These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2

Unless stated otherwise, take π = \(\frac { 22 }{ 7 }\)

Question 1.

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of n.

Solution:

Radius of cone = 1 cm. and radius of hemisphere is also = 1 cm.

Volume of solid = Volume of cone + Volume of hemisphere

Question 2.

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Solution:

Volume of air contained in the model = Total volume of the solid

Diameter of base of each cone = 3 cm

∴ Radius of base of each cone = \(\frac { 3 }{ 2 }\)

Height of each cone = 2 cm

Volume of the air inside the model = Volume of air inside = Volume of cone + volume of cylinder + volume of other cone.

Volume of the model = 66 cm³

Question 3.

A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).

Solution:

Volume of one piece of gulab jamun = Volume of the cylindrical portion + Volume of the two hemispherical ends 1 2 8

Radius of each hemispherical portion = \(\frac { 2.8 }{ 2 }\) = 1.4 cm

Radius of gulab jamun = r = \(\frac { 2.8 }{ 2 }\) = 1.4 cm and

height = 5 cm so, height of cylinder (h) = 5 – (2.8) = 2.2.

Volume 45 gulab jamuns = 25.05 x 45 = 1127.279

30% of its volume = \(\frac{1127.279 \times 30}{100}\)

= 338.18 = 338 cm³.

Question 4.

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm.

Find the volume of wood in the entire stand (see figure).

Solution:

Pen stand is made in the shape of a cuboid whose length, breadth and height are respectively = 15 cm, 10 cm, 3.5 cm.

So, the volume of pen stand = length x breadth x height

= 15 cm x 10 cm x 3.5 cm = 525 cm²

Each hole, is in the shape of a cone, so the volume of cone

= \(\frac { 1 }{ 3 }\)r²h

Radius of hole = 0 0.5 cm, and height = 1.4 cm.

∴ Volume of 4 (holes) cone = 4 x \(\frac { 1 }{ 3 }\)r²h

= \(\frac { 4 }{ 3 }\) x \(\frac { 22 }{ 7 }\) 0.5 x 0.5 x 1.4 cm³

= \(\frac{4 \times 4.4 \times 0.25}{3}\) = \(\frac { 4 }{ 3 }\) = 1.466 cm³

Volume of wood in the entire stand = Volume of cuboid – Volume 4 holes.

= (525 – 1.466) cm³.

= 523.533 cm³.

Question 5.

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

Solution:

When lead shots are dropped into the vessel, then

Volume of water flows out = Volume of leads shots

Number of lead shots dropped in the vessel = 100.

Question 6.

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use π = 3.14)

Solution:

Given: radius of 1^{st} cylinder = 12 cm

and height of 1^{st} cylinder = 220 cm

∴ Volume of 1^{st} cylinder = πr²h

= π(12)² (220) cm³

= 144 x 220π cm³

= 144 x 220 x 3.14 cm³

= 99475.2 cm³ … (i)

Given: radius of 2^{nd} cylinder = 8 cm

and height of 2^{nd} cylinder = 60 cm

∴ Volume of 2^{nd} cylinder = πr²h

= π(8)² (60) cm³ = 64 x 60π cm³

= 64 x 60 x 3.14 cm³

= 12057.6 cm³ … (ii)

Total volume of solid = Volume of 1^{st} cylinder + Volume of 2^{nd} cylinder

= 99475.2 cm³ + 12057.6 cm³ = 111532.8 cm³

Given: mass of 1 cm³ of iron = 8 g

∴ Mass of 111532.8 cm³ of iron = 111532.8 x 8 g

= 892262.4 g = 892.262 kg

Question 7.

A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Solution:

Height of right circular cone = 120 cm.

and radius of hemisphere and cone = 60 cm.

Radius of cylinder = 60 cm, and height = 180 cm.

Volume of water, left in the cylinder = Volume of cylinder – (Volume of cone + volume hemisphere)

So the left in the cylinder = 1,31 m³. (approx).

Question 8.

A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter, the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Solution:

Volume of water the glass vessel can hold = 345 cm³ (Measured by the child)

Radius of the cylindrical part = \(\frac { 2 }{ 2 }\) = 1 cm

Height of the cylindrical part = 8 cm

∴ Volume of the cylindrical part = πr²h

= 3.14 x (1)² x 8 cm³

Diameter of the spherical part Radius = 8.5 cm

∴ Radius = \(\frac { 8.5 }{ 2 }\) cm

= \(\frac{7713.41}{24}\)

= 321.392 cm³.

Total volume of the glass vessel = Volume of the cylindrical part + Volume of the spherical part

= 25.12 cm³ + 321.39 cm³ = 346.51 cm³

Volume measured by child is 345 cm³, which is not correct. Correct volume is 346.51 cm³.