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Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 9th Lesson Gravitation Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 9th Lesson Gravitation

Very Short Answer Questions

Question 1.
State the unit and dimension of the universal gravitational constant (G).
Answer:
F = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}^2}\)
units of G = Nm² Kg^-2
dimensional formula of G = \(\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}{[\mathrm{M}][\mathrm{M}]}\) = [M^-1 L³ T^-2]

Question 2.
State the vector form of Newtons’s law of gravitation.
Answer:
Vector form of Newton’s law of gravitation is
F = \(\frac{-G m_1 m_2}{r^3} \hat{r}\) where \(\hat{r}\) is unit vector.

Question 3.
If the gravitational force of Earth on the Moon is F, what is the gravitational force of moon on earth ? Do these forces to attraction-reaction pair ?
Answer:
F. Yes, they form action and reaction pair.

Question 4.
What would be the change in acceleration due to gravity (g) at the surface, if the radius of Earth decreases by 2% keeping the mass of Earth constant ?
Answer:
g₁r₁², l = g₂r₂², r₂ = \(\frac{98}{100}\) r₁
\(\frac{g_2}{g_1}=\frac{r_1^2}{r_2^2}=\frac{r_1^2}{\left(\frac{98}{100}\right) r_1^2}=\frac{100 \times 100}{98 \times 98}\)
\(\frac{g_2}{g_1}\) = 1.04
\(\frac{g_2}{g_1}\) – 1 = 1.04 – 1
\(\frac{g_2-g_1}{g_1}\) = 0.04

Question 5.
As we go from one planet to another, how will
a) the mass and
b) the weight of a body change ?
Answer:
a) The mass remains constant.
b) The weight (w = mg), changes from one planet to another planet.

Question 6.
Keeping the length of a simple pendulum constant, will the time period be the same on all planets ? Support your answer with reason.
Answer:
No, Time period depends on acceleration due to gravity (g).
T = 2π \(\sqrt{\frac{l}{g}}\)
g value varies from planet to planet. So time period changes.

Question 7.
Give the equation for the value of g at a depth ‘d’ from the surface of Earth. What is the value of ‘g’ at the centre of Earth ?
Answer:

1. g_d = g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\)) where d = Depth
   R = Radius of the Earth
2. At the centre of the Earth g = 0.

Question 8.
What are the factors that make ‘g’ the least at the equator and maximum at the poles ? > .
Answer:

1. g value is maximum at poles due to
   a) Rotation of the Earth
   b) Earth is flattened at the poles
   c) The equatorial radius is less at the poles.
2. g value minimum at equator due to
   a) Rotation of the earth
   b) Bulging near the equator.

Question 9.
“Hydrogen is in abundance around the sun but not around Earth”. Explain.
Answer:
The escape velocity on the sun is 620 km/s and escape velocity on the Earth is 11.2 km/s. The r.m.s velocities of hydrogen (2 km/s) is less than escape velocity on the Sun. So hydrogen is more abundant around the Sun and less around the Earth.

Question 10.
What is the time period of revolution of a geostationary satellite ? Does it rorate from West to East or from East to West ?
Answer:
Time period of geo-stationary satellite is 24 hours. It can rotate from west to east.

Question 11.
What are polar satellites ?
Answer:
Polar satellites are low altitude satellites (500 to 800 km), but they go around the poles of the earth in a north-south direction. Its time period is around 100 minutes.

Short Answer Questions

Question 1.
State Kepler’s laws of planetary motion.
Answer:
The three laws of Kepler can be stated as follows.

1. Law of orbits : All planets move in elliptical orbits with the sun situated at one of the foci.
   
2. Law of areas : The line that joins any planet to the sun sweeps equal areas in equal intervals of time.
3. Law of periods : The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.
   T² ∝ R³

Question 2.
Derive the relation between acceleration due to gravity (g) at the surface of a planet and Gravitational constant (G).
Answer:
Consider a body of mass m on the surface of the planet. Let R be the radius of the Earth and M be the mass of the Earth.
Force acting on the body due to gravitational pull of the planet is
F = m g → (1)
According to Newton’s gravitational law, Force on the body is F = \(\frac{\mathrm{GMm}}{\mathrm{R}^2}\) → (2)

From eq’s (1) and (2), we have
m g = \(\frac{\mathrm{GMm}}{\mathrm{R}^2}\)
g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
This is the relation between g and G
Mass of the earth (M) = Volume × density of the earth
M = \(\frac{4}{3}\) π R² × ρ
g = \(\frac{4}{3}\) π G R ρ

Question 3.
How does the acceleration due to gravity (g) change for the same values of height(h) and depth (d).
Answer:
a) g_h = g(1 – \(\frac{2 \mathrm{~h}}{\mathrm{R}}\)), g_d = g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\))
Same values of height and depth, h = d
g_h = g (1 – \(\frac{2 \mathrm{~d}}{\mathrm{R}}\)) and g_d = g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\))
∴ g_d > g_h

b) For large height and large depth
g_h = \(\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}\) and g_d = g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\))
If h = d = R
g_h = \(\frac{\mathrm{g}}{\left(1+\frac{\mathrm{R}}{\mathrm{R}}\right)^2}\) and g_d = g(1 – \(\frac{\mathrm{R}}{\mathrm{R}}\)) = 0
∴ g_h > g_d

Question 4.
What is orbital velocity ? Obtain an expression for it. [Mar. 14]
Answer:
Orbital velocity (V₀) : The horizontal velocity required for an object to revolve around a planet in a circular orbit is called orbital velocity.

Expression for orbital velocity :
Consider a body (satellite) of mass m, revolves round the earth in a circular orbit. Let h be the height of the satellite from the surface of the earth. Then (R + h) is the radius of the orbit.
The Gravitational force of attraction of the earth on the body is given by F = \(\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^2}\) ………….. (1)
Where M = Mass of the earth, R = Radius of the earth, G = universal gravitational constant. If V₀ is the orbital velocity of the body.
The centripetal force on the body is given by F = \(\frac{\mathrm{mv}_{\mathrm{o}}^2}{(\mathrm{R}+\mathrm{h})}\) …………… (2)
In order to make the body revolve in the same orbit, its centripetal force must be equal to the gravitational force

Question 5.
What is escape velocity ? Obtain an expression for it.
Answer:
Escape velocity : It is the minimum velocity with which a body should be projected, so that it moves into the space by overcoming the earth’s gravitational field.

Expression for escape velocity :
Consider a body of mass m thrown with a velocity v₂
Then K.E = \(\frac{1}{2}\) m v_e² …………. (1)
The gravitational force of attraction of the earth of mass M and Radius R on a body of mass m at its surface is F = \(\frac{\mathrm{GMm}}{\mathrm{R}^2}\) ……………… (2)
Gravitational P. E. = work done on the body
∴ P. E. = F × R = \(\frac{\mathrm{GMm}}{\mathrm{R}^2}\) × R
P.E. = \(\frac{\mathrm{GMm}}{\mathrm{R}}\) …………….. (3)
A body just escapes when its K. E. = P. E
\(\frac{1}{2}\) m v_e² = \(\frac{\mathrm{GMm}}{\mathrm{R}^2}\)
v_e² = \(\frac{2 \mathrm{GM}}{\mathrm{R}}\) (∵ g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\))
v_e = \(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\)
v_e = \(\sqrt{2 g R}\) (gR = \(\frac{\mathrm{GM}}{\mathrm{R}}\))
v_e = \(\sqrt{2} \times \sqrt{g R}\) (∵ v₀ = \(\sqrt{g R}\))
v_e = \(\sqrt{2}\) × v₀
∴ Escape velocity is \(\sqrt{2}\) times the orbital velocity.

Question 6.
What is a geostationary satellite ? State its uses. [T.S. Mar. 18, 15; A.P. Mar. 16]
Answer:
Geo-stationary satellite : If the period of revolution of an artificial satellite is equal to the period of rotation of earth, then such a satellite is called geo-stationary satellite.
Time period of geo-stationary satellite is 24 hours.
Uses :

1. Study the upper layers of atmosphere
2. Forecast the changes in atmosphere
3. Know the shape and size of the earth.
4. Identify the minerals and natural resources present inside and on the surface of the earth.
5. Transmit the T. V. programmes to distant objects
6. Under take space research i.e. to know about the planets, satellites, comets etc.

Question 7.
If two places are at the same height from the mean sea level; One is a mountain and other is in air at which place will ‘g’ be greater ? State the reason for your answer.
Answer:
The acceleration due to gravity on mountain is greater than that of air.
g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\) ………….. (1)
Mass (M) = volume × density (ρ)
M = \(\frac{4}{3}\)π R³ × ρ
g = \(\frac{\mathrm{G}}{\mathrm{R}^2}\) × \(\frac{4}{3}\) π R³ ρ
g = – \(\frac{4}{3}\) π R G ρ …………….. (2)
g ∝ ρ
So density is more at mountains. So g is more on mountain.

Question 8.
The weight of an object is more at the poles than at the equator. At which of these can we get more sugar for the same weight ? State the reason for your answer.
Answer:
Weight of the object at poles = m_p g_p (∵ w = mg)
Weight of the object at equator = m_e g_e
Given weight of the object at poles > weight of the object at equator
m_p g_p > m_eg_e
We know that g_p > g_e
Then m_p < m_e
Hence we can get more sugar at equator.

Question 9.
If a nut becomes loose and gets detached form a satellite revolving around the earth, will it fall down to earth or will it revolve around earth ? Give reasons for your answer.
Answer:
When a nut is detached from a satellite revolving around the earth, the nut is also moving with the speed of the satellite as the orbit of a satellite does not depend upon its mass. Hence nut is moving in the same orbit under centripetal force.

Question 10.
An object projected with a velocity greater than or equal to 11.2 kms it will not return to earth. Explain the reason.
Answer:
The escape velocity on the surface of the earth (v_e) = 11.2 km/s. Any object projected with the velocity greater then (or) equal to 11.2 km/s it will not come back. Because it has overcome the earth’s gravitational pull.
So an object never come back to earth.

Long Answer Questions

Question 1.
Define gravitational potential energy and derive an expression for it associated with two particles of masses m₁ and m₂.
Answer:
Gravitational potential energy : Gravitational potential energy of a body at a point in a gravitational field of another body is defined as the amount of work done in brining the given body from infinity to that point without acceleration.

Expression for gravitational potential energy : Consider a gravitational field due to earth of mass M, radius R. The mass of the earth can be supposed to be concentrated at its centre 0. Let us calculate the gravitational the potential energy of the body of mass m placed at point p in the gravitational field, where OP = r and r > R. Let OA = x and AB = dx.
The gravitational force on the body at A will be
F = \(\frac{\mathrm{GMm}}{\mathrm{X}^2}\) ……………… (1)
Small amount of work done in bringing the body without acceleration through a small distance dx is given by
dw = Force × displacement
dw = F × dx
dw = \(\frac{\mathrm{GMm}}{\mathrm{X}^2}\) × dx ……………… (2)
Total work done in bringing the body from infinity to point P is given by

This work done is stored in the body as its gravitational potential energy (u)
∴ Gravitational potential energy (u) = \(\frac{\mathrm{GMm}}{\mathrm{r}}\) ……………….. (4)
Gravitational potential energy associated with two particles of masses m, and m2 separated by a distance r is given by
u = –\(\frac{G m_1 m_2}{r}\) ……………….. (5) (if we choose u = 0 as r → ∞).

Question 2.
Derive an expression for the variation of acceleration due to gravity (a) above and (b) below the surface of the Earth.
Answer:
i) Variation of g with height:
When an object is on the surface of the earth, it will be at a distance r = R radius of the earth, then we have g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
Where G = universal gravitational constant, M = Mass of the earth
When the object is at a height h above the surface of the earth, Then r = R + h

g value decreases with altitude.

ii) Variation of g with depth :
Let us assume that the earth to be a homogeneous uniform sphere of radius R, mass M and of uniform density ρ.
We know that g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\) = \(\frac{4}{3}\) π ρ G R ………………… (1)
Consider a body of mass m be placed at a depth d.

The value of g decreases with depth.

Question 3.
State Newton’s Universal Law of gravitation. Explain how the value of the Gravitational constant (G) can be determined by Cavendish method.
Answer:
Newton’s law of gravitation :
“Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversly proportional to the square of the distance between them”
Determination of G value by cavendish method :

1. In 1798 Henry Cavendish determined the value of G experimentally.
2. The bar AB has two small lead spheres attached at its ends.
3. The bar is suspended from a rigid support by a fine wire.
4. Two large lead spheres are brought close to the small ones but on opposite sides as shown in figure.
5. The big spheres attract the nearby small ones by equal and opposite force as shown in figurer.
6. There is no net force on the bar but only a torque which is clearly equal to F times the length of the bar. When F is the force of attraction between a big sphere and its neighbouring small sphere.
7. Due to this torque, the suspended wire gets twisted till such time as the restoring torque of the wire equals the gravitational torque.
   Restoring torque = τ θ ………………… (1)
   Where τ is restoring couple per unit twist 0 is the angle
8. If d is the seperation between big and small balls having masses M and m.
   Gravitational force (F) = \(\frac{\mathrm{GMm}}{\mathrm{d}^2}\) ……………… (2)
   ix) If L is the length of the bar A B, then the torque arising out of F is F multiplied by L. At equilibrium, this is equal to the restoring torque.
   \(\frac{\mathrm{GMm}}{\mathrm{d}^2}\) = τ θ
   observations of θ thus enables one to calculate G.
   The measurement of G = 6.67 × 10^-11 Nm²/ Kg²

Problems

(Gravitational Constant ‘G’ = 6.67 × 10^-11 Nm²/ Kg^-2; Radius of earth ‘R’ = 6400 km; Mass of earth ‘ME’ = 6 × 10²⁴ kg)

Question 1.
Two spherical balls each of mass 1 kg are placed 1 cm apart. Find the gravitational force of attraction between them.
Solution:
m₁ = m₂ = 1 kg, d = 1 cm = 1 × 10^-2 m
F = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}^2}\)
F = \(\frac{6.67 \times 10^{-11} \times 1 \times 1}{\left(1 \times 10^{-2}\right)^2}\) = 6.67 × 10^-7N

Question 2.
The mass of a ball is four times the mass of another ball. When these balls are separated by a distance of 10 cm, the force of gravitation between them is 6.67 × 10^-7 N. Find the masses of the two balls.
Solution:
m₁ = m, m₂ = 4m, d = 10 cm = 10 × 10^-2 m,
F = 6.67 × 10^-7 N
G = 6.67 × 10^-11 Nm²/kg ²
F = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}^2}\)
6.67 × 10^-7 = \(\frac{6.67 \times 10^{-11} \times \mathrm{m} \times 4 \mathrm{~m}}{\left(10 \times 10^{-2}\right)^2}\)
4 m² = 10²
m² = \(\frac{100}{4}\) = 25
m = 5 kg
∴ m₁ = m = 5 kg
m₂ = 4m = 4 × 5 = 20 kg

Question 3.
Three spherical balls of masses 1 kg, 2kg and 3 kg are placed at the corners of an equilateral triangle of side 1 m. Find the magnitude of gravitational force exerted by the 2 kg and 3kg masses on the 1 kg mass.
Solution:
The force of attraction at 2 kg on the 1 kg particle
F₂ = \(\frac{\mathrm{Gmn}{\mathrm{~d}^2}\) = \(\frac{\mathrm{G} \times 1 \times 2}{1^2}\)
F₂ = 2 G

Question 4.
At a certain height above the earth’s surface, the acceleration due to gravity is 4% of its value at the surface of earth. Determine the height.
Solution:
g_h = 4% of g = \(\frac{4}{100}\)g, R = 6400 km
g_h = \(\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}\)
\(\frac{4 \mathrm{~g}}{100}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}\)
\(\left(1+\frac{h}{R}\right)^2=\frac{100}{4}\) = 25
1 + \(\frac{h}{R}\) = 5
\(\frac{h}{R}\) = 4
h = 4 × R = 4 × 6400 = 25,600 km.

Question 5.
A satellite is orbiting the earth at a height of 1000km. Find its orbital speed.
Solution:
h = 1000 km
Oribital velocity (v₀) = \(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}}\)
G = 6.67 × 10^-11 Nm²/kg ², M = 6 × 10²⁴ kg
R + h = 6400 + 1000 = 7400 km
= 7400 × 10³m
v₀ = \(\sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{7400 \times 10^3}}\)
v₀ = \(\sqrt{0.5408 \times 10^{10}}\) = 73.54 × 10³ m/s
v₀ = 7.354km/s

Question 6.
A satellite orbits the earth at a height equal to the radius of earth. Find it’s
(i) orbital speed and
(ii) Period of revolution
Solution:
Height h = R

Question 7.
The gravitational force of attraction between two objects decreases by 36% when the distance between them is increased by 4 m. Find the original distance between them.
Solution:
F₁ = F, F₂ = \(\frac{64}{100}\) F
d₁ = d, d₂ = (d + 4) m

5d = 4d + 16
d = 16 m.

Question 8.
Four identical masses of m are kept at the corners of a square of side a. Find the gravitational force exerted on one of the masses by the other masses.
Solution:

Question 9.
Two spherical balls of 1 kg and 4kg are separated by a distance of 12 cm. Find the distance of a point from the 1 kg mass at which the gravitational force on any mass becomes zero.
Solution:
m₁ = 1 kg, m₂ = 4kg, r = 12 cm
∴ x = \(\frac{r}{\sqrt{\frac{m_2}{m_1}}+1}\) from m₁
= \(\frac{12}{\sqrt{\frac{4}{1}}+1}=\frac{12}{2+1}=\frac{12}{3}\) = 4 cm
At x = 4 cm the gravitational force is zero.

Question 10.
Three uniform spheres each of mass m and radius R are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any one of the spheres due to the other two.
Solution:

Question 11.
Two satellites are revolving round the earth at different heights. The ratio of their orbital speeds ¡s 2 : 1. If one of them is at a height of 100 km, what is the height of the other satellite ?
Solution:

4R + 400 = R + h₂
h₂ = 3R + 400 = 3 × 6400 + 400
= 19200 + 400
h₂ = 19,600 km.

Question 12.
A satellite is revolving round in a circular orbit with a speed of 8 km s-1 at a height where the value of acceleration due to gravity is 8 m s-2. How high is the satellite from the Earth’s surface ? (Radius of planet = 6000 km)
Solution:
v₀ = 8 km/s = 8000 m/s
g_h = 8 m/s², R = 6000 km = 6000 × 10³ m
∴ v₀ = \(\sqrt{\frac{G M}{R+h}}=\sqrt{g(R+h)}\)
v₀² = g(R + h)
(8000)² = 8(6000 × 10³ + h)
6000 × 10³ + h = 8 × 10⁶
h = (8 – 6) 10⁶
h = 2 × 10⁶m
h = 2000 × 10³ = 2000 km.

Question 13.
(a) Calculate the escape velocity of a body from the Earth’s surface, (b) If. the Earth were made of wood, its mass would be 10% of its current mass. What would be the escape velocity, if the Earth were made of wood ?
Solution:
R = 6400 × 10³m,

Additional Problems

Question 1.
Answer the following :
a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ?
b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ?
c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull, (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why ?
Solution:
a) We cannot shield a body from the gravitational influence of nearby matter because the gravitational force on the body due to near by matter is independent of the presence of other matter, whereas it is not so in the case of electrical forces it means the gravitational screens are not possible.

b) Yes, if the size of the spaceship orbiting around the earth is large enough, an astronaut inside the spaceship can detect the variation in g.

c) Tidal effect depends inversly on the cube of the distance, unlike force which depends inversly on the square of the distance. Since the distance of moon from the ocean water is very small as compared to the distance of sun from the ocean water on earth. Therefore, the tidal effect of moon’s pull is greater than the tidal effect of the sun.

Question 2.
Choose the correct alternative :
a) Acceleration due to gravity increase^ decreases with increasing altitude.
b) Acceleration due to gravity increases/decreases with increas¬ing depth (assume the earth to be a sphere of uniform density).
c) Acceleration due to gravity is independent of mass of the earth/ mass of the body.
d) The formula – G Mm (1/r₂ – 1/r₁) is more/less accurate than the formula mg (r₂ – r₁) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth.
Solution:
a) decreases
b) decreases
c) mass of the body
d) more

Question 3.
Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ?
Solution:
Here, T_e = 1 year; T_p = \(\frac{T_c}{2}=\frac{1}{2}\) year; r_e = 1
A.U.; r_p = ?
Using Kepler’s third law, we have
r_p = r_e\(\left(\frac{T_p}{T_e}\right)^{2 / 3}\) = \(1\left(\frac{1 / 2}{1}\right)^{2 / 3}\)
= 0.63 AU

Question 4.
Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 10⁸m. Show that the mass of Jupiter is about-one-thousandth that of the sun.
Solution:
For a satellite of Jupiter, orbital period,
T₁ = 1.769 days = 1.769 × 24 × 60 × 60 s
Radius of the orbit of satellite,
r₁ = 4.22 × 10⁸ m
mass of Jupiter, M₁ is given by M₁
= \(\frac{4 \pi^2 \times\left(4.22 \times 10^8\right)^3}{G \times(1.769 \times 24 \times 60 \times 60)^2}\)
= \(\frac{4 \pi^2 r_1^3}{\mathrm{GT}_1^2}\) ……………. (1)
We know that the orbital period of earth around the sun,
T = 1 year = 365.25 × 24 × 60 × 60 s
Oribital radius, r = 1 A.U = 1.496 × 10¹¹ m

Question 5.
Let us assume that our galaxy consists of 2.5 × 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000ly from the galactic centre take to complete one revolution ? Take the diameter of the Milky Way to be 105 ly.
Solution:
Here, r = 50,000 ly =50,000 × 9.46 × 10¹⁵m
= 4.73 × 10²⁰m.
M = 2.5 × 10¹¹ solar, mass = 2.5 × 10¹¹ × (2 × 10³⁰) kg
= 5.0 × 10⁴¹ kg
We know that, M = \(\frac{4 \pi^2 r^3}{\mathrm{GT}^2}\)
or T = \(\left(\frac{4 \pi^2 r^3}{G M}\right)^{1 / 2}\)
= \(\left[\frac{4 \times(22 / 7)^2 \times\left(4.73 \times 10^{20}\right)^3}{\left(6.67 \times 10^{11}\right) \times\left(5.0 \times 10^{41}\right)}\right]^{1 / 2}\)
= 1.12 × 10¹⁶S.

Question 6.
Choose the correct alternative :
a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potentia! energy.
b) The energy required to launch an orbiting .satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.
Solution:
a) Kinetic energy
b) Less.

Question 7.
Does the escape speed of a body from the earth depend on (a) the mass of the body, (b) the Ideation from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched ?
Solution:
The escape velocity is independent of mass of the body and the direction of projection it depends upon the gravitational potential at the point from where the body is launched. Since this potential depends slightly on the latitude and height of the point, therefore, the escape velocity depends slightly on these factors.

Question 8.
A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit ? Neglect any mass loss of the comet when it comes very close to the Sun.
Solution:
A comet while going on elliptical orbit around the sun has constant angular momentum and total energy at all locations but other quantities vary with locations.

Question 9.
Which of the following symptoms is likely to afflict an astronaut in space
(a) swollen feet,
(b) swollen face,
(c) headache,
(d) orientational problem.
Solution:
a) We know that the legs carry the weight of the body in the normal position due to gravity pull. The astronaut in space is in weightless state. Hence, swollen feet may not affect his working.

b) In the conditions of weightless, the face of the astronaut is expected to get more supply. Due to it, the astronaut may develop swollen face. As eyes, ears, nose, mouth etc. are all embedded in the face, hence, swollen face may affect to great extent the seeing / hearing / eating / smelling capabilities of the astronaut in space.

c) Headache is due to metal strain it will persist whether a person is an astronaut in space or he is on earth it means headache will have the same effect on the astronaut in space as on a person on earth.

d) Space also has orientation. We also have the frames of reference in space. Hence, orientational problem will affect the astronaut in space.

Question 10.
In the following two exercises, choose the correct answer from among the given ones : The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig) (i) a, (ii) b, (iii) c, (iv) 0.

Solution:
We know that the gravitational potential is constant at all points upside a spherical shell. Therefore, the gravitational potential gradient at all points inside the spherical shell is zero [i.e as v is constant, \(\frac{\mathrm{dv}}{\mathrm{dr}}\) = 0].

Since gravitational intensity is equal to negative of the gravitational potential gradient, hence the gravitational intensity is zero at all points inside a hollow spherical shell. This indicates that the gravitational forces acting on a particle at any point inside a spherical shell, will be symmetrically placed. Therefore if we remove the upper hemispherical shell, the net gravitational forces acting on the particle at the centre Q or at some other point P will be acting downwards which will also be the direction of gravitational intensity it is so because, the gravitational intensity at a point is the gravitational force per unit mass at that point. Hence the gravitational intensity at the centre Q will be along c, i.e., option (iii) is correct.

Question 11.
For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g. .
Solution:
As per explanation given in the answer of Q. 10, the direction of gravitational intensity at P will be along e i.e., option (ii) is correct.

Question 12.
A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero ? Mass of the sun = 2 × 10³⁰ kg, mass of the earth 6 × 10²⁴ kg. Neglect the effect of other planets etc. (orbital radius 1.5 × 10¹¹ m).
Solution:
Here M_s = 2 × 10³⁰ kg ; M_c = 6 × 10²⁴ kg ; r = 1.5 × 10¹¹ m .
Let x be the distance of a point from the earth where gravitational forces on the rocket due to sun and earth become equal and opposite. Then distance of rocket from the sun
= (r – x). If m is the mass of rocket then
\(\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{m}}{(\mathrm{r}-\mathrm{x})^2}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{x}^2} \text { or } \frac{(\mathrm{r}-\mathrm{x})^2}{\mathrm{x}^2}=\frac{\mathrm{M}_{\mathrm{s}}}{\mathrm{M}_{\mathrm{e}}}\)

Question 13.
How will you ‘weigh the sun1, that is estimate its mass ? The mean orbital radius of the earth around the sun is 1.5 × 10⁸ km.
Solution:
To estimate the mass of the sun, we require, the time period of revolution T of one of its planets (say the earth). Let M_s, M_e be the masses of sun and earth respectively and r be the mean orbital radius of the earth around the sun. The gravitational force acting on earth due to sum is
F = \(\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{M}_{\mathrm{e}}}{\mathrm{r}^2}\)
Let, the earth be moving in circular orbit around the sun, with a uniform angular velocity ω, the centripetal force acting on earth is.
F¹ = M_erω² = M_er \(\frac{4 \pi^2}{T^2}\)
As this centripetal force is provided by the gravitational pull of sun on earth, So
\(\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{M}_{\mathrm{e}}}{\mathrm{r}^2}=\mathrm{M}_{\mathrm{e}} \mathrm{r} \frac{4 \pi^2}{\mathrm{~T}^2} \text { or } \mathrm{M}_{\mathrm{s}}=\frac{4 \pi^2 \mathrm{r}^3}{G \mathrm{~T}^2}\)
Knowing r and T, mass M_s of the sun can be estimated.
In this Question, we are given, r = 1.5 × 10⁸ km
= 1.5 × 10¹¹ m
T = 365 days = 365 × 24 × 60 × 60 s
∴ M_s = \(\frac{4 \times(22 / 7)^2 \times\left(1.5 \times 10^{11}\right)^3}{\left(6.67 \times 10^{-11}\right) \times(365 \times 24 \times 60 \times 60)^2}\)
= 2 × 10³⁰ kg.

Question 14.
A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 10⁸ km away from the sun ?
Solution:
Here, T_s = 29.5 T_e; R_e = 1.5 × 10⁸ km; R_s =?
Using the relation, \(\frac{\mathrm{T}_{\mathrm{s}}^2}{\mathrm{R}_{\mathrm{s}}^3}=\frac{\mathrm{T}_{\mathrm{e}}^2}{\mathrm{R}_{\mathrm{e}}^3}\)
or R = Re \(\left(\frac{\mathrm{T}_{\mathrm{s}}}{\mathrm{T}_{\mathrm{e}}}\right)^{2 / 3}\)
= 1.5 × 10⁸ \(\left(\frac{29.5 \mathrm{~T}_{\mathrm{e}}}{\mathrm{T}_{\mathrm{e}}}\right)^{2 / 3}\)
= 1.43 × 10⁹ km.

Question 15.
A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Solution:
Weight of the body = mg = 63N
At height h, the value of g is given by
g’ = \(\frac{g R^2}{(R+h)^2}=\frac{g R^2}{(R+R / 2)^2}\) = 4/9 g
Gravitational force on body at height h is
F = mg’ = m × \(\frac{4}{9}\) g = \(\frac{4}{9}\) mg
= \(\frac{4}{9}\) × 63 = 28N

Question 16.
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface ?
Solution:
wt. of body at a depth d = mg¹
= m × g \(\left(1-\frac{d}{R}\right)\)
= 250 \(\left(1-\frac{R / 2}{R}\right)\)
= 125 N

Question 17.
A rocket is fired vertically with a speed of 5 km s^-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = 6.0 × 10²⁴ kg; mean radius of the earth = 6.4 × 10⁶ m; G = 6.67 × 10^-11 N m² kg^-2.
Solution:
Let the rocket be fired with velocity v from the surface of earth and it reaches a height h from the surface of earth where its velocity becomes zero.
Total energy of rocket at the surface of energy

or h = \(\frac{\mathrm{Rv}^2}{2 \mathrm{gR}-\mathrm{v}^2}\)
= \(\frac{\left(6.4 \times 10^6\right) \times\left(5 \times 10^3\right)^2}{2 \times 9.8 \times\left(6.4 \times 10^6\right)-\left(5 \times 10^3\right)^2}\)
= 1.6 × 10⁶m

Question 18.
The escape speed of a projectile on the earth’s surface is 11.2 km s-1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth ? Ignore the presence of the sun and other planets.
Solution:
Here, v_e = 11.2 kms^-1, velocity of projection of the body v = 3ve. Let m be the mass of the projectile and v₀ be the velocity of the projectile when far away from the earth (i.e) out of gravitational field of earth) then from the law of conservation of energy
\(\frac{1}{2}\) mv₀² = \(\frac{1}{2}\) mv² – \(\frac{1}{2}\) mv_e²
or v₀ = \(\sqrt{v^2-v_e^2}\)
= \(\sqrt{(3 v e)^2-v_e^2}\)
= \(\sqrt{8} v_e=\sqrt{8}\) × 11.2 = 31.68 kms^-1

Question 19.
A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence ? Mass of the satellite = 200 kg; mass of the earth = 6.0 × 1024 kg; radius of the earth = 6.4 × 10⁶ m; G = 6.67 × 10^-11 N m² kg^-2.
Solution:
Total energy of orbiting satellite at a hight h.
= – \(\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}+\frac{1}{2} \mathrm{mv}^2\)
= – \(\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}+\frac{1}{2} m \frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}\)
= \(\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}\)
energy expended to rocket the satellite out of the earth’s gravitational field.
= – (total energy of orbiting satellite)
= \(\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}\)
= \(\frac{\left(6.67 \times 10^{-11}\right) \times\left(6 \times 10^{24}\right) \times 200}{2\left(6.4 \times 10^6+4 \times 10^5\right)}\)
= 5.9 × 10⁹J

Question 20.
Two stars each of one solar mass (= 2 × 10^30< kg) are approaching each other for a head on collision. When they are a distance i09 km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10^4< km. Assume the stars to remain undistorted until they collide. (Use the known value of G).
Solution:
Here, mass of each star, M = 2 × 10^30< kg
initial distance between two stars, r = 10^9<
km = 10^12< m.
initial potential energy of the system = – \(\frac{\text { GMM }}{r}\)
Total K.E. of the stars = \(\frac{1}{2}\) mv^2< + \(\frac{1}{2}\) mv^2<
= Mv^2<
Where v is the speed of stars with which they collide. When the stars are about to collide, the distance between their centres, r^1< = 2R.
∴ Final potential energy of two starts = \(\frac{-\mathrm{GMM}}{2 \mathrm{R}}\)
since gain in K.E. is at the cost of loss in P.E

Question 21.
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium ? If so, is the equilibrium stable or unstable ?
Solution:
Gravitational field at the mid – point of the line joining the centres of the two spheres.
= \(\frac{\mathrm{GM}}{(r / 2)^2}(-\hat{r})+\frac{\mathrm{GM}}{(r / 2)^2} \hat{r}=0\)
Gravitational potential at the mid point of the list joining the centres of the two spheres is
v = \(\frac{-\mathrm{GM}}{r / 2}+\left(\frac{-\mathrm{GM}}{r / 2}\right)=\frac{-4 \mathrm{GM}}{r}\)
\(\frac{-4 \times 6.67 \times 10^{-11} \times 100}{1.0}\) = -2.7 × 10^-8< J/kg
As the effective force on the body placed at mid-point is zero, so the body is in equilibrium. If the body is displaced a little towards either mass body from its equilibrium position, it will not return back to its initial position of equilibrium. Hence, the body is in unstable equilibrium.

Question 22.
As you have learnt in the text, a geo-stationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0 × 10²⁴ kg, radius = 6400 km.
Solution:
Gravitational potential at height h from the surface of earth is
v = \(\frac{-\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}\)
= \(\frac{-6.67 \times 10^{-11} \times\left(6 \times 10^{24}\right)}{\left(6.4 \times 10^6+36 \times 10^6\right)}\)
= -9.4 × 10⁶ J/kg.

Question 23.
A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity ? (mass of the sun = 2 × 10^30< kg).
Solution:
The object will remain struck to the surface of star due to gravity, if the accerlation due to gravity is more than the centrifugal accerlation due to its rotation.
Accerlation due to gravity, g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
= \(\frac{6.67 \times 10^{-11} \times 2.5 \times 2 \times 10^{30}}{(12000)^2}\)
= 2.3 × 10¹² m/s²
centrifugal accerlation = rw²
= r(2πv)²
= 12000 (2π × 1.5)²
= 1.1 × 10⁶ ms^-2
since g > rω² , therefore the body will remain struck with the surface of star.

Question 24.
A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system ? Mass of the space ship = 1000 kg; mass of the sun = 2 × 10³⁰ kg; mass of mars = 6.4 × 10^23< kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 10^8< km; G = 6.67 × 10^-11< N m² kg ² .
Solution:
Let R, be the radius of the orbit of mars and R be the radius of the mars. M be the mass of the sun and M’ be the mass of mars. If m is the mass of the space ship, then potential energy of space-ship due to gravitational attraction of the sun = \(\frac{-\mathrm{GMm}}{\mathrm{R}}\)
potential energy of space – ship due to gravitational attraction of mars = – \(\frac{\mathrm{GM}^1 \mathrm{~m}}{\mathrm{R}^1}\)
since K.E of space ship is zero, therefore total energy of spaceship
= \(\frac{-\mathrm{GMm}}{\mathrm{R}}\) – \(\frac{\mathrm{GM}^1 \mathrm{~m}}{\mathrm{R}^1}\)
= – Gm \(\left(\frac{M}{R}+\frac{M^1}{R^1}\right)\)
∴ energy required to rocket out the spaceship from the solar system = – (total energy of space ship)

Question 25.
A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s^-1. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it ? Mass of mars = 6.4 × 10^23< kg; radius of mars = 3395 km; G = 6.67 × 10^-11< N m² kg^-2.
Solution:
Let m = mass of the rocket, M = mass of the mars and
R = radius of mars. Let v be the initial velocity of rocket.
Initial K.E = \(\frac{1}{2}\) mv²; Initial P.E = – \(\frac{-\mathrm{GMm}}{\mathrm{R}}\)
Total initial energy = \(\frac{1}{2}\) mv² – \(\frac{-\mathrm{GMm}}{\mathrm{R}}\)
since 20% of K.E is lost, only 80% is left behind to reach the height. Therefore
Total energy available = \(\frac{80}{100} \times \frac{1}{2}\) mv²
– \(\frac{-\mathrm{GMm}}{\mathrm{R}}\) = 0.4 mv² – \(\frac{-\mathrm{GMm}}{\mathrm{R}}\)
If the rocket reaches the higher point which is at a height h from the surface of Mars, its
K.E. is zero and P.E. = \(\frac{-\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}\)
using principle of conservation of energy, we have
0.4 mv² – \(\frac{\mathrm{GMm}}{\mathrm{R}}=-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}\)
or \(\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}=\frac{\mathrm{GM}}{\mathrm{R}}\) – 0.4 v²

Textual Examples

Question 1.
Let the speed of the planet at the perihelion P in Fig. be υ_p and the Sun- planet distance SP be r_p. Relate {r_p, υ_p} to the corresponding quantities at the aphelion {r_A, υ_A}. Will the planet take equal times to traverse BAC and CPB ?

(a) An ellipse traced out by a planet around the sun. The colsest point is P and the farthest point is A. P is called the perihelion and A the aphelion. The semimajor axis (a) is half the distance AP
Answer:
The magnitude of the angular momentum at P is L_p = m_p r_p υ_p. Similarly, L_A = m_p r_A υ_A. From angular momentum conservation
m_p r_p υ_p = m_p r_A υ_A
or \(\frac{v_p}{v_A}=\frac{r_A}{r_p}\)

Question 2.
Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC. (a) What is the force acting on a mass,2m placed at the centroid O of the triangle ? (b) What is the force if the mass at the vertex A is doubled ?
Take AO = BO = CO = 1 m (see Fig)

Three equal masses are placed at the three vertices of the ∆ABC. A mass 2m is placed at the centroid O.
Answer:
(a) The angle between OC and the positive x- axix is 30° and so is the angle between OB and the negative x-axis. The individual forces a vector notation are

From the principle of superposition and the law of vector addition, the resultant gravitational force F_R on (2m) at O is
F_R = F_OA + F_OB + F_OC
F_R = 2Gm² \(\hat{\mathrm{j}}\) + 2Gm² \(-\hat{\mathrm{i}}\) cos 30° – \(\hat{\mathrm{j}}\) sin 30°) + 2Gm² (\(\hat{\mathrm{i}}\) cos 30° – \(\hat{\mathrm{j}}\) sin 30°) = 0
Alternatively, one expects on the basis of symmetry that the resultant force ought to be zero.

(b) By symmetry the x-component of the force cancels out. The y-component survives.
F_R = 4Gm² \(\hat{\mathrm{j}}\) – 2Gm² \(\hat{\mathrm{j}}\) = 2Gm² \(\hat{\mathrm{j}}\)

Question 3.
Find the potential energy of a system of four particles placed at the vertices of a square of side l. Also obtain the potential at the centre of the square.
Answer:
We have four mass pairs at distance l and two diagonal pairs at distance \(\sqrt{2}\)1 Hence,

= \(\frac{2 \mathrm{Gm}}{1}\left(2+\frac{1}{\sqrt{2}}\right)\) = -5.41 \(\frac{\mathrm{Gm}^2}{l}\)
The gravitational potential U(r) at the centre of the square (r = \(\sqrt{2}\) l / 2) is
U(r) = \(-4 \sqrt{2} \frac{\mathrm{Gm}}{\mathrm{l}}\)

Question 4.
Two uniform solid spheres of equal radii R, but mass M and 4 M have a centre to centre separation 6 R, as shown in Fig. The two spheres are held fixed A projeetile of mass m is projected from the surface of the sphere of mass M directly towards the centre of the second sphere. Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere.

Answer:
If ON = r, we have
\(\frac{\mathrm{GMm}}{\mathrm{r}^2}=\frac{4 \mathrm{GMm}}{\left(6 \mathrm{R}-\mathrm{r}^2\right)}\)
(6R – r)² = 4r²
6R – r = ±2r
r = 2R or – 6R.
The neutral point r = -6R does not concern us in this example. Thus ON = r = 2R.
Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface of M is
E_i = \(\frac{1}{2} \mathrm{~m} v^2-\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{4 \mathrm{GMm}}{5 \mathrm{R}}\)
The mechanical energy at N is purely potential.
E_N = \(-\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{4 \mathrm{GMm}}{4 \mathrm{R}}\)
From the principle of conservation of mechanical energy
\(\frac{1}{2} v^2-\frac{G M}{R}-\frac{4 G M}{5 R}=-\frac{G M}{2 R}-\frac{G M}{R}\)
υ² = \(\frac{2 G M}{R}\left(\frac{4}{5}-\frac{1}{2}\right)\)
υ² = \(\left(\frac{3 \mathrm{GM}}{5 R}\right)^{1 / 2}\)

Question 5.
The planet Mars has two moons, phobos and delmos. (i) phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4 × 10³ km. Calculate the mass of Mars, (ii) Assume that Earth and Mars move in circular orbits around the sun, with the Martian orbits being 1.52 times the orbital radius of the earth. What is the length of the Martain year in days ?
Answer:
(i) We employ T² = K (R_E + h)³ (where K = 4π² / GM_E) with the Earth’s mass replaced by the Martian mass M_m

(ii) Once again Kepler’s third law comes to our aid,
\(\frac{T_M^2}{T_E^2}=\frac{R_{M S}^3}{R_{E S}^3}\)
Where R_MS is the Mars-Sun distance and R_ES is the Earth-Sun distance.
∴ T_M = (1.52)^3/2 × 365
= 684 days
For example. the ratio of the semi-minor to semi-major axis for our Earth is, b/a = 0.99986.

Question 6.
Weighing the Earth : You are given the following data g = 9.81 ms² R_E = 6.37 x106 m the distance to the moon R = 3.4 × 10⁸ m and the time period of the moons revolution is 27.3 days. Obtain the mass of the Earth M_E in two different ways.
Answer:
(1) From g = \(\frac{F}{m}=\frac{G M_E}{R_E^2}\)
M_E = \(\frac{g R_E^2}{G}\)
= \(\frac{9.81 \times\left(6.37 \times 10^6\right)^2}{6.67 \times 10^{-11}}\)
= 5.97 × 10²⁴kg. (by Method – 1)

(2) The moon is a satellite of the Earth. From the derivation of Kepler’s third law

= 6.02 × 10²⁴kg (by Method – 2)
Both methods yield almost the same answer the difference between them being less than 1%.

Question 7.
Express the constant k T² = K (R_E + h)² where K = 4π²/GM_E of in days and kilometres. Given k = 10^-13 s² m^-3. The moon is at a distance of 3.84 × 10⁵ km from the earth. Obtain its time period of revolution in days.
Answer:
Given
k = 10^-13 s² m^-3 (d = day)
= 10^-13 \(\left[\frac{1}{(24 \times 60 \times 60)^2} d^2\right]\)
\(\left[\frac{1}{(1 / 1000)^3 \mathrm{~km}^3}\right]\) = 1.33 × 10^-14 d² km^-3
Using T² = K (R_E + h)³ (where k = 4π²/ GM_E) and the given value of k the time period of the moon is
T² = (1.33 × 10^-14) (3.84 × 10⁵)³
T = 27.3 d

Question 8.
A 400 kg satellite is in a circular orbit of radius 2R_E about the Earth. How much energy is required to transfer it to a circular orbit of radius 4R_E? What are the changes in the kinetic and potential energies ?
Answer:
Initially,
E₁ = \(\)
While finally
E_f = \(\)
The change in the total energy is
∆E = E_f – E_i
= \(\frac{\mathrm{GM}_E \mathrm{~m}}{8 R_E}=\left(\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}^2}\right) \frac{\mathrm{mR} \mathrm{R}_{\mathrm{E}}}{8}\)
∆E = \(\frac{\mathrm{gm} \mathrm{R}_{\mathrm{E}}}{8}=\frac{9.81 \times 400 \times 6.37 \times 10^6}{8}\)
= 3.13 × 10⁹J
The kinetic enegy is reduced and it mimics ∆E, namely, ∆K = K_f – K_i = -3.13 × 10⁹ J.
The change in potential energy is twice the change in the total energy, namely
∆V = V_f – V_i= -6.25 × 10⁹ J

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 10th Lesson Mechanical Properties of Solids Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 10th Lesson Mechanical Properties of Solids

Very Short Answer Questions

Question 1.
State Hooke’s law of elasticity.
Answer:
“With in the elastic limit stress directly proportional to the strain”.
Stress ∝ strain
Stress = k × strain
k = \(\frac{\text { Stress }}{\text { Strain }}\)
Where k is modulus of elasticity.

Question 2.
State the units and dimensions of stress.
Answer:

1. Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{A}\)
   S.l units → N/m² (or) Pascal
2. Dimensional formula
   Stress = \(\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}\) = [ML^-1T^-2].

Question 3.
State the units and dimensions of modulus of elasticity.
Answer:
Modulus of elasticity (k) = \(\frac{\text { Stress }}{\text { Strain }}\)
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2]

Question 4.
State the units and dimensions of Young’s modulus.
Answer:
Young’s modules (y) = \(\frac{\text { LongitudinalStress }}{\text { Longitudinal Strain }}=\frac{\frac{F}{A}}{\frac{e}{L}}\)
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2]

Question 5.
State the units and dimensions of modulus of rigidity.
Answer:
Modulus of rigidity (G) = \(\frac{F}{A \theta}=\frac{\text { Shearing Stress }}{\text { Shearing Strain }}\)
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2].

Question 6.
State the units and dimensions of Bulk modulus.
Answer:
Bulk modulus (B) = \(\frac{\text { Bulk Stress }}{\text { Bulk Strain }}=\frac{-P V}{\Delta V}\)
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2].

Question 7.
State the examples of nearly perfect elastic and plastic bodies.
Answer:

• Nearly perfect elastic bodies are quartz fibre.
• Nearly perfect plastic bodies are dough and day.

Short Answer Questions

Question 1.
Define Hooke’s Law of elasticity, proportionality, permanent set and breaking stress.
Answer:
Hooke’s law : “With in the elastic limit stress is directly proportional to the strain”.
Stress ∝ strain
Stress = k × strain
Where k is modulus of elasticity.
Proportionality limit: The maximum stress developed in a body till it obeys Hookes law is called proportionality limit.
Permanent Set : Permanent deformation produced when a body is stretched beyond its elastic limit.
Breaking stress : The maximum stress a body can bear before it breaks.

Question 2.
Define modulus of elasticity, stress, strain and Poisson’s ratio.
Answer:
Modulus of elasticity : It is the ratio stress applied on a body to the strain produced in the body.
k = \(\frac{\text { Stress }}{\text { Strain }}\)
S.I unit → N/m² (or) Pascal
Stress : When a body is subjected to an external force, the force per unit area is called stress.
Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{A}\)
S.I unit → N/m² (or) Pascal
Strain : When deforming forces act on a body, the fractional deformation produced in the body. It has no units

Poisson’s ratio (σ) : The ratio between lateral strain to longitudinal strain of a body is called poisson’s ratio.
σ = \(\frac{\text { Lateral Strain }}{\text { Longitudinal Strain }}=\frac{\frac{-\Delta \mathrm{r}}{\mathrm{r}}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}\)

Question 3.
Define Young’s modulus. Bulk modulus and Shear modulus.
Answer:
Young’s modulus (y) : With in the elastic limit, the ratio of longitudinal stress to longitudinal strain is called young’s modulus.
y = \(\frac{\text { Longitudinal Stress }}{\text { Longitudinal Strain }}=\frac{\frac{F}{A}}{\frac{e}{L}}\)
y = \(\frac{\mathrm{FL}}{\mathrm{A} \cdot \mathrm{e}}\)
S.I unit → N/m² (or) Pascal

Bulk modulus (B) : With in the elastic limit, it is defined as the ratio of Bulk stress to Bulk strain
B = \(\frac{\text { Bulk Stress }}{\text { Bulk Strain }}\)
B = \(\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{-\Delta \mathrm{V}}{\mathrm{V}}}=\frac{-\mathrm{PV}}{\Delta \mathrm{V}}\) (∵ -ve sign indicates volume decreases)
S.I unit → N/m² (or) Pascal

Rigidity modulus (G) : With in the elastic limit, it is defined as the ratio of shearing stress to shearing strain.
G = \(\frac{\text { Shearing Stress }}{\text { Shearing Strain }}\)
G = \(\frac{\frac{F}{A}}{\theta}=\frac{F}{A \theta}\)
S.I unit → N/m² (or) Pascal

Question 4.
Define stress and explain the types of stress. [T.S. Mar. 16]
Answer:
Stress : The restoring force per unit area is called stress
∴ Stress = \(\frac{\text { Restoring Force }}{\text { Area }}=\frac{F}{A}\)
Stress is classified into three types.

1. Longitudinal stress
2. Volume (or) Bulk stress
3. Tangential (or) shearing stress

1. Longitudinal stress (or) Linear stress : When a normal stress changes the length of a body, then it is called longitudinal stress.
Longitudinal stress = \(\frac{F}{A}\)

2. Volume (or) Bulk stress : When a normal stresschanges the volume of a body, then it is called volume stress.
Volume stress = \(\frac{\text { Force }}{\text { Area }}\) = pressure.

3. Tangential (or) shearing stress : When the stress is tangential to the surface due to the application of forces parallel to the surface, then the stress is called tangential stress.
Tangential stress = \(\frac{F}{A}\).

Question 5.
Define strain and explain the types of strain.
Answer:
Strain : It is the ratio of change in dimension to its original dimension.
Strain = \(\frac{\text { Changes in dimension }}{\text { Original dimension }}\)
Strain is of three types.
1. Longitudinal strain : It is the ratio of change in length to its original length.
Longitudinal strain = \(\frac{\text { Changes in length }}{\text { Original length }}=\frac{e}{L}\)

2. Shearing strain (or) Tangential strain : When simultaneous compression and extension in mutually perpendicular direction takes place in a body, the change of shape it under goes, is called shearing strain.

Shearing strain (θ) = \(\frac{1}{L}\)

3. Bulk (or) volume strain : It is the ratio of change in volume to its original volume is called bulk strain. It is called Bulk (or) volume strain.
Bulk strain = \(\frac{\text { Change in Volume }}{\text { Original Volume }}=\frac{\Delta V}{V}\)

Question 6.
Define strain energy and derive the equation for the same. [Mar. 14]
Answer:
The potential energy stored in a body when stretched is called strain energy.
Let us consider a wire of length L and cross – sectional area A. Let x be the change in length of the wire by the application of stretching force F.

Strain energy per unit volume = \(\frac{1}{2} \times \frac{F}{A} \cdot \frac{x}{L}\)
= \(\frac{1}{2}\) × stress × strain.

Question 7.
Explain why steel is preferred to copper, brass, aluminium in heavy-duty machines and in structural designs.
Answer:
The elastic behavior of materials plays an important role in everyday life. Designing of buildings, the structural design of the columns, beams and supports require knowledge of strength of material used. The elasticity of the material is due to stress developed with in the body, when extenal force acts on it. A material is of more elastic nature if it develops more stress (or) restoring force. Steel develops more stress than copper, brass, aluminium for same strain. So steel is more elastic.
y = \(\frac{\text { Stress }}{\text { Strain }}\)

Question 8.
Describe the behaviour of a wire under gradually increasing load. [A.P. – Mar. ’18, ’16, ’15; TS – Mar. ’18, ’15, ’13]
Answer:
When the load is increased in steps, a graph is drawn between stress on y-axis and corresponding strain on x-axis.

1. Proportionality limit : In the linear position OA, stress is proportional to strain, i.e. Hookes law is obeyed by the wire upto point A. The graph is a straight lint. When ever the stretching force at A is removed, the wire regains its original length.
A is called proportionality limit.

2. Elastic limit : In the graph B is the elastic limit.
Through the wire doesnot obey Hooke’s law at B. The wire regains its original length after removing the stretching force at B. upto point B the wire is under elastic behaviour.

3. Permanent set (or) yield point: In the graph c is the yield point. If the stretching force at c is removed, the wire doesnot regain its original length and the length of the wire changes permanently. In this position the wire flows like a viscous liquid. After the point c, the wire is under plastic behavior, c is called permanent set (or) yield point.

4. Breaking point: When the stress increased, the wire becomes thinner and thinner. When the stress increases to a certain limit the wire breaks. The stress at which the wire breaks is called breaking stress and the point D is called breaking point.

5. Elastic fatigue : The state of temperary loss of elastic nature of a body due to continuous strain is called elastic fatigue. When a body is subjected to continuous strain with in the elastic limit, it appears to have lost Hastic property temporarily to some extent and becomes weak.

Question 9.
Two identical solid balls, one of ivory and the other of wet-day are dropped from the same height onto the floor. Which one will rise to greater height after striking the floor and why ?
Answer:
Ivory ball rise to greater height after striking the floor. The ivory ball regain its original shape after striking the floor. The elastic property of ivory ball is more. Where as wet-day ball does not regain its original shape after striking the floor.

So wet-day ball acts like plastic body.

Question 10.
While constructing buildings and bridges a pillar with distributed ends is preferred to a pillar with rounded ends. Why ?
Answer:
Use of pillars (or) columns is also very common in buildings and bridges. A pillar with rounded ends supports less load than that with a distributed shape at the ends. The precise design of a bridge (or) a building has to take into account the conditions under which it will function, the cost and long period, reliability of usable materials.

Question 11.
Explain why the maximum height of a mountain on earth is approximately 10 km ?
Answer:
The maximum height of a mountain on earth is 10 km, can also be provided by considering the elastic properties of rocks. A mountain base is not under uniform compression and this provides some shearing stress to rocks under which they can flow. The stress due to all the material on the top should be less than the critical shearing stress at which the rocks flow.

At the bottom of a mountain of height h, the force per unit area due to the weight of the mountain is hρg. Where ρ is the density of the mountain. The material at the bottom experiences this force in the vertical direction and the sides of the mountain are free.
There is a shear component, approximately hρg itself.
Elastic limit for 3 typical rock is 30 × 10⁷ N/m²
hρg = 30 × 10⁷ (ρ = 3 × 10³ kg/m³)
h = \(\frac{30 \times 10^7}{3 \times 10^3 \times 10}\)
h = 10 km.

Question 12.
Explain the concept of Elastic Potential Energy in a stretched wire and hence obtain the expression for it.
Answer:
“When a wire is put under a tensile stress, work is done against the inter-atomic forces. The work is stored in the wire in the form of elastic potential energy”.

Expression for elastic potential energy : Consider a wire of length L and area of cross section A is subjected to a deforming force F along the length of the wire. Let the length of the wire is elongated by l.
Young’s modulus (y) = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
F = \(\frac{\mathrm{yAl}}{\mathrm{L}}\) ……………. (1)
Work done due to further elongation of small length dl
Work done (dw) = F × dl = (\(\frac{\mathrm{yAl}}{\mathrm{L}}\))dl ……………… (2)
Total work done in increasing the length of the wire from L to (L + l)
w = \(\int_0^1 \frac{\mathrm{yAl} }{\mathrm{L}} \mathrm{dl}=\frac{\mathrm{yA}}{2} \times \frac{l^2}{\mathrm{~L}}\)
w = \(\frac{1}{2} \times \mathrm{y} \times\left(\frac{l}{\mathrm{~L}}\right)^2 \times \mathrm{Al}\)
= \(\frac{1}{2}\) y × stress² × volume of the wire
w = \(\frac{1}{2}\) × stress × strain × volume of the wire.
This work is stored in the wire in elastic potential energy (u).

Long Answer Question

Question 1.
Define Hooke’s law of elasticity and describe an experiment to determine the Young’s modulus of the material of a wire.
Answer:
Hooke’s law : With in the elastic limit, stress is directly proportional to the strain.
Stress ∝ strain
Stress = k × strain
Where k is modulus of elasticity.
Determination of young’s modulus of the material of a wire:

Young’s Modulus of the Material of a wire

1. It consists of two long straight wires of same length and same area of cross-section suspended side by side from a rigid .support.
2. The wire A (reference wire) carries a metre scale M and a pan to place a weight.
3. The wire B (experimental wire) carries a pan in which known weights can be placed.
4. A vernier scale v is attached to a pointer at the bottom of the experimental wire B and the main scale M is fixed to the wire A.
5. The weights placed in the pan, the elongation of the wire is measured by the vernier arrangement.
6. The reference wire is used to compensate for any change in length that may occur due to change in room temperature.
7. Both the reference and experimental wires are given an initial small load to keep the wires straight and the vernier reading is noted.
8. Now the experimental wire is gradually loaded with more weights, the vernier reading is noted again.
9. The difference between two vernier readings gives the elongation produced in the wire.
10. Let r and L be the radius and initial length of the experimental wire. Let M be the mass that produced an elongation ∆L in the wire.
    Young’s modulus of the material of the experimental wire is given by
    y = \(\frac{\text { Longitudinal Stress }}{\text { Longitudinal Strain }}=\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}\)
    y = \(\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}\)
    y = \(\frac{\mathrm{MgL}}{\pi r^2 \times \Delta \mathrm{L}}\)
    From above equation young’s modulus of the material of the wire is determined.

Problems

Question 1.
A copper wire of 1mm diameter is stretched by applying a force of 10 N. Find the stress in the wire.
Solution:
D = 1 m.m = 10^-3m, r = \(\frac{D}{2}\) = 0.5 × 10^-3 m.
F = 10 N
Stress = \(\frac{F}{A}=\frac{F}{\pi r^2}\)
= \(\frac{10}{3.14 \times\left(0.5 \times 10^{-3}\right)^2}\)
= 1.273 × 10⁷ N/m².

Question 2.
A tungsten wire of length 20 cm is stretched by 0.1 cm. Find the strain on the wire.
Solution:
L = 20 × 10^-2 m, ∆L = 0.1 × 10^-2 m
Strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{0.1 \times 10^{-2}}{20 \times 10^{-2}}\) = 0.005.

Question 3.
If an iron wire is stretched by 1 %, what is the strain on the Wire ?
Solution:
Strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}\) = 1 %
Strain = \(\frac{1}{100}\) = 0.01

Question 4.
A brass wire of diameter 1mm and length 2 m is streched by applying a force of 20N. If the increase in length is 0.51 mm. find
(i) the stress,
(ii) the strain and
(iii) the Young’s modulus of the wire.
Solution:
D = 1 m.m, r = \(\frac{D}{2}\) = 0.5 × 10^-3 m
L = 2 m, F = 20 N, ∆L = 0.51 m.m = 0.51 × 10^-3 m

= 9.984 × 10¹⁰ N/m²

Question 5.
A copper wire and an aluminium wire have lengths in the ratio 3 : 2, diameters in the ratio 2 : 3 and forces applied in the ratio 4: 5. Find the ratio of increase in length of the two wires. (Y_cu = 1.1 × 10¹¹ Nm^-2, Y_Al = 0.7 × 10¹¹ Nm^-2).
Solution:

= \(\frac{4}{5} \times \frac{3}{2} \times\left(\frac{0.7 \times 10^{11}}{1.1 \times 10^{11}}\right) \times\left(\frac{3}{2}\right)^2\)
\(\frac{\Delta \mathrm{L}_1}{\Delta \mathrm{L}_2}=\frac{189}{110}\)

Question 6.
A brass wire of cross-sectional area 2 mm2 is suspended from a rigid support and a body of volume 100 cm3 is attached to its other end. If the decrease in the length of the wire is 0.11 mm, when the body is completely immersed in water, find the natural length of the wire.
(Y_brass = 0.91 × 10¹¹ Nm^-2, ρ_water = 10³ kg m^-3).
Solution:
A = πr² = 2 × 10^-6 m², V = 100 × 10^-6 = 10^-4 m³
∆L = 0.11 × 10^-3 m, y_Brass = 0.91 × 10¹¹ N/m², ρ = 10³ kg/m³
y = \(\frac{M g L}{A \times \Delta L}=\frac{v \rho g L}{A \times \Delta L}\)
L = \(\frac{\mathrm{yA} \Delta \mathrm{L}}{\mathrm{v \rho g}}=\frac{0.91 \times 10^{11} \times 2 \times 10^{-6} \times 0.11 \times 10^{-3}}{10^{-4} \times 10^3 \times 9.8}\)
L = 2.04 m.

Question 7.
There are two wires of same material. Their radii and lengths are both in the ratio 1:2. If the extensions produced are equal, what is the ratio of loads ?
Solution:

Question 8.
Two wires of different material have same lengths and areas of cross¬section. What is the ratio of their increase in length when forces applied are the same ?
(Y₁ = 0.9 × 10¹¹ Nm^-2, Y₂ = 3.6 × 10¹¹ Nm^-2)
Solution:
y₁ = 0.9 × 10¹¹ Nm^-2
y₂ = 3.6 × 10¹¹ Nm^-2
y = \(\frac{F L}{A \times \Delta L}\)
∆L ∝ \(\frac{1}{y}\)
\(\frac{(\Delta L)_1}{(\Delta L)_2}=\frac{y_2}{y_1}=\frac{3.6 \times 10^{11}}{0.9 \times 10^{11}}=\frac{4}{1}\)

Question 9.
A metal wire of length 2.5 m and area of cross-section 1.5 × 10^-6 m² is stretched through 2 mm. if its Young’s modulus is 1.25 × 10¹¹ N.m², find the tension in the wire.
Solution:
L = 2.5 m, A = 1.5 × 10^-6 m²
∆L = 2 × 10^-9 m
y = 1.25 × 10¹¹ N.m²
y = \(\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}\)
F = \(\frac{\mathrm{yA} \Delta \mathrm{L}}{\mathrm{L}}\)
= \(\frac{1.25 \times 10^{11} \times 1.5 \times 10^{-6} \times 2 \times 10^{-3}}{2.5}\)
F = 150 N

Question 10.
An aluminium wire and a steel wire of the same length and cross-section are joined end-to-end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is 1.35 mm, find the ratio of the
(i) stress in the two wires and
(ii) strain in the two wires.
(Y_Al = 0.7 × 10¹¹ N.m², Y_Steel = 2 × 10¹¹ Nm²).
Solution:
Y_Al = 0.7 × 10¹¹ N.m², Y_Steel = 2 × 10¹¹ Nm²
∆L₁ + ∆L₂ = 1.35 mm ……………… (1)

Question 11.
A 2 cm cube of some substance has its upper face displaced by 0.15 cm due to a tangential force of 0.3 N while keeping the lower face fixed, Calculate the rigidity modulus of the substance.
Solution:
L = 2 × 10^-2 m, A = L² = 4 × 10^-4 m²
∆x = 0.15 × 10^-2 m
F = 0.3 N
G = \(\frac{\frac{F}{A}}{\frac{\Delta x}{L}}=\frac{F L}{A \Delta x}\) (∵ θ = \(\frac{\Delta x}{L}\))
G = \(\frac{0.3 \times 2 \times 10^{-2}}{4 \times 10^{-4} \times 0.15^6 \times 10^{-2}}\)
G = 10⁴ N/m²

Question 12.
A spherical ball of volume 1000 cm³ is subjected to a pressure of 10 Atmosphere. The change in volume is 10^-2 cm³. If the ball is made of iron, find its bulk modulus.
(1 atmosphere = 1 × 10⁵ Nm^-2).
Solution:
v = 1000 cm³ = 1000 × 10^-6 = 10^-3 m³
p = 1 atm = 1 × 10⁵ = 10⁵ N/m²
-∆v = 10^-2 cm³ = 10^-2 × 10^-6 = 10^-8 m³
Bulk modulus (B) = \(\frac{-p v}{\Delta v}\)
= \(\frac{10^5 \times 10^{-3}}{10^{-8}}\)
B = 10¹⁰ N/m².

Question 13.
A copper cube of side of length 1 cm is subjected to a pressure of 100 atmosphere. Find the change in its volume if the bulk modulus of copper is 1.4 × 10¹¹ Nm^-2. (1 atm = 1 × 10⁵ Nm^-2).
Solution:
l = 1 cm = 10^-2 m
V = Volume of the cube = l³ = 1cm³
= 10^-6 m³
P = 100 atm = 100 × 10⁵ = 10⁷ N/m²
B = 1.4 × 10¹¹ N/m²
B = \(\frac{-P V}{\Delta V}\)
-∆V = \(\frac{P V}{B}=\frac{10^7 \times 10^{-6}}{1.4 \times 10^{11}}\)
-∆V = 0.7143 × 10^-10 m³.

Question 14.
Determine the pressure required to reduce the given volume of water by 2%. Bulk modulus of water is 2.2 × 10⁹ Nm^-2.
Solution:
\(\frac{-\Delta V}{V}\) = 2 % = \(\frac{2}{100}\)
B = 2.2 × 10⁹ Nm²
B = \(\frac{-P V}{\Delta V}\)
P = -B × \(\frac{\Delta V}{V}\)
= 2.2 × 10⁹ × \(\frac{2}{100}\)
P = 4.4 × 10⁷ N/m².

Question 15.
A steel wire of length 20 cm is stretched to increase its length by 0.2 cm. Find the lateral strain in the wire if the Poisson’s ratio for steel is 0.19.
Solution:
L = 20 cm = 20 × 10^-2 m
∆L = 0.2 × 10^-2 m
σ = 0.19
σ = \(\frac{\text { Lateral strain }}{\text { Longitudinal strain }\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)}\)
Lateral strain = σ × \(\frac{\Delta L}{L}\)
= \(\frac{0.19 \times 0.2 \times 10^{-2}}{20 \times 10^{-2}}\)
= 0.0019

Additional Problems

Question 1.
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10^-5 m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10^-5 m²,
Solution:
Given, for steel wire, a₁ = 3.0 × 10^-5 m², l₁ = 4.7 m, ∆l₁ = ∆l, F₁ = F
For copper wire, a₂ = 4.0 × 10^-5 m², l₂ = 3.5 m, ∆l₂ = ∆l, F₂ = F .
Let y₁, y₂ be the young modulus of steel wire and copper wire respectively.
∴ y₁ = \(\frac{F_1}{a_1} \times \frac{l_1}{\Delta l_2}=\frac{F}{3.0 \times 10^{-5}} \times \frac{4.7}{\Delta l}\) ………….. (i)
and y₂ = \(\frac{F_2 \times l_2}{a_2 \times \Delta l_2}=\frac{F \times 3.2}{4 \times 10^{-5} \times \Delta l}\)
\(\frac{\mathrm{y}_1}{\mathrm{y}_2}=\frac{4.7 \times 4 \times 10^{-5}}{3.5 \times 3.0 \times 10^{-5}}\) = 1.8
Here y₁ : y₁ = 1.8 : 1.

Question 2.
Figure shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material ?

Solution:
a) From graph, for stress = 150 × 10⁶ Nm^-2 the corresponding strain = 0.002
young’s modulus y = \(\frac{\text { Stress }}{\text { Strain }}=\frac{150 \times 10^6}{0.002}\)
= 7.5 × 10¹⁰ Nm^-2

b) Approximate yeild strength will be equal to the maximum stress it can substain with out crossing the elastic limit. Therefore, the approximate yeild strength
= 300 × 10⁶ Nm^-2
= 3 × 10⁸ Nm^-2

Question 3.
The stress-strain graphs for materials A and B are shown in Fig.

The graphs are drawn to the same scale.
a) Which of the materials has the greater Young’s modulus ?
b) Which of the two is the stronger material ?
Solution:
a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence young’s modulus (= stress/ strain) is greater for A than that of B.

b) A is stronger than B. Strength of a material is measured by the amount of stress required to cause fracture, corresponding to the point of fracture.

Question 4.
Read the following two statements below carefully and state, with reasons, if it is true or false.
a) The Young’s modulus of rubber is greater than that of steel;
b) The stretching of a coil is determined by its shear modulus.
Solution:
a) False, because for a given stress there is more strain in rubber than steel and modulus of elasticity is inversly proportional to strain.

b) True because the strecting of coil simply changes its shape without any change in the length of the wire used in the coil due to which shear modulus of elasticity is involved.

Question 5.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. The unloaded length of steel wire is 1.5 m and that of brass Wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Solution:
For steel wire, total force on steel wire,
F₁ = 4 + 6 = 10 kg, f = 10 × 9.8 N
l₁ = 1.5 m, ∆l₁ = ?, 2r₁ = 0.25cm or r₁ =(0.25/2)cm = 0.125 × 10^-2 m
y₁ = 2.0 × 10¹¹ pa
For brass wire, F₂ = 6.0 kg, f = 6 × 9.8 N
2r₂ = 0.25 cm or r₂ = (0.25/2) cm = 0.125 × 10^-2 m,
y₂ = 0.91 × 10¹¹ pa, l₂ = 1.0 m, ∆l₂ = ?

Question 6.
The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a verticle wall. A mass of 100 kdis then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Solution:
A = 0.10 × 0.10 = 10^-2 m², F = mg = 100 × 10 N
Shearing strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\left(\frac{\mathrm{F}}{\mathrm{A}}\right)}{\eta}\)
or ∆L = \(\frac{F L}{A \eta}\)
= \(\frac{(100 \times 10) \times(0.10)}{10^{-2} \times\left(25 \times 10^9\right)}\) = 4 × 10^-7 m.

Question 7.
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the corn pressional strain of each column.
Solution:
Load on each column, F = \(\frac{50,000}{4}\) kgwt
= \(\frac{50,000 \times 9.8}{4}\) N
A = π(r₂² – r₁²) = \(\frac{22}{7}\)(0.60)² – (0.30)²]
= \(\frac{22}{7}\) 0.27 m²
Compression strain = \(\frac{\frac{F}{A}}{y}=\frac{F}{A y}\)
= \(\frac{50,000 \times 9.8}{4 \times\left(\frac{22}{7} \times 0.27\right) \times 2.0 \times 10^{11}}\)
= 7.21 × 10^-7.

Question 8.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Solution:
Here, A = 15.2 × 19.2 × 10^-6 m², F = 44,
500 N, η = 42 × 10⁹ Nm^-2
Strain = \(\frac{\text { Stress }}{\text { Modulus of elasticity }}\)
= \(\frac{\frac{F}{A}}{\eta}=\frac{F}{A \eta}=\frac{44500}{\left(15.2 \times 19.2 \times 10^{-6}\right) \times 42 \times 10^9}\)
= 3.65 × 10^-3.

Question 9.
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress Is not to exceed 10⁸ N m^-2, what is the maximum load the cable can support ?
Solution:
Maximum load macimum stress × area of cross-section
= 10⁸πr²
= 10⁸ × \(\frac{22}{7}\) × (1.5 × 10^-2)²
= 7.07 × 104 N.

Question 10.
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
Solution:
As each wire has same tension F, so each wire has same extansion due to mass of rigid bar. As each wire is of same length, hence each wire has same strain, if D is the diameter of wire, then
y = \(\frac{4 \mathrm{~F} / \pi \mathrm{D}^2}{\text { Strain }}\) or D² ∝ 1/y
\(\frac{D_{\mathrm{cu}}}{\mathrm{D}_{\mathrm{iron}}}=\sqrt{\frac{\mathrm{y}_{\mathrm{iron}}}{\mathrm{y}_{\mathrm{cu}}}}\)
= \(\sqrt{\frac{190 \times 10^9}{110 \times 10^9}}=\sqrt{\frac{19}{11}}\) = 1.31.

Question 11.
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.
Solution:
Here, m = 14.5 kg, l = r = 1m, v = 2 rps, A = 0.065 × 10^-4 m²
Total pulling force on mass, when it is at the lowest position of the vertical circle is
F = mg + mrω²
= mg + mr4πv²
= 14.5 × 9.8 + 14.5 × 1 × 4 × (\(\frac{22}{7}\))² × 2²
= 142.1 +2291.6
= 2433.7 N
y = \(\frac{F}{A} \times \frac{l}{\Delta l}\) or ∆l = \(\frac{F l}{A y}\)
= \(\frac{2433.7 \times 1}{\left(0.065 \times 10^{-4}\right) \times\left(2 \times 10^{11}\right)}\)
= 1.87 × 10^-3 m
= 1.87 mm.

Question 12.
Compute the bulk modulus of water from the following data : Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Solution:
Here, V = 100 lit = 100 × 10^-3 m³, P = 100 atm = 100 × 1.013 × 10⁵ Pa
V + ∆V = 100.5 litre or ∆V= (V + ∆V) – V
= 100.5 – 100
= 0.5 litre = 0.5 × 10^-3 m³
We known that bulk modulus, B = \(\frac{\mathrm{PV}}{\Delta \mathrm{V}}\)
= \(\frac{100 \times 1.013 \times 10^5 \times 100 \times 10^{-3}}{0.5 \times 10^{-3}}\)
= 2.026 × 10⁹ Pa
Bulk modulus of air = 1.0 × 10⁵ Pa
\(\frac{\text { Bulk modulus of water }}{\text { Bulk modulus of air }}=\frac{2.026 \times 10^9}{1.0 \times 10^5}\)
= 2.026 × 10¹⁴.
It is so because gases are much more compressible than those of liquids. The molecules in gases are very poorly coupled to their neighbours as compared to those of gases.

Question 13.
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m^-3 ?
Solution:
Here, P = 80.0 atm = 80.0 × 1.013 × 10⁵ pa,
compressibility = \(\left(\frac{1}{B}\right)\) = 45.8 × 10^-11 pa^-1
Density of water at surface,
ρ = 1.03 × 10³ kg m^-3
Let p be the density of water at the given depth, if v and v’ are volumes of certain mass M of ocean water at surface and at a given

Putting this value in (i) we get
1 – \(\frac{1.03 \times 10^3}{\rho^{\prime}}\) = 3.712 × 10^-3 or
ρ’ = \(\frac{1.03 \times 10^3}{1-3.712 \times 10^{-3}}\) = 1.034 × 10³ kg m^-3.

Question 14.
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Solution:
Here, P = 10 atm = 10 × 1.013 × 10⁵ pa,
B = 37 × 10⁹ Nm^-2
Volumetric strain = \(\frac{\Delta V}{V}=\frac{P}{B}\)
= \(\frac{10 \times 1.013 \times 10^5}{37 \times 10^9}\) = 2.74 × 10^-5
∴ Fractional change in volume = \(\frac{\Delta V}{V}\)
= 2.74 × 10^-5

Question 15.
Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ Pa.
Solution:
Here, L = 10 cm = 0.10m; P = 7 × 10⁶ pa B = 140 Gpa = 140 × 10⁹ pa
As B = \(\frac{\mathrm{PV}}{\Delta \mathrm{V}}=\frac{\mathrm{Pl}^3}{\Delta \mathrm{V}}\) or ∆V = \(\frac{\mathrm{Pl}^3}{\mathrm{~B}}\)
= \(\frac{\left(7 \times 10^6\right) \times(0.10)^3}{140 \times 10^9}\) = 5 × 10^-8 m³
= 5 × 10^-2 mm³

Question 16.
How much should be pressure on a litre of water be changed to compress it by 0.10% ?
Solution:
Here, V = 1 litre = 10^-3m³;
∆V/V = 0,10/100 = 10^-3
B = \(\frac{P V}{\Delta V}\) or P = B \(\frac{\Delta V}{V}\)
= (2.2 × 10⁹) × 10^-3 = 2.2 × 10⁶pa

Question 17.
Anvils made of single crystals of diamond, with the shape as shown in Fig. are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil ?

Solution:
Here, D = 0.5 mm = 0.5 × 10^-3m = 5 × 10^-4m
F = 50,000 N = 5 × 10⁴N
Pressure at the tip of anvil.
P = \(\frac{F}{\pi D^2 / 4}=\frac{4 F}{\pi D^2}\)
P = \(\frac{4 \times\left(5 \times 10^4\right)}{(22 / 7) \times\left(5 \times 10^{-4}\right)^2}\) = 2.5 × 10¹¹pa.

Question 18.
A rod of length 1.05 m having negligible mass is supported at its end by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. The cross-sectional areas of wires A and B are 1.0 mm² and 2.0 mm², respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

Solution:
For Steel wire A, l₁ = l, A₁ = 1 mm²
Y₁ = 2 × 10¹¹Nm^-2
For aluminium wire B, l₂ = l;
A₂ = 2mm²; y₂ = 7 × 10¹⁰ Nm^-2
a) Let mass m be suspended from the rod at the distance × from the end where wires A is connected. Let F₁ and F₂ be the tension in two wires and there is equal stress in two wires, then
\(\frac{F_1}{A_1}=\frac{F_2}{A_2} \text { or } \frac{F_1}{F_2}=\frac{A_1}{A_2}=\frac{1}{2}\) …………………. (i)
Taking moment of forces about the point of suspension of mass from the rod, we have
F₁x = F₂ (1.05 – x) or \(\frac{1.05-x}{x}=\frac{F_1}{F_2}=\frac{1}{2}\)
or 2.10 – 2x = x or x = 0.70m = 70 cm

b) Let mass m be supended from the rod at distance × from the end where wire A is connected. Let F₁ and F₂ be the tension in the wires and there is equal strain in the two wires i.e.
\(\frac{F_1}{A_1 Y_1}=\frac{F_2}{A_2 Y_2}\) or \(\frac{F_1}{F_2}=\frac{A_1}{A_2} \frac{Y_1}{Y_2}\)
= \(\frac{1}{2} \times \frac{2 \times 10^{11}}{7 \times 10^{10}}=\frac{10}{7}\)
As the rod is stationary, so F₁x = F₂(1.05 – x)
or \(\frac{1.05-x}{x}=\frac{F_1}{F_2}=\frac{10}{7}\)
or 10 x = 7.35 – 7x
or x = 0.4324 m
x = 43.2cm.

Question 19.
A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10^-2 cm² is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the- mid-point.
Solution:
Refer the figure, let x be the depression at the mid point i.e CD = x
In fig. AC = CB = Z = 0.5m
m = 100g = 0.100 kg
AD = BD = (l² + x²)^1/2
Increase in length, ∆l = AD + DB – AB
= 2 AD – AB
= 2 (l² + x²)^1/2 – 2l
= 2l(1 + \(\frac{x^2}{l^2}\))^1/2 – 2l
= 2l (1 + \(\frac{x^2}{2 l^2}\)) – 2l = \(\frac{x^2}{l}\)
Strain = \(\frac{\Delta l}{2 l}=\frac{x^2}{2 l^2}\)
If T is the tension in the wire, then 2T cos θ
= mg or T = \(\frac{\mathrm{mg}}{2 \cos \theta}\)

Question 20.
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 10⁷ Pa ? Assume that each rivet is to carry one quarter of the load.
Solution:
Here, r = 6/2 = 3mm = 3 × 10^-3 m, max.
stress = 6.9 × 10⁷ Pa
Max . load on a rivet = Max stress × area of cross section
= 6.9 × 10⁷ × (22/7) × (3 × 10^-3)²
∴ Maximum tension
= 4 (6.9 × 10⁷ × \(\frac{22}{7}\) × 9 × 10^-6)
= 7.8 × 10³N.

Question 21.
The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom ?
Solution:
Here, P = 1.1 × 10⁸ Pa, V = 0.32 m³,
B = 16 × 10¹¹Pa
∆V = \(\frac{\mathrm{PV}}{\mathrm{B}}\)
= \(\frac{\left(1.1 \times 10^8\right) \times 0.32}{1.6 \times 10^{11}}\)
= 2.2 × 10^-4m³

Textual Examples

Question 1.
A structural steel rod has a radius of 10 mm and a length of 10 m. A 100 KN force stretches it along its length. Calculate (a) stress, (b) elongation, and (c) strain on the rod. Young’s modulus, of structural steel ¡s 2.0 × 10¹¹ Nm².
Answer:
a) Given Stress = \(\frac{F}{A}=\frac{F}{\pi r^2}\)
= \(\frac{100 \times 10^3 \mathrm{~N}}{3.14 \times\left(10^{-2} \mathrm{~m}\right)^2}\) = 3.18 × 10⁸ Nm^-2

b) The elongation
∆L = \(\frac{(\mathrm{F} / \mathrm{A}) \mathrm{L}}{\mathrm{Y}}\)
= \(\frac{\left(3.18 \times 10^8 \mathrm{Nm}^2\right)(1 \mathrm{~m})}{2 \times 10^{11} \mathrm{Nm}^{-2}}\)
= 1.59 × 10^-3 m
= 1.59 mm

c) The strain is given by
Strain = ∆L/L = (1.59 × 10^-3) km
= 1.59 × 10^-3 = 0.16%

Question 2.
A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.
Answer:
From y = \(\frac{\sigma}{\varepsilon}\)
we have stress = strain × Young’s modulus.
W/A = Y_c × (∆L_c/L_c) = Y_s × (∆L_s/L_s)
where the subscripts c and s refer to copper and stainless steel respectively,
∆L_c/∆L_s = (Y_s/Y_s) () (L_c/L_s)
Given L_c = 2.2 m, L_s = 1.6 m,
Y_c = 1.1 × 10¹¹ N.m^-2 and Y_s = 2.0 × 10¹¹ N.m^-2.
∆L_c/∆L_s = \(\frac{2.0 \times 10^{11}}{1.1 \times 10^{11}}=\frac{2.2}{1.6}\) = 2.5
∆L_c + ∆L_s = 7.0 × 10^-4 m
Solving the above equations,
∆L_c = 5.0 × 10^-4 m and ∆L_s = 2.0 × 10^-4 m.
∴ W = (A × Y_c × ∆L_c)L_c
= π(1.5 × 10^-3)² × \(\left[\frac{\left(5.0 \times 10^{-4} \times 1.1 \times 10^{11}\right)}{2.2}\right]\) = 1.8 × 10² N.

Question 3.
In a human pyramid in a circus, the entire weight of the balanced group is supported by the legs of a performer who is lying on his back (as shown in Fig.) The combined mass of all the persons performing the act and the tables, planks etc. involved is 280 kg. The mass of the performer lying on his back at the bottom of the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm and an effective radius of 2.0 cm. Determine the amount by which each thighbone gets compressed under the extra load.

Answer:
Total mass of all the performers, tables, plaques = 280 kg
Mass of the performer = 60 kg
Mass supported by the legs of the performer at the bottom of the pyramid = 280 – 60
= 220 kg
Weight of this supported mass = 220 kg wt.
= 220 × 9.8 N = 2156 N
Weight supported by each thighbone of the performer = \(\frac{1}{2}\) (2156) N = 1078 N.
The Young’s modulus for bone is Y = 9.4 × 10⁹ Nm^-2 (compressive)
Length of each thighbone L = 0.5 m the radius of thigbone = 2.0 cm
Thus the cross-sectional area of the thighbone
A = π × (2 × 10^-2)² m²
= 1.26 × 10^-3 m²
Using \(\frac{(F \times L)}{(A \times \Delta L)}\) the compression in each thigbone (∆L) can be computed as
∆L = \(\frac{F \times L}{Y \times A}\)
= [(1078 × 0.5)/(9.4 9 × 10⁹ × 1.26 × 10^-3)]
= 4.55 × 10^-5 m or 4.55 × 10^-3 cm.
This is a very small change ! The fractional decrease in the thighbone is ∆L/L = 0.000091 or 0.0091%.

Question 4.
A square slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of 9.0 × 10⁴ N. The lower edge is riveted to the floor. How much will the upper edge be displaced ?
Answer:
The area (A) = 50 cm × 10 cm
= 0.5 m × 0.1 m
= 0.05 m²
Therefore, the stress appIid is
= (9.4 × 10⁴N/0.05 m²)
= 1.80 × 10⁶ N.m²
We know that shearing strain = (∆x/L)
= Stress/G.
Therefore the displacement
∆x = (Stress × L)/G = \(\frac{\left(1.8 \times 10^6 \mathrm{Nm}^{-2} \times 0.5 \mathrm{~m}\right)}{5.6 \times 10^9 \mathrm{Nm}^{-2}}\)
= 1.6 × 10^-4 m = 0.16 mm.

Question 5.
The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression. ∆V/V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 109 Nm^-2. (Take g = 10 ms^-2)
Answer:
The pressure exerted by a 3000 m column of water on the bottom layer
ρ = hρg
= 3000 m × 1000 kg m^-3× 10 ms^-2
= 3 × 10⁷ Nm^-2
Fractional compression ∆V/V, is
∆V/V = stress \(\frac{\left(3 \times 10^7 \mathrm{Nm}^{-2}\right)}{2.2 \times 109 \mathrm{Nm}^{-2}}\)
= 1.36 × 10^-2 or 1.36%

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material 8th Lesson Ecology and Environment Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material 8th Lesson Ecology and Environment

Very Short Answer Type Questions

Question 1.
Define the term “ecology” and its branches.
Answer:
The word ‘ecology’ was derived from the Greek terms (‘Oikos house and ‘logos’ – ‘study’) and it can be defined as “the study of the relationship of organisms with their environment”.

Question 2.
What is an ecological population?
Answer:
A population is a group of organisms of the same species living in a specific area at a specific time.

Question 3.
Define a community.
Answer:
It is an association of the interacting members of populations of different autotrophic and heterotrophic species in a particular area.

Question 4.
What is an ecosystem?
Answer:
An ecosystem is a functional unit of the biosphere in which members of the community interact among themselves and with the surrounding environment.

Question 5.
Distinguish between ecosystem and biome.
Answer:

Ecosystem	Biome
1. Level of organization above the level of the biological community landscape.	1. Level of organization above the level.
2. It can be as small as an aquarium/tiny puddle.	2. It occupies a vast region.
3. Functional unit of the biosphere.	3. Large community of plants and animals.

Question 6.
What is a biome? Name any two biomes you studied.
Answer:
A ‘biome’ is a large community of plants and animals that occupies a vast region.
Ex: Tropical rain forest, desert, tundra (terrestrial biomes) Freshwater biomes, marine biome (aquatic biomes).

Question 7.
What is meant by ecosphere?
Answer:
It is the part of the Earth that supports ‘life’. It extends several kilometers above the Earth’s surface into the atmosphere and extends several kilometers below the ocean’s surface.

Question 8.
Explain the difference between the ‘nich’ of an organism and its ‘habitat’.
Answer:

Habitat	Nich
1. It is the place in which an organism lives.	1. Functional role of an organism in an ecosystem.
2. It is comparable to the address of a person.	2. It is comparable to the profession of a person.

Question 9.
A population has more genetically similar organisms than a biotic community. Justify the statement.
Answer:
A population is a group of organisms of the same species, living in a specific area at a specific time.
Ex: The fish belongs to the species Catla. catla living at a given time.

Question 10.
How do the fish living in Antarctic waters manage to keep their body fluids from freezing?
Answer:
During the course of millions of years of their existence, many species (fish) would have evolved a relatively constant internal (within body) environment so it permits all biochemical reactions and physiological reactions to proceed with maximal efficiency and thus, enhance the overall “fitness” of the species.

Question 11.
How does your body solve the problem of altitude sickness, when you ascend tall mountains?
Answer:
The body compensates for low oxygen availability by increasing red blood cell production and increasing the rate of breathing.

Question 12.
What is the effect of light on body pigmentation?
Answer:
Light influences the colour of the skin. The animals which live in the regions of low intensity of light have less pigmentation than the animals exposed to light.

Question 13.
Distinguish the terms phototaxis and photokinesis.
Answer:
Phototaxis is the oriented locomotion of an organism towards or away from the direction of light.
Ex: As seen in Euglena
Photokinesis is the influence of light on the non-directional movement of organisms.
Ex: Mussel crab

Question 14.
What are circadian rhythms?
Answer:
Biological rhythms that occur in a time period of 24 hours are called circadian rhythms.

Question 15.
What is photoperiodism?
Answer:
The response of organisms to the photoperiod is called photoperiodism.
Ex: Reproduction of flowers, migration of birds.

Question 16.
Distinguish between photoperiod and critical photoperiod.
Answer:
Photoperiod: The duration of light hours is known as a photoperiod.
Critical photoperiod: The specific day length which is essential for the initiation of seasonal events is called critical photoperiod.

Question 17.
Mention the advantages of some UV rays to us.
Answer:

• UV radiation kills the microorganisms present on the body surface of animals.
• UV radiation helps in the conversion of sterols present in the skin into vitamin D in mammals.

Question 18.
What is cyclomorphosis? Explain its importance in Daphnia.
Answer:
The cyclic seasonal morphological variations among certain organisms are called “Cyclomorphosis”. In the case of Daphnia, it is an adaptation to “stabilize the movement” in water and can “resist the water currents better” to stay in the water rich in food materials.

Question 19.
What are ‘regulators’?
Answer:
Organisms that are able to maintain homeostasis by physiological means which ensure constant body temperature, and constant osmotic concentration are called, “regulators”.
Ex: Mammals, Birds

Question 20.
What are conformers?
Answer:
Living organisms that cannot maintain 3 constant internal environments are described as “conformers”.

Question 21.
Define commensalism. Give one example.
Answer:
This is the interaction in which one species benefits and the other is neither harmed nor benefited.
Ex: Barnacles growing on the back of a whale benefit while the whale derives no noticeable benefit.

Question 22.
Define mutualism. Give one example.
Answer:
This is the interaction that benefits both the interacting species.
Ex: Lichens represent an intimate mutualistic relationship between a fungus and photosynthesizing algae.

Question 23.
Define amensalism. Give one example.
Answer:
Amensalism is an interaction in which one species is harmed whereas the other one is unaffected.

Question 24.
What is meant by interspecific competition? Give one example.
Answer:
A process in which the fitness of one species is significantly lower in the presence of another species is called interspecific competition.
Ex: Competition between visiting flamingos and resident fishes in shallow South American lakes.

Question 25.
What is camouflage? Give its significance.
Answer:
Some species of insects and frogs are cryptically coloured to avoid being detected easily by the predator. This phenomenon is called “Camouflage”.

Question 26.
What is Gause’s principle? When does it applicable?
Answer:
When the resources are limited, the competitively superior species will eventually eliminate the other species. It is relatively easy to demonstrate in laboratory experiments.

Question 27.
Name the association that exists in mycorrhiza.
Answer:
The association that exists in mycorrhiza is called “Mutualism”.

Question 28.
Distinguish between lotic and lentic habitats.
Answer:

Lotic habitat	Lentic habitat
The still water bodies fall under the lotic community. Ex: Lakes, ponds	Flowing water bodies are called lentic habitats. Ex: River, canals, streams

Question 29.
What is a zone of compensation in an aquatic ecosystem?
Answer:
The imaginary line that separates the limnetic zone from the profundal zone in a lake is called the zone of compensation (or) compensation point.

Question 30.
Distinguish between phytoplankton and zooplankton.
Answer:

Phyto plankton	Zooplankton
Microscopic organisms bear chlorophyll and form producers in the lakes. Ex: Diatoms, Green algae, Euglenoids	Microscopic organisms that do not bear chlorophyll move through water currents and form primary consumer levels in the lakes. Ex: Daphnia, rotifers and ostracods

Question 31.
Distinguish between neuston and nekton.
Answer:

Neuston	Nekton
The animals living at the air-water interface constitute the “neuston”. Ex: Water strides beetles, the larva of mosquitoes.	The animals capable of swimming constitute the “nekton”. Ex: Water scorpion, back swimmer, diving beetles.

Question 32.
What is periphyton?
Answer:
The animals that are attached to/creeping on the aquatic plants are called “periphyton”.
Ex: Water snails, hydras, nymphs of insects, etc.

Question 33.
Write three examples of man-made ecosystems.
Answer:
Cropland ecosystems, Aquaculture ponds, Aquaria.

Question 34.
What is meant by osmotrophic nutrition?
Answer:
The state of pre-digested food material through the body surface is known as osmotrophic nutrition.

Question 35.
Explaining the process of “leaching”.
Answer:
When the water-soluble inorganic nutrients go down into the soil and get precipitated as unavailable salts that entire process is called “leaching”.

Question 36.
What is PAR?
Answer:
PAR means “Photosynthetically Active Radiation”.

Question 37.
What is the percentage of PAR, in the incident solar radiation?
Answer:
Of the incident solar radiation, less than 50% of it is PAR.

Question 38.
Define entropy.
Answer:
As per the second law of thermodynamics, the energy dispersed is in the form of unavailable heat energy and constitutes entropy.

Question 39.
What is a standing crop?
Answer:
Each trophic level has a certain mass of living material at a particular time and it is called the “Standing crop”.

Question 40.
Explain the terms GPP and NPP.
Answer:
GPP means Gross primary productivity.
NPP means Net primary productivity.

Question 41.
Distinguish between upright and inverted ecological pyramids.
Answer:

Upright Pyramid	Inverted Pyramid
Producers (I – trophic level Biomass) are more in number than other trophic levels. Ex: Grazing food chain	Producers are less in number biomass than other trophic levels. Ex: Parasitic food chain

Question 42.
Distinguish between litter and detritus.
Answer:
Litter: Litter is just like manure that is formed by dead (Either plant (or) Animal clusters) organic matter. It is the primary food source in the detritus food chain.
Detritus: It is a decaying organic matter being decomposed into detritivores organisms.

Question 43.
Distinguish between primary and secondary productivity.
Answer:

Primary Productivity	Secondary Productivity
1. The amount of biomass produced per unit area over a period of time by plants.	1. The rate formation of new organic matter by consumers.
2. It can be divided into gross primary productivity (GPP), net primary productivity (NPP)	2. It also can be divided into two types cross secondary productivity, and net secondary productivity.

Question 44.
Which air pollutants are chiefly responsible for acid rains?
Answer:
Sulphur dioxide (SO₂) and nitrogen oxides are the major causes of acid rain.

Question 45.
What is BOD?
Answer:
BOD means Biological Oxygen Demand. It is a measure of the content of biologically degradable substances in sewage.

Question 46.
What is biological magnification?
Answer:
An increase in the concentration of the pollutant (or) toxicant at successive trophic levels in an aquatic food chain is called ‘Bio-magnification’.

Question 47.
Why are incinerators used in hospitals?
Answer:
Disposal of hospital wastes that contain disinfectants, harmful chemicals, and also pathogenic micro-organisms incinerators are used in hospitals.

Short Answer Type Questions

Question 1.
Considering the benefits of a constant internal environment to the organism we tend to ask ourselves why the conformers had not evolved to become regulators.
Answer:
Thermoregulation is energetically expensive for many organisms. This is particularly true in small animals like shrews and hummingbirds. Heat loss or heat gain is a function of the surface area. Since small animals have a larger surface area relative to their volume, they tend to lose body heat very fast when it is cold outside. Then they have to spend much energy to generate body heat through metabolism. This is the main reason why very small animals are rarely found in polar regions. During the course of evolution, the costs and benefits of maintaining a constant internal environment are taken into consideration. Some species have evolved the ability to regulate, but only over a limited range of environmental conditions, beyond which they simply conform.

Question 2.
The individuals who have fallen through the ice and been submerged under cold water for long periods can sometimes be revived – explain.
Answer:
Temperature variations occur with seasonal changes. These differences in the temperature form thermal layers in water. Water shows maximum density at 4°C decrease its density. Generally, during the winter season the surface water cools down in the upper water phases in the temperature reaches 0°C. below the upper icy layer, the cool water occupies the lake. The aquatic animals continue their life below the icy layer at lower temperatures the activity of bacteria and the rate of oxygen consumption by aquatic animals decrease. Hence organisms can survive below the frozen (icy) upper water without being subject to hypoxia.

Question 3.
What is summer stratification? Explain.
Answer:
During summer in temperate lakes, the density of the surface water decreases because of an increase in its temperature (21-25°C). This ‘uppermost warm layer’ of a lake is called the epilimnion. Below the epilimnion, there is a zone in which the temperature decreases at the rate of 1°C per meter in-depth, and it is called thermocline or metalimnion. The bottom layer is the hypolimnion, where water is relatively cool, stagnant, and with low oxygen content (due to the absence of photosynthetic activity).

During autumn (also called fall). The epilimnion cools down, and the surface water becomes heavy when the temperature is 4°C and sinks to the bottom of the lake overturns bringing about a ‘uniform temperature’ in lakes during that period, this circulation during the autumn is known as fall. The upper oxygen-rich water reaches the hypolimnion and the nutrient-rich bottom water comes to the Surface. Thus there is a uniform distribution of nutrients and oxygen in the lake.

Question 4.
What is the significance of stratification in lakes?
Answer:
Temperature variations occur with seasonal changes in temperature regions. These differences in the temperature form thermal layers in water. These phenomena are called thermal stratification.

During autumn (also called fall), the epilimnion cools down and the surface water becomes heavy when the temperature is 4°C and sinks to the bottom of the lake. Overturns bring about uniform temperatures in lakes during that period. This circulation during autumn is known as the fall or autumn overturn. The upper oxygen-rich water reaches the hypolimnion and the nutrient-rich bottom water comes to the surface. Thus there is a uniform distribution of nutrients and oxygen in the lake. In the spring season the temperatures start rising when it reaches 4°C, the water becomes more dense and heavy and sinks to the bottom, taking oxygen-rich sinks down and bottom nutrient-rich water reaches the surface.

Question 5.
Explain Vant’ Hoff’s rule.
Answer:
Van’t Hoff, a Nobel Laureate in thermo chemistry proposed that, with the increase of every 10°C, the rate of metabolic activities doubles. This rule is referred to as Van’t Hoff’s rule. Van’t Hoffs rule can also be stated in reverse saying that the reaction rate is halved with the decrease of every 10°C. The effect of temperature on the rate of a reaction is expressed in terms of the temperature coefficient of the Q10 value. Q10 values are estimated by taking the ratio between the rate of a reaction at X°C and the rate of reaction at (X – 10°C). In the living systems, the Q10 value is about 2.0. If the Q10 value is 2.0, it means, for every 10°C increase, the rate of metabolism doubles.

Question 6.
Unlike mammals, reptiles cannot tolerate environmental fluctuations in temperature. How do they adapt to survive in desert conditions?
Answer:
Some organisms show behavioural responses to cope with variations in their environment. Desert lizards manage to keep their body temperature fairly constant by behavioural means. They ‘bask’ (staying in the warmth of sunlight) in the sun and absorb heat when their body temperature drops below the comfort zone, but move into the shade when the temperature starts increasing. Some species are capable of burrowing into the soil to escape from the excessive heat above the ground level.

Question 7.
How do terrestrial animals protect themselves from the danger of being dehydration of bodies?
Answer:
In the absence of an external source of water, the kangaroo rat of the North American deserts is capable of meeting all its water requirements through oxidation of its internal fat (in which water is a by-product – of metabolic water). It also has the ability to concentrate its urine, so that minimal volume of water is lost in the process of removal of its excretory products.

Question 8.
How do marine animals adapt to hypertonic seawater?
Answer:
To overcome the problem of water loss, marine fishes have glomerular kidneys with less number of nephrons. Such kidneys minimize the loss of water through urine. To compensate for water loss marine fish drink more water and along with this water, salts are added to the body fluids and disturb the internal equilibrium. To maintain salt balance (salt homeostasis) in the body they have salt-secreting chloride cells in their gills. Marine birds like seagulls and penguins eliminate salts in the form of salty fluid that drips through their nostrils. In turtles, the ducts of chloride-secreting glands open near the eyes. Some cartilaginous fishes retain urea and trimethylamine oxide (TMO) in their blood to keep the body fluids isotonic to the seawater and avoid dehydration of the body due to exosmosis.

Question 9.
Discuss the various type of adaptations in freshwater animals.
Answer:
Animals living in fresh waters have to tackle the problem of endosmosis. The osmotic pressure of freshwater is very low and that of the body fluids of freshwater organisms is much higher. So water tends to enter into bodies by endosmosis. To maintain the balance of water in the bodies, the freshwater organisms acquired several adaptations such as contractile vacuoles in the freshwater protozoans, and large glomerular kidneys in fishes, etc., They send out large quantities of urine along which some salts are also lost. To compensate for the ‘salt loss’ through urine freshwater fishes have ‘salt absorbing chloride cells’ in their gills.

The major problem in freshwater ponds is in summer most of the ponds dry up. To overcome this problem most of the freshwater protists undergo encystment. The freshwater sponges produce asexual reproductive bodies, called gemmules, to tide over the unfavourable conditions of the summer. The African lungfish Protopterus burrows into the mud and forms a gelatinous cocoon around it, to survive, in summer.

Question 10.
Compare the adaptations of animals with fresh water and seawater mode of life.
Answer:

Adaptations in freshwater	Adaptations in seawater
1. Freshwater fishes have glomerular kidneys with more nephrons.	1. Marine water fishes have glomerular kidneys with a number of nephrons.
2. They send out large quantities of urine.	2. They minimize the loss of water through urine.
3. To compensate for the salt loss through urine freshwater fishes have salt-absorbing chloride cells in their gills.	3. To maintain salt balance in the body they have salt-secreting chloride cells.
4. They undergo encystment to overcome the problems.	4. Some fishes retain urea in their blood to keep the body fluid isotonic to the seawater.

Question 11.
Distinguish between euryhaline and stenohaline animals.
Answer:

Euryhaline	Stenohaline
Organisms that are adapted to stand wide fluctuations in salinity are called Euryhaline animals. Ex: Salmon fish, eel fish, etc.	Those that cannot stand wild fluctuations in salinity are known as steno haline animals. Ex: Aromatic insects, Aromatic insects

Question 12.
Many tribes living at high altitudes in the Himalayas normally have higher red blood cell count (or) total haemoglobin than the people living in the plains. Explain?
Answer:
Some organisms possess adaptations that are physiological and allow them to respond quickly to a stressful situation. If you had ever been to any high-altitude place (e.g. > 3,500 M Rohtang pass near Manali and Manasarovar, in Tibet) you must have experienced what is called altitude sickness. Its symptoms include nausea (vomiting sense), fatigue (tiredness), and heart palpitations (abnormality in heartbeat). This is because, in the low atmospheric pressure of high altitudes, the body does not get enough oxygen. But, you gradually get acclimatized and overcome the altitude sickness. How did your body solve this problem? The body compensates for low oxygen availability by increasing red blood cell production and increasing the rate of breathing.

Question 13.
An orchid plant is growing on the branch of the mango tree. How do you describe this interaction between the orchid and mango tree?
Answer:
An orchid growing as an epiphyte on a mango branch gets the benefit of exposure to light, while the mango tree does not drive any noticeable benefit. So in this interaction, one species (arched) get benefitted the other (Mango) is neither armed nor benefitted. So the interactions between orchids and mango trees are commensalism.

Question 14.
Predation is not an association. Support the statement.
Answer:
Predation is not an association (it is a feeding strategy), it is an interaction between two different species. The predator gets benefits at the cost of the prey. Besides acting as) pipelines for energy transfer across trophic levels predators play other important roles. They keep the prey populations under control. In the absence of predators, the prey species could achieve very high population densities and cause instability in the ecosystem.

Question 15.
What is the biological principle behind the biological control method of managing pest insects?
Answer:
The prickly pear cactus introduced in Australia in the early 1920s caused havoc by spreading rapidly into millions of hectares of Rangel and (vast natural grasslands). Finally, the invasive cactus was brought under control only after a cactus-feeding predator (a moth) was introduced into the country. Biological control methods adopted in agricultural pest control are based on the ability of the predators to regulate prey populations.

Question 16.
Discuss competitive release.
Answer:
Another evidence for the occurrence of competition in nature comes from what is called competitive releases. Competitive release occurs when one of the two competing species is removed from an area, thereby releasing the remaining species from one of the factors that limited its distributional range dramatically when the competing species is experimentally removed. This is due to the phenomenon called competitive release, Connells field experiments showed that on the rocky sea coasts of Scotland the larger and competitively superior barnacle Balanus dominates the intertidal area, and excludes the smaller barnacle chathamalus from that zone. When the dominant one is experimentally removed, the population of the smaller one’s increases. In general, herbivores and plants appear to be more adversely affected by competition than carnivores.

Question 17.
Write a short note on the parasitic adaptations.
Answer:
In order to lead successful parasitic life, parasites evolved special adaptations such as:

• Loss of sense organs (which are not necessary for most parasites).
• Presence of adhesive organs such as suckers, and hooks to cling to the host’s body parts.
• Loss of digestive system and presence of high reproductive capacity.
• The life cycles of parasites are often complex, involving one or two intermediate hosts or vectors to facilitate the parasitization of their primary hosts.

Examples:

• The human liver fluke depends on two intermediates (secondary) hosts (a snail and a fish) to complete its life cycle.
• The malaria parasite needs a vector (mosquito) to spread to other hosts. The majority of the parasites harm the host: they may reduce the survival, growth, and reproduction of the host and reduce its population density. They might render the host more vulnerable to predation by making it physically weak.

Question 18.
Explain brood parasitism with a suitable example.
Answer:
Certain birds are fascinating examples of a special type of parasitism, in which the parasitic bird lays its eggs in the nest of its host and lets/allows the host incubates them. During the course of evolution, the eggs of the parasitic bird have evolved to resemble the host’s egg in size and colour to reduce the chances of the host bird detecting the foreign eggs and ejecting them from the nest.

Question 19.
How do predators act as biological control?
Answer:
The prickly pears cactus introduced in Australia in the early 1920s caused havoc by spreading rapidly into millions of hectares of Rangel and (vast natural grasslands) Finally, the invasive cactus was brought under control only after a cactus-feeding predator (a moth) was introduced into the country. Biological control methods adopted in agricultural pest control are based on the ability of the predators to regulate prey populations.

Question 20.
Write notes on the structure and functioning of an ecosystem.
Answer:
‘An ecosystem’ is a functional unit of nature, where living organisms interact among themselves and also with the surrounding physical environment.

Ecosystem varies greatly in size from a small pond to a large forest or a sea. Many ecologists regard the entire biosphere as a global ecosystem as a composite of all local ecosystems on Earth. Since this system is too big and complex to be studied at one time it is convenient to divide it into two basic categories, namely natural and artificial. The natural ecosystems include aquatic ecosystems of water and terrestrial ecosystems of the land. Both types of natural and artificial ecosystems have several subdivisions.

The Natural Ecosystem: These are naturally occurring ecosystems and there is no role of humans in the formation of such types of ecosystems. These are categorized mainly into two types – aquatic and terrestrial ecosystems. These are man-made ecosystems such as agricultural or agroecosystems. They include cropland ecosystems, aquaculture ponds, and aquaria.

Question 21.
Explain the different types of aquatic ecosystems.
Answer:
Based on the salinity of water, three types of aquatic ecosystems are identified marine, fresh water, and estuarine.

• The marine ecosystem: It is the largest of all the aquatic ecosystems. It is the most stable ecosystem.
• Estuarine ecosystem: Estuary is the zone where the river joins the sea, and seawater ascends up into the river twice a day (effect of high tides and low tides). The salinity of water in an estuary also depends on the seasons. During the rainy season outflow of river water makes the estuary saline and the opposite occurs during the summer. Estuarine organisms are capable of withstanding the fluctuations in salinity.
• The freshwater ecosystem: The freshwater ecosystem is the smallest aquatic ecosystem. It includes rivers, lakes, ponds, etc., It is divided into two groups the lentic and lotic. The still water bodies like ponds, lakes reservoirs, etc., fall under the category of lentic ecosystems, whereas streams, rivers, and flowing water bodies are called lotic ecosystems. The communities of the above two types are called lentic and lotic communities respectively. The study of freshwater ecosystems is called limnology.

Question 22.
Explain the different types of terrestrial ecosystems.
Answer:
The ecosystems of land are known as terrestrial ecosystems. Some examples of terrestrial ecosystems are the forest, grass, and desert.

• The forest ecosystem: The two important types of forests seen in India are tropical rain forests and tropical deciduous forests.
• The grassland ecosystems: These are present in the Himalayan region of India. They occupy large areas of sandy and saline soils in western Rajasthan.
• Desert ecosystem: The areas having less than 25 cm of rainfall per year are called deserts. They have characteristic flora and fauna. The deserts can be divided into two types – hot type and cold type deserts, the desert in Rajasthan is an example of the hot type of desert. Cold-type desert is seen in Ladakh.

Question 23.
Discuss the main reason for the low productivity of the ocean.
Answer:
The Primary productivity is very low in the ocean ecosystem compared with the terrestrial ecosystem.

Unlike terrestrial ecosystems, the majority of primary production in the ocean is performed by feel living microscopic organisms called Phytoplankton, large autotrophs such as the seagrasses and macroalgae or seaweeds are generally confined to the littoral zone.

The sunlight zone of the ocean is called the photic zone or euphotic zone, it is a thin layer upto 10 to 100 m near the Ocean’s surface where there is sufficient light for photosynthesis to occur. Light is attenuated down the water column by its absorption or scattering by the water itself. Net photosynthesis in the water column is determined by the interaction between the photic zone and the mixed layer. In the deep water of the ocean (Demersal) there is no light penetration for photosynthesis.

Another factor relatively recently discovered to play a significant role in oceanic primary production is the micronutrient iron. The factors limiting primary production in the ocean are also very different from those on land. However, the availability of light, the source of energy for photosynthesis, and mineral nutrients, building the blocks for new growth, play a crucial role in regulating primary production in the ocean.

Question 24.
Explain the terms saprotrophs detritivores and mineralizers.
Answer:
Saprotrophs are microorganisms such as fungi and bacteria which live on dead organic matter. Detritivores ingest small fragments of decomposing organic materials, termed detritus mineralizers affect the mineralization of humans.

Question 25.
Define decomposition and describe the process and products of decomposition.
Answer:
When organisms die, their bodies and the waste materials passed from the bodies of living organisms form a source of energy and nutrient for the decomposer organisms like saprotrophs detritivores, and mineralizers. Saprotrophs absorb substances through the general body surface of the dead bodies. Detritivores ingest detritus as food. Mineralized mineralize humus these decomposers are referred to as micro consumers of the ecosystem. The decomposition of organic matter includes three phases. In the first phase, particulate detritus is formed by the action of saprotrophs. The second stage is the rapid action of saprotrophs and detritivores to convert detritus into humic substances. The third process is the slower mineralization of the hummus.

Decomposers also play an important role in an ecosystem by converting complex molecules of dead organisms into simpler and reusable molecules. The breakdown products of the dead organisms and waste materials are recycled in the ecosystem and are made available to the producers. The producers cannot continue to exist forever in the absence of the decomposers (as minerals are not returned to the environment).

Question 26.
Write a note on DFC. Give its significance in a terrestrial ecosystem.
Answer:
The detritus food chain (DFC) begins with dead organic matter (such as leaf litter, and bodies of dead organisms). It is made up of decomposers which are heterotrophic organisms ‘mainly’ the fungi’ and ‘bacteria’. They meet their energy and nutrient requirements by degrading dead organic matter to detritus. These are also known as saprotrophs.

Decomposers secrete digestive enzymes that break down dead and waste materials (such as feces i) into simple absorbable substances. Some examples of detritus food chains are:

• Detritus – Earthworm – Frogs – Snakes
• Dead animals – Flies and maggots – Frogs – Snakes

In an aquatic ecosystem. GFC is the major conduit for the energy flow. As against this, in a terrestrial ecosystem, a much larger fraction of energy flows through the detritus food chain than through the GFC. The Detritus food chain may be connected with the grazing food chain at some levels. Some of the organisms of DFC may form the prey of the GFC animals. For example, in the detritus food chain given above, the earthworms of the DFC may become the food of the birds of the GFC. It is to be understood that food chains are not ‘isolated1 always.

Question 27.
What is primary productivity? Give a brief description of the factors that affect primary productivity.
Answer:
Primary productivity is defined as the amount of biomass or organic matter produced per unit area over a period of time by plants, during photosynthesis. It can be divided into Gross Primary Productivity (GPP) and Net Primary Productivity (NPP).
(a) Gross Primary Productivity: Of an ecosystem is the rate of production of organic matter
during photosynthesis. A considerable amount of GPP is utilized by plants for their catabolic process (respiration).
(b) Net Primary Productivity: Gross Primary Productivity minus respiratory loss (R), is the Net Primary Productivity (NPP). On average about 20-25 percent of GPP is used for catabolic (respiratory) activity.
GPP – R = NPP
The Net Primary productivity is the biomass available for the consumption of the heterotrophs (herbivores and decomposers).

Question 28.
Define ecological pyramids and describe with examples, pyramids of numbers and biomass.
Answer:
It is a graphical representation of the trophic structure and function of an ecosystem. The base of each pyramid represents the producers of the first trophic level, while the apex represents the tertiary or top-level/top-order consumers. The three types of ecological pyramids that are usually studied are (a) pyramid of numbers (b) pyramid of biomass and (c) pyramid of energy. These pyramids were first represented by Elton, hence the name ELTONIANpyramids/Ecological pyramids.

Any calculations of energy content, biomass, or numbers have to include all organisms at that trophic level. No generalizations we make will be true if we take only a few individuals of any trophic level into account. In most ecosystems, all the pyramids – of numbers, energy, and biomass are uprights. i.e., producers are more in number and biomass than the herbivores, and herbivores are more in number and biomass than the carnivores. Also, energy (available) at a lower trophic level is always more than that at a higher level.

There are exceptions to this generalization. In the case of a parasitic food chain, the pyramid of numbers is inverted. A large tree (single producer) may support many herbivores like squirrels and fruit-eating birds. On these herbivores, many ectoparasites such as ticks, mites, and lice (secondary consumers) may live. These secondary consumers may support many more top-level consumers and also the hyper-parasites. Thus in each trophic level from the bottom to the top, the numbers of organisms increase and form an ‘inverted pyramid’ of numbers.

The pyramid of biomass in the sea is also generally inverted because the biomass of fishes far exceeds that of phytoplankton.

Question 29.
What are the deleterious effects of depletion of ozone in the stratosphere?
Answer:
The depletion of ozone is particularly marked over the Antarctic region. This has resulted in the formation of a large area of thinned ozone layer commonly called the ‘ozone hole.

UV radiation with wavelengths shorter than that of UV-B is almost completely absorbed by Earth’s atmosphere, provided that the ozone layer is intact. But IJV-B damages DNA and may induce mutations. It causes aging of the skin, damage to skin cells, and various types of skin cancers. In the human eye, the cornea absorbs UV-B radiation, and a high dose of UV-B causes inflammation of the cornea called snow-blindness, cataract, etc., such exposure may permanently damage the cornea.

Question 30.
Describe the ‘Green House’ Effect.
Answer:
The term Green House effect’ has been derived from a phenomenon that occurs in a greenhouse. The greenhouse is a small glasshouse and is used for growing plants, especially during winter. In a greenhouse, the glass panel allows the passage of light into it but does not allow heat to escape (as it is reflected back). Therefore, the greenhouse warms up, very much like inside a car that has been parked in the sun for a few hours.

The greenhouse effect is a naturally occurring phenomenon that is responsible for heating the Earth’s surface and atmosphere. It would be surprising to know that without the greenhouse effect the average temperature of the Earth’s surface would have been chilly – 18°C rather than the present average of 15°C.

When sunlight reaches the outermost layer of the atmosphere, clouds and gases reflect about one-fourth of the incoming solar radiation and absorb some of it. Almost half of the incoming solar radiation falls on the Earth’s surface and heats it up. While a small proportion is reflected back.

Question 31.
Discuss briefly the following:
(a) Greenhouse gases
(b) Noise pollution
(c) Organic farming
(d) Municipal solid wastes
Answer:
(a) Greenhouse gases: The Earth’s surface re-emits heat in the form of infrared radiation but part of this does not escape into space as atmospheric gases (e.g. carbon dioxide, methane, etc.) absorb a major fraction of it. The molecules of these gases radiate heat energy, a major part of which again comes back to the Earth’s surface, thus heating it up once again. The above-mentioned gases- Carbon dioxide and methane are commonly known as greenhouse gases.

(b) Noise pollution: Undesirably high sounds constitute noise pollution. Sound is measured in units called decibels. The human ear is sensitive to sounds ranging from 0 to 180 dB. 0 dB is the threshold limit of hearing and 120 dB is the threshold limit for the sensation of pain in the ear. Any noise above 120 dB is considered to be noise pollution. Brief exposure to the extremely high sound level. 150 dB or more generated by jet planes while taking off may damage eardrums causing permanent hearing impairment. Even long-term exposure to a relatively higher level of noise in cities may also cause hearing impairment. Noise also causes auditory fatigue, anxiety, sleeplessness/msommaj, increased heartbeat, and altered breathing pattern thus causing considerable stress to humans.

(c) Organic farming: Integrated organic farming is a zero waste procedure, where the recycling of waste products is efficiently carried out. This allows the maximum utilization of resources and increases the efficiency of production. A method practiced by Ramesh Chandra Dagar, all these processes support one another and allow an extremely economical and sustainable venture. Natural – biogas generated in the process can be used for meeting the energy needs of the farm. Enthusiastic about spreading information and helping in the practice of integrated organic farming, Dagar has created the Haryana Kisan welfare club.

(d) Municipal Solid waste: Anything (substance/material/articles/goods) that is thrown out as waste in solid form is referred to as solid waste. The municipal solid wastes generally consist of paper, food wastes, plastics, glass, metals, rubber, leather, textile, etc., The wastes are burnt to reduce the volume of the waste. As a substitute for open-burning dumps, sanitary landfills are adopted. There is a danger of seepage of chemicals and pollutants from these landfills, which may contaminate the underground water resources.

Question 32.
Discuss the causes and effects of global warming. What measures need to be taken to control ‘Global warming’?
Answer:
An increase in the level of greenhouse gases has led to considerable heating of the “Earth leading to global warming. During the past century, the temperature of the earth has increased by 0.6°C most of it during the last three decades. Scientists believe that this rise in temperature is leading to severe changes in the environment. Global warming is causing climatic changes (e.g. as El Nino effect) and is also responsible for the melting of polar ice caps and other snow caps of mountains such as the Himalayas. Over many years, this will result in a rise in sea levels, all over the world, that can submerge many coastal areas. The total spectrum of changes that global warming can bring about is a subject that is still under active research.

Global warming: Control measures

• The measures include cutting down the use of fossil fuels.
• Improving the efficiency of energy usage.
• Planting trees and avoiding deforestation.

Question 33.
Write critical notes on the following:
(a) Eutrophication
(b) Biological magnification
Answer:
(a) Eutrophication: Natural ageing of a lake by nutrient enrichment of its water is known as eutrophication. In a young lake, the water is cold and clear supporting little life. Gradually nutrients such as nitrates and phosphates are carried into the lake via streams, in course of time. This encourages the growth of aquatic algae and other plants. Consequently, animal life proliferates and organic matter gets deposited on the bottom of the lake. Over centuries, as silt and organic debris pile up the lake grows shallower and warmer. As a result, the aquatic organisms thriving in the cold environment are gradually replaced by warm-water organisms. Marsh plants appear by taking root in the shallow regions of the lake. Eventually, the lake gives way to large masses of floating plants (bog) and is finally converted into land.

(b) Biological magnification: Increase in the concentration of the pollutant or toxicant at successive trophic levels in an aquatic food chain is called biological magnification or Bio-magnification. This happens in the instance where a toxic substance accumulated by an organism is not metabolized or excreted and thus passes on to the next higher trophic level. This phenomenon is well known regarding DDT and mercury pollution.

As shown in the above example, the concentration of DDT is increased at successive trophic levels. Starting at a very low concentration of 0.003 PPb (PPb part per billion) in water, which ultimately reached an alarmingly high concentration of 25 ppm (ppm = parts per million) in fish-eating birds through biomagnification. High concentrations of DDT disturb calcium metabolism in birds, which causes thinning of eggshells and their premature breaking, eventually causing a decline in bird populations.

(c) Groundwater depletion and ways for its replenishment: Sewage arising from homes and hospitals may contain undesirable pathogenic microorganisms. If it is released untreated into water courses, there is a likelihood of an outbreak of serious diseases, such as dysentery, typhoid, jaundice, cholera, etc.

Untreated industrial effluents released into water bodies pollute most of the rivers, freshwater streams, etc. Effluents contain a wide variety of both inorganic and organic pollutants such as oils, greases, plastics, metallic wastes, suspended solids, and tonics. Most of them are non-degradable. Arsenic, Cadmium, Copper, Chromium, Mercury, Zinc, and Nickel are the common heavy metals discharged from industries.

Effects: Organic substances present in the water deplete the dissolved oxygen content in water by increasing the BOD (Biological Oxygen Demand) and COD (Chemical Oxygen Demand). Most of the inorganic substances render the water unit for drinking.

Removal of dissolved salts such as nitrates, phosphates, and other nutrients and toxic metal ions and organic compounds is much more difficult. Domestic sewage primarily contains biodegradable organic matter, which will be readily decomposed by the action of bacteria and other microorganisms.

Water-logging and soil salinity: Irrigation without proper drainage of water leads to water logging in the soil. Besides affecting the crops, water logging draws salt to the surface of the soil (salinization of the topsoil). The salt then is deposited as a thin crust on the land surface or starts collecting at the roots of the plants. This increased salt content is inimical (unfavourable) to the growth of crops and is extremely damaging to agriculture. Water logging and soil salinity are some of the problems that have come in the wake of the Green Revolution.

Long Answer Type Questions

Question 1.
Write an essay on temperature as an ecological factor.
Answer:
Temperature is a measure of the intensity of heat. The temperature on land or in water is not uniform. On land, the temperature variations are more pronounced when compared to the aquatic medium because land absorbs or loses heat much more quickly than water. The temperature on land depends on seasons and the geographical area on this planet. Temperature decreases progressively when we move from the equator to the poles. Altitude also causes variations in temperature. For instance, the temperature decreases gradually as we move to the top of the mountains.

Biological effects of Temperature:
Temperature Tolerance: A few organisms can tolerate and thrive in a wide range of temperatures they are called eurythermal, but, a vast majority of organisms are restricted to a narrow range of temperatures such organisms are called stenothermal. The levels of thermal tolerance of different species determine their geographical distribution.

Temperature and Metabolism: Temperature affects the working of enzymes and through it, the basal metabolism, and other physiological functions of the organism. The temperature at which the metabolic activities occur at the climax level is called the optimum temperature. The lowest temperature at which an organism can live indefinitely is called minimum effective temperature. It an animal or plant is subjected to a temperature below the minimum effective limit, enters into a condition of inactivity called a chill coma. The metabolic rate increases with the rise of temperature from the minimum effective temperature to the optimum temperature.

The maximum temperature at which a species can live indefinitely in an active state is called maximum effective temperature, the animals enter into a ‘heat coma’. The maximum temperature varies much in different animals.

Vant Hoffs’s rule: Vant Hoff, a Nobel Laureate the thermochemistry proposed that, with the increase of every 10°C, the rate of metabolic activities doubles. This rule is referred to as the Vant Hoffs rule. Vant Hoffs’s rule can also be stated in reverse saying that the reaction rate is halved with the decrease of every 10°C. The effect of temperature on the rate of a reaction is expressed in terms of temperature coefficient or Q10 value. Q10 values are estimated by taking the ratio between the rate of a reaction at X°C and the rate of reaction at (X-10°C). In the ‘living systems’ the Q10 value is about 2.0. If the Q10 value is about 20, it means, for every 10°C increase, the rate of metabolism doubles.

Cyclomorphosis: The cyclic seasonal morphological variations among certain organisms are called cyclomorphosis. This phenomenon has been demonstrated in the Cladoceran (a subgroup of Crustacea) and Daphnia (water flea). In the winter season, the head of Daphnia is ’round’ in shape (typical or non helmet morph). With the onset of the spring season, a small ‘helmet’/’hood’ starts developing on it. The helmet attains the maximum size in summer. In ‘autumn’ the helmet starts receding. By the winter season, the head becomes round. Some scientists are of the opinion that Cyclomorphosis is a seasonal adaptation to changing densities of the water in lakes. In summer as the water is less dense Daphnia requires a larger body surface to keep floating easily. During winter the water is denser, so it does not require a larger surface area of the body to keep floating.

Temperature adaptations: Temperature adaptations in animals can be dealt with under three heads:
(a) Behavioural adaptations
(b) Morphological and Anatomical adaptations and
(c) Physiological adaptations.

(a) Behavioural adaptations: Some organisms show behavioural responses to cope with variations in their environment. Desert lizards manage to keep their body temperature fairly constant by behavioural means. They bask (staying in the warmth of sunlight) in the sun and absorb heat when their body temperature drops below the comfort zone, but move into the shade when the temperature starts increasing. Some species are capable of burrowing into the soil to escape from the excessive heat above the ground level.

(b) Morphological and anatomical adaptations: In the polar seas, aquatic mammals such as the seals have a thick layer of fat (blubber) that acts as an insulator and reduces the loss of body heat, underneath their skin. The animals which inhabit the colder regions have larger body sizes with greater mass. The body mass is useful to generate more heat. As per Bergmann’s rule mammals and other warm-blooded living in colder regions have less surface area to body volume ratio. Then their counterparts live in the tropical regions. The small surface area helps to conserve heat. For instance, the body size of American moose/Eurasian elk (Alces alces), increases with the latitudes in which they live. Moose in the northern part of Sweden show 15-20% more body moss than the same species (counterparts) living in southern Sweden.

Mammals from colder climates generally have shorter earlobes and limbs (extremities of the body) to minimize heat loss. Large earlobes and long limbs increase the surface area without changing the body volume. This is known as Allen’s rule. For instance, the polar fox, Vulpes lagopus (formerly called Alopex lagopus) has short extremities to minimize the heat loss from the body. In contrast, the desert fox has short extremities to minimize heat loss from the body. In contrast, the desert fox, Vulpes zerda, has large ear lobes and limbs to facilitate better heat loss from the body.

(c) Physiological adaptations: In most animals, all the physiological functions proceed ‘optimally’ in a narrow temperature range (in humans, it is 37°C). But there are microbes (archaebacteria) that flourish in hot springs and in some parts of deep seas, where temperatures far exceed 100°C. Many fish thrive in Antarctic waters where the temperature is always below zero. Having realized that the abiotic conditions of many habitats may vary over a time period, we now ask. How do the organisms living in such habitats manage stressful conditions?

One would expect that during the course of millions of years of their existence, many species would have evolved a relatively constant internal (within the body) environment. It permits all biochemical reactions and physiological functions to proceed with maximal efficiency and thus, enhance the overall fitness of the species. This constancy could be chiefly in terms of optimal temperature and osmotic concentration of body fluids. So the organism should try to maintain the constancy of its internal environment (homeostasis) despite varying external environmental conditions that tend to upset its homeostasis. This is achieved by the processes described below.

Thermal migration: The organism can move away temporarily from the stressful habitat to a more hospitable (comfortable) area and return when the stressful period is over. In human analogy comparison, this strategy is comparable to a person moving from Delhi to Shimla for the duration of summer. Many animals, particularly birds, during winter undertake long-distance migrations to more hospitable areas. Every winter, many places in India including the famous Keoladeo Ghana National Park (Formerly – Bharatpur bird sanctuary) in Rajastan and Pulicat Lake in Andhra Pradesh host thousands of migratory birds coming from Siberia and other extremely cold northern regions.

Diapause: Certain organisms show a delay in development, during periods of unfavourable environmental conditions and spend periods in a state of inactiveness called diapause. This dormant period in animals is a mechanism to survive extremes of temperature drought, etc. It is seen mostly in insects and embryos of some fish. Under unfavourable conditions, many zooplankton species in Lakes and ponds are known to enter diapause.

Question 2.
Write an essay on water as an ecological factor.
Answer:
Water is another important factor influencing the life of organisms. Life is unsustainable without water. Its availability is so limited in deserts that only certain special adaptations make it possible for them to live there. You might think that organisms living in oceans, lakes, and rivers should not face any water-related problems, but It is not true. For aquatic organisms the quality (chemical composition, pH, etc.) of water becomes important. The salt concentration is less than 5 percent in inland waters and 30-35 percent in seawater. Some organisms are tolerant to a wide range of salinities (euryhaline) but others are restricted to a narrow range (stenohaline) Many freshwater animals cannot live for long in seawater and vice versa because of the osmotic problems, they would face.

Adaptations in freshwater habitat: Animals living in freshwater have to tackle the problem of endosmosis. The osmotic pressure of freshwater is very low and that of the body fluids of freshwater organisms is much higher. So water tends to enter into bodies by endosmosis. To maintain the balance of water in the bodies, the freshwater organisms acquired several adaptations such as contractile vacuoles in the freshwater protozoans, large glomerular kidneys in fishes, etc… They send out large quantities of urine freshwater fishes have salt-absorbing ‘chloride cells’ in their gills. The major problem in freshwater ponds is – in summer most of the ponds dry up. To overcome this problem, most of the freshwater protists undergo encystment. The freshwater sponges produce asexual reproductive bodies, called gemmules to tide over the unfavorable conditions of the summer. The African lungfish, Protopterus, burrows into the mud and forms a gelatinous cocoon around it, to survive, in summer.

Adaptations in marine habitat: Seawater is high in salt content compared to that body fluids. So, marine animals continuously tend to lose water from their bodies by exosmosis and face the problem of dehydration. To overcome the problem of water loss marine fishes have glomerular kidneys with less number of nephrons. Such Kidneys minimize the loss of water through urine. To compensate for water loss marine fish drink more water, and along with this water, salts are added to the body fluids and disturb the internal equilibrium.

To maintain salt balance (salt homeostasis) in the body they have salt-secreting chloride cells in their gills. Marine birds like seagulls and penguins eliminate salts in the form of salty fluid that drips through their nostrils. In turtles, the ducts of chloride-secreting glands open near the eyes. Some cartilaginous fishes retain urea and trimethylamine oxide (TMO) in their blood to keep the body fluid isotonic to the seawater and avoid dehydration of the body due to exosmosis.

Water-related adaptations in brackish water animals: The animals of brackish water are adapted to withstand wide fluctuations in salinity. Such organisms are called euryhaline animals and those that cant with stand is known as stenohaline. The migratory fishes such as salmon and Hilsa are anadromous fishes i.e. they ‘migrate from the sea to freshwater, for breeding; Anguilla bengalensis is a catadromous fish i.e. it migrates from the river to sea, for breeding. In these fishes, their glomerular kidneys are adjusted to changing salinities. The chloride cells are adapted to excrete or absorb salts depending on the situation. On entering the river they drink more freshwater to maintain the concentration of body fluids equal to that of the surrounding water.

Water-related adaptations for terrestrial life: In the absence of an external source of water, the Kangaroo rat of the North American deserts is capable of meeting all its water requirements through oxidation of its internal fat (in which water is a by-product – of metabolic water). It also has the ability to concentrate its urine, so that minimal volume of water is lost in the process of removal of its excretory products.

Question 3.
Describe the lake as an ecosystem giving examples for the various zones and the biotic components in it.
Answer:
Deep water lakes contain three distinct zones namely

• Littoral zone
• Limnetic zone
• Profundal zone

Littoral Zone: It is the shallow part of the lake closer to the shore. Light penetrates up to the bottom. It is euphotic (having good light) and has rich vegetation and a higher rate of photosynthesis, hence rich in oxygen.

Limnetic Zone: It is the open water zone away from the shore. It extends up to the effective light penetration level, vertically. The imaginary line that separates the limnetic zone from the profundal zone is known as the zone of compensation/compensation point light compensation level. It is the zone of effective light penetration. Here the rate of photosynthesis is equal to the rate of respiration. The limnetic zone has no contact with the bottom of the lake.

Profundal Zone: It is the deep water area present below the limnetic zone and beyond the depth of effective light penetration. Light is absent. Photosynthetic organisms are absent and so the water is poor in oxygen content. It includes mostly the anaerobic organisms which feed on detritus. The organisms living in lentic habitats are classified into pedantic forms, which live at the bottom of the lake and those living in the open waters of lakes, away from the shore vegetation are known as limnetic forms.

Biota (animal and plant life of a particular region) of the littoral zone: Littoral zone is rich with pedantic flora (especially up to the depth of the effective light penetration). At the shore, proper emergent vegetation is abundant with firmly fixed roots at the bottom of the lake, and shoots and leaves are exposed above the level of water. These are amphibious plants. Certain emergent rooted plants of the littoral zone are the cattails (Typha), bulrushes (Scirpus) arrowheads (Sagittaria),. Slightly deeper are the rooted plants with floating leaves, such as the water lilies (Nymphaea), Nelumbo, Trapa, etc., still deeper are the submerged plants such as Hydrilla – Chara, Potamogeton, etc… The free-floating vegetation includes pistia, Wolffia, Lemna (duckweed), Azolla, Eichhornia, etc.

The phytoplankton of the littoral zone is composed of diatoms (Coscinodiscus, Nitzschia, etc) green algae (Volvox, spirogyra, etc), euglenoids (Euglena, phacus, etc), and dinoflagellates (Gymnodinium, Cystodinium, etc ….)

Animals, the consumers of the littoral zone, are abundant in this zone of the lake, these are categorized into zooplankton, neuston, nekton, periphyton, and benthos. The Zoo-plankton of the littoral zone consists of water fleas such as Daphnia, rotizers, and ostracods.

The animals living at the air-water interface constitute the neuston. They are of two types. The epineuston and hyponeuston. Water striders (Gerris), beetles, and water bugs (Dineutes) form the epineuston/ supraneuston and the hyponeuston/infraneuston includes the larvae of mosquitoes.

The animals such as fishes, amphibians, water-snakes, terrapins, insects like water scorpion (Ranatra), back swimmer (Notonecta), diving beetles (Dytiscus), capable of swimming constitute the nekton.

The animals that are attached to/creeping on the aquatic plants, such as the ‘water snails’, nymphs of insects, bryozoans, turbellarians, hydras, etc, constitute the periphyton. The animals that rest on or move on the bottom of the lake constitute the ‘benthos’, e.g.: red annelids, chironomid larvae, crayfishes, some isopods amphipods, clams, etc.

Biota of the limnetic zone: Limnetic zone is the largest zone of a lake. It is the region of rapid variations of the level of the water, temperature, oxygen availability, etc., from time to time. The limnetic zone has autotrophs (photosynthetic plants) in abundance. The chief autotrophs of this region are the phytoplankton such as the euglenoids, diatoms, cyanobacteria, dinoflagellates, and green algae. The consumers of the limnetic zone are the zooplanktonic organisms such as the copepods, Fisher frogs, water snakes, etc., which form the limnetic nekton.

Biota of the profundal zone: It includes the organisms such as decomposers (bacteria), chironomid larvae, Chaoborus (phantom larva), red annelids, clams, etc., that are capable of living in low oxygen levels. The decomposers of this zone decompose the dead plants and animals and release nutrients that are used by the biotic communities of both littoral and limnetic zones.

The lake ecosystem performs all the functions of any ecosystem and of the biosphere as a whole, i.e., conversion of inorganic substances into organic material, with the help of the radiant solar energy by the autotrophs, consumption of the autotrophs by the heterotrophs; decomposition and mineralization of the dead matter to release them back for reuse by the autotrophs (recycling of minerals).

Question 4.
Describe different types of food chains that exist in an ecosystem.
Answer:
The food energy passes from one trophic level to another trophic level mostly from the lower to higher trophic levels. When the path of food energy is ‘linear’ the components resemble the ‘links’ of a chain and it is called a ‘food chain’. Generally, a food chain ends with decomposers. The three major types of food chains in an ecosystem are the Grazing Food Chain, Parasitic Food Chain, and Detritus Food Chain.

(i) Grazing food chain: It is also known as the predatory food chain, it begins with the green plants (producers), and the second third, and fourth trophic levels are occupied by the herbivores, primary carnivores, and secondary carnivores respectively. In some food chains, these are yet another trophic level – the climax carnivores. The number of trophic levels in food chains varies from 3 to 5 generally. Some examples from the grazing food chain (GFC) are given below.

I trophic level	II trophic level	III trophic level	IV trophic level	V trophic level
Rosebush →	Aphids →	Spiders →	Small birds →	Hawk
Grass →	Grasshopper →	Frog →	Snake →	Hawk
Plants →	Caterpillar →	Lizard →	Snake
Phytoplankton →	Zooplankton →	Fish →	Bird
Grass →	Goat →	Man

(ii) Parasitic food chain: Some authors included the Parasitic Food Chains as a part of the GFC. As in the case of GFC’s, it also begins with the producers, the plants (directly or indirectly). However, the food energy passes from large organisms to small organisms in the parasitic chains. For instance, a tree that occupies the 1st trophic level provides shelter and food for many birds. These birds host many ectoparasites and endo parasites. Thus, unlike in the predator food chain, the path of the flow of energy includes fewer, large-sized organisms in the lower trophic levels and numerous, small-sized organisms in the successive higher trophic levels.

(iii) Detritus Food Chain: The detritus food chain (DFC) begins with dead organic matter (such as leaf litter, and bodies of dead organisms). It is made up of decomposers which are heterotrophic organisms, mainly fungi’ and ‘bacteria’. They meet their energy and nutrient requirements by degrading dead organic matter or detritus. These are also known as saprotrophs (sappro: to decompose).

Decomposers: Secrete digestive enzymes that break down dead and waste materials (such as feces) into simple absorbable substances. Some examples of detritus food chains are:

• Detritus (formed from leaf litter) – Earthworms – Frogs – Snakes
• Dead animals – Flies and maggots – Frogs – Snakes.

In an aquatic ecosystem, GFC is the major ‘conduit for the energy flow. As against this in a terrestrial ecosystem, a much larger fraction of energy flows through the detritus food chain than through the GFC. The Detritus food chain may be connected with the grazing food chain at some levels. Some of the organisms of DFC may form the prey of the GFC animals. For example, in the detritus food chain given above, the earthworms of the DFC may become the food of the birds of the GFC. It is to be understood that food chains are not ‘isolated’ always.

Question 5.
Give an account of the flow of energy in an ecosystem.
Answer:
Except for the deep sea hydro-thermal ecosystem, the sun is the only source of energy for all ecosystems on Earth. Of the incident solar radiation, less than 50 percent of it is photosynthetically active radiation (PAR). We know that plants and photosynthetic bacteria (autotrophs), fix the sun’s radiant energy to synthesize food from simple inorganic materials. Plants capture only 2-10 percent of the PAR and this small amount of energy sustains the entire living world. So, it is very important to know how the solar energy captured by plants flows through different organisms of an ecosystem. All heterotrophs are dependent on the producers for their food, either directly or indirectly. The law of conservation of energy is the first law of thermodynamics. It states that energy may transform from one form into another form, but it is neither created nor destroyed. The energy that reaches the earth is balanced by the energy that leaves the surface of the earth as invisible heat radiation.

The energy transfers in an ecosystem are essential for sustaining life. Without energy transfers, there could be no life and ecosystem. Living beings are the natural proliferations that depend on the continuous inflow of concentrated energy. Further, ecosystems are not exempted from the Second Law of thermodynamics. It states that no process involving energy transformation will spontaneously occur unless there is the degradation of energy. As per the second law of thermodynamics – the energy dispersed is in the form of unavailable heat energy and constitutes the entropy (energy lost or not available for work in a system).

The organisms need a constant supply of energy to synthesize the molecules they require. The transfer of energy through a food chain is known as energy flow. A constant input of mostly solar energy is the basic requirement for any ecosystem to function. The important point to note is that the amount of energy available decreases at successive trophic levels. When an organism dies, it is converted to detritus or dead biomass that serves as a source of energy for the decomposers. Organisms at each trophic level depend on those at the lower trophic level, for their energy demands.

Each trophic level has a certain mass of living material at a particular time and it is called the standing crop. The standing crop is measured as the mass of living organisms (biomass) or the number of organisms per unit area. The biomass of a species is expressed in terms of fresh or dry weight (dry weight is more accurate because water contains no usable energy).

The 10 percent Law: The 10 percent law for the transfer of energy from one trophic level to the next was introduced by Lindeman (the Founder of modern Ecosystem Ecology). According to this law, during the transfer of energy from one trophic level to the next, only about 10 percent of the energy is stored/converted as body mass/biomass. The remaining is lost during the transfer or broken down in catabolic activities (Respiration). Lindeman’s rule of trophic efficiency/Gross ecological efficiency is one of the earliest and most widely used measures of ecological efficiency. For example: If the NPP (Net Primary Production) in a plant is 100 kJ, the organic substance converted into the body mass of the herbivore which feeds on it is 10 kJ only. Similarly, the body mass of the carnivore-I is 1 kJ only.

Question 6.
List out the major air pollutants and describe their effects on human beings.
Answer:
Air pollutants cause injury to all living organisms. They reduce the growth and yield of crops. They are harmful to the respiratory system of humans and animals. An increase in the concentration of pollutants or duration of exposure increases the harmful effects on the organisms.
The major air pollutants:
1. Carbon monoxide (CO): It is produced mainly due to incomplete combustion of fossil fuels. Automobiles are a major cause of CO pollution in larger cities and towns. Automobile exhausts, fuels from factories, emissions from power plants, forest fires and even burning of firewood contribute to CO pollution. Haemoglobin has a greater affinity for CO and SO, and CO competitively interferes with oxygen transport. CO symptoms such aS headache and blurred vision at lower concentrations. In higher concentrations, it leads to coma and death.

2. Carbon Dioxide (CO₂): Carbon dioxide is the main pollutant that is leading to global warming. Plants utilize CO₂ for photosynthesis and all living organisms emit carbon dioxide in the process of respiration. With rapid urbanization, automobiles, aeroplanes, power plants, and other human activities that involve the burning of fossil fuels such as gasoline, carbon dioxide is turning out to be an important pollutant of concern.

3. Sulphur Dioxide (SO₂): It is mainly produced by burning fossil fuels. Melting of sulphur ores is another important source of SO2 pollution. The metal smelting and other industrial processes also contribute to SO₂ pollution. Sulfur dioxide and nitrogen oxides are the major causes of acid rains, which cause acidification of soils, lakes, and streams and also accelerated corrosion of buildings and monuments. High concentrations of sulphur dioxide (SO₂) can result in breathing problems in asthmatic children and adults. Other effects associated with long-term exposure to sulphur dioxide, include respiratory illness, alterations in the lungs defenses, and aggravation of existing cardiovascular problems.

To control SO₂ pollution, the emissions are filtered through scrubbers. Scrubbers are devices that are used to clean the impurities in exhaust gases. Gaseous pollutants such as SO₂ are removed by scrubbers.

4. Nitrogen Oxides: Nitrogen oxides are considered to be major primary pollutants. The source is mainly automobile exhaust. The air polluted by nitrogen oxide is not only harmful to humans and animals but also dangerous for the life of plants. Nitrogen oxide pollution also results in acid rain and the formation of photochemical smog. The effect of nitrogen oxides on plants includes the occurrence of necrotic spots on the surface of leaves. Photosynthesis is affected in crop plants and the yield is reduced. Nitrogen oxides combine with volatile organic compounds by the action of sunlight to form secondary pollutants called Peroxyacetyl nitrate (PAN) which are found especially in photochemical smog. They are powerful irritants to the eyes and respiratory tract.

5. Particulate matter/Aerosols: Tiny particles of solid matter suspended in a gas or liquid constitute the particulate matter. Aerosols refer to particles and /or liquid droplets and the gas together (a system of colloidal particles dispersed in a gas) Combustion of “fossil fuels” (petrol, diesel, etc) fly ash produced in thermal plants, forest fires, cement factories, asbestos mining, and manufacturing units, spinning and ginning mills, etc., are the main sources of particulate matter pollution. According to the Central Pollution Control Board (CPCB) particles of 2.5 micrometers or less in diameter are highly harmful to man and other air-breathing organisms.

An electrostatic precipitator is a widely used filter’ for removing particulate matter from the exhaust of thermal power plants. It can remove 99% of particulate matter. It has high voltage electrodes which produce a ‘corona’ that releases electrons. These are collected by collecting plates that attract charged particles. The air flowing between the plates is kept at a low velocity so as to allow the dust particles to fall. Thus clean air is released into the atmosphere.

Question 7.
What are the causes of water pollution and suggested measures for control of water pollution?
Answer:
Domestic Sewage: Sewage is the major source of water pollution in large cities and towns. It mainly consists of human and animal excreta and other waste materials. It is usually released into freshwater bodies or the sea directly. As per the regulations the sewage has to be passed through treatment plants before it is released into the water sources. Only 0.1 percent of impurities from domestic sewage are making these water sources unfit for human consumption. In the treatment of sewage, solids are easy to remove. Removal of dissolved salts such as nitrates, phosphates, and other nutrients and toxic metal ions and organic compounds is much more difficult. Domestic sewage primarily contains biodegradable organic matter, which will be readily decomposed by the action of bacteria and other microorganisms.

Biological Oxygen Demand (BOD): BOD is a measure of the content of biologically degradable substances in sewage. The organic degradable substances are broken down by microorganisms using oxygen. The demand for oxygen is measured in terms of the oxygen consumed by microorganisms over a period of 5 days (BOD 5) or seven days (BOD 7). BOD forms an index for measuring pollution load in the sewage. Microorganisms involved in the biodegradation of organic matter in water bodies consume a lot of oxygen and as a result, there is a sharp decline in dissolved oxygen causing the death of fish and other aquatic animals.

Algal blooms: The presence of large amounts of nutrients in waters also causes excessive growth of plankton algae and the phenomenon is commonly called “algal blooms1′. Algal blooms impart distinct colour to the bodies and deteriorate the quality of water. It also causes the mortality of fish. Some algae which are involved in algal blooms are toxic to human beings and animals.

Excessive growth of aquatic plants such as the common water hyacinth (Eichhornia crassipes), the world’s most problematic aquatic weed which is also called “Terror of Bengal” causes blocks in our waterways. They grow faster than our ability to remove them. They grow abundantly in eutrophic water bodies (water bodies rich in nutrients) and lead to an imbalance in the ecosystem dynamics of the water body.

Sewage arising from homes and hospitals may contain undesirable pathogenic microorganisms. If it is released untreated into water courses, there is a likelihood of an outbreak of serious diseases, such as dysentery, typhoid, jaundice, cholera, etc.

2. Industrial Effluents: Untreated industrial effluents released into water bodies pollute most of the rivers, freshwater streams, etc. Effluents contain a wide variety of both inorganic and organic pollutants such as oils, greases, plastics, metallic wastes, suspended solids, and toxins. Most of them are non-degradable. Arsenic, Cadmium, Copper, Chromium, Mercury, Zinc, and Nickel are the common heavy metals discharged from industries.

Effects: Organic substances present in the water deplete the dissolved oxygen content in water by increasing the BOD (Biological Oxygen Demand) and COD (Chemical Oxygen Demand). Most of the inorganic substances render the water unfit for drinking. Outbreaks of dysentery, typhoid, jaundice, cholera, etc., are caused by sewage pollution.

Biomagnification: Increase in the concentration of the pollutant or toxicant at successive trophic levels in an aquatic food chain is called Biological Magnification or Bio-magnification. This happens in the instances where a toxic substance accumulated by an organism is not metabolized or excreted and thus passes on to the next higher trophic level. This phenomenon is well known regarding DDT and mercury pollution.

As shown in the above example, the concentration of DDT is increased at successive trophic levels starting at a very low concentration of 0.003 ppb (ppb = parts per billion) in water, which ultimately reached an alarmingly high concentration of 25 ppm (ppm = parts per million) in fish-eating birds, through biomagnification. High concentrations of DDT disturb calcium metabolism in birds, which causes thinning of eggshells and their premature breaking, eventually causing a decline in bird populations.

Eutrophication: Natural ageing of a lake by nutrient enrichment of its water is known as eutrophication. In a young lake, the water is cold and clear, supporting little life. Gradually nutrients such as nitrates and phosphates are carried into the lake via streams, in course of time. This encourages the growth of aquatic algae and other plants. Consequently, animal life proliferates and organic debris piles up, the lake grows shallower and warmer. As a result, the aquatic organisms thriving in the cold environment are gradually replaced by warm-water organisms. Marsh plants appear by taking root in the shallow regions of the lake. Eventually, the lake gives way to large masses of floating plants (bog) and is finally covered in land.

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 11th Lesson Mechanical Properties of Fluids Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 11th Lesson Mechanical Properties of Fluids

Very Short Answer Questions

Question 1.
Define average pressure. Mention it’s unit and dimensional formula. Is it a scalar or a vector ? [A.P. Mar. 17]
Answer:
Average pressure (P_av) : Average power is defined as the normal force acting per unit area.
P_av = \(\frac{F}{A}\)
units → N/m² (or) pascal
Dimensional formula → [ML^-1 T^-2]
Pressure is a scalar quantity.

Question 2.
Define Viscosity. What are it’s units and dimensions ?
Answer:
Viscosity : The property of a liquid which opposes the relative motion between its layers is called viscosity. .
G.G.S unit poise
S.I unit → Nm^-2s
Dimensional formula = [M¹L^-1T^-1]

Question 3.
What is the principle behind the carburetor of an automobile ? [AP – Mar. 15; TS – Mar. ’18, ’17]
Answer:
The carburetor of automobile has a venturi channel (nozzle) through which air flows with a large speed. The pressure is then lowered at the narrow neck and the petrol is sucked up in the chamber to provide the correct mixture of air to fuel necessary for combustion. ”

Question 4.
What is magnus effect ? [T.S. Mar. 16; A.P. Mar. 15]
Answer:
The difference in the velocities of air results in the pressure difference between the lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spinning is called magnus effect.

Question 5.
Why are drops and bubbles spherical ? [A.P. Mar. 18, 17, 16, 14]
Answer:
The surface tension of a liquid tends to have minimum surface area. For a given volume, the surface area is minimum for a sphere. Hence rain drops are spherical shape.

Question 6.
Give the expression for the excess pressure in a liquid drop. [T.S. Mar. 17]
Answer:
Excess pressure in a liquid drop, p_i – p₀ = \(\frac{2 s}{r}\)
where s = Surface tension; r = Radius of the liquid drop.

Question 7.
Give the expression for the excess pressure in an air bubble inside the liquid.
Answer:
Excess pressure in an air bubble inside the liquid, p_i – p₀ = \(\frac{2 s}{r}\)
where s = Surface tension
r = Radius of the air bubble.
Air bubble forms inside the liquid, hence it has one liquid surface.

Question 8.
Give the expression for the excess pressure soap bubble in air. [T.S. Mar. 16]
Answer:
Soap bubble have two interfaces, hence excess pressure inside a soap bubble is p_i – p₀ = \(\frac{4 s}{r}\)
where s = Surface tension
r = Radius of the soap bubble.

Question 9.
What are water proofing agents and water wetting agents ? What do they do ?
Answer:
Water proofing agents are added, to create a large angle of contact between water and fibres.
Soaps, detergents and dying substances are wetting agents. When they are added, the angle of contact becomes small. So that they may penetrate well and become effective.

Question 10.
What is angle of contact ? [A.P. Mar. 16]
Answer:
The angle between tangent to the liquid surface at the point of contact and solid surface inside the liquid is termed as angle of contact (0).

Question 11.
Mention any two examples that obey Bernoullis theorem and justify them. [A.P. Mar. 18]
Answer:

1. In heavy winds house roof’s are blown off. When the velocity of the wind is greater on the roof top than inside the house, then the pressure on the roof top becomes less than that inside the house. This pressure difference causes the dynamic lift.
2. When a fan is rotating, papers are blown off from .the table top. The velocity of wind on the paper increases due to fan and hence pressure decreases. Due to this pressure difference papers are blown off.

Question 12.
When water flows through a pipe, which of the layers moves fastest and slowest ?
Answer:
Water flows through a pipe, the layers near the axis of the tube are fastest and at the walls of the tube are slowest

Question 13.
“Terminal velocity is more if surface area of the body is more”. Give reasons in support of your answer.
Answer:
Surface area (A) = 4πr² and terminal velocity (υ_t) αr²
As surface area increases, r2 is also increases. Then terminal velocity is also increases.
∴ Terminal velocity is more, if surface area of the body is more.

Short Answer Questions

Question 1.
What is atmospheric pressure and how is it determined using Barometer ?
Answer:
Atmospheric pressure : Atmospheric pressure at any point is equal to the weight of a column of air of unit cross sectional area extending from that point to the top of the earth’s atmosphere.
1 atm = 1.013 × 10⁵ pa
Determination of atmospheric pressure using barometer: A long glass tube closed at one end and filled with mercury is inverted into a trough of mercury. This device is known as mercury barometer.

The space above the mercury column in the tube contains only mercury vapour whose pressure Pis so small, that it may be neglected. The pressure inside the column at point A must equal the pressure at B.
∴ Pressure at B = Atmospheric pressure = P_a
P_a = ρgh = Pressure at A …………………… (1)
Where ρ is density of mercury and h is the height of the mercury column in the tube. In the experiment it is found that the mercury column in the barometer has a height of about 76cm at the sea level equivalent to one atmosphere.

Question 2.
What is guage pressure and how is a manometer used for measuring pressure differences ?
Answer:
Guage pressure : The guage pressure is the difference of the actual pressure and the atmospheric pressure. P_g = P – P_a
measurement of pressure differences :

1. The manometer consists of a U-shaped tube, which is filled with a low density liquid (oil) for measuring small pressure difference and high density liquid
   (mercury) for measuring large pressure difference.
2. One end of the tube is connected to the vessel D whose pressure of air measure and the other end of the tube is open.
3. If pressure of air in vessel D is more than the earth’s atmosphere, the level of liquid in arm I will go down up to point A and level of liquid in arm II rises up to C.
4. Then the pressure of air in vessel is equal to pressure at point A.
   
5. Note the difference of liquid levels in the two arms of U-tube.(Say h). p be density of the liquid. Pa is the atmospheric pressure.
6. Pressure at point A (P_A) = Pressure at point B = Pressure at point C + Pressure due to column of liquid.
   P_A = P_c + hρg (or) P_A – P_c = hρg
   Here P_c = P_a, P_A = P, ∴ P – P_a = hρg
   P – P_a = P_g = guage pressure = hρg

Question 3.
State Pascal’s law and verify it with the help of an experiment.
Answer:
Pascal’s law : It states that if gravity effect is neglected, the pressure at every point of liquid in equilibrium of rest is same.
Proof :

1. Imagine a circular cylinder of uniform cross-sectional area A, such that points C and D lie on flat faces of the cylinder.
2. The liquid inside the cylinder is in equilibrium under the action of forces exerted by the liquid outside the cylinder.
3. These forces are acting every where perpendicular to the surface of the cylinder.
4. Thus the forces on the flat faces of the cylinder at C and D will be perpendicular to the forces on the curved surface of the cylinder.
5. Since liquid is in equilibrium, the sum of the forces acting on the curved surface of the cylinder must be zero.
   
6. If P₁ and P₂ are the pressures at points C and D respectively. F₁ and F₂ are the forces acting on the flat surfaces of the cylinder due to liquid, then
   F₁ = P₁ A and F₂ = P₂A Since liquid is in equilibrium, therefore
   F₁ = F₂
   P₁ A = P₂A
   (or) P₁ = P₂
   It means the pressures at C and D are the same. This proves the pascal’s law.

Question 4.
Explain hydraulic lift and hydraulic brakes.
Answer:
Hydraulic lift and hydraulic brakes are based on the Pascal’s law.

Hydraulic lift : Here C and D are two cylinders of different areas of cross section. They are connected to each other with a pipe E. Each cylinder is provided with airtight frictionless piston. Let a, A be the area’s of cross-sections of the piston at C and D (a < < A)

The cylinders are filled with an incompressible liquid.
Let f be the applied force at C. Pressure exerted on the liquid
P = \(\frac{f}{a}\) ……………. (1)
According to pascal’s law, this pressure is transmitted to piston of cylinder D. Upward force at D will be
F = PA = \(\frac{f}{a}\) A = f \(\frac{A}{a}\) …………… (2)
As A > > a
∴ F > > f .
∴ Heavy load placed on the larger piston is easily lifted.

Hydraulic Brakes :
When the brake pedal is pressed, the piston (P) of the master cylinder is pushed inwards. There will be increased pressure on liquid at P, which is transmitted equally to P₁ and P₂ of wheel cylinder in accordance with pascal’s law. Due to which P₁ and P₂ move outwards. Breakshoes to move away from each other which in turn press against the inner rim of the wheel. The brake becomes operative.

Question 5.
What is hydrostatic paradox ?
Answer:
Consider three vessels A, B and C of different shapes. They are connected at the bottom by a horizontal pipe. On filling with the level in the three vessels is the same, though they hold different amounts of water. This is so because water at the bottom has the same pressure below each section of the vessel. It means the liquid pressure at a point is independent of the quantity of liquid but depends upon the depth of point below the liquid surface. This is known as hydrostatic paradox.

Question 6.
Explain how pressure varies with depth.
Answer:
Consider a fluid at rest in a container. In figure point 1 is at a height h above a point 2. The pressure at points 1 and 2 are P₁ and P₂. As the fluid is at rest, the horizontal forces should be zero. The resultant vertical forces should balance the weight of the element. Pressure at top (P₁A) acting downward, pressure at bottom (P₂A) acting upward.

(P₂ – P₁) A = mg ………………… (1)
Mass of the fluid (m) = ρv = ρhA
(P₂ – P₁) = ρgh ………………. (2)
Pressure difference depends on the vertical distance h.
If the point 1 under discussion is shifted to the top of the fluid (water), which is open to the atmosphere, P₁ may be replaced by atmospheric pressure (P_a) and P₂ by P. Then eq (2) gives
P – P_a = ρgh
P = P_a + ρgh ……………… (3)
Thus the pressure P, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh.

Question 7.
What is Torricelli’s law ? Explain how the speed of efflux is determined with an experiment.
Answer:
Torricelli’s theorem : The velocity of efflux i.e., the velocity with which the liquid flows out of an orifice (i.e., a narrow hole) is equal to that which is freely falling body would acquire in falling through a vertical distance equal to the depth of orifice below the free surface of liquid.

Consider ideal liquid of density p contained in a tank provided with a narrow hole.
Let h = Height of free surface of liquid above O.
P = Atmospheric pressure
v = Velocity of efflux
Applying Bernoulli’s theorem at A and O
(P + ρgh + O)_atA = [P + 0 + \(\frac{1}{2}\)ρv²]_atO
P + ρgh = P + \(\frac{1}{2}\) ρv² = ρgh = \(\frac{1}{2}\)ρv²
v = \(\sqrt{2 g h}\)

Question 8.
What is Venturi-meter? Explain how it is used.
Answer:
Venturi meter: The Venturi-meter is a device to measure the flow speed of incompressible fluid.

1. It consists of a tube with a broad diameter and a small constriction at the middle.
2. A manometer in the form of a U-tube is also attached to it, with one of arm at the broad neck point of the tube and the other at constriction as shown in figure.
   
3. The manometer contains a liquid of density ρ_m.
4. The pressure difference causes the fluid in the U-tube connected at the narrow neck to rise in comparison to the other arm.
5. Filter pumps, sprayers used for perfumes, carburetor of automobile has used on this principle.

Question 9.
What is Reynolds number ? What is it’s significance ?
Answer:
Reynolds number : “Reynold number is a pure number which determines the nature of flow of liquid through a pipe”.
Reynold number (R_e) = \(\frac{\rho \mathrm{vd}}{\eta}\)
Where ρ is density of the fluid
v is speed of the fluid
d is diameter of the pipe

1. If the flow is stream line (or) laminor R_e < 1000
2. If the flow is turbulent, R_e > 2000
3. If the flow becomes unsteady, 1000 < R_e < 2000

Physical significance of Reynolds number : Reynold’s number describes the ratio of the inertial force per unit area to the viscous force per unit area for a flowing fluid.

Question 10.
Explain dynamic lift with examples.
Answer:
Dynamic lift: Dynamic lift is the force that act on a body, by virtue of its motion through a fluid.

e.g.1 : Fig (a) shows, ball moving without spin. Stream lines are equally distributed above and below the ball. The velocity above and below the ball is same resulting zero pressure difference. There no upward (or) downward force on the ball.

Fig (b) shows, ball moving with spin stream lines are more crowded above the ball than below. The velocity of air above the ball is large (v + v_r) and below it is smaller (v – v_r). As a result, there is a pressure difference between lower and upper faces. Pressure is less at top of the ball and pressure is morebelow the ball. There is a net upward force on the ball.
e.g. 2 : Dynamic lift also acts on the an aeroplane wing.

Question 11.
Explain Surface Tension and Surface energy. [Mar. 13]
Answer:
Surface tension (S): The force acting per unit length of an imaginary line drawn on the surface of a liquid, normal to it and parallel to the surface is called surface tension.
T = \(\frac{\mathrm{F}}{l}\)
S.l unit → N/m
D.F → [MT^-2]
Surface energy (E): The additional potential energy due to molecular forces per unit surface area is called surface tension.
Surface energy = \(\)
S.l Unit → J/m²
D.F → (MT^-2 )

Consider a horizontal liquid film ending in bar free to slide over parallel guides. We move the bar by a small distance d. The area of the surface increases, the system now has more energy, this means – that some work has to be done against an internal force F.
Work done (W) = F.d
If the surface energy of the film is S per unit area, the extra area is 2dl. A film has two sides and the liquid in between them.
So there are two surfaces and the extra energy is
S (2dl) = Fd
S = \(\frac{\mathrm{F}}{2 l}\)
Surface tension is equal to the surface energy and is also equal to the force per unit length exerted by the fluid on the movable bar.

Question 12.
Explain how surface tension can be measured experimentally.
Answer:
A flat vertical glass plate, below which a vessel of some liquid is kept, forms one arm of the balance. The plate is balanced by weights on the otherside, with its horizontal edge just over water. The vessel is raised slightly till the liquid just touches the glass plate and pulls it down a little because of surface tension. Weights are added till the plate just clears water. Suppose the additional weight requires is W.

Surface tension of liquid air interface is
S_la = \(\frac{\mathrm{W}}{2 l}=\frac{\mathrm{mg}}{2 l}\)
Where l is length of the plate edge, m is extra mass.

Long Answer Questions

Question 1.
State Bernoulli’s principle. From conservation of energy in a fluid flow through a tube, arrive at Bernoulli’s equation. Give an application of Bernoulli’s theorem.
Answer:
Bernoulli’s principle : Bernoulli principle states that in a stream line flow, the sum of the pressure, the K.E per unit volume and the P.E per unit volume remains a constant.
P + \(\frac{1}{2}\) ρv² + ρgh = constant

1. Consider a non-viscous, incompressible fluid is flowing through the pipe in a steady flow.
2. Consider the flow at two regions BC and DE. Initially the fluid lying between B and D.
   
3. During short time interval At, this fluid would have moved. Suppose V₁ is the speed at B and V₂ is the speed at D.
4. In time ∆t, distance moved from B to C is V₁ ∆t in the same interval (∆t) distance moved from D to E is V₂∆t.
5. Let P₁ and P₂ be pressure act at area’s of cross¬sections A₁ and A₂ of the two regions.
6. The work done on the fluid at left end (BC)
   = Force × displacement
   = Pressure × Area × displacement
   = P₁A₁ × V₁∆t (∆V = A₁V₁∆t) = P₁∆V ………………. (1)
7. Similarly work done by the fluid at right end (DE)
   = P₂A₂ × V₂∆t = P₂∆V …………………. (2)
8. Work done on the fluid is taken as positive and workdone by the fluid is taken as negative.
   ∴ Total work done (W) = (P₁ – P₁) ∆V ………………. (3)
   Part of this work goes into changing the K.E of the fluid and part goes into changing gravitational P.E.
9. Mass of the fluid(∆m) passing through the pipe in time (∆t) is ∆m = ρA₁V₁∆t
   where ρ is the density of the fluid.
   ∆m = ρ∆V ………………….. (4)
10. Gravitational P.E = ρg∆V (h₂ – h₁) ………………… (5)
    Change in K.E (∆K) = \(\frac{1}{2}\) ρ∆V (V₂² – V₁²) …………….. (6)
11. According to law of conservation of energy
    (P₁ – P₂) ∆V = \(\frac{1}{2}\) ρ ∆V (V₂² – V₁²) + ρg∆V (h₂ – h₁)
    P₁ – P₂ = \(\frac{1}{2}\) ρ (V₂² – V₁²) + ρg (h₂ – h₁)
    P₁ + \(\frac{1}{2}\) ρ V₁² + ρgh₁ = P₁ + \(\frac{1}{2}\) ρ V₂² + ρgh₂
    P + \(\frac{1}{2}\) ρ v² + ρgh = constant ………………. (7)

∴ Sum of the pressure, K.E per unit volume and P.E per unit volume remains constant. Application of Bernoulli’s theorem :

1. It explains the dynamic lift on the wings of aeroplanes.
2. It explains the dynamic lift experienced by a spinning cricket ball.

Question 2.
Define coefficient of viscosity. Explain Stoke’s law and explain the conditions under which a rain drop attains terminal velocity, υ_t. Give the expression for υ_t.
Answer:
Coefficient of viscosity (η): The coefficient of viscosity is defined as the tangential force per unit area of the layer, required to maintain unit velocity gradient.
η = \(\frac{F}{A\left(\frac{\Delta V}{\Delta x}\right)}\)
S.l. unit → Nm^-2 s (or) P_aS
C.G.S unit → Poise
Dimensional formula = [ML^-1T^-1]
Stoke’s law : According to this law the viscous force acting on a moving body which is spherical in shape is directly proportional to

1. Coefficient of viscosity of fluid (η)
2. Radius of the spherical body (r)
3. Velocity of the body (v)

∴ F ∝ ηrv
F = Kηrv
Where K is a constant of proportionality. Experimentally it was found to be 6π.
∴ F = 6πηrv
When rain drops falling through air from the clouds reach the surface with almost constant speed through they are moving under gravitational force. This velocity is called terminal velocity. After attaining the terminal velocity, net force acting on the rain drop is zero.
According to stokes law, F ∝ ηrv
F = 6πηrv (∵ 6π = K = Proportionality constant)
Let ρ, r be the density and radius of the sphere.
The fluid density be σ.
The forces acting on the sphere are

1. Weight of the sphere W = mg
   W = Vρg = \(\frac{4}{3}\)πr³ρg …………….. (1)
2. The force of buoyancy (B) = V σ g = \(\frac{4}{3}\)πr³σg ……………. (2)
3. Viscous force (f) = 6πηv ………………… (3)
   When the sphere attains terminal velocity (v_t), the net force on the body becomes zero.
   
   ∴ At terminal velocity
   Net downward force = Net upward force
   W = B + f, W – B = f
   \(\frac{4}{3}\)πr³ρg – \(\frac{4}{3}\)πr³σg = 6πηrv_t;
   ∴ \(\frac{4}{3}\)πr³g (ρ – σ) = 6πηrv_t
   V_t = \(\frac{2}{9} \frac{r^2 g(\rho-\sigma)}{\eta}\) ………………. (4)

Problems

Question 1.
Calculate the work done in blowing a soap bubble of diameter 0.6 cm against the surface tension force. (Surface tension of soap solution = 2.5 × 10² Nm^-1).
Solution:
D = 0.6 cm = 0.6 × 10² m
r = \(\frac{D}{2}=\frac{0.6 \times 10^{-2}}{2}\) = 0.3 × 10² m
S = 2.5 × 10² N/m
W = 8πr²s
= 8 × 3.14 × (0.3 × 10^-2)₂ × 2.5 × 10^-2
W = 5.652 × 10^-6 J

Question 2.
How high does methyl alcohol rise in a glass tube of diameter 0.06 cm ? (Surface tension of methyl alcohol = 0.023 Nm^-1 and density = 0.8 gmcm^-3. Assume that the angle of contact is zero)
Solution:
D = 0.06cm, θ = 0°
r = \(\frac{D}{2}=\frac{0.06}{2}\) = 0.03 cm = 0.03 × 10^-2 m
= 3 × 10^-4 m
S = 0.023 N/m, Density = 0.8 gm/c.c.
= 0.8 × \(\frac{10^{-3}}{10^{-6}}\)
ρ = 0.8 × 10³ kg/m³
S = \(\frac{\mathrm{hr \rho g}}{2 \cos \theta}\)
h = \(\frac{2 \cos \theta}{r \rho g}\)
= \(\frac{2 \times 0.023}{3 \times 10^{-4} \times 0.8 \times 10^3 \times 9.8}\) (∵ cos 0° = 1)
= 0.0019 × 10^-1 m
≈ 0.002 m
h = 2 cm .

Question 3.
What should be the radius of a capillay tube if water has to rise to a height pf 6 cm in it ? (Surface tension of water = 7.2 × 10^-2 Nm^-1).
Solution:
h = 6 × 10^-2 m
S = 7.2 × 10^-2 N/m
Density of water (ρ) = 10³ kg/m³
S = \(\frac{\mathrm{hr \rho g}}{2}\)
r = \(\frac{2 \mathrm{~S}}{\mathrm{h \rho g}}=\frac{2 \times 7.2 \times 10^{-3}}{6 \times 10^{-2} \times 10^3 \times 9.8}\)
r = \(\frac{14.4}{58.8} \) × 10^-3
r = 0.24 × 10^-3 m
r = 0.24 mm

Question 4.
Find the depression of the meniscus in the capillary tube of diameter 0.4 mm dipped in a beaker containing , mercury. (Density of mercury = 13.6 × 10³ Kg m^-3 and surface tension of mercury = 0.49 Nm^-1 and angle of contact = 135°).
Solution:
D = 4 m.m
r = \(\frac{D}{2}=\frac{4}{2}\) = 2m.m = 2 × 10^-3 m
ρ = 13.6 × 10³ kg/m³
θ = 135°, S = 0.49 Nm
cosθ = cos 135°
= – sin 45° = – \(\frac{1}{\sqrt{2}}\)
S = \(\frac{\mathrm{hr \rho g}}{2 \cos \theta}\)
h = \(\frac{2 \mathrm{~s} \cos \theta}{\mathrm{r \rho g}}\)
= 2 × 0.49 \(\left(\frac{-1}{\sqrt{2}}\right) \frac{1}{2 \times 10^{-3} \times 13.6 \times 10^3 \times 9.8}\)
h = \(\frac{-0.49}{13.6 \times 9.8 \times \sqrt{2}}\)
h = -0.024m.

Question 5.
If the diameter of a soap bubble is 10 mm and its surface tension is 0.04 Nm^-1, find the excess pressure inside the bubble. [Mar. 14]
Solution:
D = 10 m.m
r = \(\frac{D}{2}=\frac{10}{2}\) = 5 m.m
= 5 × 10^-3
S = 0.04 N/m
P_i – P₀ = \(\frac{4 S}{r}=\frac{4 \times 0.04}{5 \times 10^{-3}}\)
= 0.032 × 10³
P_i – P₀ = 32 N/m² (or) Pascal.

Question 6.
If work done by an agent to form a bubble of radius R is W, then how much energy is required to increase its radius to 2R ?
Solution:
R₁ = R, R₂ = 2R
Initial work done (W) = 8πR²S
Final work done (W) = 8π[R₂² – R₁²]S
= 8π [4R² – R²]S
= 3 × 8π R²S
W = 3W.

Question 7.
If two soap bubbles of radii R₁ and R₂ (in vacuum) coalasce under isothermal conditions, what is the radius of the new bubble. Take T as the surface tension of soap solution.
Solution:
R₁, R₂ and R be the radii of first, second and resultant bubble. The soap bubbles coalesce in vacuum, so surface energy do not change
E = E₁ + E₂
8π R²T = 8π R₁²T + 8π R₂² T
R² = R₁² + R₂²
R = \(\sqrt{R_1^2+R_2^2}\)

Additional Problems

Question 1.
Explain why
a) The blood pressure in humans is greater at the feet than at the brain.
b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km.
c) Hydrostatic pressure is a scalar quantities even though pressure is force divided by area.
Answer:
a) The height of the blood column in the human body is more at feet than at the brain. That is why, the blood exerts more pressure at the feet than at the brain (∴ pressure = hρg)

b) We know that the density of air is maximum near the surface of earth and decreases rapidly with height and at a height of about 6 km, it decreases to near by half its value at the sea level. Beyond 6 km height the density of air decreases very slowly with height. Due to this reason, the atmospheric pressure at a height of about 6 km decreases to nearby half of its value at sea level.

c) Since due to applied force on liquid, the pressure is transmitted equally in all directions inside the liquid. That is why there is no fixed direction for the pressure due to liquid. Hence hydrostatic pressure is a scalar quantity.

Question 2.
Explain why
a) The angle of contact of mercurywith glass is obtuse, while that of water with glass is acute.
b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.) [T.S. Mar. 15]
c) Surface tension of a liquid is independent of the area of the surface.
d) Water with detergent disolved in it should have small angles of contact.
e) A drop of liquid under no external forces is always spherical in shape.
Answer:
a) When a small quantity of liquid is poured on solid, three interfaces, namely liquid- air, solid-air and solid-liquid are formed. The surface tensions corresponding to these three interfaces are S_LA, S_SA and S_SL respectively. Let 0 be the angle of contact between the liquid and solid. The molecules in the region, where the three interfaces meet are in equilibrium. It means net force acting on them is zero. For the molecule at 0 to be in equilibrium, we have.

In case of mercury glass, S_SA < S_SL, therefore cosθ is negative or θ > 90° i.e. obtuse. In case of water-glass, S_SA > S_SL, therefore cosθ is positive or θ < 90° i,e. acute.

b) For mercury glass, angle of contact is obtuse. In order to achieve this obtuse value of angle of contact, the mercury tends to form a drop. In case of water glass, the angle of contact is acute. To achieve this acute value of angle of contact, the water tends to spread.

c) Surface tension of liquid is the force acting per unit length on a line drawn tangentially to the liquid surface at rest. Since this force is independent of the area of liquid surface, therefore surface tension is also independent of the area of the liquid surface.

d) We know that the cloth has narrow spaces in the form of capillaries. The rise of liquid in a capillary tude is directly proportional to cos0. if 0 is small cos0 will be large. Due to which capillary rise will be more and so the detergent will penetrate more in cloth.

e) In the absence of external forces, the surface of the liquid drop tends to acquire the minimum surface area due to surface tension. Since for a given volume, the surface area of sphere is least, hence the liquid drop takes the spherical shape.

Question 3.
Fill in the blanks using the word (S) from the list appended with each statement ;
a) Surface tension of liquids generally … with temperatures (increases / decreases)
b) Viscosity of gases… with temperature, whereas viscosity of liquids … with temperature (increases / decreases)
c) For solids with elastic modulus of rigidity, the shearing force is proportional to…. while for fluids it is proportional to … (shear strain / rate of shear strain)
d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle)
e) For the model of a plane in a wind tunnel, turbulence occurs at a… speed for turbulence for an actual plane (greater / smaller)
Answer:
a) Decreases
b) increases; decreases
c) Shear strain; rate of shear strain.
d) Conservation of mass; Bernoullis principle.
e) Greater.

Question 4.
Explain why
a) To keep a piece of paper horizontal, you should blow over, not under, it.
b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.
d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel.
e) A spinning cricket ball in air does not follow a parabolic trajectory.
Answer:
a) When we blow over the paper, the velocity of air blow increases and hence pressure of air on it decreases (according to Bernoullis theorem), whereas pressure of air below the paper is atmosphere. Hence the paper stays horizontal.

b) By doing so the area of the outlet of water jet is reduced, so velocity of water increases according to equation of continuity av = a constant.

c) When a fluid is flowing out of a small hole in a vessel it acquires a large velocity and hence possesses large momentum. Since no external force is acting on the system, a backward velocity must be attained by the vessel (according to law of conservation of momentum). As a result of it, impulse (backward thrust) is experienced by the vessel.

d) There, size of the needle controls velocity of flow and thumb pressure controls pressure. According to Bernoulli’s theorem, P + ρgh + \(\frac{1}{2}\) ρV² = a constant shows that P occurs with power one and V occurs with power two, hence the velocity has more influence. That is why the needle has a better control over flow.

e) If the ball is spinning well as moving linearly, the streamlines at the top of ball due to two types of motion are opposed to each other and those below are in the same direction. As a result of it, the velocity of air flow is greater below than above the ball. Now, according to Bernoullis principle, the pressure on the upper side of the ball becomes more than the pressure on the lower side of ball. Due to it, a resultant force F acts upon the ball at right angle to linear motion in the downward, direction, resulting the ball to move along a curved path.

Question 5.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor ?
Answer:
Here, m = 50kg; r =D/2 = 1/2 cm = \(\frac{1}{200}\) m
Pressure = \(\frac{\text { Force }}{\text { area }}=\frac{m g}{\pi r^2}=\frac{50 \times 9.8}{(22 / 7)(1 / 200)^2}\)
= 6.24 × 10⁶ Nm^-2.

Question 6.
Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^-3. Determine the height of the wine column for normal atmosphere pressure.
Answer:
P = 0.76 × (13.6 × 10³) × 9.8
= h × 984 × 9.8 or
h = \(\frac{0.76 \times 13.6 \times 10^3 \times 9.8}{984 \times 9.8}\)
= 10.5 m.

Question 7.
A vertical off-shore structure is built to withstand a maximum stress of 10⁹ Pa. Is the structure suitable for putting up on top of an oil well in the ocean ? Take the depth of the ocean to be roughly 3 km and ignore ocean currents.
Answer:
Here, maximum stress = 10⁹Pa,
h = 3km = 3 × 10³m;
ρ(water) = 10³kg/m³ and g = 9.8 m/s²
The structure will be suitable for putting upon top of an oil well provided the pressure exerted by sea water is less than the maximum stress it can bear.
Pressure due to sea water, P = hρg
= 3 × 10³ × 10³ × 9.8
= 2.94 × 10⁷ Pa
Since the pressure of sea water is less than the maximum stress of 10⁹ Pa, the structure will be suitable for putting upon top of the oil well.

Question 8.
A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm². What maximum pressure would the smaller piston have to bear ?
Answer:
The maximum force, which the bigger piston can bear,
F = 3000 kgf = 3000 × 9.8 N
∴ Area of piston, A = 425 cm²
= 425 × 10^-4 m²
∴ maximum pressure on the bigger piston.
P = \(\frac{F}{A}=\frac{3000 \times 9.8}{425 \times 10^{-4}}\) = 6.92 × 10⁵ Pa
Since the liquid transmits pressure equally, therefore the maximum pressure the smaller piston can bear is 6.92 × 10⁵ Pa.

Question 9.
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit ?
Answer:
For water column in one arm of U-tube,
h₁ = 10.0 cm; ρ₁ (density) = 1g cm^-3
For spirit column in other arm of U-tube, h₂ = 12.5 cm; ρ₁ = ?
As the mercury column in the two arms of U-tube are in level, therefore pressure exerted by each is equal. Hence h₁ρ₁g = h₂ρ₂g or
P₂ = \(\frac{h_1 \rho_1}{h_2}=\frac{10 \times 1}{12.5}\) = 0.8 g cm^-3
Therefore, relative density of spirit = ρ₂/ρ₁
= \(\frac{0.8}{1}\) = 0.8

Question 10.
In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific, gravity of mercury = 13.6)
Answer:
On pouring 15.0 cm of water and spirit each into the respective arms of U-tube, the mercury level will rise in the arm containing spirit. Let h be the difference in the levels of mercury in two arms of U-tube and p be the density of mercury.
∴ The pressure exerted by h cm of mercury column = difference in pressure exerted by water and spirit.
∴ h₁ρ₁g = h₂ρ₂g ……………. (1)
Here h = ? ρ =13.6 g cm^-3
ρ₁ =1 cm^-3
ρ₂ = 0.8 g cm^-3
h₁ = 15 + 10 = 25 cm
h₂ = 15 + 12.5 = 27.5 cm
Putting values in (i) we get h × 13.6 × g
= 25 × 1 × g – 27.5 × 0.8 × g = 3g
or h = 3/13.6 = 0.22cm

Question 11.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river ? Explain.
Answer:
No, Bernoulli’s theorem is used only for stream – line flow.

Question 12.
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation ? Explain.
Answer:
No, it does not matter if one uses gauge instead of absolute pressures in applying Bernoulli’s’ equation, provided the atmospheric pressure at the two points where Bernoulli’s equation is applied are significantly different.

Question 13.
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10^-3 kg s^-1, what is the pressure difference between the two ends of the tube ? (Density of glycerine = 1.3 × 10³ kg m^-3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].
Answer:
Here, l = 1.5 m, r = 1.0 cm = 10^-2 m, ρ = 1; 1.3 × 10^-3 kg/m3; η = 0.83 Nsm^-2.
Mass of glycerine flouring per sec, M = 4 × 10^-3 kg/s
Volume of glycerine flouring per sec, V = \(\frac{M}{\rho}\)
V = \(\frac{4 \times 10^{-3}}{1.3 \times 10^3} \mathrm{~m}^3 \mathrm{~s}^{-1}\) m³s^-1
If ρ is the difference of pressure between two ends of the tube,then using poisevilles formula we have
V = \(\frac{\pi \rho r^4}{8 \eta l}\) or P = \(\frac{V \times 8 \eta l}{\pi r^4}\)
P = \(\left(\frac{4 \times 10^{-3}}{1.3 \times 10^3}\right) \times \frac{8 \times 0.83 \times 1.5}{3.142 \times\left(10^{-2}\right)^4}\)
= 975.37 Pa

Question 14.
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s^-1 and 63 m s^-1 respectively. What is the lift on the wing if its area is 2.5 m² ? Take the density of air to be 1.3 kg m^-3.
Answer:
Let V₁, V₂ be the speeds on the upper and lower surfaces of the wing of aeroplane and P₁ and P₂ be the pressures on upper and lower surfaces of the wing respectively.
Then V₁ = 70ms^-1; V₂ = 63ms^-1,
P = 1.3kg m^-3.
This difference of pressure provides the lift to the aeroplane.
\(\frac{P_1}{\rho}\) + gh + \(\frac{1}{2} V_1^2\) = \(\frac{P_2}{\rho}\) + gh + \(\frac{1}{2} V_2^2\)
∴ \(\frac{P_1}{\rho}-\frac{P_2}{\rho}=\frac{1}{2}\left(V_2^2-V_1^2\right)\)
or P₁ – P₂ = \(\frac{1}{2} \rho\left(V_2^2-V_1^2\right)\)
= \(\frac{1}{2}\) × 1.3 [(70)² – (63)²]
= 605.15 Pa
This difference of pressure provides the lift to the aeroplane.
So, lift on the aeroplane = Pressure difference × area of wings
= 605.15 × 2.5
= 1512.875 N
= 1.51 × 10³N.

Question 15.
Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?

Answer:
Figure a is incorrect. According to equation of continuity i.e. av = a constant, where area of cross – section of tube is less, the velocity of liquid flow is more than the other portion of tube. According to Bernoulli’s theorem,
P + \(\frac{1}{2}\) ρv² = a constant i.e. where V is more P is less and avice versa.

Question 16.
The cylindrical tube of a spray pump has a cross-section of 8.0 cm² one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min^-1, what is the speed of ejection of the liquid through the holes ?
Answer:
Area of cross – section of tube, a₁ = 8.0 cm² = 8 × 10^-4 m²
No. of holes = 40, Diameter of each hole, D = 1 mm = 10^-3 m
∴ Radius of hole, r = \(\frac{D}{2}=\frac{1}{2}\) × 10^-3 m
= 5 × 10^-4 m
Area of cross – section of each hole = πr²
= π(5 × 10^-4)²m²
Total area of cross – section of 40 holes,
a₂ = 40 × π (5 × 10^-4)²m²
Speed of liquid inside the tube,
V₁ = 1.5m/min
= \(\frac{1.5}{60}\) ms^-1
If V₂ is the velocity of ejection of the liquid through the holes, then
a₁V₁ = a₂V₂ or V₂ = \(\frac{a_1 V_1}{a_2}\)
V₂ = \(\frac{\left(8 \times 10^{-4}\right) \times 1.5}{60 \times 40 \times \pi \times\left(5 \times 10^4\right)^2}\)
= 0.637 ms^-1

Question 17.
A U-shaped wire is dipped in a soap solution and removed. The thin shaped film formed between the surfaces and the light slider supports a weight of 1.5 × 10^-2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film ?
Answer:
We know that soap film has two free surfaces, so total length of the film to be supported,
= 2l = 2 × 30 = 60 cm
= 0.6 m
Total force on the slider due to surface tension will be,
F = S × 2l
= S × 0.6 N
In equilibrium position, the force F on slider due to surface tension must balance the weight mg
(1.5 × 10^-2)N i.e. F = 1.5 × 10^-2
= S × 0.6 Or
S = \(\frac{1.5 \times 10^{-2}}{0.6}\)
2.5 × 10^-2Nm^-1

Question 18.
Figure (a) shows a thin liquid film supporting a small weight = 4.5 × 10^-2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c) ? Explain your answer physically.

Answer:
a) Here, length of film supporting the weight
= 40cm = 0.4 m
Total liquid film supported (or force)
= 4.5 × 10^-2 N
film has two free surfaces, ∴ surface tension,
S = \(\frac{4.5 \times 10^{-2}}{2 \times 0.4}\)
S = 5.625 × 10^-2 Nm^-1
Since the liquid is same for all the cases (a), (b), (c) and temperature is also same, therefore surface tension for cases (b) and (c) will also be the same = 5.625 × 10^-2. In figure (b) and (c) the length of the film supporting the weight is also the same as that of(a), hence the total weight supported in each case is 4.5 × 10^-2 N.

Question 19.
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ? Surface tension of mercury at that temperature (20°C) is 4.65 × 10^-1 Nm^-1. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop.
Answer:
Here, r = 3.0 mm = 3 × 10^-3 m;
S = 4.65 × 10^-1 Nm^-1;
P = 1.01 × 10⁵ Pa
Excess of pressure inside the drop of mercury is given by
P = \(\frac{2 S}{r}=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}\)
= 310 Pa
Total pressure inside the drop = P + ρ
= 1.01 × 10⁵ + 310
= 1.01 31 × 10⁵ Pa

Question 20.
What is the excess pressure inside a bubble of soap solution of radius 5.0 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50 × 10^-2 Nm^-1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is 1.01 × 10⁵ Pa).
Answer:
Here, S = 2.5 × 10^-2 Nm^-1, r = 5.00 mm = 5 × 10^-3 m.
Density of soap solution, ρ = 1.2 × 10³ kg m^-3
Excess pressure inside the soap bubble,
P = \(\frac{4 s}{r}=\frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\) = 20 Pa
Excess pressure inside the air bubble, P’ = \(\frac{2S}{r}\)
= \(\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\) = 10 Pa
Total pressure inside the air bubble at depth h in soap solution – ρ’ + atmospheric pressure + hρg
= 10 + 1.01 × 10⁵ + 0.4 × 1.2 × 10³ × 9.8
= 1.06 × 10³ Pa

Question 21.
A tank with a square base of area 1.0 m² is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm². The tank is filled with water in one compartment and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.
Answer:
For compartment containing water,
h₁ = 4m, ρ₁ = 10³ kg m^-3
The pressure exerted by water at the door provided at bottom

P₁ = h₁ρ₁g
= 4 × 10³ × 9.8
= 3.92 × 10⁴ Pa
For compartment containing acid,
h₂ = 4m,
ρ₂ = 1.7 × 1.0³ kg/m³
The pressure exerted by acid at the door provided at bottom.
P₂ = h₂ρ₂g
= 4 × 1.7 × 10³ × 9.8
= 6.664 × 10⁴ Pa
∴ Difference of pressure = P₂P₁
= 6.664 × 10⁴ – 3.92 × 10⁴
= 2.774 × 10⁴Pa
Given, area of door, A = 20cm²
= 20 × 10^-4m²
Force on the door = difference in pressure × area
= (P₂ – P₁) × A
= (2.774 × 10⁴) × (20 × 10⁴)
= 54.88N
≈ 55N
To keep the door closed, the force equal to 55 N should be applied horizontally on the door from compartment containing water to that containing acid.

Question 22.
A manometer reads the pressure of a gas in an enclosure as shown in Fig. (a) when a pump removes some of the gas, the manometer reads as in Fig. (b) The liquid used in the manometers is mercury and the atmo-spheric pressure is 76 cm of mercury.
a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and b) in units of cm of mercury.
b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury, are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).

Answer:
a) Here, atomospheric presure, p = 76 cm of mercury
In Fig (a) Pressure head
h = + 20 cm
∴ Absolute pressure = p + h
= 76 + 20
= 96 cm of mercury
Gauge pressure = h = 20 cm of mercury.
In fig (b) pressure head, h = -18 cm
Absolute pressure = p + h
= 76+ (-18)
= 58 cm of mercury
Gauge pressure = h = -18cm of mercury.

b) Here 13.6 cm of water added in right limb is equiralent to
\(\frac{13.6}{13.6}\) = 1 cm of mercury column.
i.e., h¹ = 1 cm of mercury column.
Now pressure at A, PA = P + h¹ = 76 + 1
= 77 cm
Let the difference in mercury levels in the two lumbs be h₁, then pressure at B,
P_B = 58 + h₁ or
As P_A = P_B = 77 = 58 + h₁
h₁ = 77 – 58 = 19 cm of mercury column.

Question 23.
Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular comon height. Is the force exerted by the water on the base of the vessel the same in the two cases ? If so, why do the vessels filled with water to that same height give different readings on a weighing scale ?
Answer:

Since the pressure depends upon the height of water column and the height of the water column in the two vessels of different shapes is the same.hence there will be same pressure due to water on the base of each vessel. As the base area of each vessel is same, hence there will be equal force acting on the two base areas due to water pressure. The water exerts force on the walls of the vessel also. In case, the walls of the vessel are not perpendicular to base, the force exerted by water on the walls has a net non-zero vertical component which is more in first vessel than that of second vessel. That is why, the two vessel:; filled with water to same vertical height show different readings on a weighing machine.

Question 24.
During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein ? Use the density of whole blood from Table 1.
Answer:
h = \(\frac{p}{p g}=\frac{200}{1.06 \times 10^3}\) × 9.8 = 0.1925 m
The blood may just enter the vein if the height at which the blood container be kept must be slightly greater than 0.1925 m i.e. 0.2 m.

Question 25.
In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy.
a) What is the largest average velocity of blood flow in an artery of diameter 2 × 10^-3 m if the flow must remain laminar ?
b) Do the dissipative forces become more important as the fluid velocity increases ? Discuss qualitatively.
Answer:
a) If dissipative forces are present, then some forces in liquid flow due to pressure different is spent against dissipative forces. Due to which the pressure drop becomes large.

b) The dissipative forces become more important with increasing flow velocity, because of tubulence.

Question 26.
a) What is the largest average velocity of blood flow in an artery of radius 2 × 10^-3m if the flow must remain lanimar ?
Answer:
Here, r = 2 × 10^-3m ;
D = 2r = 2 × 2 × 10^-3 = 4 × 10^-3m
η = 2.084 × 10^-3 Pa s;
p = 1.06 × 10³ kgm³
For flow tobe laminar, N_r = 2000
a) Now, V_c = \(\frac{N_r \eta}{\rho D}\)
= \(\frac{2000 \times\left(2.084 \times 10^{-3}\right)}{\left(1.06 \times 10^3\right) \times\left(4 \times 10^{-3}\right)}\) = 0.98m/s.

b) What is the corresponding flow rate ? (Take viscosity of blood to be 2.084 × 10^-3 Pa s).
Answer:
Volume flowing per second = πr²Vc
= \(\frac{22}{7}\) × (2 × 10^-3)² × 0.98
= 1.23 × 10^-5 m³s^-1

Question 27.
A plane is in level flight at constant speed and each of its two wings has an area of 25m². If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m^-3)
Answer:
Here, V₁ = 180 km/h = 50m/s, V₂ = 234 km/ h = 65 m/s;
A = 2 × 25 = 50m²; P = 1kg/m³
P₁ – P₂ = \(\frac{1}{2}\) p (V₂² – V₁²)
= \(\frac{1}{2}\) × 1 × (65² – 50²)
Upward force = (P₁ – P₂) A = \(\frac{1}{2}\) × (65² – 50²) × 50N
As the plane is in level flight, so
mg = (P₁ – P₂)A
or m = \(\frac{\left(P_1-P_2\right) A}{g}=\frac{1 \times\left(65^2-50^2\right) \times 50}{2 \times 9.8}\)
= 4.4 × 10³N

Question 28.
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10^-5 m and density 1.2 × 10³ kg m^-3. Take the viscosity of air at the temperature of the experiment to be 1.8 × 10⁵ Pa s. How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop due to air.
Answer:
Here, r = 2.0 × 10^-5m; ρ = 1.2 × 10³ kg m^-3;
η = 1.8 × 10⁵ Ns m^-2
P₀ = 0, V = ?, F = ?
Terminal velocity V = \(\frac{2 r^2\left(\rho-\rho_0\right) g}{9 \eta}\)
= \(\frac{2 \times\left(2.0 \times 10^{-5}\right)^2 \times\left(1.2 \times 10^3-0\right) \times 9.8}{9 \times 1.8 \times 10^{-5}}\)
= 5.8 × 10^-2ms^-1 = 5.8 cms^-1
Viscous force on the drop, F = 6πηrv
= 6 × \(\frac{22}{7}\) × (1.8 × 10^-5) × (2.0 × 10^-5) × (5.8 × 10^-2)
= 3.93 × 10^-10N.

Question 29.
Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.456 N m^-1. Density of mercury = 1.36 × 10³ kg m^-3.
Answer:
Here, θ = 140°, r = 1 × 10^-3 m;
S = 0.465 Nm^-1, ρ = 13.6 × 10³ kg, h = ?]
Cos = 140° = – cos40° = -0.7660
Now h = \(\frac{2 S \cos \theta}{r \rho g}\)
= \(\frac{2 \times 0.465 \times \cos 140^{\circ}}{10^{-3} \times 13.6 \times 10^3 \times 9.8}\)
= \(\frac{2 \times 0.465 \times(-0.7660)}{10^{-3} \times 13.6 \times 10^3 \times 9.8}\)
= -5.34 × 10^-3m
= -5.34mm
Negative value of h shows that the mercury levels is depressed in the tube.

Question 30.
Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to forma a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube ? Surface tension of water at the temperature of the experiment is 7.3 × 10^-2 N m^-1. Take the angle of contact to be zero and density of water to be 1.0 × 10³ kg m^-3 (g = 9.8 m^-2).
Answer:
Here, S = 7.3 ^-2 Nm^-1, ρ = 1.0 × 10³ kg m^-3; θ = 0°
For narrow tube, 2r₁ = 3.00 mm = 3 × 10^-3 m or r₁ =1.5 × 10^-3 m .
For wider tube, 2r₂ = 6.00 mm = 6 × 10^-3 m or r₂ = 3 × 10^-3 m
let h₁, h₂ be the heights to which water rises in narrow tube and wider tube respectively.
Then, h₁ = \(\frac{2 s \cos \theta}{r_1 \rho g}\) and h₂ = \(\frac{2 s \cos \theta}{r_2 \rho g}\)
∴ Difference in levels of water in two limbs of U-tube is

Question 31.
a) It is known that density of air decreases with height y as
ρ = ρ₀e^-y/y₀
Where p₀ = 1.25 kg m^-3 is the density at sea level, and y₀ is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of g remains constant.
Answer:
We know that the rate of decrease of density p of air is directly proportional to height y i.e.
\(\frac{-d \rho}{d y}\) ∝ p or \(\frac{d \rho}{d y}\) = – Kρ
Where K is a constant of proportionality. Here – ve sign shows that ρ decreases as y increases.
\(\frac{d y}{\rho}\) = – Kρ
Integrating it with in the conditions, as y changes fromotoy density changes from ρ₀ to ρ, we have
\(\left.\int_{P_0}^p \frac{d P}{P}=-\int_0^y k d y=\left[\log e^p\right)\right]_{\rho_0}^p=k y\)
\(\frac{\rho}{\rho_0}=e^{-k y}=\rho=\rho_0 e^{-\mathrm{y} / y_0}\)

b) A large He balloon of volume 1425 m³ is used to lift a payload of 400 kg. Assum that the balloon maintains constant radius as it rises. How high does it rise ?
(Take y₀ = 8000 m and P_He = 0.18 kgm^-3)
Answer:
The balloon will rise to aheight, where its density becomes equal to the air at that height.

y = 8000 × 1 = 8000 m
= 8 km.

Textual Examples

Question 1.
The two thigh bones (femurs), each of cross-sectional area 10 cm² mass 40 kg. Estimate the average pressure sustained by the femurs.
Answer:
Total cross-sectional area of the femurs is A = 2 × 10 cm² 20 × 10^-4 m². The force acting on them is F = 40 kg wt = 400 N (taking g 10 ms^-2). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is
P_av = \(\frac{F}{A}\) = 2 × 10⁵ N m^-2

Question 2.
What is the pressure on a swimmer 10m below the surface of a lake ?
Answer:
Here
h = 10 m and p = 1000 kg m^-3.
Take g 10 m S^-2
From Equation = P = P_a + ρgh
P = P_a + ρgh
= 1.01 × 10⁵ Pa + 1000 kgm^-3 × 10m S^-2 × 10 m
= 2.10 × 10⁵ Pa
≈ 2 atm
This is a 100% increase in pressure from surface level. At a depth of 1 km the increase in pressure is 100 atm! Submarines are designed to withstand such enormous pressures.

Question 3.
The density of the atmosphere at sea level is 1.29 kg/m³. Assume that it does not change with altitude. Then how high would the atmosphee extend ?
Answer:
We use P – P_a = ρgh
ρgh = 1.29 kg m^-3 × 9.8 ms² × hm
= 1.01 × 10⁵ pa
∴ h = 7989 m ≈ 8 km
In reality the density of air decreases with height. So does the value of g. The atmospheric cover extends with decreasing press-ure over 100 km. We should also note that the sea level atmospheric pressure is not always 760 mm of Hg. A drop in the Hg level by 10 mm or more is a sign of an approaching storm.

Question 4.
At a depth of 1000 m in an ocean
(a) What is the absolute pressure ?
b) What is the guage Pressure ?
c) Find the force acting on the window of area 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sealevel atmospheric pressure. (The density of sea water is 1.03 × 10³ kg m^-3, g = 10ms^-2.
Answer:
Here h = 1000 m and ρ = 1.03 × 10³ kg m^-3
a) From P₂ – P₁ = ρgh absolute pressure
P = P_a + ρgh
= 1.01 × 10⁵ Pa + 1.03 × 10³ kg m^-3 × 10 m s^-2 × 1000 m
= 104.01 × 10⁵ Pa .
= 104 atm

b) Gauge pressure is P – P_a = ρgh = P_g
P_g = 1.03 × 10³ kg m^-3 × 10 m s² × 1000 m
= 103 × 10⁵ Pa
≈ 103 atm

c) The pressure outside the submarine is P = P_a + ρgh and the pressure inside it is P_a. Hence, the net pressure acting on the window is gauge pressure, P_g = ρgh. Since the area of the window is A = 0.04 m², the force acting on it is
F = P_gA = 103 × 10⁵ Pa × 0.04m²
= 4.12 × 10⁵ N
≈ 103 atm

Question 5.
Two syringes of different cross sections A₁, A₂ and lengths L₁, L₂ (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively.
a) Find the force exerted on the larger piston when a force of ION is applied to the smaller piston, b) if the smaller piston is pushed in through 6.0 cm,’how much does the larger piston move out ?
Answer:
a) Since pressure is transmitted undiminished throughout the fluid.
F₂ = \(\frac{A_2}{A_1} F_1=\frac{\pi\left(3 / 2 \times 10^{-2} m\right)^2}{\pi\left(1 / 2 \times 10^{-2} \mathrm{~m}\right)^2} \times 10 \mathrm{~N}\)
= 90 N

b) Water is considered to be perfectly incompressible. Volume covered by the move-ment of smaller piston inwards is equal to volume moved outwards due to the larger piston.
L₁A₁ = L₂A₂
L₂ = \(\frac{A_1}{A_2} L_1=\frac{\pi\left(1 / 2 \times 10^{-2} \mathrm{~m}\right)^2}{\pi\left(3 / 2 \times 10^{-2} \mathrm{~m}\right)^2}\)
= × 6 × 10^-2m
≃ 0.67 × 10^-2m = 0.67 cm.
Note, atmospheric pressure is common to both pistons and has been ignored.

Question 6.
In a car lift compressed air exerts a force F₁ on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston or radius 15 cm (Fig). If the mass of the car to be lifted is 1350 kg, calculate F₁. What is the pressure necessary to accomplish this task ? (g = 9.8 ms^-2).

Answer:
Since pressure is transmitted undiminished throughout the fluid.
F₁ = \(\frac{A_1}{A_2} F_2=\frac{\pi\left(5 \times 10^{-2} \mathrm{~m}\right)^2}{\pi\left(15 \times 10^{-2} \mathrm{~m}\right)^2}\)
= (1350 N × 9.8m s^-2) = 1470 N
= 1.5 × 10³N
The air pressure that will produce this force is
P = \(\frac{F_1}{A_1}=\frac{1.5 \times 10^{-3} \mathrm{~N}}{p\left(5 \times 10^{-2}\right)^2 \mathrm{~m}}\) = 1.9 × 10⁵ Pa
This is almost double the atmospheric pressure.

Question 7.
Blooc velocity : The flow of blood in a large artery of an anesthetised dog is diverted through a Venturi meter. The wider part of the meter has a cross-sectional area equal to that of the artery. A = 8 mm². The harrower part has an area = 4mn². The pressure drop in the artery is 24 Pa. What is the speed of the blood in the artery ?
Answer:
We take the density of blood from whole blood to be 1.06 × 10³ kg m³. The ratio of the areas is \(\left(\frac{\mathrm{A}}{\mathrm{a}}\right)\) = 2
Using v₁ = \(\left(\sqrt{\frac{2 \rho_m g h}{\rho}}\right)\left(\left(\frac{A}{2}\right)^2-1\right)^{-1 / 2}\)
we obtain
v₁ = \(\sqrt{\frac{2 \times 24 \mathrm{pA}}{1060 \mathrm{~kg} \mathrm{~m}^{-3} \times\left(2^2-1\right)}=0.125 \mathrm{~m} \mathrm{~s}^{-1}}\)

Question 8.
A fully loaded Boeing aircraft has a mass of 3.3 × 10⁵ kg. Its total wing area is 500 m2. It is in level flight with a speed of 960 km/h. a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is ρ = 1.2 kgm^-3].
Answer:
a) The weight of’the Boeing aircraft is balanced by the upward force due to the pressure difference
∆P × A – 3.3 × 10⁵ kg × 9.8 = mg.
∆P = (3.3 × 10⁵ kg × 9.8 m s^-2) / 500 m²
= 6.6 × 10³ N m^-2

b) We ignore the small height difference between the top and bottom sides in
P₁ + (\(\frac{1}{2}\)) ρv₁² + ρgh₁ = P₂ + (\(\frac{1}{2}\))ρv₂² + ρgh₂
The pressure difference between them is then
∆P = \(\frac{\rho}{2}\left(v_2^2-v_1^2\right)\)
Where v₂ is the speed of air over the upper surface and v₁ is the speed under the bottom surface.
(v₂ – v₁) = \(\frac{2 \Delta P}{\rho\left(v_2+v_1\right)}\)
Taking the average speed v_av (v₂ + v₁)/2 – 960 km/h = 267 m s^-1, we have
(v₂ – v₁)/v_av = \(\frac{\Delta \mathrm{P}}{\rho v_{\mathrm{av}}^2}\) ≈ 0.08
The speed of air above the wing needs to be only 8% higher than that below.

Question 9.
A metal block of area 0.10 m² is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in Fig. A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released, the block moves to the right with a constant speed of 0.085 ms^-1. Find the co-efficient of viscosity of the liquid.

Answer:
The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m. Thus the shear force F is
F = T = mg = 0.010 kg × 9.8 ms^-2
= 9.8 × 10^-2 N
Shear stress on the fluid = F/A = \(\frac{9.8 \times 10^{-2}}{0.10}\)
Strain rate = \(\frac{v}{l}=\frac{0.085 \mathrm{~ms}^{-1}}{0.3 \times 10^{-3} \mathrm{~m}}\)
η = \(\frac{\text { stress }}{\text { strain rate }}\)
= \(\frac{\left(9.8 \times 10^{-2} \mathrm{~N}\right)\left(0.30 \times 10^{-3} \mathrm{~m}\right)}{\left(0.085 \mathrm{~ms}^{-1}\right)\left(0.10 \mathrm{~m}^2\right)}\)
= 3.45 × 10^-3 Pa s

Question 10.
The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20°C is 6.5 cm s^-1. Compute the viscosity of the oil at 20°C. Density of oil is 1.5 × 10³ kg m^-3, density of copper is 8.9 × 10³ kg m^-3.
Answer:
We have v_t = 6.5 × 10^-2 ms^-1,
a = 2 × 10^-3m,
g = 9.8 ms^-2, ρ = 8.9 × 10³ kg m^-3,
σ = 1.5 × 10³ kg m^-3. From
v_t = \(\frac{2 a^2(\rho-\sigma) g}{(9 \eta)}\)
η = \(\frac{2}{9} \times \frac{\left(2 \times 10^{-3}\right) \mathrm{m}^2 \times 9.8 \mathrm{~ms}^{-2}}{6.5 \times 10^{-2} \mathrm{~ms}^{-1}}\) × 7.4 × 10³ kg m^-3
= 9.9 × 10^-1 kg m^-1s^-1

Question 11.
a) The flow rate of water from a tap of diameter 1.25 cm is L7min. The co-efficient of viscosity of water is 10^-3 Pa s.
b) After sometime the flow rate is increased to 3L / min. Characterise the flow for both the flow rates.
Answer:
a) Let the speed of the flow be v and the diameter of the tap be d = 1.25 cm. The volume of the water flowing out per second is
Q = v × π d²/4
v = 4Q / d² π
We then estimate the Reynolds number to be
R_e = 4vQ / πdη
= 4 × 10³ kg m^-3 × Q / (3.14 × 1.25 × 10^-2 m × 10^-3 Pa S)
= 1.019 × 10⁸ m^-3 SQ
Since initially (a)
Q = 0.48 L/min = 8cm³/s
= 8 × 10^-6 m³ s^-1, we obtain,
R_e = 815
Since this is below 1000, the flow is steady.
After some time

b) When Q = 3L/ min = 50 cm³, s = 5 × 10^-5 m³ s^-1 we obtain,
R_e = 5095
The flow will be turbulent. You may carry out an experiment in your washbasin to determine the ransition from laminar to turbulent flow.

Question 12.
The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water ? The surface tension of water at temperature of the experiments is 7.30 × 10^-2 Nm¹. 1 atmospheric pressure = 1.01 × 10⁵ Pa, density of water = 1000 kg/m³, g = 9.8 × ms^-2. Also calculate the excess pressure.
Answer:
The excess pressure in a bubble of gas in a liquid is given by 2S / r, where S is the surface tension of the liquid gas interface. You should note there is only one liquid surface in this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for excess pressure in that case is 4S / r.) The radius of the bubble is r. Now the pressure outside the bubble, within water, P₀ equals atmospheric pressure plus the pressure due to 8.00 cm of water column. That is
P₀ = (1.01 × 10⁵ Pa + 0.08 m × 1000 kg m^-3 × 9.80 m s^-2)
= 1.01784 × 10⁵ Pa. .
Therefore, the pressure inside the bubble is
P₁ = P₀ + 2S / r (as r = 1 mm)
= 1.01784 × 10⁵ Pa + (2 × 7.3 × 10^-2 Pa m/10^-3 m)
= (1.01784 + 0.00146) × 10⁵ Pa
= 1.02 × 10⁵ Pa
where the radius of the bubble is taken to be equal to the radius of the capillary tube, since the bubble is hemispherical ! (The answer has been rounded off to three significant figures). The excess pressure in the bubble is 146 Pa from (1.0178 + 0.00146) × 10⁵ Pa.