Class 10 Maths Chapter 12 Surface Areas and Volumes Important Questions with Answers (English Medium)

Q. Write the formula for the curved surface area of a hemisphere of radius $r$.

Answer: CSA of hemisphere $=\frac12(4\pi r^2)=2\pi r^2$.

Q. Two cubes, each of edge 6 cm, are joined end to end. Find the surface area of the resulting cuboid.

Answer: Cuboid $12\times6\times6$: TSA $=2(72+36+72)=360$ cm².

Q. What is the greatest diameter of a hemisphere that can be placed on a cubical block of edge 9 cm?

Answer: 9 cm (it cannot be wider than a face of the cube).

Q. A solid is a cylinder with a hemisphere at each end. Which surfaces make up its total surface area?

Answer: CSA of the cylinder + CSA of the two hemispheres (the joining circles are hidden).

Q. Find the slant height of a cone of radius 6 cm and height 8 cm.

Answer: $l=\sqrt{6^2+8^2}=\sqrt{100}=10$ cm.

Q. Find the volume of a hemisphere of radius 3 cm in terms of $\pi$.

Answer: $\frac23\pi(3)^3=18\pi$ cm³.

Q. Why is the total surface area of a toy made of a cone and a hemisphere not equal to the sum of their total surface areas?

Answer: Because the flat circular faces where they are joined are hidden inside the toy, so they are not part of its surface.

Q. A toy is a cone of radius 5 cm and height 12 cm mounted on a hemisphere of the same radius. Find its total surface area in terms of $\pi$.

Answer: $l=\sqrt{25+144}=13$ cm.
TSA $=\pi rl+2\pi r^2=\pi(5)(13)+2\pi(25)=65\pi+50\pi=115\pi$ cm².

Q. A vessel is a hollow hemisphere of diameter 14 cm with a hollow cylinder of height 5 cm mounted on it. Find the inner surface area of the vessel ($\pi=\frac{22}{7}$).

Answer: $r=7$ cm, $h=5$ cm. Inner surface $=$ CSA of hemisphere $+$ CSA of cylinder $=2\pi r^2+2\pi rh=2\pi r(r+h)$
$=2\times\frac{22}{7}\times7\times(7+5)=44\times12=528$ cm².

Q. A solid is a cone standing on a hemisphere, both of radius 3 cm, and the height of the cone is 4 cm. Find its volume in terms of $\pi$.

Answer: Volume $=\frac13\pi r^2h+\frac23\pi r^3=\frac13\pi(9)(4)+\frac23\pi(27)=12\pi+18\pi=30\pi$ cm³.

Q. A medicine capsule is a cylinder with two hemispheres stuck to its ends. The whole capsule is 21 mm long and 7 mm in diameter. Find its surface area ($\pi=\frac{22}{7}$).

Answer: $r=3.5$ mm, length of cylindrical part $=21-7=14$ mm.
SA $=2\pi rh+4\pi r^2=2\pi r(h+2r)=2\times\frac{22}{7}\times3.5\times(14+7)=22\times21=462$ mm².

Q. A cubical block of edge 10 cm is surmounted by a hemisphere of diameter 7 cm. Find the surface area of the solid ($\pi=\frac{22}{7}$).

Answer: SA $=6a^2-\pi r^2+2\pi r^2=6a^2+\pi r^2=600+\frac{22}{7}\times3.5\times3.5=600+38.5=638.5$ cm².

Q. A glass is a cylinder of inner diameter 7 cm and height 12 cm, with a raised hemisphere of diameter 7 cm at its bottom. Find its actual capacity ($\pi=\frac{22}{7}$).

Answer: Apparent capacity $=\pi r^2h=\frac{22}{7}\times3.5\times3.5\times12=462$ cm³.
Hemisphere $=\frac23\pi r^3=\frac23\times\frac{22}{7}\times3.5^3\approx89.83$ cm³.
Actual capacity $\approx462-89.83=372.17$ cm³.

Q. A shed is a cuboid 20 m long, 14 m wide and 6 m high, surmounted by a half cylinder of diameter 14 m running along its length. Find the volume of the shed ($\pi=\frac{22}{7}$).

Answer: Cuboid $=20\times14\times6=1680$ m³.
Half cylinder $=\frac12\pi r^2h=\frac12\times\frac{22}{7}\times7\times7\times20=1540$ m³.
Volume $=1680+1540=3220$ m³.

Q. Explain why we add volumes directly for a combined solid, but do not add total surface areas directly.

Answer: On joining, the faces in contact go inside, so part of each surface area disappears; we add only the outside surfaces. But no material is lost on joining, so the volume of the combined solid is exactly the sum of the volumes of its parts.

Q. A hemispherical depression of diameter 4.2 cm is cut out from one face of a cube of edge 5 cm. Find the surface area of the remaining solid ($\pi=\frac{22}{7}$).

Answer: The top loses a circle $\pi r^2$ but gains the inside of the hemisphere $2\pi r^2$, $r=2.1$ cm.
SA $=6a^2+\pi r^2=150+\frac{22}{7}\times2.1\times2.1=150+13.86=163.86$ cm².

Q. A tent is a cylinder of diameter 28 m and height 4 m surmounted by a conical top of slant height 15 m. Find the area of canvas used (base not covered) and its cost at ₹120 per m² ($\pi=\frac{22}{7}$).

Answer: $r=14$ m, $h=4$ m, $l=15$ m.
Canvas $=2\pi rh+\pi rl=\pi r(2h+l)=\frac{22}{7}\times14\times(8+15)=44\times23=1012$ m².
Cost $=1012\times120=₹121440$.

Q. A toy rocket is a cone of radius 3 cm and height 4 cm mounted on a cylinder of radius 2 cm and height 10 cm. The cone (with the visible ring of its base) is painted red and the cylinder (with its bottom) blue. Find each area ($\pi=3.14$).

Answer: Cone: $l=\sqrt{9+16}=5$ cm.
Red $=\pi rl+\pi(r^2-r'^2)=3.14\times[15+(9-4)]=3.14\times20=62.8$ cm².
Blue $=2\pi r'h'+\pi r'^2=\pi r'(2h'+r')=3.14\times2\times22=138.16$ cm².

Q. A wooden article is made by scooping out a hemisphere from each end of a solid cylinder of height 15 cm and radius 3.5 cm. Find the total surface area of the article ($\pi=\frac{22}{7}$).

Answer: Outside surfaces: CSA of cylinder $+$ two hollow hemispheres.
TSA $=2\pi rh+2(2\pi r^2)=2\pi r(h+2r)=2\times\frac{22}{7}\times3.5\times(15+7)=22\times22=484$ cm².

Q. A solid iron pole is a cylinder of height 100 cm and base diameter 20 cm, surmounted by another cylinder of height 50 cm and radius 5 cm. Find the mass of the pole if 1 cm³ of iron has mass 8 g ($\pi=3.14$).

Answer: Volume $=\pi(10^2)(100)+\pi(5^2)(50)=3.14\times(10000+1250)=3.14\times11250=35325$ cm³.
Mass $=35325\times8=282600$ g $=282.6$ kg.

Q. A wooden pen stand is a cuboid 20 cm × 12 cm × 4 cm with six conical depressions, each of radius 0.7 cm and depth 1.5 cm. Find the volume of wood in the stand ($\pi=\frac{22}{7}$).

Answer: Cuboid $=20\times12\times4=960$ cm³.
One depression $=\frac13\times\frac{22}{7}\times0.7\times0.7\times1.5=0.77$ cm³; six $=4.62$ cm³.
Wood $=960-4.62=955.38$ cm³.

Q. A vessel is an inverted cone of height 12 cm and top radius 6 cm, filled with water to the brim. Lead shots, each a sphere of radius 0.5 cm, are dropped in, and one-fourth of the water flows out. Find the number of lead shots.

Answer: Volume of water $=\frac13\pi(6^2)(12)=144\pi$ cm³; water out $=\frac14\times144\pi=36\pi$ cm³.
One shot $=\frac43\pi(0.5)^3=\frac{\pi}{6}$ cm³.
Number $=36\pi\div\frac{\pi}{6}=216$.

Q. A spherical glass vessel has a cylindrical neck 7 cm long and 2 cm in diameter; the spherical part has diameter 14 cm. A student measures its volume as 1460 cm³. Check whether her reading agrees with the calculated volume to the nearest 10 cm³, taking these as inside measurements ($\pi=\frac{22}{7}$).

Answer: Sphere: $r=7$ cm, volume $=\frac43\times\frac{22}{7}\times343\approx1437.33$ cm³.
Neck: $r=1$ cm, $h=7$ cm, volume $=\frac{22}{7}\times1\times7=22$ cm³.
Total $\approx1459.33$ cm³. To the nearest 10 cm³ this is 1460 cm³, so her reading agrees.

Q. From a solid cylinder of radius 3 cm and height 4 cm, a conical cavity of the same radius and height is hollowed out. Find, in terms of $\pi$, (i) the total surface area and (ii) the volume of the remaining solid.

Answer: (i) $l=\sqrt{9+16}=5$ cm. TSA $=2\pi rh+\pi r^2+\pi rl=\pi r(2h+r+l)=3\pi(8+3+5)=48\pi$ cm².
(ii) Volume $=\pi r^2h-\frac13\pi r^2h=\frac23\pi(9)(4)=24\pi$ cm³.

Q. A solid made of a cone of height 12 cm and radius 6 cm standing on a hemisphere of radius 6 cm is placed upright in a right circular cylinder full of water so that it touches the bottom. The cylinder has radius 6 cm and height 18 cm. Find the volume of water left in the cylinder ($\pi=\frac{22}{7}$). Also show that the cone and the hemisphere have equal volumes.

Answer: Cylinder $=\pi(36)(18)=648\pi$ cm³.
Cone $=\frac13\pi(36)(12)=144\pi$ cm³; hemisphere $=\frac23\pi(216)=144\pi$ cm³. So the cone and hemisphere have equal volumes.
Solid $=288\pi$ cm³.
Water left $=648\pi-288\pi=360\pi=360\times\frac{22}{7}\approx1131.43$ cm³.

Q. A model is a cylinder with a cone attached at each end, made of thin sheet. The model has diameter 6 cm and total length 20 cm, and each cone is 4 cm high. Find (i) the volume of air inside the model and (ii) the area of sheet used ($\pi=\frac{22}{7}$).

Answer: $r=3$ cm, cylinder length $=20-2\times4=12$ cm, cone slant height $l=\sqrt{9+16}=5$ cm.
(i) Volume $=\pi r^2h+2\times\frac13\pi r^2h_1=\pi(9)(12)+2\times\frac13\pi(9)(4)=108\pi+24\pi=132\pi\approx414.86$ cm³.
(ii) Sheet $=2\pi rh+2\pi rl=\pi(72)+\pi(30)=102\pi\approx320.57$ cm².

Q. A solid toy is a hemisphere of radius 3.5 cm surmounted by a cone of height 12 cm. Find (i) the volume of the toy, (ii) its total surface area, and (iii) the difference between the volume of the cylinder circumscribing it and the volume of the toy ($\pi=\frac{22}{7}$).

Answer: (i) Volume $=\frac23\pi r^3+\frac13\pi r^2h=\frac{22}{7}\left(\frac23\times42.875+\frac13\times12.25\times12\right)\approx89.83+154=243.83$ cm³.
(ii) $l=\sqrt{3.5^2+12^2}=12.5$ cm. TSA $=\pi r(l+2r)=\frac{22}{7}\times3.5\times19.5=214.5$ cm².
(iii) Cylinder: $r=3.5$ cm, height $=3.5+12=15.5$ cm, volume $=\frac{22}{7}\times12.25\times15.5=596.75$ cm³. Difference $\approx596.75-243.83=352.92$ cm³.