## AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.4

### 10th Class Maths 10th Lesson Mensuration Ex 10.4 Textbook Questions and Answers

Question 1.

A metallic sphere of radius 4.2 cm. is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Answer:

Given, sphere converted into cylinder.

Hence volume of the sphere = volume of the cylinder.

Sphere:

Radius, r = 4.2 cm

Volume V = \(\frac{4}{3}\)πr^{3}

= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 4.2 × 4.2 × 4.2

= 4 × 22 × 0.2 × 4.2 × 4.2

= 4 x 22 x 0.2 x 4.2 x 4.2

= 310.464

Cylinder:

Radius, r = 6 cm

Height h = h say

Volume = πr^{2}h

= \(\frac{22}{7}\) × 6 × 6 × h

= \(\frac{22 \times 36}{7} h\)

= \(\frac{792}{7} h\)

Hence, \(\frac{792}{7} h\) = 310.464

h = \(\frac{310.464 \times 7}{792}\) = 2.744cm

!! π can be cancelled on both sides i.e., sphere = cylinder

Question 2.

Three metallic spheres of radii 6 cm., 8 cm. and 10 cm. respectively are melted together to form a single solid sphere. Find the radius of the resulting sphere.

Answer:

Given : Radii of the three spheres r_{1} = 6 cm r_{2} = 8 cm r_{3} = 10 cm

These three are melted to form a single sphere.

Let the radius of the resulting sphere be ‘r’.

Then volume of the resultant sphere = sum of the volumes of the three small spheres.

∴ 1728 = (2 × 2 × 3) × (2 × 2 × 3) × (2 × 2 × 3)

r^{3} = 12 × 12 × 12

r^{3} = 12^{3}

∴ r = 12

Thus the radius of the resultant sphere = 12 cm

Question 3.

A 20 m deep well with diameter 7 m. is dug and the earth got by digging is evenly spread out to form a rectangu¬lar platform of base 22 m. × 14 m. Find the height of the platform.

Answer:

Volume of earth taken out = πr^{2}h

= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 20

= 770 m

Let height of plot form = H m.

∴ 22 × 14 × H = \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 20

H = \(\frac{35}{14}\) = \(\frac{5}{2}\) = \(2 \frac{1}{2} \mathrm{~m}\)

∴ The height of the plat form is \(2 \frac{1}{2} \mathrm{~m}\)

Question 4.

A well of diameter 14 m. is dug 15 m. deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 7 m to form an embankment. Find the height of the embankment

Answer:

Volume of the well = Volume of the embank

Volume of the cylinder = Volume of the embank

Cylinder :

Radius r = \(\frac{d}{2}\) = \(\frac{14}{2}\) = 7 cm

Height/depth, h = 15 m

Volume V = πr^{2}h

= \(\frac{22}{7}\) × 7 × 7 × 15

= 22 × 7 × 15

= 2310 m^{3}

Embank:

Let the height of the embank = h m

Inner radius ‘r’ = Radius of well = 7 m

Outer radius, R = inner radius + width

= 7m + 7m = 14 m

Area of the base of the embank = (Area of outer circle) – (Area of inner circle)

= πR^{2} – πr^{2}

= π(R^{2} – r^{2})

= \(\frac{22}{7}\)\(\left(14^{2}-7^{2}\right)\)

= \(\frac{22}{7}\) × (14+7) × (14-7)

= \(\frac{22}{7}\) × 21 × 7

= 462 m^{2}

∴ Volume of the embank = Base area × height

= 462 × h = 462 h m^{3}

∴ 462 h m^{3} = 2310 m^{3}

h = \(\frac{2310}{462}\) = 5 m.

Question 5.

A container shaped like a right circular cylinder having diameter 12 cm. and height 15 cm. is full of ice-cream. The ice-cream is to be filled into cones of height 12 cm. and diameter 6 cm., having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.

Answer:

Let the number of cones that can be filled with the ice-cream be ‘n’.

Then total volume of all the cones with a hemi spherical top = Volume of the ice-cream

Ice-cream cone = Cone + Hemisphere = πr^{2}h

Cone:

Radius = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm

Height, h = 12 cm

Volume V = \(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 12

= \(\frac{22}{7}\) × 36

= \(\frac{792}{7}\)

Hemisphere:

Radius = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm

Volume V = \(\frac{2}{3}\)πr^{3}

= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 3

= \(\frac{44 \times 9}{7}\)

= \(\frac{396}{7}\)

∴ Volume of each cone with ice-cream = \(\frac{792}{7}\) + \(\frac{396}{7}\) = \(\frac{1188}{7}\) cm^{3}

Cylinder:

Radius = \(\frac{d}{2}\) = \(\frac{12}{2}\) = 6 cm

Height, h = 15 cm

Volume V = πr^{2}h

= \(\frac{22}{7}\) × 6 × 6 × 15

= \(\frac{22 \times 36 \times 15}{7}\)

= \(\frac{11880}{7}\)

∴ \(\frac{11880}{7}\) = n × \(\frac{11880}{7}\)

⇒ n = \(\frac{11880}{7}\) × \(\frac{7}{1188}\) = 10

∴ n = 10.

Question 6.

How many silver coins, 1.75 cm in diameter and thickness 2 mm., need to be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Answer:

Let the number of silver coins needed to melt = n

Then total volume of n coins = volume of the cuboid

n × πr^{2}h = lbh [∵ The shape of the coin is a cylinder and V = πr^{2}h]

∴ 400 silver coins are needed.

Question 7.

A vessel is in the form of an inverted cone. Its height is 8 cm. and the radius of its top is 5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, 1/4 of the water flows out. Find the number of lead shots dropped into the vessel.

Answer:

Let the number of lead shots dropped = n

Then total volume of n lead shots = \(\frac{1}{4}\) volume of the conical vessel.

Lead shots:

Radius, r = 0.5 cm

Volume V = \(\frac{4}{3}\)πr^{3}

= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.5 × 0.5 × 0.5

Total volume of n – shots

= n × \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.125

Cone:

Radius, r = 5 cm;

Height, h = 8 cm

Volume, V = \(\frac{1}{3}\) πr^{2}h

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5 × 5 × 8

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 200

∴ Number of lead shots = 100.

Question 8.

A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter 4 \(\frac{d}{2}\) cm and height 3 cm. Find the number of cones so formed.

Answer:

Let the no. of small cones = n Then,

total volume of n cones = Volume of sphere Diameter = 28 cm.

Cones:

Radius r = \(\frac{d}{2}\)

Height, h = 3 cm

Total volume of n-cones = n . \(\frac{154}{9}\) cm^{3}

Sphere:

Radius = \(\frac{d}{2}\) = \(\frac{28}{2}\) = 14 cm

No. of cones formed = 672.