Class 10 NCERT Solutions
Chapter 12: Surface Areas and Volumes
Master the visualization of combined solids, the calculation of total surface areas and volumes, and the logic of 3D transformations with our step-by-step logic.
Exercise 12.1
1. 2 cubes each of volume $64\text{ cm}^3$ are joined end to end. Find the surface area of the resulting cuboid.
Side of each cube $= \sqrt[3]{64} = 4$ cm. When joined end to end: $l = 8$ cm, $b = h = 4$ cm.
$$\text{Surface area} = 2(lb+bh+lh) = 2(32+16+32) = 160 \text{ cm}^2$$
2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Radius $r = 7$ cm, height of cylinder $h = 13-7 = 6$ cm.
$$\text{Inner surface area} = 2\pi r^2+2\pi rh = 2\pi r(r+h) = 2\times\frac{22}{7}\times7\times13 = 44\times13 = 572 \text{ cm}^2$$
3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
$r = 3.5$ cm, height of cone $= 15.5-3.5 = 12$ cm.
$$l = \sqrt{12^2+3.5^2} = \sqrt{144+12.25} = \sqrt{156.25} = 12.5 \text{ cm}$$ $$\text{Total surface area} = 2\pi r^2+\pi rl = \pi r(2r+l) = \frac{22}{7}\times\frac{7}{2}(7+12.5) = 11\times19.5 = 214.5 \text{ cm}^2$$
4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Greatest diameter $= 7$ cm (side of cube), so $r = 3.5$ cm.
$$\text{Surface area} = 6(7)^2+2\pi r^2-\pi r^2 = 6\times49+\pi r^2 = 294+\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}$$ $$= 294+38.5 = 332.5 \text{ cm}^2$$
5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter $l$ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
The remaining surface = 6 faces of cube + curved surface of hemisphere $-$ circular base of hemisphere:
$$= 6l^2+2\pi\left(\frac{l}{2}\right)^2-\pi\left(\frac{l}{2}\right)^2 = 6l^2+\frac{\pi l^2}{4} = \frac{l^2}{4}(24+\pi)$$
6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
$r = \dfrac{5}{2}$ mm, cylindrical height $h = 14-2\times2.5 = 9$ mm.
$$\text{Surface area} = 2\times2\pi r^2+2\pi rh = 2\pi r(2r+h) = 2\times\frac{22}{7}\times\frac{5}{2}(5+9) = \frac{110}{7}\times14 = 220 \text{ mm}^2$$
7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas at the rate of ₹500 per m².
$r = 2$ m, $h = 2.1$ m, $l = 2.8$ m.
$$\text{Canvas area} = 2\pi rh+\pi rl = \pi r(2h+l) = \frac{22}{7}\times2(4.2+2.8) = \frac{22}{7}\times2\times7 = 44 \text{ m}^2$$ $$\text{Cost} = ₹(44\times500) = ₹\,22{,}000$$
8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
$r = 0.7$ cm, $h = 2.4$ cm, $l = \sqrt{2.4^2+0.7^2} = \sqrt{6.25} = 2.5$ cm.
$$\text{Total surface area} = \pi rl+2\pi rh+\pi r^2 = \pi r(l+2h+r)$$ $$= \frac{22}{7}\times0.7(2.5+4.8+0.7) = \frac{22}{7}\times0.7\times8 = \frac{22\times0.8}{1}\approx 17.6 \approx \mathbf{18} \text{ cm}^2$$
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
$$\text{Total surface area} = 2\times2\pi r^2+2\pi rh = 2\pi r(2r+h) = 2\times\frac{22}{7}\times3.5(7+10) = 22\times17 = 374 \text{ cm}^2$$
Exercise 12.2
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 $\pi$.
$$\text{Volume} = \frac{2}{3}\pi(1)^3+\frac{1}{3}\pi(1)^2(1) = \frac{2\pi}{3}+\frac{\pi}{3} = \pi \text{ cm}^3$$
2. Rachel was asked to make a model shaped like a cylinder with two cones attached at its two ends. The diameter of the model is 3 cm and its length is 12 cm. If each cone has height of 2 cm, find the volume of air contained in the model.
$r = \dfrac{3}{2}$ cm, cone height $h_1 = 2$ cm, cylinder height $h = 12-4 = 8$ cm.
$$\text{Volume} = 2\times\frac{1}{3}\pi r^2 h_1+\pi r^2 h = \pi r^2\left[\frac{2h_1}{3}+h\right] = \frac{22}{7}\times\frac{9}{4}\times\left[\frac{4}{3}+8\right] = \frac{22}{7}\times\frac{9}{4}\times\frac{28}{3} = 66 \text{ cm}^3$$
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.
$r = 1.4$ cm, cylindrical height $h = 5-2.8 = 2.2$ cm.
$$\text{Volume of 1 gulab jamun} = \pi r^2\left[\frac{4r}{3}+h\right] = \frac{22}{7}\times1.96\times\left[\frac{5.6}{3}+2.2\right] \approx 25.07 \text{ cm}^3$$ $$\text{Sugar syrup in 45 jamuns} = \frac{30}{100}\times45\times25.07 \approx 338 \text{ cm}^3$$
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.
$$\text{Volume of cuboid} = 15\times10\times3.5 = 525 \text{ cm}^3$$
$$\text{Volume of 4 depressions} = 4\times\frac{1}{3}\times\frac{22}{7}\times(0.5)^2\times1.4 = 1.47 \text{ cm}^3$$
$$\text{Volume of wood} = 525-1.47 = 523.53 \text{ cm}^3$$
5. 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 brim. When lead shots, each 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 water that flows out $= \dfrac{1}{4}\times\dfrac{1}{3}\pi\times25\times8$. Each lead shot has volume $\dfrac{4}{3}\pi\times(0.5)^3$.
$$x = \frac{\frac{1}{4}\times\frac{1}{3}\pi\times25\times8}{\frac{4}{3}\pi\times0.125} = \frac{25\times8\times3}{4\times4\times3\times0.5\times0.5\times0.5} = 100$$$\therefore$ 100 lead shots should be dropped.
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 $1\text{ cm}^3$ of iron has approximately 8 g mass. (Use $\pi = 3.14$)
$$\text{Volume} = \pi(12)^2\times220+\pi(8)^2\times60 = \pi(31680+3840) = 35520\pi \text{ cm}^3$$
$$\text{Mass} = 35520\times3.14\times\frac{8}{1000} = 892.26 \text{ kg}$$
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. If the radius of the cylinder is 60 cm and its height is 180 cm, find the volume of water left in the cylinder.
$$\text{Volume of water left} = \pi(60)^2(180)-\left[\frac{2}{3}\pi(60)^3+\frac{1}{3}\pi(60)^2(120)\right]$$
$$= 648000\pi – \frac{4}{3}\pi\times216000 = 648000\pi – 288000\pi = 360000\pi \text{ cm}^3 \approx 1.131 \text{ m}^3$$
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 $345\text{ cm}^3$. Check whether she is correct, taking the above as the inside measurements and $\pi = 3.14$.
$$\text{Volume} = \frac{4}{3}\pi\left(\frac{8.5}{2}\right)^3+\pi(1)^2\times8 = 3.14\left[\frac{4}{3}\times(4.25)^3+8\right]$$
$$= 3.14[102.35+8] = 3.14\times110.35 = 346.51 \text{ cm}^3$$
The child measured $345\text{ cm}^3$, but the correct value is approximately $346.51\text{ cm}^3$. Therefore, the child is not correct.
