to be examined by the doctor.
So, I appreciate the role of spherical mirrors in our daily life.

Question 22.
How do doctors use concave mirrors?
Answer:

Dentists and ENT specialist doctors use concave mirrors to get a magnified clearer view of the teeth, skin and ear cavities.
A concave mirror converges the rays of light falling on it in a direction parallel to the principal axis, on to its focus. This is the reason why ENT specialists use it as a head mirror.

Question 23.
Why do we prefer a convex mirror as a rear-view mirror in the vehicles?
Answer:

Convex mirror always gives an erect and diminished image.
The image distance will be smaller than the object distance.
A convex mirror has a wider field of view than a plane mirror.
Hence we prefer convex mirror as rear-view mirror in the vehicles, as we have to observe the vehicles which are at the back of our vehicle.

Question 24.
Complete the table-1 which is related to experiment done by a concave mirror.

Object distance	Image distance	Magnification
1. 10 cm
2. 20 cm
3. 30 cm
4.	40 cm	1

Answer:

Object distance	Image distance	Magnification
1. 10 cm	20 cm	2
2. 20 cm	Infinity	Cannot measure
3. 30 cm	60 cm	2
4. 40 cm	40 cm	1

From the given data, m = $\frac{40 \mathrm{~cm}}{40 \mathrm{~cm}}$ cm = 1.
Hence, R = 40 cm, f = 20 cm.

Question 25.
How can you show the diverging and the converging of light by using laser lights?
Answer:
Aim :
To show diverging and converging of light by using laser lights.

Material required :
Concave mirror, convex mirror, laser lights-2, screen, V-stand, Agarbathi.

Procedure:

Place a concave mirror on a V-stand and place the V-stand on a table.
Take two laser lights.
Focus the light rays of laser lights parallel to the axis of the concave mirror.
The light rays (beams) incident on the concave mirror are reflected back and converge at one point.
Place the screen to catch that converging point of reflected light rays.
Light a Agarbathi near the table.

Observation :
We can observe path of the incident and reflected rays clearly in the smoke of Agarbathi.

Now place a convex mirror on the stand.
Again pass the laser light rays parallel to the principal axis at the convex mirror.

Observation :
We can observe diverging light rays. We cannot catch any converging point on the screen.

Question 26.
Collect information about the history of spherical mirrors in human civilization. Display it in your classroom.
Answer:

The idea of mirror came into existence long back when people saw their images in water, on poluted surfaces, etc.
The history of mirrors starts in the 6000 BC. The earliest man made mirrors were pieces of poluted stone.
The first glass mirrors were produced in 1A.D. by Romans.
The first modern silver-glass mirror was created by Justus Von Liebig, a German chemist in 1835.
The invention of glass blowing method during the 14th century led to the discovery of spherical mirrors, which increased the popularity of glass mirrors.
By the end of 18^th century, decorative mirrors have widely used.
New, cheaper techniques of mirror production in the 19th century led to a great proliferation in their use.

Question 3.
Think about the objects which act as a concave or convex mirrors in your surroundings. Make a table and display it in your classroom.
(OR)
Collect some objects from your surroundings that act as concave, convex mirrors and display them your classroom.
Answer:
Object used as convex and concave mirror in our surroundings.

Convex mirror	Concave mirror
1. Spoon bulged outwards.	1. Spoon bulged inwards.
2. Outer surface of cooking vessel.	2. Inner surface of cooking vessel.
3. Sink outwards.	3. Sink (wash basin) inwards.
4. Glass outwards.	4. Glass inwards.

Question 4.
How will our image be in concave and convex mirrors? Collect photographs and display in your classroom.
Answer:

In concave mirrors our image is thin and enlarged.
As we move away from mirror the image will be diminished and become pointed at the pole.
In convex mirror, our image is bulged and size of image is diminished.
As we move away from the mirror, the image is further diminished.

Question 5.
How do you appreciate the use of reflection of light by a concave mirror in making of TV antenna dishes?
Answer:

The parabolic shape of dish reflects the signals to the dish’s focal point.
Mounted on brackets at the dish’s focal point is a device called a feed horn.
This feed horn is a wave guide that gathers the signals at or near the focal point and conducts them to a LNB (Low-Noise Block down converter).
The LNB converts these electromagnetic waves into electrical signals and shifts to the receiver i.e. T.V. set.
This is all possible only with the help of parabolic dish antennas (concave shape).
We know, the T.V. plays an important role in our daily life. So the inventions of the utilities of concave shapes are more helpful for us.

Question 6.
A convex mirror with a radius of curvature of 3 m is used as rear view in an automobile. If a bus is located at 5 m from this mirror, And the position, nature and size of the image.
Answer:

Position = 1.15 m behind the mirror. Nature = Virtual and erected.
∴ Image is erect and smaller in size by a factor of 0.23.

Question 7.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
(OR)
An object 5 cm high is placed at a distance of 10 cm from a convex mirror, the focal length of the mirror is 15 cm. Find the nature, position and size of the image.
(v = +6m ; m = 0.6, h₂ = 3 cm)
Answer:
Given, u = – 10 cm ; f = 15 cm ; v =?

So the Image is virtual and seen in the mirror.
Magnification m = $-\frac{v}{u}=\frac{-6}{-10}=\frac{6}{10}$ = 0.6
Position : 6 cm behind the mirror.
Nature : Virtual and erect image.
So the image is erect and diminished.

Question 8.
Write answers to the following questions based on the data given in the table – 3.
1) What changes will come gradually in the size of an image, when we move an object away from the concave mirror?
2) In which situation inverted images formed by a concave mirror?
3) If the centre of curvature of a concave mirror is 10 cm, where the object should be placed to get an image at centere of curvature?
Answer:
1) a) If we move an object from a concave mirror to its focus (F), we will get a virtual image.
b) If we place the object at ‘F, the size of the image is infinity.
c) If we move, the object from ‘F to centre of curvature the size of the image is decreases gradually but the size of the image is bigger than that of the size of the object.
d) At ‘C the size of the image is same as the size of the object.
e) From ‘C’ to infinity distance the size of the image is smaller than the siz4 of$he object and decreases gradually.
f) Hence, the size of the image decreases if we move an object away from the focus fo the concave mirror.

2) If we place an object beyond focus (F) inverted image will be formed.

3) At 10 cm distance (or) at u = 10 cm.

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces InText Questions and Answers

9th Class Physical Science Textbook Page No. 111

Question 1.
Did you burn a paper by using a magnifier?
Answer:
Yes.

Question 2.
What causes burning of the paper, while doing it?
Answer:
Converging light rays through magnifier at one point on the paper.

Question 3.
Can you use a plane mirror to burn a paper instead of a magnifier lens?
Answer:
Cannot.

Question 4.
Can’t? Why?
Answer:
Plane mirror cannot converge light rays.

Question 5.
Which type of mirrors may be used to converge light rays?
Answer:
Concave mirrors.

Question 6.
Which type of reflected surface used by Archimedes?
Answer:
Curved reflected surfaces.

9th Class Physical Science Textbook Page No. 112

Question 7.
Can we burn a paper by using light rays of the torch light?
Answer:
Cannot.

Question 8.
Can’t? Why?
Answer:
The light rays coming from the torch light are not parallel rays. Hence, they cannot converge at one point on the paper with high intensity.

Question 9.
Which type of incident rays are required to get high intensity and point sized converging point?
Answer:
Parallel light rays.

Question 10.
Which line is taken as a base to draw a reflected ray to the given incident ray?
Answer:
Normal taken as base.

9th Class Physical Science Textbook Page No. 114

Question 11.
Does a concave mirror from image at focus in all situations?
Answer:
No

9th Class Physical Science Textbook Page No. 114

Question 12.
See the given figure, a set of parallel rays are falling on a convex mirror. What conclusions can you draw from this?

Answer:
The parallel rays are diverging after reflection. If we extend the reflected rays backwards, they meet at F which is focal point of convex mirror. The image formed is virtual, straight and point sized.

Question 13.
Will you get a point image if you place a paper at the focal point?
Answer:
We can get highly enlarged image at infinity when we place paper at focal point.

9th Class Physical Science Textbook Page No. 118

Question 14.
How can you find the focal length of a convex mirror?
Answer:

The focal length of a convex mirror can be determined by introducing a convex lens between the object and the convex mirror.
An image can be obtained with the help of a convex lens side by side with object when the convex mirror reflects the rays along the same path.
Then, the radius of curvature (R) of the mirror is the distance between the screen and the mirror.

The focal length ‘f of the convex mirror is calculated using the formula R (Radius of curvature)

Question 15.
Can we get a magnified image by a convex mirror?
Answer:
No, convex mirror cannot form magnified image.

9th Class Physical Science Textbook Page No. 119

Question 16.
How can one see an image formed on the object itself?
Answer:
When an object placed at centre of curvature the image should form at C in inverted position. So the image formed on the object itself.

Question 17.
Do you get an image when object is placed at F? Draw a ray diagram. Do the experiment.
Answer:

When an object placed at F, we can get the image at an infinite distance.

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces Activities

Activity – 1

1. Observation the light rays on concave and convex surfaces.
Answer:
Procedure :

Take a rectangular shaped (3″ x 6″) Acrylic sheet.
Hold the Acrylic sheet with your palm without bending it.
Ask your friend to focus light with a torch light on the sheet.
Now adjust the sheet as reflected rays are caught on a wall.
Now bend slowly your palm as the sheet bends inwards in the shape of concave.
Again ask your friend to focus the light.
Now bend slowly your palm as the sheet bends outwards in the shape of convex.
Again ask your friend to focus the light.

Observation :

In the first case, when without bending the sheet, reflected light is not converged at one place. Because the sheet was acted as a plane mirror.
In the second case, reflected light converged at one place. Because the acrylic sheet is bent inwards and it is acted as a concave mirror.
In the third case, reflected light rays are not converged, of course they diverged with low intensity. Because the sheet was acted as a convex mirror.

Conclusion :
Concave surfaces converge light rays and convex surfaces diverge.

Activity – 2

Question 2.
Show that a powerful source far away is needed to get parallel rays of light.
Answer:

Stick two pins on a thermocole block.
The pins are exactly parallel to each other.
When a source of light is kept very near, we see the shadows diverging (from the base of the pins).
As we move the source away from the pins, the angle of divergence gets reduced.
If we move the source far away we will get parallel shadows. But the light intensity reduces.
That means to get a beam of parallel rays all that we need is a powerful source far away.

Activity – 3

Question 3.
Identify the focus or focal point of concave mirror.
(OR)
How do you able to find the focal point of a given concave mirror? Explain it with an activity.
Answer:

Hold a concave mirror perpendicular to the direction of sunlight.
Take a small paper and slowly move it in front of the mirror and find out the point where we get smallest and brightest spot image of the sun.
The rays coming from the sun parallel to the concave mirror are converging at a point.
This point is called Focus or focal point (F) of the concave mirror.

Activity – 4

Question 4.
Find the normal to a curved surface.
(OR)
How do you find normal to a curved surface ? What is the name given to wavering point of normals, explain.
Answer:

Take a small piece of thin foam or rubber (like the sole of a slipper).
Put some pins in a straight line on the foam.
All these pins are perpendicular to the plane of foam.
If the foam is considered as a mirror, each pin would represent the normal at that point.
Any ray incident at the point where the pin makes contact with the surface will reflect at the same angle as the incident ray made with the pin-normal.
If we bend the foam piece inwards, we will notice that all the pins tend to converge at a point.
If we bend the foam piece outwards, the pins seem to move away from each other or they diverge.
This gives us an idea about spherical mirror.
A concave mirror will be like the rubber sole bent inwards (fig. b) and the convex mirror will be like the rubber sole bent outwards (fig. c).

For a concave mirror, like these pins all normals will converge towards a point. This point is called centre of curvature (C) of the mirror.
For the ray R, the incident angle is the angle it makes with the radius (normal) shown as Zi and the reflected angle is shown as Zr. We know by first law of reflection Zi = Zr.

Lab Activity

Question 5.
Write the experimented method in measure the distances of object and image using concave mirror. And write the table for observations.
(OR)
Write an experimental activity which gives the information about types of images and measuring the object distance and image distance.
Answer:
Aim :
Observing the types of images formed by a concave mirror and measuring the object distance and image distance.

Material required :
A candle, paper, concave mirror, V-stand, measuring tape or meter scale.

Procedure:

1) Place the concave mirror on V-stand, a candle and meter scale as shown in figure.
2) Keep the candle at different distances from the mirror (10 cm to 80 cm) along the axis and by moving the paper find the position where we get sharp image on paper.

3) Note down the observations in table -1

4) Since we know the focal point and centre of curvature, we can re-classify the above observations as shown in the table – 2.

5) Then note down the observations in table – 2

AP Board 9th Class Physical Science Solutions Chapter 6 Chemical Reactions and Equations

November 6, 2023November 5, 2023 by AP Board

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 6 Chemical Reactions and Equations Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 6th Lesson Chemical Reactions and Equations

9th Class Physical Science 6th Lesson Chemical Reactions and Equations Textbook Questions and Answers

Improve Your Learning

Question 1.
What is a balanced chemical equation? Why should chemical equations be balanced?
Answer:
1) A chemical equation in which the number of atoms of different elements on the reactants side (left side) are same as those on product side (right side) is called a balanced chemical equation.
Ex : Zn + 2HCl → ZnCl₂ + H₂ ↑

2) All the chemical equations must balance because atoms are neither created nor destroyed in chemical reactions.

3) The number of atoms of each element before and after reaction must be the same.

4) According to the law of conservation of mass, the total mass of the substances that are taking part in a chemical reaction must be the same before and after the reaction.

Question 2.
Balance the following chemical equations.
a) NaOH + H₂SO₄ → Na₂SO₄ + H₂O
b) Hg (NO₃)₂ + KI → Hgl₂ + KNO₃
c) H₂ + O₂ → H₂O
d) KClO₃ → KCl + O₂
e) C₃H₈ + O₂ → CO₂ + H₂O
Answer:
a) 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
b) Hg (NO₃)₂ + 2 KI → Hgl₂ + 2KNO₃
c) 2H₂ + O₂ → 2H₂O
d) 2KClO₃ → 2KCl + 3O₂
e) C₃Hg + 5O₂ → 3CO₂ + 4H₂O

Question 3.
Write the balanced chemical equations for the following reactions.
a) Zinc + Silver nitrate → Zinc nitrate + Silver
b) Aluminium + Copper chloride → Aluminium chloride + Copper
c) Hydrogen + Chlorine → Hydrogen chloride
d) Ammonium nitrate → Nitrous Oxide + Water
Answer:
a) Zn + 2AgNO₃ → Zn(NO₃)₂ + 2Ag
b) 2Al + 3CuCl₂ → 2AlCl₃ + 3Cu
c) H₂ + Cl₂ → 2HCl
d) NH₄NO₃ → N₂O + 2H₂O

Question 4.
Write the balanced chemical equations for the following and identify the type of reaction in each case.
a) Calcium hydroxide(aq) + Nitric acid(aq) → Water(l) + Calcium nitrate(aq)
b) Magnesium(sJ + Iodine → Magnesium Iodide(s)
c) Magnesium(s) + Hydrochloric acid(aq) → Magnesium chloride(aq) + Hydrogen^
d) Zinc(s) + Calcium chloride(aq) → Zinc Chloride(aq) + Calcium(s)
Answer:
a) Ca(OH₂) + HNO₃ → H₂O + Ca(NO₃)₂
This is double decomposition reaction.

b) Mg + I₂ → Mgl₂
This is chemical combination reaction.

c) Mg + 2HCl → MgCl₂ + H₂ ↑
This is chemical displacement reaction.

d) Zn + CaCl₂ → ZnCl₂ + Ca
This is chemical displacement reaction. This reaction is not possible, because calcium is more reactive than zinc.

Question 5.
Write an equation for decomposition reaction where energy is supplied in the form of heat / light / electricity.
Answer:
Chemical decomposition reaction : A chemical reaction in which a single substance splits into two or more substances is called chemical decomposition.
For decomposition reaction energy is supplied in the form of :

Question 6.
What do you mean by precipitation reaction?
Answer:
A reaction in which insoluble substance in water is formed as product is called precipitation reaction.

Question 7.
How does chemical displacement reactions differ from chemical decomposition reaction? Explain with an example for each.
Answer:

Question 8.
Name the reactions taking place in the presence of sunlight.
Answer:
1) In the presence of sunlight plants prepare their food by taking C02 from the air and H20 from the soil with their chloroplasts of the green leaves. This reaction is called photosynthesis.

Question 9.
Why does respiration considered as an exothermic reaction? Explain.
Answer:

We need energy to stay alive.
We get this energy from food we eat.
During digestion, food is broken down into simpler substances.
For example, rice and potato contains starch.
The starch breaks down to form glucose.
This glucose combines with oxygen in the cells of our body and releases energy, which helps to do the various works.
During this process, energy is given out. Hence this reaction can be called exothermic reaction.
The special name of this reaction is respiration.
So respiration is considered as exothermic reaction.
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + Q (Energy)

Question 10.
What is the difference between displacement and double displacement reactions? Write equations for these reactions.
Answer:

Chemical displacement reaction	Chemical double displacement reaction
1. In a reaction when an active element displaces less active element from its compound is called displacement reaction.	1. If two reactants exchange their constituents chemically and form two products, then the reaction is called as double displacement reaction.
2. Zn + 2HCl → ZnCl₂ + H₂In this example Zn displaces Hydrogen from HCl.	2. Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl In this reaction SO₄^2- and Cl^– are mutually exchanged.
3. General formula to the reaction is A + BC → AC + B	3. General formula to the reaction is AB + CD → AD + BC

Question 11.
MnOz + 4HCl → MnCl₂ + 2H₂O + Cl₂ ↑
In the above equation, name the compound which is oxidized and which is reduced.
Answer:
In the above equation, HCl compound is oxidized and MnO₂ is reduced.

Question 12.
Give two examples for oxidation-reduction reaction.
Oxidation :
Oxidation is a reaction that involves the addition of oxygen or loss of hydrogen or electrons.

Reduction :
The process in which a substance loses oxygen or gains hydrogen or electrons is known as reduction.

Question 13.
In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write the reaction involved.
Answer:

Cu_(s) + 2 AgNO_3(aq) > Cu(NO₃)_2(aq) + 2Ag_(s)
This is redox reaction, copper is the reducing agent and the silver is reduced.
Electrons from the copper metal are transferred to the silver.
This reaction can also be called a displacement reaction because copper displaces silver as it is more reactive.

Question 14.
What do you mean by corrosion? How can you prevent it?
Answer:

When some metals are exposed to moisture, acids, etc. they tarnish due to the formation of respective metal oxide on their surface. This process is called “corrosion”.
Corrosion can be prevented by painting, oiling, greasing, galvanizing, chrome plating or making alloys.
Galvanizing is a method of protecting iron from rusting by coating them a thin layer of zinc.

Question 15.
Explain rancidity.
Answer:

Oxidation reactions in food material that were left for a long period are responsible for spoiling of food. This process is called “rancidity”.
When these processes occur in food, undesirable odours and flavours can result.
Rancidity is an oxidation reaction.

Question 16.
Balance the following chemical equations including the physical states,
a) C₆H₁₂O₆ → C₂H₅OH + CO₂
b) Fe + O₂ → Fe₂O₃
c) NH₃ + Cl₂ → N₂H₄ + NH₄Cl
d) Na + H₂O → NaOH + H₂
Answer:
a) C₆H₁₂O_6(s) → C₂H₅OH_(l) + CO_2(g)
b) Fe_(s) + O_2(g) → Fe₂O_3(s)
c) NH_3(aq) + Cl_2(g) → N₂H_4(l) + NH₄Cl_(aq)
d) 2Na_(s) + 2H₂O_(l) → NaOH + H₂

Question 17.
Balance the chemical equation by including the physical states of the substances for the following reactions.
a) Barium chloride and Sodium sulphate aqueous solutions react to give insoluble Barium sulphate and aqueous solution of Sodium chloride.
b) Sodium hydroxide reacts with Hydrochloric acid to produce Sodium chloride and water.
c) Zinc pieces react with dilute Hydrochloric acid to liberate Hydrogen gas and forms Zinc chloride.
Answer:
a) BaCl_2(aq) + Na₂SO_4(aq) → BaSO_4(s)↓ + NaCl_(aq)
b) NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)
c) Zn_(s) + 2HCl_(aq) → ZnCl_2(aq) + H_2(g)

Question 18.
A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Can you predict the element ‘X’ and the black coloured substance formed? How do you support your predictions?
Answer:
The brown coloured element is copper (Cu). On heating copper reacts with oxygen
present in the atmosphere to form copper oxide which is black in colour.
The reaction is shown below.

If we pass hydrogen gas over hot copper oxide we will notice that black coating on copper turns brown because copper oxide loses oxygen to form copper.
This will support our prediction.

Question 19.
Why do we apply paint on iron articles?
Answer:

Ferrous reacts with oxygen in the air and form iron oxide.
2 Fe + O₂ → 2 FeO
This reaction is called corrosion. It spoils the iron articles by rusting.
Corrosion of iron articles can be prevented or minimized by shielding the metal surface from oxygen and moisture.
It can be prevented by applying paint on the articles.

Question 20.
What is the use of keeping food in air tight containers?
Answer:

Oxidation is defined as the interaction of oxygen molecules with all the different substances from metal to living tissue which may come into contact with it.
When fats and oils are oxidized they become rancid. Their smell and taste change.
Keeping food in airtight containers helps to slow down oxidation process.
So, manufacturers of potato chips usually flush bags of chips with gas such as nitrogen to prevent the chips from getting oxidized.

Fill in the Blanks

1. The decomposition of vegetable into compost is an example of ……………. reaction.
2. The chemical reaction in which energy is absorbed to form a new compound is called ………….
3. The reaction 2N₂O → 2N₂ + O₂ is an example for ………….. reaction.
4. The reaction Ca + 2H₂O → Ca(OH)₂ + H₂ ↑ is an example for ………….. reaction.
5. The substances that are present on left side of a chemical equation are called
6. The arrow mark between the products and reactants of a chemical equation shows of the reaction.
Answer:

oxidation
endothermic reaction
decomposition
displacement
reactants
direction

7. Match the following :

1) 2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃	a) combination reaction
2) 2NH₃ → N₂ + 3H₂	b) decomposition reaction
3) C₂H₄ + H₂O → C₂H₆O	c) displacement reaction
4) Fe₂O₃ + 3CO → 2Fe + 3CO₂	d) double displacement reaction

Answer:

1) 2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃	d) double displacement reaction
2) 2NH₃ → N₂ + 3H₂	b) decomposition reaction
3) C₂H₄ + H₂O → C₂H₆O	a) combination reaction
4) Fe₂O₃ + 3CO → 2Fe + 3CO₂	c) displacement reaction

Multiple Choice Questions:

1. Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
The above reaction is an example of:
A) Combination reaction
B) Decomposition reaction
C) Displacement reaction
D) Double decomposition reaction
Answer:
C) Displacement reaction

2. What happens when dil. hydrochloric acid is added to iron filings? Choose the correct answer.
A) Hydrogen gas and iron chloride are produced.
B) Chlorine gas and iron hydroxide are produced.
C) No reaction takes place.
D) Iron salt and water are produced.
Answer:
A) Hydrogen gas and iron chloride are produced.

3. 2PbO_(s) + C_(s) → 2Pb_(s) + CO_2(g)
Which of the following statements are correct for the above chemical reaction?
i) Lead is reduced
ii) Carbon dioxide is oxidized
iii) Carbon is oxidized
iv) Lead oxide is reduced
A) (i) and (ii)
B) (i) and (iii)
C) (i), (ii) and (iii)
D) all
Answer:
B) (i) and (iii)

4. The chemical equation
BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl represents following type of chemical reaction.
A) displacement
B) combination
C) decomposition
D) double-displacement
Answer:
D) double-displacement

5. The reaction of formation hydrogen chloride from hydrogen and chlorine repre¬sents following type of chemical reaction i
A) decomposition
B) displacement
C) combination
D) double-displacement
Answer:
C) combination

9th Class Physical Science 6th Lesson Chemical Reactions and Equations InText Questions and Answers

9th Class Physical Science Textbook Page No. 90

Question 1.
What changes do you notice generally?
Answer:
I noticed two types of changes generally. They are :

Physical change,
Chemical change.

Question 2.
“Coal is burnt”, “crackers are burnt” ………. changes Are they physical changes (or) chemical changes?
Answer:
They are all chemical changes.

Question 3.
Are they (coal, crackers) temporary changes or permanent changes?
Answer:
They are permanent changes.

Question 4.
How do we know a chemical reaction has taken place?
Answer:
While we observe the following things, we can conclude that a chemical reaction has taken place.

A change that changes state and colour of substance.
A change that releases heat energy.
A change which forms an insoluble substance as precipitate.
A change that liberates a gas.

9th Class Physical Science Textbook Page No. 92

Question 5.
Do the atoms of each element on left side equal to the atoms of the element on the right side of the equation?
Answer:
Yes. The atoms of each element in left side are equal to the atoms of their corresponding element.

9th Class Physical Science Textbook Page No. 94

Question 6.
Are the atoms of all elements of reactants present in products?
Answer:
Yes. The atoms of all elements of reactants are present in products.

Question 7.
2 C₃H₈ + 10 O₂ → 6 CO₂ + 8 H₂O.
Is it a well balanced equation as per rules ? How do you say?
Answer:
Yes, it is a balanced equation.

L.H.S.	R.H.S.
Number of ‘C’ atoms = 6	Number of ‘C’ atoms = 6
Number of ‘H’ atoms = 16	Number of ‘H’ atoms = 16
Number of ‘O’ atoms = 20	Number of ‘O’ atoms = 20

9th Class Physical Science Textbook Page No. 105

Question 8.
Did you notice the colour coating on silver and copper articles?
Answer:
Yes. I noticed. To prevent from corrosion they are colour coated like that.

9th Class Physical Science Textbook Page No. 106

Question 9.
How can we prevent the spoiling of food?
Answer:
The spoilage of food can be prevented by adding preservatives like vitamin C and vitamin E.

9th Class Physical Science 6th Lesson Chemical Reactions and Equations Activities

Activity – 1

Question 1.
Write an activity when calcium oxide reacts with water.
What type of reaction is this ? Write balanced chemical equation.
Answer:

Take about 1 gm of quick lime (CaO) in a beaker.
Add 10 ml of water to this.
Touch the beaker with your finger.
The beaker is hot.
The reason is that the calcium oxide reacts with water and releases heat energy.
Calcium oxide dissolves in water producing colourless solution [Ca(OH)2],
Dip the red litmus paper in it.
Red litmus paper changes into blue colour.
We conclude that the solution obtained is a basic solution.
It is a chemical combination reaction.
CaO_(s) + H₂O_(l) → Ca(OH)₂

Activity – 2

Question 2.
What change do you observe by mixing of sodium sulphate solution and barium chloride solution?
(OR)
Which type of reaction is this (mixing of sodium sulphate and barium chloride solutions)?
Answer:

Take about 100 ml of water in a beaker.
Dissolve a small quantity of sodium sulphate (Na₂SO₄) in it.
Take about 100 ml of water in another beaker.
Dissolve a small quantity of barium chloride (BaCl₂) in it.
Add these (Na₂SO₄, BaCl₂) two solutions.
We will get a white precipitate of barium sulphate.
Na₂SO₄ + BaCl₂ → BaSO₄↓ + 2NaCl
It is a double displacement reaction.

Activity – 3

Question 3.
Formation of H₂ gas by the action of dil. HCl and Zn pieces.
(OR)
What happens if dilute HCl is added to zinc granules. Explain the process with an experiment. What type of reaction is this? Write the balanced chemical equation for this process.
Answer:

Hydrogen gas experiment:

Take a few zinc granules in a conical flask.
Add about 5 ml of dilute hydrochloric acid to the conical flask.
Observe the changes in the conical flask.
Keep a burning matchstick near the mouth of the conical flask.
The light of burning matchstick is put off with ‘pop’ sound.
This indicates the H₂ gas has released in this reaction.
When we touch the bottom of the conical flask with our finger, we feel hot.
So, this reaction produces heat.
It is a displacement reaction
Zn_(s) + 2HCl_(aq) → ZnCl_2(aq) + H_2(g)

Activity – 4 Chemical Combination

Question 4.
Write an activity on burning of magnesium ribbon in the presence of air.
(OR)
Write an activity which shows burning of magnesium ribbon is a chemical combination reaction.
Answer:

1) Take about 3 cm long piece of magnesium ribbon.
2) Rub the magnesium ribbon with sand paper.
3) Hold it with a pair of tongs.
4) Burn it with a spirit lamp or burner.
5) Magnesium burns in oxygen by producing dazzling white flame and changes into white powder (magnesium oxide).

7) In this reaction magnesium and oxygen combine to form a new substance magnesium oxide.
8) A reaction in which single product is formed from two or more reactants is known as chemical combination reaction.

Activity – 5 Decomposition Reaction

Question 5.
How can you prove that CO₂ is released on heating CaCO₃?
(OR)
When calcium carbonate is heated it releases certain gas. What is the gas that has been released? How do you identify that gas? Write the activity of this reaction.
Answer:

Take a pinch of Calcium Carbonate (lime stone) in a boiling tube.
Heat the boiling tube over the flame of spirit lamp or burner.
Now take a burning matchstick near the mouth of delivery tube.

Heating of calcium carbonate and testing the gas evolved with burning matchstick

We can observe that the matchstick kept near the mouth of the boiling tube will be put off.
In the above activity, on heating CaC03, it decomposes to Calcium oxide (CaO) and Carbon dioxide (CO₂).
When a decomposition reaction is carried out by heating, it is called thermal decomposition reaction.

Activity – 6

Question 6.
Write an activity to show that nitrogen dioxide (NO^ is released by heating of lead nitrate.
(OR)
Write an activity which shows thermal decomposition reaction.
Answer:

Take about 0.5 g of lead nitrate powder in a boiling test tube.
Hold the boiling tube with a test tube holder.
Heat the boiling tube over a flame.
We observe that on heating of lead nitrate brown fumes of nitrogen dioxide are released.
On heating lead nitrate, it decomposes and gives lead oxide, oxygen and nitrogen dioxide.

6) Reaction :

7) This is also an example for thermal decomposition reaction.

Activity – 7

Question 7.
Write an activity to show dissociate of water into hydrogen and oxygen.
(OR)
Write an activity showing decomposition of a compound in the presence of electricity.
(OR)
Draw a neat diagram representing electrolysis of water.
How do you prove that water contains hydrogen and oxygen in the ratio of 2 : 1?
(OR)
Explain electrolysis of water.
(OR)
What are the materials required for the experiment to show the chemical decomposition of water ? Write the procedure of the experiment. Name the products which we get in this reaction.
(OR)
Draw a neat diagram showing the electrolytic decomposition reaction of water. Write the balanced chemical equation of the above reaction.
Answer:

Take a plastic mug.
Drill two holes at its base.
Fit two ‘one holed rubber stoppers’ in these holes.
Insert two graphite electrodes in these rubber stoppers.
Connect the electrodes to 9V battery.
Fill the mug with water, so that the electrodes are immersed.
Add few drops of dilute sulphuric acid to water.
Take two test tubes filled with water and invert them over the two graphite electrodes.
Switch on the current and leave the apparatus undisturbed for sometime.
We will notice that the liberation of gas bubbles at both the electrodes.
These bubbles displace the water in the test tubes.
We also observe that the volume of gas collected in the both test tubes is different.
Once the test tubes are filled with gases take them out carefully.
Test both the gases separately by bringing a burning candle near the mouth of each test tube.
The gas which occupies high volume makes to burn with blue flame and put off candle flame with ‘puf sound is hydrogen gas.
The gas which occupies low volume and makes to burn candle brightly is oxygen gas.
In the above activity on passing the electricity, water dissociates to hydrogen and oxygen gases. This is called electrolytic composition reaction.

Activity – 8

Question 8.
Write an activity to observe silver bromide decomposes in the presence of sunlight.
(OR)
Write an activity for photochemical reaction.
Answer:

Take some quantity of silver bromide on a watch glass.
When we observe the colour of silver bromide it is light yellow in colour.
Place the watch glass in sunlight for sometime.
Silver bromide decomposes to silver and bromine in sunlight.
After sometime we observe that the colour of silver bromide changes into grey colour.

Light yellow coloured AgBr changes into Ag and Br.
This decomposition reaction occurs in the presence of sunlight and such reactions are called photochemical reactions.
The above decomposition reactions require some energy to convert reactants to products.
This type of reactions are called endothermic reactions.

Activity – 9

Question 9.
Write an activity to show that zinc displaces hydrogen from dil. hydrochloric acid.
Answer:

Take a small quantity of zinc dust in a conical flask.
Add dilute hydrochloric acid slowly.
Now take a balloon and tie it to the mouth of the conical flask.
We can observe that the gas bubbles coming out from the solution and the balloon bulges out.
Zinc pieces react with dilute hydrochloric acid and liberate hydrogen gas.
The element zinc has displaced hydrogen from dilute hydrochloric acid. This is one of the examples for displacement reaction.
Equation : Zn_(s) + 2 HCl_(aq) → ZnCl_2(aq) + H_2(g) ↑

Activity – 10

Question 10.
Write an activity to show that iron displaces copper from copper sulphate.
(OR)
Write an activity about how you conduct an experiment to show that more reactive metals replace less reactive metals from their compounds. (OR)
Why is an iron nail kept in a solution brownish? Explain the activity.
Answer:

Take two iron nails and clean them by rubbing with sand paper.
Take two test tubes and mark them as A and B.
Take about 10 ml of copper sulphate solution in each test tube.
Dip one iron nail in copper sulphate solution of test tube A.
Keep the test tube with iron nail undisturbed for 20 minutes.
Keep the other iron nail and test tube aside. ;
Compare the colours of the solutions in the test tubes.
Now take out the iron nail from copper sulphate solution.
Keep the iron nail and test tubes A and B side by side.
Compare with the other iron nail that has been kept aside.
We will observe that the iron nail dipped in copper sulphate solution becomes brownish.
Blue colour of copper sulphate solution in test tube ‘A’ fades.
Iron is more reactive than copper, so it displaces copper from copper sulphate.
This is one of the examples for displacement reaction.
Equation : Fe_(s) + CuSO_4(aq) → FeSO_4(aq) + Cu_(s)

Activity – 11 Double displacement reaction

Question 11.
Write an activity for the formation of lead iodide and potassium nitrate.
(OR)
Your friend has a doubt about chemical double displacement reaction. How can you clarify his/her doubt by showing an experiment? Explain.
(OR)
What happens if lead nitrate solution is added to potassium iodide solution? Explain the activity.
Answer:

It forms yellow precipitate of lead iodide.

Take a pinch of lead nitrate and dissolve in 5.0 ml of distilled water in a test tube.
Take a pinch of potassium iodide in another test tube and dissolve in distilled water.
Mix lead nitrate solution with potassium iodide solution.
We observe that a yellow coloured substance which is insoluble in water, is formed.
This insoluble substance is known as precipitate.
The precipitate is Lead Iodide.
Equation : Fb(NO₃)_2(aq) + 2 KI_(aq) → Pbl_2(s) + 2KNO_3(aq)
In the above reaction, lead ion and potassium ion exchange their places each other.
Lead ion combines with iodide ion and forms Pbl2 as precipitate and KN03 remains in the solution.
Such reaction is double displacement reaction.

Activity – 12

Question 12.
Write an activity to show that the oxidation of copper to copper oxide.
(OR)
Write an activity for example of oxidation and reduction.
Answer:

Take about 1.0 gram of copper powder in a China dish.
Keep the China dish on a tripod stand containing wire gauze.
Heat it with a bunsen burner or with a spirit lamp.
We can find that the surface layer of copper becomes black.
On heating copper reacts with oxygen present in the atmosphere to form copper oxide.

Here copper combines with oxygen to form copper oxide.
Here oxygen is gained and the process is called oxidation.
Now pass hydrogen gas over hot copper oxide.

We can observe that the black coating on copper turns brown because copper oxide loSes oxygen to form copper.
In this process oxygen is lost and the process is called ”Reduction”.

AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies

November 4, 2023November 3, 2023 by AP Board

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 9th Lesson Floating Bodies

9th Class Physical Science 9th Lesson Floating Bodies Textbook Questions and Answers

Improve Your Learning

Question 1.
A solid sphere has a radius of 2 cm and a mass of 0.05 kg. What is the relative density of the sphere? (AS 1)
Answer:
Radius of the sphere = 2 cm

Question 2.
A small bottle weighs 20 g when empty and 22 g when filled with water. When it is filled with oil it weighs 21.76 g. What is the density of oil? (AS 1)
Answer:
Weight of water = 22 – 20 = 2 gm
Weight of oil = 21.76 – 20 = 1.76 gm

Question 3.
An ice cube floats on the surface of a glass of water (density of ice = 0.9 g/cm3). When the ice melts will the water level in the glass rise? (AS 1)
Answer:
Yes, the water level rises.

Reason :
The ice cube floats on water, because its density is less than the density of water. When ice cube melts, it becomes water, so that the water level rises.

Question 4.
The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g/cm³, will the substance sink or float when placed on the surface of water? What will be the mass of water displaced by the substance? (AS 1)
Answer:

∴ Weight of the water displaced by the substance = 20 g

Question 5.
Find the pressure at a depth of 10 m in water if the atmospheric pressure is 100 kPa. [1Pa = 1 N/m²] [100kPa = 10⁵Pa = 10⁵ N/m² = 1 atm.] (AS 1)
Answer:
Depth ‘h’ = 10 m ; Atmospheric pressure P₀ = 100 kPa
Density of water p = 1 gm/cm³ = 1 kg/m³
Pressure at a depth ‘h’ is P = P₀ + hρg
= 100 + 10 x 1 x 9.8
= 100 + 98
= 198 kPa

Question 6.
Why do some objects float on the water? And some sink? (AS 1)
Answer:

Floating or sinking of objects on water depends on two factors.
a) Relative density
b) Weight of the water displace by the object
If the relative density of an object is greater than 1, the object sinks otherwise it floats.
Eventhough the relative density is greater than 1, if the weight of the water displaced by the object equal to the weight of the object itself, the object floats on the water.

Question 7.
Explain density and relative density and write their formulae. (AS 1)
Answer:
Density :
Density is defined as mass per unit volume.
Density = $\frac{\text { Mass }}{\text { Volume }}$

Unit of density is gm/cm³ or kg/m³.

Relative density :
Relative density of an object is the ratio of density of the object to the density of water.

Question 8.
What is the value of density of water? (AS 1)
Answer:
Value of density of water = 1 gm/cm³ (or) 1 kg/cm³.

Question 9.
Find the relative density of wood. Explain the process. (Lab Activity 1) (AS 3)
Answer:
Aim :
To find the relative density of wood.

Materials required :
Overflow vessel, 50 ml measuring cylinder, spring balance, wooden block, water.

Procedure:

Weigh the 50 ml measuring cylinder and note its weight.
Weigh the wooden block and note its weight.
Pour water in the overflow vessel until it starts dripping from its beak.
When water stops dripping from the beak, place the 50 ml measuring cylinder under it.
Slip the wooden block gently into the overflow vessel, ensuring that the water does not splash out.
Once the wooden block is in the overflow vessel, water flows out of the beak and collects in the 50 ml cylinder.
Wait till the flow of water from beak, stops.
Weigh the cylinder with the water that overflowed and record the weight.

Question 10.
Which is denser, water or milk? (AS 2)
Answer:
The density of water is 1 gm/cc and that of milk is 1.02 gm/cc. Hence milk is slightly denser than water.

Question 11.
What is buoyancy? (AS 1)
Answer:
Buoyancy is the upward force that a fluid exerts on an object less denser than itself.
(or)
Buoyancy is the ability of an object to float in a liquid.

Question 12.
Classify the following things into substances having relative density > 1 and relative density < 1. Wood, iron, rubber, plastic, glass, stone, cork, air, coal, ice, wax, paper, milk, kerosene, groundnut oil, soap. (AS 1)
Answer:

Relative density > 1	Relative density < 1
Iron	Wood
Glass	Rubber
Stone	Plastic
Milk	Cork
Soap	Air
	Coal
	Ice
	Wax
	Paper
	Kerosene
	Groundnut oil

Question 13.
How can you appreciate the technology of making ships float, using the material which sink in water? (AS 6)
Answer:

We know that a piece of iron sinks in water.
The relative density of iron is 8.5 which is many times more than water.
A ship that floats on water is made up of tonnes of iron. It is really wonderful.
According to the Archimedes principle of buoyancy, any object can float when its weight is equal to the weight of water displaced by it.
Hence ships are made with a larger surface area so that it displaces the water, whose weight is equal to the weight of loaded ship.
It involves very sharp scientific calculations and a large engineering technology.
Really such type of technologies are highly appreciable and the scientists who formulates all these are also really great.

Question 14.
Can you make iron float? How? (AS 3)
Answer:
Yes, we can make iron to float on water.
Procedure:

Take a piece of iron and drop it in a vessel of water.
We observe that the iron piece sinks in water.
Take a thin foil of iron and fold it into four folds.
Drop it in water. It sinks.
Now unfold the foil and bend it in the form of a bowl. [You can use an iron tin also]
Now drop the bowl in water.
If floats on water.

Reason:
The weight of water displaced by iron bowl (iron tin) is less than the weight of the iron bowl (iron tin).

Question 15.
How can you find the relative density of a liquid? (Lab Activity – 2) (AS 3)
Answer:
Aim :
To find the relative density of a liquid.

Materials required :
Small bottle of 50 ml capacity (the bottle should weigh not less than 10 gm), spring balance, any liquid (milk or oil or kerosene) about 50 ml.

Procedure:

Find the weight of empty 50 ml bottle.
Fill the bottle with water and weigh it.
Find the weight of 50 ml water.
Remove water from the bottle and fill it with any liquid (say milk).
Weigh the bottle with liquid.
Weight of 50 ml liquid = Weight of bottle with liquid – Weight of empty bottle

Question 16.
Find the relative density of different fruits and vegetables and make a list. (AS 3)
Answer:

Follow the procedure in Q.No (11) of A.S – (3).
Use different fruits and vegetables in the place of wooden block.
Write the observed values in the following table.

Name of the fruit or vegetable	Relative density
Cabbage	0.36
Cauliflower	0.26
Bottleguard	0.56
Potato	0.67
Onion	0.59
Chilli	0.29
Bitterguard	0.4
Apple	1.22
Grape	1.04
Orange	0.34

Question 17.
Make a lactometer with ball point refill. What would you do to make the refill stand vertically straight? (Activity – 2) (AS 5)
Answer:

Take an empty ball pen refill. It should have a metal point.
Take a boiling tube and fill it with water.
Put the refill in with metallic point inside the water.
Use a pen to mark the point on the refill to show the part which is above the water surface.
Pour out the water from the boiling tube and fill it with milk.
Float the refill in the milk.
Puf the record mark on the refill, at the point showing the part which is above the surface of the milk.
These two marks are not at the same place. This is the improvised lactometer.
We have to attach a small weight at the bottom of the refill to make it stand vertically straight.

Question 18.
Draw the diagram of a mercury barometer. (AS 5)
Answer:

Question 19.
Write a note on Pascal’s discovery in helping to make hydraulic jacks. (AS 6)
Answer:
Pascal’s principle:
External pressure applied to an enclosed body of fluid is transmitted equally in all directions throughout the fluid volume and the walls of the containing vessel.

Use :

This principle is used in designing and working of hydraulic jacks.
Hydraulic jacks are useful to lift heavy objects like cars and other vehicles in automobile work shops when the vehicles are to be required or repaired.
Here very less force is used to lift such heavy vehicles.

Appreciation :

Hydraulic jacks are used not only in automobile work shops, but also in so many industries where heavy weights are to be lifted with a little force.
All the comforts which we are enjoying now are the efforts of scientists who discovered the laws and principles.
Hence we have to appreciate the efforts of Pascal for his contribution in designing these principles.

Question 20.
Write a note on Archimedes discovery of force of buoyancy. (AS 6)
Answer:
Archimedes principle:
When a body is immersed in a fluid, it experiences an upward force of buoyancy equal to the weight of fluid displaced by the immersed portion of the body.

Use :
The principle is used to determine the purity of metals.

Appreciation :

This principle was discovered by Archimedes eventually when he was taking bath.
With this principle he could solve the problem assigned by the king to him.
So many problems in our life may be solved by so many scientific principles discovered by various scientists.
You might have heard about falling of the statue of Budha in Hussain Sagar.
That statue was lifted by using the principle of buoyancy.
Archimedes is thought to be so important as a mathematician that scientists honoured him.
a) A large hole or crator on the moon is named after Archimedes.
b) Some mountains on the moon are called the monte – Archimedes.
Hence the efforts of Archimedes in discovering such type of principle may be highly appreciated.

Question 21.
You found the relative densities of some solids and some liquids by doing some activities. List the solids and liquids in increasing order of relative density. (AS 4)
Answer:

Substance	Relative density
Kerosene	0.81
Rubber	0.94
Milk	1.02
Glass	1.29
Iron	8.5

Question 22.
Iron sinks in water, wood floats in water. If we tie an iron piece to wood piece of the same volume, buritlle and drop it in water, would bundle sink or float? Make a guess and find out whether your guess is correct or wrong with an experiment. Give reasons. (AS 2, AS 3)
Answer:
The body sinks in water.

Reason :
The combined mass of the system increases, so the combined density also increases. Hence the body sinks in water.

Question 23.
Air brakes in automobiles work on Pascal’s principle. What about air brakes? Collect the information about the working process of air brakes. (AS 4)
Answer:

Air brakes works on the principle of conversion of energy. Generally in trains, while it is moving -fit .produces kinetic energy. This kinetic energy has to be reduced to make the train stop. Here air is used to reduce kinetic energy by converting it into heat energy.
The system of air brakes in trains has been shown in the figure.
The important parts are compressor, main reservoir, driver’s brake value, brake pipe, triple value, auxiliary reservoir, brake cylinders, and brake block.

Working:

When the driver placed the value in application position, the air pressure in the brake pipe escapes.
The loss of pressure is detected by the slide value in the triple value.
Now a connection between the auxiliary reservoir and the brake cylinder has been opened and the air in the auxiliary reservoir feeds through into the brake cylinder.
The air pressure forces the piston to move against the spring pressure and causes the brakes to be applied to the wheels.

Question 24.
Where do you observe Archimedes principle in daily life? Give two examples.
Answer:
Daily life application of Archimedes principle :

Archimedes principle of buoyancy is applied in our daily life in many ways.
Fish, human swimmers, ice bergs and ships float follow Archimedes principle of buoyancy.
Rise of balloon in air also follows Archimedes principle.
While dragging water from a well, the bucket filled with water seems to be weightless till it reaches the surface of the water in the well. This is also due to buoyancy.
Swimming of duck in water is also an example of Archimedes principle.

Question 25.
Where do you observe Pascal’s principle in daily life? Give a few examples.
(OR)
Write any one application of Pascal’s principle in daily life.
Answer:
Daily life application of Pascal’s principle :
Pascal’s principle is applied in the working of

Hydraulic jacks
Hydraulic lifts
Hydraulic pumps
Hydraulic cranes
Siphons
Artesian wells
Water towers and dams

9th Class Physical Science 9th Lesson Floating Bodies InText Questions and Answers

9th Class Physical Science Textbook Page No. 141

Question 1.
a) Did kerosene float above the water or did water float above the kerosene?
Answer:
Kerosene floats above the water.

b) Which objects float in kerosene?
Answer:
Plastic buttons, match stick, tiny paper balls, wax, etc. floats in kerosene.

c) Which objects sink in kerosene but float on water?
Answer:
Wax sink in kerosene but float on water.

d) Which objects sink in water?
Answer:
Pins, small pebbles, sand, etc. sink in water.

e) Draw a diagram of the tube, showing the results of your activity.
Answer:

f) Why did different objects behave differently?
Answer:
The density of objects is the main reason to behave differently.

9th Class Physical Science Textbook Page No. 144

Question 2.
1) What is the relative density of wood?
Answer:
0.8

2) What is the relative density of glass?
Answer:
1.2

3) Which is denser, rubber or plastic?
Answer:
Rubber is denser than plastic.

4) Which is denser, wood or cork?
Answer:
Wood is denser than cork.

5) Do objects that have a relative density less than 1 Sink in water or float on it?
Answer:
Float on water.

6) Do the objects that sink in water have a relative density less than 1 or more than 1?
Answer:
More than 1.

7) Classify the above materials as denser than stone and less denser than the stone.
Answer:
Denser than the stone : iron, nails.
Less denser than the stone :
eraser, wood, glass slides, plastic cube, Aluminium, glass marbles, cork.

8) What relationship do you find between the relative density of objects and floating sinking of the objects?
Answer:
If the relative density is less than 1 the object of less density will float above the other in the water.

9th Class Physical Science Textbook Page No. 145

Question 3.
1) Which liquid will float on top if groundnut oil is poured over water?
Answer:
Groundnut oil float on the water.

2) If we put a wooden block in kerosene, will it float or sink? Give reasons for your answer.
Answer:

Wooden block will sink in the kerosene.
Because, the density of the wooden block is more than the kerosene.

3) A piece of wax floats in water but the same piece of wax sinks in a liquid say liquid ‘X’. Will the relative density of liquid . ‘X’ be less than 1 or greater than 1? How can you say?
Answer:

The relative density of liquid ‘X’ is less than 1.
The density of wax is less than 1 because it floats on the water.
The density of liquid ‘X’ is less than the density of the wax this means less than 1. because wax sinks in that liquid ‘X’.

4) If we mix some water in milk, will the relative density of the mixture be less than or more than the relative density of milk?
Answer:

We know that relative density of milk is more than 1 and density of milk is more ‘ than water.
If we mix some water in milk, the density of the mixture will decreases and the relative density will also be less than the relative density of milk.

5) If we take two bottles of equal volume and pour pure milk in one and milk mixed with water in the other, which one will be heavier?
Answer:

Pure milk is heavier than water mix milk.
This is due to density of milk is more than water.

9th Class Physical Science Textbook Page No. 151

Question 4.
Why is the height of mercury column nearly 76 cm in the tube?
Answer:
Air pressure is the weight of air in the atmosphere above the reservoir (bowl of mercury). So, the level of mercury continues to change until the weight of mercury in the glass tube is exactly equal to the weight of the air above the reservoir, which is 76 cm.

9th Class Physical Science Textbook Page No. 154

Question 5.
What happens if we replace this cylindrical liquid column with another object which is made up of a material whose density is equal to the density of liquid?
Answer:
We know that the pressure difference in the liquid,
P₂ – P₁ = hρg

Since F = PA and W = mg
We get F = W (Values of displaced liquid)

Here F’ is the force applied on the object and ‘W’ is the weight of the liquid.
So, the”force applied on the object by the liquid is equal to the weight of the displaced liquid.

9th Class Physical Science Textbook Page No. 155

Question 6.
Why does the stone lose weight when it is immersed?
Answer:

Suspend a stone from a spring balance. Note the reading of the spring balance.
Take a beaker half-filled with water.
Now immerse the stone in the water, note the reading of the spring balance.
We notice that the stone, when immersed appears to lose some weight.
The immersed stone appears to lose weight because the force of buoyancy.
Thus the apparent loss of weight must be equal to the force of buoyancy acting on the immersed stone.

9th Class Physical Science Textbook Page No. 142

Question 7.
Let us suppose you have two blocks and you do not know what material they are made of. The volume of one block is 30 cm3 while the other is 60 cm3. The second block is heavier than the first. Based on this information, can you tell which of the two blocks is denser?
Answer:
No, we cannot say which of the two blocks is denser, because any one of the quantity
i. e., either volume or weight must be same.

9th Class Physical Science Textbook Page No. 153

Question 8.
a) What would happen if Toricelli’s experiment is done on moon?
b) A stopper is inserted in the small hole of the glass tube of the mercury barometer below the top level of the mercury in it. What happens when you pull out the . stopper from the glass tube?
c) Why don’t we use water instead of mercury in Toricelli experiment? If we are ready to do this experiment, what length of tube is needed?
d) Find the weight of the atmosphere around the earth (take the radius of earth as 6400km)
Answer:
a) If Toricelli’s experiment had been done on moon, the height of mercury column will be zero. Because there is no atmosphere on the moon.

b) The mercury level does not change, because there is vacuum above the mercury level. Hence no pressure is on the mercury. Also, the weight of mercury column must be equal to the air pressure above the reservoir. Hence there will be no change in the height of the mercury column.

c) We cannot use water instead of mercury in Toricelli experiment because, if we want to use water, we have to take the glass tube of length nearby above 10 m, which is inconvenient.
If water is taken,
P₀ = ρhg
1 01 yin5
1.01 × 10⁵ = 1 × 10³ × h × 9.8 ⇒ h = = 0.1030 × 10² = 10.3 m

d) Weight of atmosphere = Atmospheric pressure x Surface area of the earth

9th Class Physical Science Textbook Page No. 157

Question 9.
a) Why is it easier for you to float in saltwater than in freshwater?
b) Why is there no horizontal buoyant force on a submerged body?
c) Two solid blocks of identical size are submerged in water. One block is iron and the other is aluminium. Upon which is the buoyant force greater?
d) A piece of iron when placed on a block of wood, this makes the wood to float lower in the water. If the iron piece is suspended beneath the wood block, would it float at the same depth? Or lower or higher?
Answer:
a) Salt water is denser than freshwater.
b) Buoyant force is the upward force only. The body is submerged means, its weight is more than the buoyant force. Here there will be no horizontal buoyant force.
c) Buoyant force on iron block is more than that of aluminium block.
d) It floats on higher depth than in the first case.

9th Class Physical Science 9th Lesson Floating Bodies Activities

Activity – 1

Question 1.
Comparing density – relative density.
Take two test tubes of the same size and fill one to the brim with water and the other with oil.
a) Which will weigh more?
Answer:
The test tube with oil will weigh more.

b) Which liquid is denser?
Answer:
Oil is denser than water.
Take two equal sized blocks made of wood and rubber.

c) Which of these two blocks is heavier?
Answer:
Wooden block is heavier.

d) Which one is denser?
Answer:
Wooden block is denser than rubber.

Activity – 3

Question 2.
Do the objects denser than water float in it? Prove it with an activity.
Answer:

Collect some objects listed in the table below.
Place them one by one in a glass of water and observe whether they sink or float in water.
Record your observations in the table.

Observations:

Find the relative densities of all the objects.
We observe that some objects floats and some objects like iron nail, glass marble, stone sinks in water.
Geometry box made of iron floats on water, though its relative density is greater than 1. This is due to its larger surface area.
Geometry box made of iron floats on water though it is made up of a substance denser than water.
We can say that the floating or sinking of an object does not depend not only on its relative density, but also its surface area which displaces the water.

Activity – 4

Question 3.
Show that for a floating object, the weight of the object is equal to the weight of water displaced by it.
Answer:

Take a beaker and weigh it. Note down its weight.
Fill water in an overflow jar, wait until the water stops dripping from the outlet of the overflow jar.
Now place the beaker below the outlet of the overflow jar.
Take a wooden block, moisten it with water and gently drop it into the overflow jar.
Water will flow out of the overflow jar and collects in the beaker kept under the overflow jar.
Measure the weight of beaker with water.
Subtract the weight of beaker from this. The value gives the weight of water displaced by wooden block. Note it.
Now remove wooden block from the overflow jar, make it cry and weight.
We can observe that the weight of wooden block is equal to weight of water displaced by it.
Do the same experiment with some other substances and record your observations in the table.

Activity – 5

Question 4.
Explain how does metal aluminium floats on the water with an activity.
Answer:

Take a small sheet of aluminium foil.
Fold it four or five times, pressing the foil tight after each fold.
We can find that the aluminium foil will sink in the water.
Now unfold the aluminium foil and make it as a small bowl.
We can find that the bowl will float on the water.
The metal bowl displace larger amount of water than a metal piece.
Weight of the displaced water is more than the metal sheet.
So, bowl will float on the water.

Activity – 6

Question 5.
Prove that the water exerts upward force on objects.
Answer:

Take an empty plastic bottle.
Put the cap on it tightly.
Place the bottle in a bucket of water.
The bottle will float.
Push the bottle into the water by your hand as shown in figure.
We feel some upward force.
Try to push it further down. We feel increase in the upward force.
Now release the bottle.
It bounce back to the surface of water.
Here the upward force of water a real, observable force.
This force acting on unit area of the surface of an object is called static pressure of the water.

Activity – 7

Question 6.
Describe an activity to observe the air pressure.
Answer:

Take a glass tumbler.
Stick some cotton at the bottom of it.
Immerse it inversely in water upto the bottom of the container as shown in figure.
Take out the tumbler from water.
We observe that the cotton attached at the bottom of the tumbler is not wet.
This is due to the force of air which is applied on water by the air present in the tumbler and stops water from entering the tumbler.
This force on unit area of water is the pressure of air.

Activity – 8

Question 7.
How can you measure the force of buoyancy and how much?
(OR)
Why does and stone lose weight when it is immersed in water?
Answer:

Suspend a stone from a spring balance.
Note that the reading of the spring balance.
The reading gives the weight of the stone.
Take a beaker half filled with water.
Now immerse the stone in the water.
Note the reading of the spring balance.
The reading gives the weight of the immersed stone.
We may notice that the stone, when immersed, appears to lose some weight.
The immersed stone appears to lose weight because the force of BUOYANCY, exerted on the stone by the water in the upward direction.
Thus the apparent loss of weight must be equal to the force of buogancy acting on the immersed stone.
The lose of weight of stone is equal to weight of the water displaced by the stone, this is the force of buoyancy.

Activity – 9

Question 8.
‘State and prove Archimedes principle of buoyancy.
Answer:

Archimedes principle :
Archimedes principle states that when a body is immersed in a fluid, it experiences an upward force of buoyancy equal to the weight of fluid displaced by the immersed portion of the body.

Proof:

Suspend a stone from a spring balance.
Note the reading on the spring balance. This reading is the weight of the stone.
Take an overflow jar with water and place a graduated beaker below the beak as shown in the figure.
Now immerse the stone in the water and note the reading on the spring balance.
Measure the volume of the water that overflows into the graduated beaker.
The reading of the spring balance gives the weight of the immersed stone.
The beaker reading gives the volume of water displaced by the stone.
The difference in the two readings of spring balance gives the apparent loss of weight of the stone.
Now weigh the water in the graduated beaker.
We observe that the apparent loss of weight of the stone is equal to the weight of the water displaced by the stone.
Hence Archimedes principle is proved.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

November 4, 2023November 3, 2023 by AP Board

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 11 Sound Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 11th Lesson Sound

9th Class Physical Science 11th Lesson Sound Textbook Questions and Answers

Improve Your Learning

Pick out the correct answer :

Question 1.
When we say sound travels in a medium (AS 1)
A) the medium travels
B) the particles of the medium travel
C) the source travels
D) the disturbance travels
Answer:
D) the disturbance travels

Question 2.
A sound wave consists of (AS 1)
A) number of compression pulses only
B) number of rarefaction pulses only
C) number of compression and rarefaction pulses one after the other
D) vacuum only
Answer:
C) number of compression and rarefaction pulses one after the other

Question 3.
Hertz stands for oscillations per (AS 1)
A) second
B) minute
C) hour
D) milli second
Answer:
A)second

Question 4.
When we increase the loudness of sound of a TV, the property of sound that changes is (AS 1)
A) amplitude
B) frequency
C) wavelength
D) speed
Answer:
A) amplitude

Question 5.
The characteristic of the sound that describes how the brain interprets the frequency of sound is called (AS 1)
A) pitch
B) loudness
C) quality
D) sound
Answer:
A) pitch

Question 6.
In a stethoscope, sound of heart beats travel through stethoscope tube (AS 1)
A) by bending along the tube
B) in a straight line
C) undergoing multiple reflections
D) all of the above
Answer:
C) undergoing multiple reflections

Question 7.
Explain the following terms : (AS 1)
a) amplitude
b) wavelength
c) frequency
Answer:
a) Amplitude :
The maximum variation in density or pressure from the mean value is called amplitude.
(or)

The maximum disturbance of particles of a medium from their mean position is called amplitude.

b) Wavelength :
The distance between two consecutive compressions or two consecutive rarefactions is called the wavelength of a sound wave, denoted by W. Wavelength is measured in ‘meters’.

c) Frequency :

The number of oscillations of the density of the medium at a place per unit time is called the frequency of the sound wave.
Frequency is denoted by ‘o’.
The S.I. unit of frequency is ‘Hertz’.

Question 8.
Deduce the relation between wavelength, frequency, and speed of sound. (AS 1)
Answer:
1) Speed of sound can be defined as the distance by which a point on the wave, such as a compression or rarefaction, travels in unit time.

2) Let the distance travelled by a wave in T seconds = X metres

3) The distance travelled by a wave in 1 second = $\frac{\lambda}{\mathrm{T}}$ meters

4) Thus by definition of speed of wave, v = $\frac{\lambda}{\mathrm{T}}$ ……….. (1)

5) We know that frequency and time period are related as υ = $\frac{1}{T}$ ……….. (2)

6) From (1) and (2) we get v = λ .υ
∴ Speed of sound = Frequency × Wavelength

Question 9.
How are multiple reflections of sound helpful to doctors and engipeeps? (AS 7)
Answer:
1) Doctors use multiple reflections of sound to hear the sounds produced with in the body using stethoscope.

2) Doctors can see the images of patient’s organs like liver, gall bladder, uterus, etc. to know the abnormalities in their functioning, using ultrasounds.

3) Engineers use the reflections of sound in designing concert halls and cinema halls.

4) Generally the ceilings of concert halls, conference halls, cinema halls are designed such that sound after reflection reaches all corners of the hall as shown in the figure.

Question 10.
Name two quantities that vary periodically at a place in air as a sound wave travels through it. (AS 1)
Answer:
The two quantities that vary periodically at a place in air as a sound wave travels through it are density and pressure of particles.

Question 11.
Which has larger frequency – infrasonic sound or ultrasonic sound? (AS 7, AS 2)
Answer:

Infrasonics are the sounds of frequency less than 20 Hz.
Ultrasonics are the sounds of frequency greater than 20 kHz.
Hence the ultrasonics have larger frequency.

Question 12.
The grandparents and parents of two-year-old girl are playing with her in a room. A sound source produces a 28 kHz sound. Who in the room is most likely to hear the sound? (AS 2, AS 7)
Answer:

The two-year-old girl is able to hear the sound.
Children can hear sounds of somewhat higher frequencies up to 30 kHz.

Question 13.
Does the sound follow same laws of reflection as light does? (AS 1)
Answer:

Reflection of sound follows the same laws as the reflection of light when sound is reflected.
The directions in which the sound is incident and reflected make equal angles with the normal to the reflecting surface.

Question 14.
Why is soft furnishing avoided in concert halls? (AS 7)
Answer:

Sound reflects like the reflection of light.
But unlike to light, sound reflects more on rough surfaces than soft surfaces.
In concert halls, sound must undergo multiple reflections, so as to reach all corners of the hall.
Hence for better reflection, soft furnishing is avoided in concert halls.

Question 15.
Two sources A and B vibrate with the same amplitude. They produce sounds of frequencies 1 kHz and 30 kHz respectively. Which of the two waves will have larger power? (AS 1)
Answer:
Frequency of source A = 1 kHz; Frequency of source B = 30 kHz

As the speed of wave increases with frequency and both the waves have same amplitude, the sound produced from source B has larger power.

Question 16.
What do you understand by a sound wave? (AS 1)
Answer:

Sound is produced from a vibrating body.
It travels through air in the form of a wave.
Sound waves are longitudinal.

Question 17.
Define the wavelength of a sound wave. How is it related to the frequency and the wave speed? (AS 1)
Answer:
Wavelength :
The distance between two consecutive compressions or rarefactions is called wavelength.

Relation between wavelength, frequency, and wave speed :

Speed of sound can be defined as the distance by which a point on the wave, such as a compression or rarefaction, travels in unit time.
Let the distance travelled by a wave in T seconds = λ metres
The distance travelled by a wave in 1 second = $\frac{\lambda}{\mathrm{T}}$ meters
Thus by definition of speed of wave, v = $\frac{\lambda}{\mathrm{T}}$ ………… (1)
We know that frequency and time period are related as o = $\frac{1}{T}$ ……….. (2)

Question 18.
Explain how echoes are used by bats to judge the distance of an obstacle in front of them. (AS 1)
Answer:

Bats search out prey and fly in dark night by emitting and detecting reflections of ultrasonic waves.
The high pitched ultrasonic squeaks of the bat are reflected from the obstacles or prey and returned to bat’s ear.
The nature of reflections tells the bat where the obstacle or prey is and what it is like.
The bats use ultrasound for navigation and location of the food in dark.

Question 19.
With the help of a diagram describe how compression and rarefaction pulses are produced in air near a source of sound. (AS 5)
Answer:

Consider a vibrating membrane of a musical instrument like a drum or tabla.
As it moves back and forth, it produces a sound.
The figure shows the membrane at different instants and the condition of the air near it at those instants.
As the membrane moves forward, it pushes the particles of air in the layer in front of it.

So, the particle of air in the layer get closer to each other, hence the density increases.
This layer of air pushes and compresses the layer next to it and so on.
We call this disturbance as compression pulse.
When the membrane moves backward, it drags back the layer of air near it. Hence the density decreases.
The particles of air in the next layer on the right move into fill this less dense area.
This is a rarefaction pulse moves to right.
As the membrane moves back and forth repeatedly, compression and rarefaction pulses are produced, one after the other.
These two pulses travel one behind the other, carrying the disturbance with it.

Question 20.
How do echoes in a normal room affect the quality of the sounds that we hear? (AS 7)
Answer:

Echo is a reflected sound, arriving at the position of listener more than 0.1s after the direct sound.
Quality is the characteristic of a sound which enables us to distinguish between musical notes emitted by different musical instruments.
In a normal room, if echo is formed, we can hear multiple sounds, at same time.
Our ear cannot perceive and judge the sound from where it is coming.
So, quality of sound does not work here.

Question 21.
Explain the working and applications of SONAR. (AS 1)
Answer:

SONAR stands for Sonographic Navigation And Ranging.
This is a method for detecting and finding the distance of objects under water by means of reflected ultrasonic waves.

Working of SONAR:

SONAR system consists of a transmitter and a detector in the “Observation Centre” on board of a ship.
From the observation centre, ultrasonic waves of high frequency are sent in all directions under the water through transmitter.
These waves travel in straight lines till they hit an object such as a submarine, a sunken ship, etc.
The waves are then reflected and are received back by the receiver at the observation centre.
The study of these reflected waves gives information about the direction of the object located.
The time between sending ultrasonic wave and receiving its echo, the distance of the object is calculated.
Reflections from various angles can be utilized to determine the shape and size of the object.

Mathematical expression :

Let’d’ be the distance between SONAR and an underwater object.
‘t’ be the time between sending an ultrasonic wave and receiving its echo.
‘u’ be the speed of sound in water.
The total distance covered by the wave from the SONAR to the object and back is 2d.
From the equation s = ut ⇒ 2d = ut ⇒ d = $\frac{ut}{2}$.

Application:
Marine geologists use this method to determine the depth of the sea and to locate underwater hills and valleys.

Question 22.
Find the time period of a source of a sound wave whose frequency is 400 Hz. (AS 1)
Answer:
Frequency υ = 400 Hz
Time period T =?

Question 23.
A sound wave travels at a speed of 340 m/s. If its wavelength is 2 cm, what is the frequency of the wave? Will it be in the audible range? (AS 1)
Answer:
Speed of sound v = 340 m/s.; Wavelength λ = 2 cm = 0.02 m.
Frequency υ =?

The audible range of sound wave is 20 Hz to 20 kHz.
Hence this is in the audible range.

Question 24.
Given that sound travels in air at 340 m/s, find the wavelength of the waves in air produced by a 20 kHz sound source. If the same source is put in a water tank, what would be the wavelength of the sound waves in water? (AS 7)
Speed of sound in water = 1,480 m/s.
Answer:
In air:
Speed of sound wave (v) = 340 m/s ; Frequency of source of sound (o) = 20 kHz
Wavelength of the sound wave λ =?
v = υλ

∴ Wavelength of the sound wave in air = 17 m

Same source is kept in water :
∴ Speed of sound in water (v) = 1480 m/s
Frequency of sound wave (p) = 20 kHz
Wavelength of sound wave λ =?
v = υλ

∴ Wavelength of the sound wave in water = 74 m

Question 25.
A man is lying on the floor of a large, empty hemispherical hall, in such a way that his head is at the centre of the hall. He shouts “Hello!” and hears the echo of his voice after 0.2 s. What is the radius of the hall ? (Speed of sound in air 340 m/s) (AS 7)
Answer:
Let the distance travelled by the sound wave = 2d m

As the head of the man is at the centre of hemispherical room, then ‘d’ is the radius of the hall.
∴ Radius of the hall = 34 m

Question 26.
“We know that sound is a form of energy. So, the large amount of energy produced due the sound pollution in cosmopolitan cities can be used to our day-to-day needs of energy. It also helps us to protect biodiversity in urban areas”. Do you agree with this statement? Explain.
Answer:

Sound is a form of mechanical energy.
So, the mechanical energy can be converted into electrical energy.
Experiments are going on this concept.
If this is successful, we have the following benefits.
a) Sound pollution can be controlled.
b) Conventional methods of producing electrical energy from coal or water will cause in loss of biodiversity. This can be avoided.
c) Natural resources like water can be protected.
d) Increasing needs of energy can be overcome by this method.

Question 27.
How do you appreciate efforts of a musician to produce melodious sound using a musical instrument by simultaneously controlling frequency and amplitude of the sounds produced by it.
Answer:

The sounds which produce pleasing effect on the ear are called musical sounds.
Any instrument which produces musical sound is called musical instrument.
The person who plays a musical instrument to produce melodious sound is called a musician.
The musician must have control on breathing, concentration on the output of the sound, which is a very hard task.
For this the musician needs a lot of practise.
With the musician’s practise and knowledge over musical notes only we can hear melodious sound otherwise it could only be a noise.
Hence the efforts of a musician are highly appreciable.

Question 28.
You might have observed that sometimes your pet dog starts barking though no one is seen near in its surroundings or no disturbance heard nearby. Does this observation raise any doubts in your mind about the peculiar behaviour of dog after your understanding about ‘range of hearing the sound’. If yes, write them.
Answer:

Dogs can hear sounds of frequencies up to about 50 kHz, which is ultrasonic.
After hearing this ultrasonics, a dog will bark panicly, though no one is seen near.
I understood this after studying about ‘range of hearing the sound’.
Before the knowledge of ‘range of hearing the sound’, I felt that the dogs are barking by seeing some devils, which is a misconception.
Scientific knowledge helps us to know reasons for many misconcepts.

Question 29.
Find out the names of animals (and their photographs from internet) which communicate using infra-sonic or ultra-sonic sound and prepare a scrap book.
Answer:
Scrap book

9th Class Physical Science 11th Lesson Sound InText Questions and Answers

9th Class Physical Science Textbook Page No. 184

Question 1.
How does sound reach our ears from the source of its production?
Answer:
Soufid travels in the form of waves. It reaches our ears from the source of its production, in the form of waves.

Question 2.
Does it travel by itself or is there any force bringing it to our ears?
Answer:
Sound does not travel by itself. When a sound is produced, the kinetic energy of the source, vibrates the nearest particles in the medium. These particles transfers energy and finally, it reaches our ears.

Question 3.
What is sound? Is it a force or an energy?
Answer:
Sound is a form of energy.

Question 4.
Why don’t we hear sounds when our ears are closed?
Answer:
When our ears are closed, the energy in the form of waves reaches our ear, but it cannot make the eardrum to vibrate. Hence we cannot hear the sound.

Question 5.
Why is the light ray dancing, after sound is made in the tin?
Answer:
The dancing of light ray, after the sound is made in the tin shows that the sound travels in the form of waves from the source of its production.

Question 6.
What do you infer from this?
Answer:
Sound is a form of energy which can travel in the form of waves through the medium.

Question 7.
Can we say that sound is a form of mechanical energy?
Answer:
Yes, sound is a form of mechanical energy.

9th Class Physical Science Textbook Page No. 185

Question 8.
Do you hear any sound?
Answer:
We cannot hear any sound.

Question 9.
Do you see any vibrations in the tuning fork?
Answer:
Yes, we can sense the vibrations in the tuning fork.

Question 10.
What do you conclude from the above activity?
Answer:
We observe that vibrating tuning fork produces sound.

Question 11.
Can you produce sound without vibration in the body?
Answer:
We cannot produce sound without vibration in the body.

Question 12.
Give some examples of vibrating bodies which produce sound.
Answer:
Drums, tabla, calling bell, school bell, etc.

Question 13.
What part of our body vibrates when we speak?
Answer:
When we speak, vocal cord vibrates in our body.

Question 14.
Do all vibrating bodies necessarily produce sound?
Answer:
All vibrating bodies produce sound, but we cannot hear some of them, due to the limit of audible frequency.

9th Class Physical Science Textbook Page No. 186

Question 15.
If sound travels in the form of a wave then what is the pattern?
Answer:
Sound travels in the form of longitudinal or transverse waves in the air or in the other material.

9th Class Physical Science Textbook Page No. 188

Question 16.
What do you say about sound waves in air by the above activity?
Answer:
From the above activity, we can say that there involves change in the density of medium while sound waves are travelling in air.

Question 17.
Are they longitudinal or transverse?
Answer:
Sound waves in air are longitudinal.

Question 18.
Does sound get reflected at the surface of a solid?
Answer:
Yes, sound gets reflected at the surface of a solid as in the case of reflection of light.

Question 19.
What happens if you lift your tube slightly above the table?
Answer:
If we lift the tube slightly we cannot hear the sound clearly.

Question 20.
Are able to listen to the sound? If not why?
Answer:
We are unable to hear the sound. If we lift one of the pipes then the pipe carrying incident sound, the pipe carrying reflected sound will not be in the same plane. Hence we cannot hear the sound.

9th Class Physical Science Textbook Page No. 195

Question 21.
Do hard surfaces reflect sound better than soft ones?
Answer:
Generally, hard surfaces reflect sound better than soft surfaces. But sound reflects quite well from rough surfaces than polished surfaces.

9th Class Physical Science Textbook Page No. 187

Question 22.
Do compressions and rarefactions in sound wave travel in same directions or in opposite directions? Explain.
Answer:

Compressions and rarefactions in a sound wave will be in opposite direction.
In a compression, all the particles come close, so the density and pressure increases.
In a rarefaction, all particles drag back, so the density and pressure decreases.
In a microscopic view of particle, the compression and rarefaction travel in opposite directions.

9th Class Physical Science Textbook Page No. 191

Question 23.
Does the frequency of sound waves depend on the medium in which it frawels? How?
Answer:
Yes.

2) As speed of wave differs from medium to fnedium, the frequency also changes, keeping the wavelengths constant.

Question 24.
The frequency of source of sound is 10 Hz. How many times does it vibrate in one minute?
Answer:
Number of vibrations per second = 10
Number of vibrations in one minute = 10 × 60 = 600

Question 25.
Gently strike a hanging bell (temple bell) and try to listen to the sound produced by it with a stethoscope keeping it both at bottom portion and top portion of the bell. Is the pitch and loudness of the sound same at the two portions? Why?
Answer:
No. The bob of the bell strikes at the bottom portion of the bell. Hence the pitch and loudness are high.

Question 26.
During a thunderstorm if you note a 3 second delay between the flash of lightning and sound of thunder. What is the approximate distance of thunderstorm from you ?
Answer:
Time taken to reach the sound = 3 s ; Speed of sound in air = 343.2 m/sec.
Distance of thunderstorm = 343.2 x 3 = 1029.6 m

9th Class Physical Science Textbook Page No. 194

Question 27.
Two girls are playing on identical stringed instruments. The strings of the both instruments are adjusted to give notes of same pitch. Will the quality of two notes be same? Justify your answer.
Answer:
If the two girls are playing with same instruments, then the quality is same. If the girls are playing with different stringed instruments the quality will be different.

Quality is the characteristic which enables us to distinguish between musical notes emitted by different musical instruments.

Question 28.
What change, would you expect in the characteristic of a musical sound when we increase its frequency one instance and amplitude in another instance?
Answer:
When frequency is increased, the pitch of sound increases. When amplitude is increased, its loudness increases.

9th Class Physical Science Textbook Page No. 195

Question 29.
What could be the reason for better reflection of sound by rough surfaces than polished surfaces?
Answer:

Sound reflects better on rough surfaces than polished surfaces.
The rough surface reflects sound in all directions, so it can reach in many directions.

9th Class Physical Science Textbook Page No. 196

Question 30.
Why is an echo weaker than the original sound?
Answer:

Echo is the reflected sound.
While it travels back, it gradually loses its energy.
Hence the echo is weaker than original sound.

Question 31.
In a closed box if you say hello, the sound heard will be Hellooooo ………. What does it mean?
Answer:
In a closed box, the multiple reflections of sound has no way to go out side. Hence we will hear the sound as hellooooooo ………

9th Class Physical Science Textbook Page No. 197

Question 32.
What is the advantage of having conical openings in horns, megaphones, etc?
Answer:
The conical openings in horns, megaphones, etc. will guide the reflected sound waves in forward direction and spreads towards the audience.

Question 33.
Why do we put cushions on the chairs, carpet on the floor, straw materials on the walls in cinema halls?
Answer:

These materials absorb unnecessary reflections of sound, so that we can hear more clearly.
As these material absorb reflected sound waves, the people outside the cinema halls, cannot hear the sound produced in the cinema hall.

9th Class Physical Science Textbook Page No. 199

Question 34.
What is the benefit of using ultrasound over light waves in the above applications?
Answer:

Light waves cannot penetrate in the internal organs like liver, kidney, etc.
Light wave after reflection do not form image.
Ultrasound waves can penetrate through internal organs like liver, kidney, etc.
After reflection, ultrasound waves produce image. So, in ultrasonography and surgeries, ultrasound waves are better than light waves.

9th Class Physical Science 11th Lesson Sound Activities

Activity 1

Question 1.
How can .you say that the sound is a form of energy?
Answer:

Take a tin can and remove both ends to make a hollow cylinder.
Take a balloon and stretch it over the can.
Wrap a rubber band around the balloon.
Take a small mirror and stick it on the balloon.
Take a laser light and let it fall on the mirror.
After reflection the light spot is seen on the wall.
Now shout directly into the open end of the can and observe the dancing light.

Observations:

When sound is made, the energy produced from the sound vibrates the membrane of the balloon, resulting in the dancing of light ray.
This shows that sound is a form of energy which travels in air.

Activity – 2

Question 2.
Prove that the sound is produced from a vibrating source.
Answer:

Attach a small piece of steel wire to one of the prongs of a tuning fork as shown in the figure.
Strike the tuning fork with a rubber hammer.
While it is vibrating, try to draw a straight line on a piece of smoked glass as quick as possible with, it.
Keep the end of the wire in such a way just it touches the glass.
A line is formed in the form of a wave.
Repeat the experiment when the tuning fork is not vibrating and observe the difference in the line.

Observations:

We have produced vibrations in the tuning fork by striking it with a hammer.
Thus the vibrating tuning fork produces sound.
Thus the sound is produced by vibrating bodies.

Activity – 3

Question 3.
а) How do you demonstrate the formation of compressions and rarefactions in a slinky?
Answer:

Take a slinky.
It is a spring-shaped toy which can be extended or compressed very easily.
Lay it down on a table or the floor as shown in figure.
Ask a friend to hold one end.
Pull the other end to stretch the slinky and then move it to and fro along its length.
We will see alternate compressions and rarefactions of the coil.
This is similar to the pattern of varying density produced in a medium when sound passes through it.

b) Mow do you demonstrate the formation of crests and troughs in a slinky?
Answer:

Hang a slinky from a fixed support.
Hold it gently at the lower end and quickly move your hand sideways and back.
This will cause a hump on the slinky near the lower end.
The hump travels upwards on the slinky as shown in the figure.
The humps formed alternately are known as crests and troughs.

Activity – 4

Question 4.
Describe an activity to listen the reflected sound.
Answer:

Take two long, identical tubes and place them on
Ask your friend to speak softly into one tube while you use the other tube to listen.
Adjust the tube until you hear the best sound.
You will find that you hear your friend’s voice best when the tube makes equal angles with a normal to the wall.
This shows that reflection of sound follows the laws of reflection of light.
Lift your tube slightly above the table.
You will not be able to listen the voice clearly, because the plane carrying the incident wave and reflected wave are changed.
Repeat the experiment by placing flat objects of different materials (steel and plastic trays, a card board, a tray wrapped with cloth, etc.) against the wall and observe the changes in the sound.

Observations:

Reflection of sound follows the laws of reflection of light.
When the plane carrying incident wave and reflecting wave changes, the reflected sound cannot be heard clearly.
Hard or rough surfaces reflect the sound better than soft surfaces.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

November 3, 2023 by AP Board

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 8th Lesson Gravitation

9th Class Physical Science 8th Lesson Gravitation Textbook Questions and Answers

Improve Your Learning

Question 1.
What path will the moon take when the gravitational interaction between the moon and earth disappears? (AS 2)
Answer:
The force of attraction between moon and earth is given by F = $\frac{\mathrm{GMm}}{\mathrm{R}^{2}}$
M = mass of the earth ; m = mass of the moon ; R = radius of the earth
Here the gravitational interaction between moon and earth disappears.
∴ G = 0 ⇒ F = 0

Therefore the moon neither revolves around the earth nor fall into the earth.
It takes a straight path away from the earth.

Question 2.
A Car moves with constant speed of 10 m/s in a circular path of radius 10 m. The mass of the car is 1000kg. Who or what is providing the required centripetal force for the car? How much is it? (AS 1)
Answer:
Speed of the car (v) = 10 m/s ; Radius of the path (r) = 10 m
Weight of the car (m) = 1000kg

The required centripetal force is provided by the friction between the tyres of the car and the road.

Question 3.
A small metal washer is placed on the top of a hemisphere of radius R. What minimum horizontal velocity should be imparted to the washer to detach it from the hemisphere at the initial point of motion? (AS 1, AS 7)

Answer:
Radius of hemisphere = R; Mass of hemisphere = M
Let the radius of washer = r and mass = m
Distance between hemisphere and washer = R + r
The centripetal force required to rotate the washer = $\frac{\mathrm{mv}^{2}}{\mathrm{r}}$
The gravitational force of washer due to hemisphere is = $\frac{\mathrm{GMm}}{\mathrm{R}^{2}}$
But the necessary centripetal force must be equal to the gravitational force

Question 4.
Explain why a long pole is more beneficial to the tight rope walker if the pole has slight bending. (AS 1, AS 7)
Answer:

You must have noticed the circus artists doing rope walking.
During this act, they carry a long bamboo pole in their hands.
The reason for this is that the line joining the centre of gravity and centre of equilibrium must fall within the rope for achieving the stable equilibrium.
Thus when an artist finds that he is falling towards left, he shifts bamboo pole towards right, so that his centre of gravity stay undisturbed.
Thus he can balance himself on the rope.

Question 5.
Why is it easier to carry the same amount of water in two buckets, one in each hand rather than in a single bucket? (AS 7)
Answer:
It is because in the later case centre of gravity of our body shifts towards the bucket and there is a tendency that the line joining the centre of gravity and centre of equilibrium may fall outside our feet.

However in the former case the centre of gravity not only gets lowered, but also it is at such a point that line joining the C.G. and C.E. falls within our feet.
Hence one is a stable equilibrium.

Question 6.
What is the speed of an apple dropped from a tree after 1.5 second? What distance will it cover during this time? Take g = 10 m/s². (AS 1)
Answer:
An apple is dropped from a tree.
∴ Initial velocity u = 0 ; Time t = 1.5 s
a = g = 10 m/s2 ; Final velocity v = ?
v = u + at = 0 + 10 x 1.5 = 0 + 15 = 15 m/s
Distance covered (s) = ut + $\frac{1}{2}$ at² = 0 × 1.5 + $\frac{1}{2}$ × 10 × 1.5 × 1.5 = 0 + 5 × 2.25 = 11.25 m

Question 7.
A body is projected with a speed of 40 m/s vertically up from the ground. What is the maximum height reached by the body? What is the entrie time of motion? What is the velocity at 5 seconds after the projection? Take g = 10 m/s². (AS 1)
Answer:
Initial speed u = 40 m/s ; g = 10 m/s²
Maximum height reached (h) = $\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}=\frac{40 \times 40}{2 \times 10}$ = 80 m
Entire time of motion (T) = $\frac{2 \mathrm{u}}{\mathrm{g}}=\frac{2 \times 40}{10}$ = 8 s
Entire time of motion is 8 seconds.
∴ It starts to fall down after 4 seconds.
At 5 seconds the body is in downward direction.
u = 0 m/s, a = g = 10 m/s², t = 5 – 4 = 1 sec.
v = u + at = 0 + 10 × 1 = 10 m/s
∴ The velocity at 5 seconds is 10 m/s downward.

Question 8.
A boy is throwing balls into the air one by one in such a way that when the first ball thrown reaches maximum height he starts to throw the second ball. He repeats this activity. To what height do the balls rise if he throws twice in a second? (AS 1, AS 7)
Answer:
The boy throws the second ball when the first ball reaches its maximum height. He throws twice in a second.
Time of ascent of first ball is 1/2 sec.
After 1/2 sec, the first ball starts to fall down and the second ball starts from ground.
Let the distance travelled = s meters.; Initial velocity = u m/s
Time of ascent (t₁) = 1/2 sec.

They reach a height of 15/4 meters.

Question 9.
A man is standing against a wall such that his right shoulder and right leg are in contact with the surface of the wall along his height. Can he raise his left leg at this position without moving his body away from the wall? Why? Explain. (AS 7)
Answer:

When the right leg and right shoulder are in contact with the surface of the wall along the height of a man, his weight is towards the wall.
The centre of gravity will be away from the foot. Just like in the case of carrying a bucket full of water with one hand.
The line joins the centre of gravity and centre of equilibrium is not perpendicular to the horizontal.
Hence he cannot lift his left leg without moving his body along the wall.

Question 10.
A ball is dropped from a height. If it takes 0.2 s to cross the last 6 m before hitting the ground, find the height from which it is dropped. Take g = 10 m/ s². (AS 1)
Answer:

For last 6 m, distance travelled
s = 6 m ; u = ? ; t = 0.2 sec ; a = g = 10 m/s²
s = ut + $\frac{1}{2}$ at²
6 = u (0.2) + $\frac{1}{2}$ × 10 × (0.2)²
6 = (0.2) u + (5 × 0.04)
6 – 0.2 = 0.2 u
5.8 = 0.2 u ⇒ u = $\frac{5.8}{0.2}$ = 29
∴ u = 29 m/s
This u will be the final velocity while travelling the distance x.
∴ s = x, v = 29 m/s, a = g = 10 m/s², u = 0 m/s
v² – u² = 2as ⇒ 29² – 0 = 2 × 10 × x ⇒ x = $\frac{841}{20}$ = 42.05 m
∴ Total distance = x + 6 = 42.05 + 6 = 48.05 m
∴ The ball is dropped from a height of 48.05 m.

Question 11.
A ball is dropped from a balloon going up at a speed of 5 m/s. If the balloon was at a height 60 m. At the time of dropping the ball, how long will the ball take to reach the ground? (AS 1, AS 7)
Answer:
At t = 0, the stone was going up with a velocity of 5 m/s. After that it moves as a freely falling body, with downward acceleration ‘g’.
If it reaches the ground at time t₁,

Question 12.
A ball is projected vertically up with a speed of 50 m/s. Find the maximum height, the time to reach the maximum height, and the speed at of the maximum height. (g = 10 m/ s²) (AS 1)
Answer:
Initial speed u = 50 m/s ; g = 10 m/s²

After reaching maximum height, the velocity becomes ‘zero’.

Question 13.
Two cars having masses m₁ and m₂ move in circles of radii r₁ and r₂ respectively. If they complete the circle in equal time. What is the ratio of their speeds and centripetal accelerations? (AS 1)
Answer:
Masses of cars : m₁ and m₂; Radius of circles : r₁ and r₂
Given that their time period is equal.

Question 14.
Two spherical balls of mass 10 kg each are placed with their centers 10 cm apart. Find the gravitational force of attraction between them. (AS 1)
Answer:
Masses of balls M₁ and M₂ = 10 kg each.; Distance d = 10 cm = 0.1 m
Gravitational force of attraction between them is

Question 15.
Find the free-fall acceleration of an object on the surface of the moon, if the radius of the moon and its mass are 1740 km and 7.4 × 10²² kg respectively. Compare this value with free fall acceleration of a body on the surface of the earth. (AS 1)
Answer:
Radius of the moon = 1740 km = 1740 × 10³ m
Mass of the moon = 7.4 × 10²² kg ; G = 6.67 × 10^-11 Nm² kg^-2
Free fall acceleration of a body on the surface of the moon

Question 16.
Can you think of two particles which do not exert gravitational force on each other? (AS 2)
Answer:
Two particles which do not exert gravitational force on each other will be mass less particles. But every particle has even a little mass. Hence we cannot find two particles which do not exert gravitational force on each other.

Question 17.
An apple falls from a tree. An insect in the apple finds that the earth is falling towards it with an acceleration g. Who exerts the force needed to accelerate the earth with this acceleration? (AS 7)
Answer:

According to Newton’s third law, when an apple is freely falling, the force on the apple due to earth is equal to the force on the earth due to apple.
The force is due to gravity, which causes acceleration in the body.
As the insect is inside the freely falling apple, it feels that the earth is falling towards it with an acceleration ’g’.
According to the insect, the acceleration is due to the force on the apple due to the earth.
Actually the earth is not falling towards the apple.

Question 18.
A scooter weighing 150 kg together with its rider moving at 36 kin/hr is to take a turn of radius 30 in. What force on the scooter towards the center is needed to make the turn possible ? Who or what provides this? (AS 1)
Answer:
Weight of the scooter with rider = 150 kg
Speed = 36 km/hr = 36 × $\frac{5}{18}$ = 10 m/sec.
Radius of the turning = 30 m
The force needed on the scooter towards the centre is centripetal force.

∴ A centripetal force of 500 N is required on the car.
This force is provided by the friction between the tyres of the car and road.

Question 19.
The bob of simple pendulum of length 1 m has mass 100 g and a speed of 1.4 m/s at the lowest point in its path. Find the tension in the string at this moment. (Take g = 9.8 m/sec²) (AS 1)
Answer:
Mass of the bob = 100 g = 0.1 kg ; Length of the string = 1 m
Speed of the bob v = 1.4 m/s ; Let the tension in the string be T.
The forces acting on the bob are

a) Weight of the bob mg, downwards
b) Tension in the string ‘T’ upward
Weight of bob = $\frac{\mathrm{mv}^{2}}{\mathrm{l}}$
Tension in the string T = g cos θ
∴ According to Newton’s third law

Question 20.
How can you find the centre of gravity of India map made with steel? Explain. (AS 3)
(OR)
Describe an activity to know the center of gravity of a India map.
Answer:

Take an India map made of steel.
Make holes at four different corners i.e., top, bottom, left, and right, and name them as A, B, C, and D as shown in figure.
Suspend the map through a hole A, by means of a string to a nail ‘P’ as shown in figure.
Suspend a plumb line from the nail P.
Draw the line AX along the plumb line.
Similarly suspend the map through other holes B, C and D and draw lines BY, CZ, DW along the plumb line.
These lines (more than two) intersect at a point.
This point (G) is the centre of gravity of the map.

Question 21.
Explain some situations where the center of gravity of man lies out side of the body. (AS 1)
Answer:
Centre of gravity of a human being is located interior to the second sacral vertebra.
Centre of gravity of man lies out side the body in the following situation :
a) While doing sit-ups.
b) While carrying a load like bucket full of water with one hand.
c) While walking on a narrow base like walking on a rope or pole or narrow wall, etc.
d) While walking with one leg.
e) As age increases its position changes.

Question 22.
Where does the centre of gravity of atmosphere of the earth lie? (AS 2)
Answer:
The earth’s atmosphere is about 10,000 km thick, but most of its bulk is contained in the first 11 km above the earth’s surface. Since the earth and its atmosphere are roughly spherical, the centre of the earth is also the centre of gravity of the earth’s atmosphere.

Question 23.
Where does the center of gravity lie, when a boy is doing sit-ups? Explain. (AS 7)
Answer:

The centre of gravity of a boy when he start erect, falls in the foot.
When he is doing sit-ups, the centre of gravity shifts from foot to his base.
The weight vector also move from the base.
Hence the boy stretches his hands or bends slightly towards the earth. While doing sit-ups, in order to make the weight vector pass through base, so that he acquire stability.

9th Class Physical Science 8th Lesson Gravitation InText Questions and Answers

9th Class Physical Science Textbook Page No. 126

Question 1.
What is that force?
Answer:
The force acting on these objects to make them move around another object, instead of moving in straight line in the gravitational force.

Question 2.
Is the motion of the earth around the sun uniform motion?
Answer:
The earth takes 365.25 days to complete one rotation around the sun. Hence it is uniform motion.

Question 3.
Is the motion of the moon around the earth uniform motion?
Answer:
Moon takes 27.3 days to complete one rotation around the earth, which does not change. Hence it is uniform motion.

9th Class Physical Science Textbook Page No. 127

Question 4.
Does the velocity of the body change in uniform circular motion? Why?
Answer:
The velocity of the body in uniform circular motion is constant.

If the velocity is not constant, the time period changes from time to time and it cannot be treated as uniform circular motion.

Question 5.
Does the body in uniform circular motion have an acceleration? What is the direction of acceleration?
Answer:
The body in uniform circular motion have an acceleration, which is directed towards the centre of the circle, known as centripetal acceleration.

9th Class Physical Science Textbook Page No. 129

Question 6.
Do you know what questions arose in his mind from this observation?
Answer:
When apple fell to the ground, Newton might have thought like this:
a) Why did the apple fall on the earth?
b) Why doesn’t the apple go up in the sky?
c) Whether all the objects fall to ground ?
d) What makes them to fall on the ground?

Question 7.
Why did the apple fall to the ground?
Answer:
The apple fell to the ground due to the gravitational attraction of the earth.

Question 8.
Why does the moon not fall to the ground?
Answer:
The gravitational force in the moon due to earth is equal to the gravitational force on the earth due to moon. Hence the moon does not fall to the ground.

Question 9.
What makes the moon to move in a circular orbit around the earth?
Answer:
The gravitational force between moon and earth act as centripetal force and makes the moon to revolve around the earth in uniform circular motion.

9th Class Physical Science Textbook Page No. 135

Question 10.
How do you feel during the free-fall of your body from the height?
Answer:

I feel weightlessness during the free-fall of my body from the height.
This is due to the gravitational force of earth on the body.

9th Class Physical Science Textbook Page No. 136

Question 11.
Try to balance ladder on your shoulder. When does it happen?
Answer:
When the centre of gravity of the ladder and the centre of gravity of our body lies in the same line then we can balance the ladder on our shoulders.

9th Class Physical Science Textbook Page No. 128

Question 12.
Can an object moVe along a curved path if no force acts on it?
Answer:

If a body has to move in a curved path, its direction of velocity must change continuously.
This is done by the centripetal force.
Hence an object cannot move along a curved path if no force acts on it.

Question 13.
As a car speeds up when rounding a curve, does its centripetal acceleration increase? Use an equation to defend your answer.
Answer:
Centripetal acceleration $a_{c}=\frac{v^{2}}{r}$
As v increases, its centripetal acceleration also increases.

Question 14.
Calculate the tension in a string that whirls a 2 kg – toy in a horizontal circle of radius 2.5 iv when it moves at 3 m/s.
Answer:
Mass of the toy m = 2 kg ; Radius of the circle = 2.5 m ; Speed of the toy = 3 m/s
As the toy is moving in a horizontal circle, the necessary centripetal force is provided due to the tension in the string.

9th Class Physical Science Textbook Page No. 131

Question 15.
In figure, we see that the moon falls around earth rather than straight into it. If the magnitude of velocity were zero, how would it move?

Answer:
If the magnitude of velocity were zero, the moon would move towards earth due to acceleration due to gravity.

Question 16.
According to the equation for gravitational force, what happens to the force between two bodies if the mass of one of the bodies doubled?
Answer:

∴ The gravitational force will be doubled.

Question 17.
If there is an attractive force between all objects, why we do not feel ourselves gravitating toward massive buildings in our vicinity?
Answer:

Earth is massive than the building.
Hence the gravitational force between ourselves and earth is more than that of between ourselves and building.
Hence we do not feel gravitating towards massive building in our vicinity.

Question 187.
Is the force of gravity stronger on a piece of iron than on a piece of wood if both have the same mass?
Answer:
Yes. The force of gravity is stronger on a piece of iron then on a piece of wood if both have the same mass.

Question 19.
An apple falls because of the gravitational attraction of the earth. What is the gravitational attraction of apple on the earth?
Answer:

We know that acceleration due to gravity, g = $\frac{\mathrm{GM}}{\mathrm{R}^{2}}$
Here M is the mass of the earth.
In this mass of the object which is freely falling has no effect on ‘g’.
Hence the gravitational attraction of apple on the earth is negligible.

9th Class Physical Science Textbook Page No. 133

Question 20.
Give an example for the motion of an object of zero speed and with non-zero acceleration?
Answer:
Protons and neutrons inside the nucleus.

Question 21.
Two stones are thrown into air with speeds 20 m/s, 40 m/s respectively. What accelerations are possessed by the objects?
Answer:
Stone -1 :
Initial velocity u = 20 m/s
After a time t, it reaches to ground then final velocity v = 0 m/s
Accelerahon = $\frac{0-20}{t}=\frac{-20}{t} \mathrm{~m} / \mathrm{s}^{2}$

Stone – 2 :
Initial velocity u = 40 m/s
Final velocity v = 0 m/s
Accelerahon = $\frac{0-40}{t}=\frac{-40}{t} \mathrm{~m} / \mathrm{s}^{2}$

9th Class Physical Science Textbook Page No. 135

Question 22.
When is your weight equal to mg?
Answer:
When we are on the surface of the earth, our weight is equal to ‘mg’ on earth.

Question 23.
Give example of when your weight is zero?
Answer:

When we are at the centre of the earth, our weight is zero.
When we are freely falling, we feel weightlessness.

9th Class Physical Science Textbook Page No. 138

Question 24.
Where does the centre of gravity of a sphere and triangular lamina lie?
Answer:
1) Centre of gravity of a sphere is a point where the whole mass of the sphere is assumed to be concentrated which is called centre.

2) Centre of gravity of a triangular lamina lies where the whole mass of the triangle is assumed to be concentrated which is centroid.

Question 25.
Can an object have more than one centre of gravity?
Answer:
Basing on the position or distribution of mass, an object can have more than one centre of gravity.

Question 26.
Why doesn’t the leaning tower of Pisa topple over?
Answer:

The centre of gravity of the leaning tower of Pisa is very close to earth.
Even though it is leaning, the line of action of total weight passes through the base. The base area of leaning tower of Pisa is very large. So it doesn’t topple over.

Question 27.
Why must you bend forward when carrying a heavy load on your back?
Answer:
If we bend forward, then the line of action of total weight passes through the body. So we are stable. Hence we must bend forward when carrying a heavy load on our back.

9th Class Physical Science 8th Lesson Gravitation Activities

Activity – 1 Uniform circular motion

Question 1.
Describe an activity to observe uniform circular motion.
Answer:
Uniform Circular Motion :
Uniform circular motion is a motion of the body with a constant speed in circular path.

Material required :
Electric motor, old C.D., small wooden block, battery, connecting chords, stop clock.

Procedure:

Take an electric motor and fix a disc to the shaft of the electric motor.
Place a small wooden block on the disc, as shown in figure.
Switch on the motor.
Find the time required to complete ten revolutions by the block.
Repeat the same two to three times.
We observe that the wooden block moves in a circular path with a constant speed.
So, this motion of wooden block is called uniform circular motion.

Activity – 2

Question 2.
Define centripetal acceleration and derive an expression for centripetal acceleration.
Answer:
Centripetal acceleration :
The acceleration which can change only the direction of velocity of a body is called centripetal acceleration.

Derivation :

Let a body is in uniform circular motion.
Draw velocity vectors at successive time intervals as shown in the figure.
Transfer tails of each velocity vector to coincide at a single point without changing their direction as shown in the figure.
In the above figure, the directed line joining two vectors represents change in velocity (Av).
Let us consider the change in velocity during the course of a complete revolution of a body i.e., the sum of the magnitudes of the changes in velocity during a complete revolution will be equal to the sum of the sides of the depicted polygon.
The smaller the sides of our polygon, the closer they cling to the circle of radius v, consequently the magnitude of change in velocity of the body during the course of revolution, will be equal to the circumference ’2πv’ of the circle.

9) As the vertex angle of isosceles triangle decreases, the angle between the change in velocity approaches to 90°.

10) Therefore, the acceleration of a body in uniform circular motion is directed perpendicular to its velocity.

Activity – 3 Free fall

Question 3.
Accelerahon is independent of masses.
Place a small paper on a book. Release the book with the paper from certain height from the ground.
a) What is your observation?
Answer:
The book and paper will fall on the ground at same time, if the paper is kept in the book. If the paper is kept on the book and released, then the paper will be separated from book and falls on the ground little bit later than the book.

b) Drop the book and paper separately, what happens?
Answer:
Book reaches the ground first, then the paper reaches.

Activity – 4

Question 4.
What is the direction of ‘g’?
Throw a stone vertically up. Measure the time required for it to come back to earth’s surface with stop clock.

a) What happens to its speed when it moves up and down?
Answer:

When the stone moves up, its velocity decreases gradually and finally it becomes zero, because it is moving against the direction of gravitation.
When its velocity becomes zero, it starts to fall down due to gravitational force of earth, and its velocity increases.

b) What is the direction of acceleration?
Answer:

While the stone is moving up, the direction of acceleration is in upward direction, which is against to the direction of gravitational force.
While the stone is falling down, the direction of acceleration is in downward direction, which is in the direction of gravitational force.

Activity – 5

Question 5.
Describe the method of measuring the weight of a free fail body.
Answer:

Take a spring balance and suspend it to the ceiling and put some weight to it.
Note the reading of the spring balance.
Now drop the spring balance with load from certain height to fall freely.
Carefully observe the change in the position of indicator on the spring balance scale while it is in free-fall.
We observe that the indicator of spring balance shows zero while it is in free-fall.
This is due to the weightlessness of a free-fall body.

Activity – 6

Question 6.
How do you demonstrate the changes during the free-fall of a body?
Answer:

Take a transparent tray and make holes on opposite sides.
Take two or three rubber bands and tie them tightly, close to each other between the holes.
Now place a stone on the bands as shown in the figure.
We observe that the bands bend due to the weight of the stone.
Now drop the tray with stone.
We observe that the bands do not bend. They straight.
When in equilibrium on a firm surface, weight is evidenced by a support force. Then weight equals to mg.
When the body falls freely then it experiences weightlessness.
Even in this weightless condition, there is still a gravitational force acting on the body causing downward acceleration.

Activity – 7

Question 7.
Give one activity that showing center of gravity and balancing.
(OR)
How do you balance spoon and fork?
Answer:

As shown in the figure arrange a fork, spoon and wooden match stick together.
The combination will balance nicely on the edge of the glass.
The weights of the spoon and fork will pass through the centre of gravity and the base (glass).

Activity – 8

Question 8.
Can you get up without bending?
Sit in a chair comfortably as shown in figure. Try to get up from the chair without bending your body or legs.

a) Are we able to do so? If not why?
Answer:

We cannot get up from the chair without bending our body or legs.
When we sit in a chair, we are in weightless condition.
We need a support force to get up from the chair.
This support force can be obtained by bending our body or legs.

Activity – 9

Question 9.
How do you balance a ladder on your shoulder?
Answer:

Take a ladder.
Find the mid stick by counting the sticks of the ladder.
Put that mid stick on the shoulder and balance the ladder.
That is the center of gravity of a ladder.

Activity – 10

Question 10.
How can you locate the centre of gravity of a uniform object?
Answer:

Take a meter scale.
Suspend it from various points.
We observe that it bends to one side.
Suspend it from the mid point.
We observe that the scale will be in horizontal position without bending to any side.
At this point, the scale behaves as if its entire weight is concentrated at this point.
The support given at this single point gives support to the entire scale.
This point is its centre of gravity.

Activity – 11

Question 11.
Identifying the centre of gravity of a ring.
The center of gravity of any freely suspended object lies directly beneath the point of suspension.

If a vertical line is drawn through the point of suspension, the center of gravity lies somewhere along that line. To determine exactiy where it lies along the line, we have only to suspend the object from the some other point and draw a second vertical line through that point of suspension. The center of gravity lies where the two lines intersect.

a) Where does the center of gravity of a ring lie?
Answer:
The centre of gravity of a ring lies at its geometric centre.

b) Does the center of gravity of a body exist outside the body?
Answer:
For the bodies which are in right angled triangle shape, the centre of gravity falls out side the body. Hence we can say the CG of a body may be outside or inside the body.

c) Does center of gravity of an object exist at a point where there is no mass of the object?
Answer:
The centre of gravity exist at a point where there is no mass of the object, for the objects like annual rings.

Activity – 12

Question 12.
Shift of the centre of gravity and its effects.
Try to touch your toes as shown in figure (a). Try this again when standing against a wall as shown in figure (b).

a) Are you able to touch your toes in second case as shown in figure? If not why?
Answer:

We cannot touch our toes in the second case as shown in figure.
When we stand erect with the support of wall, the centre of gravity shifts its place from second sacral vertebra to the place near to abdomen.

b) What difference do you notice in the center of gravity of your body in above two positions ?
Answer:

In the first position, the centre of gravity is at a place near to the centre of back-bone.
In the second position, the centre of gravity shifts from the place near to the centre of backbone to the place near to abdomen.

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

November 3, 2023November 2, 2023 by AP Board

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 10th Lesson Work and Energy

9th Class Physical Science 10th Lesson Work and Energy Textbook Questions and Answers

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Question 1.
Define work and write its units. (AS 1)
Answer:
Work : Work can be defined as the product of the force (F) and the distance (s) moved along the direction of the force.
Work done = Force × displacement = F × s
[This formula is used in only translatory motion of the object]

Work is a scalar quantity.
Unit of work is ‘N – m’ or ‘Joule’ (J).

Question 2.
Give few examples where displacement of an object is in the direction opposite to the force acting on the object. (AS 1)
Answer:

When a ball is thrown up, the motion is in upward direction, whereas the force due to earth’s gravity is in downward direction.
If a ball is moving on plain ground, will get stopped after sometime, due to frictional force acting on it in opposite direction.

Question 3.
Identify the wrong statement among the following. Rewrite them by making necessary corrections. (AS 1)
a) Work and energy have different units.
Answer:
Correction :
Work and energy have same units.

b) When an aeroplane takes off, the work done by its weight is positive.
Answr:
Correction : When an aeroplane takes off, the work done by its weight is negative.

c) The potential energy of spring increases when it is extended and decreases when it is compressed.
Answer:
Correct.

d) If the work done by external forces on a system is negative then the energy of the system decreases.
Answer:
Correct.

e) When a body is falling freely from a height, its kinetic energy remains constant.
Answer:
Correction : When a body is falling freely from a height, its K.E. increases and P.E. decreases, but the total energy of the body remains constant.

f) The unit of power is watt.
Answer:
Correct.

Question 4.
What is mechanical energy? (AS 1)
Answer:
The sum of the kinetic energy and the potential energy of an object is called its mechanical energy.

Question 5.
State the principle of conservation of energy. (AS 1)
Answer:
Energy can neither be created nor destroyed. It can only be changed from one form to another.

Question 6.
When you lift a box from the floor and put it on an almirah the potential energy of the box increases but there is no change in its kinetic energy. Is it violation of conservation of energy? Explain. (AS 7)
Answer:

When the box is on the floor, it has no K.E., but has some P.E.
Its P.E. increases as it is lifted and put on an almirah.
In this event, the energy produced by the person converts into K.E. and the total M.E. remains same.
Hence it is not the violation of conservation of energy.

Question 7.
One person says that potential energy of a particular book kept in an almirah is 20 J and other says it is 30 J. Is one of them necessarily wrong? Give reasons. (AS 2, AS 1)
Answer:
Two cases will arise in this situation.
Case – I :
One of them will be wrong, if both of them kept the book in the same shelf of the almirah.

Case – II :
Both of them will be correct, if the first person keeps the book in lower shelf and the second one keeps the book in upper shelf of the almirah.

Question 8.
In which of the following cases is the work done positive or zero or negative? (AS 1)
a) Work done by the porter on a suitcase in lifting it from the platform on to his head.
Answer:
Positive

b) Work done by the force of gravity on suitcase as the suitcase falls from porter’s head.
Answer:
Positive

c) Work done by the porter standing on platform with suitcase on his head.
Answer:
Zero

d) Work done by force of gravity on a ball thrown up vertically up into the sky.
Answer:
Negative

e) Work done by force applied by hands of a man swimming in a pond.
Answer:
Negative

Question 9.
What is potential energy? Derive an equation for gravitational potential energy of a body of mass ‘m’ at a height ‘h’. (AS 1)
Answer:
Potential energy:
The energy possessed by an object because of its position or shape is called its potential energy.

Potential energy of a body at height (Or) Gravitational potential energy :

The gravitational potential energy of an object at a point above the ground is defined as the work done in raising it from the ground to that point against gravity.
Consider an object of mass ‘m’ raised to height ’h’ from the ground.
A force is required to do this.
The minimum force required to raise the object is equal to the weight of the object (mg).
The object gains energy equal to the work done on it.
Let the work done on the object against gravity be ‘W’.
∴ W = Force × displacement
= mg × h
W = mgh
The object gains the energy equal to the mgh’ units.
This energy is the potential energy of the object at a height ‘h’.
∴ P.E. = mgh

Question 10.
When an apple falls from a tree what happens to its gravitational potential energy just as it reaches the ground? After it strikes the ground? (AS 7)
Answer:

Let us imagine an apple of mass ‘m’ falls from a height h.
Before starting to fall down, its velocity is zero. Hence its energy is purely potential.
When it falls down, its gravitational potential energy gradually converts into kinetic energy, but the entire energy in the system remains same.
When it strikes the ground, its energy is fully kinetic, till its velocity becomes zero.

Question 11.
Let us assume that you have lifted a suitcase and kept it on a table. On which of the following does the work done by you depend or not depend? Why?
a) The path taken by the suitcase
b) The time taken by you in doing so
c) The weight of the suitcase
d) Your weight
Answer:
We know that P.E. = mgh
a) The work done by you does not depend on the path taken by the suitcase, because the height from the ground to the top of the table is fixed in this case, (h constant)

b) The work done by you does not depend the time taken, in doing so, because the work done is same. If you take less time your power will be more otherwise the power is less.

c) The work done by you depends on the weight of the suitcase. As the weight of the suitcase increases, the work done will also increase (See the formula P.E. = mgh).

d) The work done by you depends on your weight, we know that a person with more weight can Mft the suitcase easily.

Question 12.
When you push your bicycle up an incline, the potential energy of the bicycle and yourself increase. Where does this energy come from? (AS 7)
Answer:
The energy inside our body causes increase in the P.E. of the bicycle and ourself.

As we are pushing the bicycle upon incline, the work done by the gravitational force is negative. Hence the energy loses.

Question 13.
Why does a person standing for a long time get tired when he does not appear to be doing any work? (AS 7)
Answer:
1) A person standing for a long time, does not appear to be doing any work, because there is no displacement.
2) Eventhough work has not been done, the person gets tired.
3) This is due to the utilization of energy produced in his body.

Question 14.
What is kinetic energy? Derive an expression for the kinetic energy of a body of mass ‘m’ moving at a speed ‘v’. (AS 1)
Answer:
Kinetic energy :
The energy possessed by an object due to its motion is called kinetic energy.

Numerical expression for K.E.:
1) Let us assume that an object of mass (m) is at rest on a smooth horizontal plane as shown in figure.

2) Let be displaced through a distence ‘s’ from the point A to B by a force (F) acting upon it in the direction of the displacement.
3) In the horizontal direction the net force ‘F_net‘ is equal to the force applied ‘F’.
∴ W = F_net.s = F.s …………. (1)
4) Let the work done on the object cause a change in its velocity from ‘u’ to ‘v’ anr the ‘acceleration produced be ‘a’.

8) As we Have assumed that object is at rest, then the initial velocity u = 0.
∴ W = $\frac{1}{2}$ mv²

9) We know that K.E. of a body moving with certain velocity is equal to work done on the object to acquire that velocity from rest.

10) Thus the K.E. of an object of mass’m’ and moving with velocity V is equal to
K.E = $\frac{1}{2}$ mv²

Question 15.
A free-fall object eventually stops on reaching the ground. What happens to its kinetic energy? (AS 1)
Answer:

Its kinetic energy becomes ‘zero’.
As the ball stops on reaching the ground, its final velocity ‘v’ becomes zero.
Hence the K.E. = $\frac{1}{2}$ mv² = $\frac{1}{2}$.m.0 = 0

Question 16.
A man carrying a bag of total mass 25 kg climbs up to a height of 10 m in 50 seconds. Calculate the power delivered by him on the bag. (AS 1)
Answer:
Mass of the bag = 25 kg ; Height ‘h’ = 10 m
Acceleration due to gravity g = 9.8 m/s²
The P.E. of the man = mgh = 25 × 9.8 × 10 = 2450 J
The work done by the man is equal to the P.E. possessed by the man.
∴ The work done by the man (W) = 2450 J ; Time (t) = 50 sec.
∴ Power delivered by the man on the bag = $\frac{\mathrm{W}}{\mathrm{t}}=\frac{2450 \mathrm{~J}}{50 \mathrm{sec}}$ = 49 watts

Question 17.
A 10 kg ball is dropped from a height of 10 m. Find (a) the initial potential energy of the ball, (b) the kinetic energy just before it reaches the ground and (c) the speed just before it reaches the ground.
Answer:
Mass of the ball = 10 kg ; It is dropped from a height h = 10 m
Acceleration due to gravity g = 9.8 m/sec²
a) The initial P.E. of the ball = mgh = 10 × 9.8 × 10 = 980 J
b) We know that the energy is conserved in a freely falling object at anywhere in its journey.
∴ The K.E. just before it reaches the ground = 980 J

c) K.E. just before it reaches the ground = 980 J ;
Speed = v m/s

Question 18.
Calculate the work done by a person in lifting a load of 20 kg from the ground and placing it 1 m high on a table. (AS 1)
Answer:
Mass of a load = 20 kg ; Height h = 1 m
Acceleration due to gravity = 9.8 m/s²
The work done by a person in lifting the load is equal to the P.E.
P.E. = mgh = 20 × 9.8 × 1 = 196 J

Question 19.
Find the mass of a body which has 5 J of kinetic energy while moving at a speed of 2 m/s. (AS 1)
Answer:
KE = 5 J
Speed v = 2 m/sec
Mass of the body = m kg.

Question 20.
A cycle together with its rider weighs 100 kg. How much work is needed to set it moving at 3 m/s?
Answer:
Mass of the bicycle with its rider = 100 kg ; Speed = 3 m/s Amount of work needed to set it moving is equal to its K.E.
K.E. = $\frac{1}{2}$mv² = $\frac{1}{2}$ × 100 × 3 × 3 = 450 J
∴ 450 J work is needed.

Question 21.
When the speed of a ball is doubled its kinetic energy (AS 1)
A) remains same
B) gets doubled
C) becomes half
D) becomes 4 times
Answer:
D) becomes 4 time

Question 22.
Two bodies of unequal masses are dropped from the top of a building. Which of the following is equal for both bodies at any instant?
A) Speed
B) Force of gravity
C) Potential energy
D) Kinetic energy
Answer:
B) Force of gravity

Question 23.
A man with a box on his head is climbing up a ladder. The work done by the man on the box is ……….
A) Positive
B) Negative
C) Zero
D) Undefined
Answer:
A) Positive

Question 24.
A porter with a suitcase on his head is climbing up steps with uniform speed. The work done by the “weight of the suitcase” on the suitcase is ……
A) Positive
B) Negative
C) Zero
D) Undefined
Answer:
B) Negative

Question 25.
How will the increasing energy needs and conservation of energy influence international peace, cooperation and security? Discuss.
Answer:
1) The increasing energy needs and conservation of energy influence international peace, co-operation and security.

2) For example, many great nations depend on the gulf countries and many countries for fuel (Petrol, diesel, etc.). The need of fuel brings co-operation among the countries. Also the need of energy (electricity, fuel, etc.) link all nations to a small circle so that there comes the peace, co-operation and security.

3) Atomic energy, thermal energy, chemical energy, etc. bring unity among all nations.

Question 26.
How would you assess the role of energy conversion occurring naturally in maintaining ecological balance of nature?
Answer:

The chemical changes in the sun is providing us the heat energy and light energy to the earth.
The heat energy of the sun evaporates the water into water vapour. This helps us to get heavy rainfall.
The solar energy is taken by the plant to change CO₂, water and chloroplast into starch which is the food of the plant. This process is called photosvnthesis.
In the photosynthesis process the plant releases O₂ (oxygen).
The leguminous (Pea plants) plants help the bacteria present at their roots. In return they help the plants to fix nitrogen in them The nitrogen cycle is maintained.
From the above examples we can understand that the energy conversion is helpful in maintaining ecological balance of nature.

Question 27.
Collect pictures showing various situation where potential energy possessed by an object depends on its shape and position. Prepare a scrap book.
Answer:
Note : Students can collect pictures of different objects at different heights, with different shapes.
e.g.: 1) A stone on a hill.
2) A book on a table and a pen on a table, etc.
Paste these pictures in your scrap book.

Question 28.
Draw a diagram to show conservation of mechanical energy in case of free falling body.
Answer:

9th Class Physical Science 10th Lesson Work and Energy InText Questions and Answers

9th Class Physical Science Textbook Page No. 161

Question 1.
How are these works being done?
Answer:
All the works mentioned above are being done by utilizing human muscular energy or electrical energy.

Question 2.
What do you need to do these works?
Answer:
Both human beings or machines need energy to do work.

Question 3.
Where does the energy spent go ultimately?
Answer:
The energy spent will be converted into another form to complete the task.

Question 4.
Is there any transfer of energy while work is being done?
Answer:
Yes. We can observe the transfer of energy while work is being done. For example, in fan, electrical energy is converted into mechanical energy. In heater, electrical energy is converted into heat energy, etc.

Question 5.
Can we do any work without transfer of energy?
Answer:
No, we cannot do any work without transfer of energy.

9th Class Physical Science Textbook Page No. 162

Question 6.
What is work?
Answer:
When a force is applied on an object and a displacement is observed, the work is said to be done.

Question 7.
Why is there difference between general view of work and scientific view of work?
Answer:
In general view, work is said to be done while reading, writing, eating or even standing for some time. We cannot observe any displacement in these cases. But in scientific view, work is said to be done when there is a displacement in the direction of force applied.

9th Class Physical Science Textbook Page No. 163

Question 8.
Observe the following examples.

a) Are all the people mentioned in the above examples doing work ?
Answer:
No.
In example (1) and (2) there is displacement. Hence they are doing work.
In example (3) and (4) there is no displacement. Hence they are not doing any work.

b) How do you define work?
Answer:
Work is defined as the product of force and displacement.
W = F × s

9th Class Physical Science Textbook Page No. 165

Question 9.
What would be the work done when the force on the object is zero?
Answer:
We know work done W = F × s
Here F = 0
⇒ W = 0 × s = 0
∴ The work done is zero.

Question 10.
What would be the work done when the displacement of the object is zero?
Answer:
Work done W = F × s
Here s = 0
⇒ W = F × 0 = 0
∴ The work done is zero.

Question 11.
Can you give some examples, where the displacement of the object is zero?
Answer:

When a boy pushes a wall, the displacement in the wall is zero.
When a boy stands for some time at one place, then there is no displacement in the boy.
Wind causes the swing in trees, but the tree does not have any displacement.

9th Class Physical Science Textbook Page No. 166

Question 12.
What happens to the speed of the ball while it moves up?
Answer:
The speed of the ball while it moves up decreases gradually, due to negative gravitational force acting on it.

Question 13.
What is the speed at its maximum height?
Answer:
The speed of the ball at its maximum height is zero.

Question 14.
What happens to the speed of the ball during its downward motion?
Answer:
During its downward motion, the ball starts from zero velocity*and its velocity gradually increases due to the positive gravitational force acting on it.

9th Class Physical Science Textbook Page No. 167

Question 15.
What is energy?
Answer:
Energy is the capability to do work.

Question 16.
How can we decide that an object possess energy or not?
Answer:
The capacity of doing work by an object on another object depends on position and state of the object which is doing work.

Question 17.
What will happen to the plate? Why?
Answer:
The plate breaks into pieces, because of the potential energy of the metal ball.

Question 18.
What changes do you notice? Why?
Answer:
The toy car is at rest before winding the key but the same toy gets energy to move when the key attached to it is wound up.

9th Class Physical Science Textbook Page No. 168

Question 19.
What could be the reason?
Answer:

The reason for breaking up of the plate is the potential energy of the metal ball.
The reason for moving the toy car is the potential energy of the spring in the toy.

Question 20.
Where does this energy go?
Answer:
The energy is transferred from one object to another.

Question 21.
Is there any energy transfer between the object doing the work and the object on which work has been done?
Answer:
Yes. In case of the toy car, the potential energy of the wounded spring is transferred to car and made the toy car to move.

Question 22.
Can any force do work without energy transfer?
Answer:
No.

9th Class Physical Science Textbook Page No. 169

Question 23.
Where do we get energy from?
Answer:
The Sun is the biggest natural and primary source of energy for us. Many other secondary sources are derived from the sun. We can also get energy from the interior of the earth and from tides of the sea.

Question 24.
Can you think of other sources of energy?
Answer:
The other natural sources of energy are wind energy, tidal energy, gravitational energy, etc. The artificial sources of energy are electrical energy, heat energy, muscular energy, etc.

Question 25.
Are there sources of energy which are not dependent on the Sun?
Answer:
Energy can exists in several forms like mechanical energy, light energy, thermal energy, sound energy, electrical energy, magnetic energy, etc. These energies, though they are artificial, they ultimately depend on the Sun.

Question 26.
Do you know why a person gets tired standing at a place for long time?
Answer:
Though the person standing is not doing any work externally, a lot of work is being done inside the body.

The muscles of the body become stretched when he stands for long time and heart has to pump more blood to muscles. This leads to loss of energy inside the body and hence he gets tired.

9th Class Physical Science Textbook Page No. 176

Question 27.
How do green plants produce food?
Answer:
Green plants produce food from sunlight, by means of photosynthesis.

Question 28.
How are fuels like coal and petroleum formed?
Answer:
Dead plants hurried deep below the earth’s surface for millions of years get converted to fuels like petroleum and coal which have chemical energy stored in them.

Question 29.
What kind of energy conversions sustain the water cycle in nature?
Answer:
Water cycle and conversation of energies :
1) The Sun heats water in oceans and seas. Then water evaporates as water vapour into the air.
Here potential energy converts into kinetic energy.

2) Rising air currents take the vapour up into the atmosphere. There cooler temperatures cause it to condense into clouds.
Here kinetic energy converts into potential energy.

3) Air currents move water vapour around the globe, cloud particles collide, grow and fall out as rain.
Here potential energy converts into kinetic energy.

4) Run off and ground water are stored as fresh water in lakes. Some precipitation falls as snow or hail.
Here kinetic energy converts into potential energy.

9th Class Physical Science Textbook Page No. 178

Question 30.
Do all of us do the work at the same rate?
Answer:
The rate of work done by all of us will not same, because the capacity to do work differs from person to person depending on the weight, nature of work, etc.

Question 31.
Is the energy spent by the force doing work the same every time?
Answer:
No. The energy spent by the force doing work depends on the capacity of the machine or the capacity of a person.

Question 32.
Do the machines consume or transfer energy at same rate every time while doing a particular work?
Answer:
No. For example, consider two grinders of different wattages. The grinder with high wattage can perform fast. In this the energy transfer is faster. The grinder with less wattage can perform slowly. In this the energy transfer is slow.

9th Class Physical Science Textbook Page No. 179

Question 33.
Whose argument is correct?
Answr:
Raheem’s argument is correct, in view of work done.
Second labourer’s argument is correct in view of working hours.

Question 34.
Is work done in two cases same?
Answer:
The work done in two cases is same.

Question 35.
Why is there a change in rate of doing work?
Answer:
The rate of doing work is different, as the two labourers completed in different times.

9th Class Physical Science Textbook Page No. 165

Question 37.
A wooden chair is dragged on the level floor and brought to the same place. Let the distance covered be ‘s’ and frictional force acted on the chair by the floor be T. What is the work done by the frictional force?
Answer:
The displacement of the chair is zero. Hence the work done is zero.

9th Class Physical Science Textbook Page No. 166

Question 38.
Lift an object up from the ground. Work done by the force exerted by you on the object moves it in upward direction. Thus the force applied is in the direction of displacement. However there exists a force of gravitation on the object at the same time
• Which one of these forces is doing positive work?
• Which one is doing negative work?
• Give reasons.
Answer:

Force applied by us on the object, because the object moves in the direction of force applied.
Force of gravitation on the object is doing negative work. Here the object is moving against to gravitational force.

9th Class Physical Science Textbook Page No. 169

Question 39.
What would happen if nature does not allow the transfer of energy? Discuss with few examples.
Answer:
If nature could not allow the transfer of energy, the normal life is not possible, e.g.:
1) If the solar energy is not transferred into chemical energy, in which plants prepare their food, we cannot find any plant on the earth. Life without plants is impossible. We can not get food, cloth, shelter and fresh air for our survival.

2) Water cycle is also an example of transfer of energy in nature. If this does not take place, ground water cannot be recharged, we cannot find any water further.

9th Class Physical Science Textbook Page No. 172

Question 40.
a) Why is it easier to stop a lightly loaded truck than heavier one that has equal speed?
Answer:
Due to its light weight, the truck with lighter load is easy to stop.

b) Does the kinetic energy of a car change more when it goes from 10 m/s to 20 m/s or when it goes from 20 m/s to 30 m/s?
Answer:
Case – I :
K.E._(I) = $\frac{1}{2}$ × m × 10 × 10 = 50m J
K.E._(F) = $\frac{1}{2}$ × m × 20 × 20 = 200m J
Difference in K.E. = 200m – 50m = 150m J

Case – II :
K.E._(I) = $\frac{1}{2}$ × m × 20 × 20 = 200m J
K.E._(F) = $\frac{1}{2}$ × m × 30 × 30 = 450m J
Difference in K.E. = 450m – 200m = 250m J

c) A person starts from rest and begins to run. The runner puts a certain momentum into himself. What is the momentum of ground? And the runner puts a certain amount of kinetic energy into himself. What is the kinetic energy of the ground?
Answer:
Here two cases arise.
Case – I :
In view of the observer, ground has no velocity. Hence the momentum and kinetic energy of the ground will be zero.

Case – II :
In view of the runner, ground has a velocity equal to the velocity of the runner.

The momentum of ground is equal to the momentum of the runner in magnitude but opposite in direction.
The K.E. of ground is also equal to the K.E of the runner in magnitude but opposite in direction.

9th Class Physical Science Textbook Page No. 175

Question 41.
Does the international space station have gravitational potential energy?
Answer:
International space station is situated in space where the gravitational force is zero. Hence the gravitational potential energy is zero.

9th Class Physical Science Textbook Page No. 178

Question 42.
a) Someone wanting to sell you a super ball claims’ that it will bounce to a height greater than the height from which it is dropped. Would you buy this ball? If yes explain, if not explain.
Answer:
Yes. I will buy this ball. When the applied force is more, the ball bounce to a height greater than the hefght from which it is dropped.

b) A ball, initially at the top of the inclined hill, is allowed to roll down: At the bottom its speed is 4 m/s. Next, the ball is again rolled down the hill, but this time it does not start from rest. It has an initial speed of 3 m/s. How fast is it going when it gets to the bottom?
Answer:
A ball start from the top of an inclined plane u} = 0 m/s; v, = 4 m/s.
Let the acceleration a’ and distance ’s’.
v₁² – u₁² = 2as …………. (1)
Now the ball is rolled down from the top of an inclined plane with a velocity 3 m/s.
∴ u₂ = 3 m/s; v₂ =?
Here the acceleration of the inclined plane is ‘a’ and the distance is ‘s’.

9th Class Physical Science Textbook Page No. 179

Question 43.
The work done by a force F₁ is larger than the work done by another force F₂. Is it necessary that power delivered by F₁ is also larger than that of F₂? Why?
Answer:
Power delivered by F₁ is may or may not be larger than F₂.
Two cases arise.
Case – I :
If the displacement is same, as F₁ > F₂, the power delivered by F₁ is larger than F₂.

Case – II :
If the displacement is different, power delivered by F₁ may not be larger than F₂.

9th Class Physical Science 10th Lesson Work and Energy Activities

Activity – 1

Question 1.
Read Ex : 1, 2, 3, and 4 from page 162 and 163. Now fill the following table.
Answer:

Activity – 2

Question 2.
Explain the relation between work done on an object and energy of the object with an activity.
Answer:

Take a hard spring and keep it on the table as shown in the figure.
Now compress the spring with your palm and release it after few seconds.
We will notice that when a spring is being compressed there is a change in its size.
When it is released it gains some energy and may even jump from the table.
The work done by your palm on the spring increases its energy and makes it to jump.
Thus we can conclude that the object which does work loses energy and the object on which work has been done gains energy.

Activity – 3

Question 3.
List the energy sources.
Answer:
The main source of energy is the sun. Many other secondary sources of energy are electrical energy, magnetic energy, chemical energy, muscular energy, geothermal energy, energy from fossil fuels, etc.

Activity – 4

Question 4.
Explain the energy of moving objects with an activity.
Answer:

Consider a metal ball and a hollow plastic block which are kept on a table side by side.
Now, suppose that the ball is separated from the block and brought to one end of the table and pushed to roll on the table with speed ‘v’.
We may notice that, the ball hits the plastic block and displaces it from point A to B.
Thus a moving ball is more energetic than the ball at rest.
We can conclude that a body possesses more energy when it is moving than when it is at rest.

Activity – 5

Question 5.
Explain potential energy with the help of a bow.
Answer:

Take a bamboo stick and make a bow as shown in figure.
Place an arrow made of a light stick on it with one end supported by the string of the bow and stretch the string gently and release the arrow.
We notice that the arrow gets separated from the bow and falls down on the ground.
Now place the arrow on the bow with one end supported by the string, applying more force and release the arrow.
We observe that the arrow flies with great speed into the air.
From this we can conclude that the bow in normal shape is not able to push the arrow, but when we stretch the string, it acquires energy to throw the arrow into air with great speed.
The energy acquired by the bow due to change in its shape is known as its potential energy.

Activity – 6

Question 6.
Write an activity to show the energy in stretched rubber band.
Answer:
Take a rubber band hold it at one end and pull it from the other end. Then release the rubber band at one of the ends.

When we release the rubber band, it strikes our hand with great force. This force is due to the energy acquired by the rubber band by stretching it.

Activity – 7

Question 7.
Show that the object at some height possesses energy.
Answer:

Take a heavy ball.
Drop it on a thick bed of wet sand from different heights from 25 cm to 1.5 m.
Observe the depression created by the ball on die bed of sand. Compare the depths of these depressions.
When the ball is dropped from different heights, the depressions created by the ball on the bed of sand will be different. ‘
As the height from which the ball was dropped increases, the depth of depression also increases.
Thus we can conclude that a body at some height possesses energy.

Activity – 8

Question 8.
List out the energy conversions in nature and in day to day life.
Answer:
Discuss various ways of energy conversions in nature as well as in our day to day activities and make a separate list of situations for natural conversions of energy and energy conversions in day to day life and write them in the given tables.
Table

Sl.No.	Situation of energy conversion in nature
1	Heat energy from the sun used for preparing food by plants gets converted into chemical energy.
2	Heat energy of the earth’s crust is useful in the formation of fossil fuels like petroleum, coal, etc.
3	Heat energy from the sun is converted into wind energy and wave energy.
4	Food we take will convert into energy.
5	Water cycle Heat energy → Water vapour → water (rain)

Table

Situation of energy conversion	Gadgets / appliances used for energy conversion
1. Conversion of electrical energy into mechanical energy	Electric fan
2. Conversion of mechanical energy into electrical energy	Dynamo
3. Conversion of chemical energy into electrical energy	Cell
4. Conversion of chemical energy into mechanical energy	Motor vehicles
5. Electrical energy into heat energy	Electric stove
6. Electrical energy into light energy	Bulb
7. Solar energy into electrical energy	Solar lamp
8. Wind energy to electrical energy	Wind mills
9. Gravitational potential energy to electrical energy	Hydro electric power

Activity – 9

Question 9.
Write an activity which shows the conservation of mechanical energy.
Answer:

Take a long thread say 50-60 cm long.
Attach a small heavy object like a metal ball at one end.
Tie other end to a nail fixed to the wall as shown in figure.
Now pull the bob of the pendulum to one side to the position A₁ and release it.
The bob swings towards opposite side and reaches the point A₂. It repeats the motion over and over again.
The potential energy of the bob is minimum at A and reaches maximum at A₁ because the height of the bob is maximum at that position.
When the bob is released from this point (A₁), its P.E. decreases and K.E. starts increasing slowly.
When the bob reaches the position A, its K.E. reaches maximum, and P.E. becomes minimum.
As the bob proceeds from A to A₂, its P.E. increases slowly and becomes maximum at A₂.
The total P.E. and K.E. at any point on the path of motion during the oscillation of the pendulum.
Thus the total mechanical energy in the system of pendulum remains constant.

Activity – 10

Question 10.
Calculate the total energy of free-fall at different heights.
An object of mass 20 kg is dropped from a height of 4 m. Compute the potential and kinetic energy in each case and write the values in the table. (Take g = 10 ms-2)

a) What do you say about total energy of system of freely falling body?
Answer:
In a freely falling body, the total energy of system (i.e. the sum of potential energy and kinetic energy) is same at any instance of its travel.

b) Is the mechanical energy conserved in the system?
Answer:
The sum of potential energy and kinetic energy at any instance of its travel is same. Hence we can say that the mechanical energy is conserved in the system.