NCERT Class 10 Maths – Exercise 12.2 Solutions

NCERT Class 10 Maths

Chapter 12 – Surface Areas and Volumes | Exercise 12.2

(Rationalized Syllabus 2025-26 | Formerly Ex 13.2)

💡 Concept: Volume of Combinations

The volume of a solid formed by joining two or more basic solids is the sum of the volumes of the individual parts.

Q1

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 $\pi$.

Dimensions: $r = 1$ cm, Height of cone $h = 1$ cm.
Total Volume = Vol of Cone + Vol of Hemisphere
$$V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$$
$$= \frac{1}{3}\pi(1)^2(1) + \frac{2}{3}\pi(1)^3$$
$$= \frac{1}{3}\pi + \frac{2}{3}\pi = \pi \text{ cm}^3$$
✔️ Volume = $\pi$ cm³
Q2

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.

Dimensions:
Diameter = 3 cm $\Rightarrow r = 1.5$ cm.
Total length = 12 cm. Height of each cone ($h_c$) = 2 cm.
Height of cylinder ($h_{cyl}$) = $12 – (2 + 2) = 8$ cm.
Volume:
$$V = \text{Vol Cylinder} + 2 \times \text{Vol Cone}$$
$$= \pi r^2 h_{cyl} + 2 \times \frac{1}{3}\pi r^2 h_c = \pi r^2 (h_{cyl} + \frac{2}{3}h_c)$$
$$= \frac{22}{7} \times (1.5)^2 \times (8 + \frac{4}{3})$$
$$= \frac{22}{7} \times 2.25 \times \frac{28}{3} = 22 \times 0.75 \times 4 = 66$$
✔️ Volume = 66 cm³
Q3

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.

Dimensions:
Diameter = 2.8 cm $\Rightarrow r = 1.4$ cm.
Total length = 5 cm. Length of cylinder ($h$) = $5 – (1.4 + 1.4) = 2.2$ cm.
Volume of one Gulab Jamun:
$$V = \pi r^2 h + 2 \times \frac{2}{3}\pi r^3 = \pi r^2 (h + \frac{4}{3}r)$$
$$= \frac{22}{7} \times (1.4)^2 \times (2.2 + \frac{5.6}{3})$$
$$= 6.16 \times \frac{6.6 + 5.6}{3} = 6.16 \times \frac{12.2}{3} \approx 25.05 \text{ cm}^3$$
Volume of Syrup:
Volume of 45 Jamuns = $45 \times 25.05$.
Syrup = 30% of Total Volume.
$$\text{Syrup} = 0.30 \times 45 \times 25.05 \approx 338.175$$
✔️ Approx Volume of Syrup = 338 cm³
Q4

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.

Volume of Cuboid:
$$V_{box} = 15 \times 10 \times 3.5 = 525 \text{ cm}^3$$
Volume of 4 Conical Depressions:
$r = 0.5$ cm, $h = 1.4$ cm.
$$V_{cones} = 4 \times \frac{1}{3}\pi r^2 h = 4 \times \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4$$
$$= \frac{88}{21} \times 0.25 \times 1.4 \approx 1.47 \text{ cm}^3$$
Volume of Wood:
$$525 – 1.47 = 523.53 \text{ cm}^3$$
✔️ Volume of Wood = 523.53 cm³
Q5

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.

Volume of Cone (Water):
$$V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (5)^2 (8) = \frac{200}{3}\pi$$
Volume of Water Flows Out:
$$V_{out} = \frac{1}{4} \times V_{cone} = \frac{1}{4} \times \frac{200}{3}\pi = \frac{50}{3}\pi$$
Volume of 1 Lead Shot (Sphere):
$$V_{shot} = \frac{4}{3}\pi (0.5)^3 = \frac{4}{3}\pi (0.125) = \frac{0.5}{3}\pi$$
Number of Shots ($n$):
$$n \times V_{shot} = V_{out}$$
$$n \times \frac{0.5}{3}\pi = \frac{50}{3}\pi$$
$$0.5n = 50 \Rightarrow n = 100$$
✔️ Number of Lead Shots = 100
Q6

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 8g mass. (Use $\pi = 3.14$)

Dimensions:
Large Cylinder: $H=220, R=12$. Small Cylinder: $h=60, r=8$.
Total Volume:
$$V = \pi R^2 H + \pi r^2 h = 3.14 (12^2 \times 220 + 8^2 \times 60)$$
$$= 3.14 (144 \times 220 + 64 \times 60)$$
$$= 3.14 (31680 + 3840) = 3.14 (35520) = 111532.8 \text{ cm}^3$$
Total Mass:
$$\text{Mass} = 111532.8 \times 8 \text{ g} = 892262.4 \text{ g}$$
✔️ Mass = 892.26 kg
Q7

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.

Volume of Water Left = Volume of Cylinder – Volume of Solid
Solid Volume = Vol Cone + Vol Hemisphere.
Calculation:
$$V_{cyl} = \pi r^2 H = \pi (60)^2 (180)$$
$$V_{solid} = \frac{1}{3}\pi (60)^2 (120) + \frac{2}{3}\pi (60)^3$$
$$V_{left} = \pi (60)^2 [180 – (\frac{1}{3} \times 120 + \frac{2}{3} \times 60)]$$
$$= 3600\pi [180 – (40 + 40)] = 3600\pi (100) = 360000\pi$$
$$= \frac{360000 \times 22}{7} \approx 1131428.57 \text{ cm}^3$$
✔️ Volume = 1.131 m³ (approx)
Q8

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 $\pi = 3.14$.

Volume Calculation:
Sphere: $R = 4.25$ cm. Cylinder: $r = 1$ cm, $h = 8$ cm.
$$V = \text{Vol Sphere} + \text{Vol Cylinder}$$
$$V = \frac{4}{3}\pi (4.25)^3 + \pi (1)^2 (8)$$
$$= 3.14 [\frac{4}{3} \times 76.765 + 8] = 3.14 [102.353 + 8]$$
$$= 3.14 \times 110.353 \approx 346.51 \text{ cm}^3$$
The child found 345 cm³, but the correct volume is approx 346.51 cm³.
✔️ She is incorrect. Correct Volume = 346.51 cm³
🎉 Exercise 12.2 Completed | Chapter 12 Finished!
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