NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.2#
Exercise 12.2 covers the volume of combinations of solids — finding total volume when solids are combined together.
Key Volume Formulas:
- Cylinder: $V = \pi r^2 h$
- Cone: $V = \dfrac{1}{3}\pi r^2 h$
- Sphere: $V = \dfrac{4}{3}\pi r^3$
- Hemisphere: $V = \dfrac{2}{3}\pi r^3$
- Cuboid: $V = l \times b \times h$; Cube: $V = a^3$
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 π. #
Answer
Radius = 1 cm, height of cone = 1 cm.
$$V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{\pi}{3}(r^2 h + 2r^3) = \frac{\pi}{3}(1+2) = \pi \text{ cm}^3$$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.) #
Answer
Radius = 1.5 cm, height of cylinder = $12 - 2 - 2 = 8$ cm, height of each cone = 2 cm.
$$V = \pi r^2 h_{cyl} + 2 \times \frac{1}{3}\pi r^2 h_{cone}$$$$= \pi r^2\left(h_{cyl} + \frac{2h_{cone}}{3}\right) = \pi \times 2.25 \times \left(8 + \frac{4}{3}\right) = 2.25\pi \times \frac{28}{3} = \frac{63\pi}{3} = 21\pi \approx 66 \text{ cm}^3$$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. #
Answer
Radius = 1.4 cm, diameter = 2.8 cm.
Height of cylinder = $5 - 2 \times 1.4 = 5 - 2.8 = 2.2$ cm.
Volume of one gulab jamun:
$$V = \pi r^2 h + \frac{4}{3}\pi r^3 = \pi r^2\left(h + \frac{4r}{3}\right) = \frac{22}{7} \times 1.96 \times \left(2.2 + \frac{5.6}{3}\right)$$$$= \frac{22}{7} \times 1.96 \times \frac{6.6+5.6}{3} = \frac{22}{7} \times 1.96 \times 4.0\overline{6}$$$= \frac{22}{7} \times 1.96 \times \frac{37.4}{3} \times \frac{3}{3}$…
$h + \frac{4r}{3} = 2.2 + \frac{4 \times 1.4}{3} = 2.2 + \frac{5.6}{3} = 2.2 + 1.867 = 4.067$
$V = \dfrac{22}{7} \times 1.96 \times 4.067 = 3.14 \times 1.96 \times 4.067 \approx 25.05$ cm³
Volume of 45 gulab jamuns = $45 \times 25.05 = 1127.25$ cm³
Volume of syrup = $30\% \times 1127.25 = 338.175 \approx$ 338 cm³
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. #
Answer
Volume of cuboid = $15 \times 10 \times 3.5 = 525$ cm³
Volume of one conical depression = $\dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times 0.25 \times 1.4 = \dfrac{1}{3} \times \dfrac{22}{7} \times 0.35 = \dfrac{7.7}{21} = \dfrac{11}{30} \approx 0.367$ cm³
Volume of 4 depressions = $4 \times 0.367 = 1.467$ cm³
Volume of wood = $525 - 1.467 \approx$ 523.53 cm³
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. #
Answer
Volume of cone = $\dfrac{1}{3}\pi \times 25 \times 8 = \dfrac{200\pi}{3}$ cm³
Volume of water that flows out = $\dfrac{1}{4} \times \dfrac{200\pi}{3} = \dfrac{50\pi}{3}$ cm³
Volume of one lead shot = $\dfrac{4}{3}\pi \times (0.5)^3 = \dfrac{4}{3}\pi \times 0.125 = \dfrac{\pi}{6}$ cm³
Number of shots = $\dfrac{50\pi/3}{\pi/6} = \dfrac{50}{3} \times 6 = 100$
∴ Number of lead shots = 100.
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) #
Answer
Volume of lower cylinder: $\pi \times 12^2 \times 220 = 3.14 \times 144 \times 220 = 99475.2$ cm³
Volume of upper cylinder: $\pi \times 8^2 \times 60 = 3.14 \times 64 \times 60 = 12057.6$ cm³
Total volume = $99475.2 + 12057.6 = 111532.8$ cm³
Mass = $111532.8 \times 8 = 892262.4$ g $\approx$ 892.26 kg
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. #
Answer
Volume of cylinder = $\pi \times 60^2 \times 180 = 648000\pi$ cm³
Volume of cone = $\dfrac{1}{3}\pi \times 60^2 \times 120 = 144000\pi$ cm³
Volume of hemisphere = $\dfrac{2}{3}\pi \times 60^3 = 144000\pi$ cm³
Volume of solid = $(144000 + 144000)\pi = 288000\pi$ cm³
Volume of water left = $648000\pi - 288000\pi = 360000\pi = 360000 \times \dfrac{22}{7} \approx 1131428.57$ cm³ $\approx$ 1.131 m³
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. #
Answer
Volume of cylindrical neck: $\pi r^2 h = 3.14 \times 1 \times 8 = 25.12$ cm³
Volume of sphere: $\dfrac{4}{3}\pi r^3 = \dfrac{4}{3} \times 3.14 \times (4.25)^3 = \dfrac{4}{3} \times 3.14 \times 76.77 = \dfrac{4 \times 3.14 \times 76.77}{3}$
$= \dfrac{964.22}{3} \approx 321.39$ cm³
Total volume = $25.12 + 321.39 = 346.51$ cm³
The child’s answer of 345 cm³ is approximately correct (close to 346.51 cm³, the small difference is due to rounding).