Practice MCQ Questions and Answer on Volume and Surface Area

41.

The ratio of the volume of a hemisphere and a cylinder circumscribing this hemisphere and having a common base is :

• (A) 1 : 2
• (B) 2 : 3
• (C) 3 : 4
• (D) 4 : 5

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 Let the radius of the hemisphere be be r cm Then, radius of the cylinder = r cm Height of the cylinder = r cm ∴ Required ratio : $$\eqalign{ & = \frac{{{\text{Volume of hemisphere}}}}{{{\text{Volume of cylinder}}}} \cr & = \frac{{\frac{2}{3}\pi {r^3}}}{{\pi {r^2} \times r}} \cr & = \frac{2}{3}\, Or\,2:3 \cr} $$

42.

What length of solid cylinder 2 cm in diameter must be taken to cast into a hollow cylinder of external diameter 12 cm, 0.25 cm thick and 15 cm long ?

• (A) 42.3215 cm
• (B) 44.0123 cm
• (C) 44.0625 cm
• (D) 44.6023 cm

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 External radius = 6 cm Internal radius = (6 - 0.25) = 5.75 cm Volume of material in hollow cylinder : $$\eqalign{ & = \left[ {\frac{{22}}{7} \times \left\{ {{{\left( 6 \right)}^2} - {{\left( {5.75} \right)}^2}} \right\} \times 15} \right]{\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{22}}{7} \times 11.75 \times 0.25 \times 15} \right){\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{22}}{7} \times \frac{{1175}}{{100}} \times \frac{{25}}{{100}} \times 15} \right){\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{11 \times 705}}{{56}}} \right){\text{c}}{{\text{m}}^3} \cr} $$ Let the length of solid cylinder be h Then, $$\eqalign{ & \frac{{22}}{7} \times 1 \times 1 \times h = \left( {\frac{{11 \times 705}}{{56}}} \right) \cr & \Rightarrow h = \left( {\frac{{11 \times 705}}{{56}} \times \frac{7}{{22}}} \right) \cr & \Rightarrow h = 44.0625\,{\text{cm}} \cr} $$

43.

If two cylinders of equal volumes have their heights in the ratio 2 : 3, then the ratio if their radii is :

• (A) $$\sqrt 6 :\sqrt 3 $$
• (B) $$\sqrt 5 :\sqrt 3 $$
• (C) $$2 : 3 $$
• (D) $$\sqrt 3 :\sqrt 2 $$

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 Let their heights be 2h and 3h and radii be r and R respectively Then, $$\eqalign{ & \pi {r^2}\left( {2h} \right) = \pi {R^2}\left( {3h} \right) \cr & \Rightarrow \frac{{{r^2}}}{{{R^2}}} = \frac{3}{2} \cr & \Rightarrow \frac{r}{R} = \frac{{\sqrt 3 }}{{\sqrt 2 }}\,i.e.,\sqrt 3 :\sqrt 2 \cr} $$

44.

A spherical ball of lead, 3 cm in diameter is melted and recast into three spherical ball. The diameter of two of these are 1.5 cm and 2 cm respectively. The diameter of the third ball is :

• (A) 2.5 cm
• (B) 2.66 cm
• (C) 3 cm
• (D) 3.5 cm

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 Let the radius of the third ball be R cm Then, $$\frac{4}{3}\pi \times {\left( {\frac{3}{4}} \right)^3} + \frac{4}{3}\pi \times {\left( 1 \right)^3}$$   $$ + \frac{4}{3}\pi \times {R^3}$$   $$ = \frac{4}{3}\pi \times {\left( {\frac{3}{2}} \right)^3}$$ $$\eqalign{ & \Rightarrow \frac{{27}}{{64}} + 1 + {R^3} = \frac{{27}}{8} \cr & \Rightarrow {R^3} = \frac{{125}}{{64}} = \frac{{{{\left( 5 \right)}^3}}}{{{{\left( 4 \right)}^3}}} \cr & \Rightarrow R = \frac{5}{4} \cr} $$ ∴ Diameter of the third ball : $$ = 2R = \frac{5}{2}cm = 2.5\,cm$$

45.

If the radius of the base and height of a cylinder and cone are each equal to r, and the radius of a hemisphere is also equal to r, then the volumes of the cone, cylinder and hemisphere are in the ratio ?

• (A) 1 : 2 : 3
• (B) 1 : 3 : 2
• (C) 2 : 1 : 3
• (D) 3 : 2 : 1

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 Required ratio : = Volume of cone : Volume of cylinder : Volume of hemisphere $$\eqalign{ & = \frac{1}{3}\pi {r^2}:\pi {r^2}r:\frac{2}{3}\pi {r^3} \cr & = \frac{1}{3}:1:\frac{2}{3} \cr & = 1:3:2 \cr} $$

46.

Length of each edge of a regular tetrahedron is 1 cm. It volume is :

• (A) $$\frac{{\sqrt 3 }}{{12}}{\text{ }}c{m^3}$$
• (B) $$\frac{1}{4}\sqrt 3 {\text{ }}c{m^3}$$
• (C) $$\frac{{\sqrt 2 }}{6}{\text{ }}c{m^3}$$
• (D) $$\frac{1}{{12}}\sqrt 2 {\text{ }}c{m^3}$$

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 Length of each edge of a regular tetrahedron = 1 cm Volume of regular tetrahedron : $$\eqalign{ & = \frac{{{a^3}}}{{6\sqrt 2 }}{\text{ c}}{{\text{m}}^3} \cr & = \frac{1}{{6\sqrt 2 }} \cr & = \frac{{\sqrt 2 }}{{6\sqrt 2 \times \sqrt 2 }}{\text{ c}}{{\text{m}}^3} \cr & = \frac{{\sqrt 2 }}{{12}}{\text{ Or }}\frac{1}{{12}}\sqrt 2 {\text{ c}}{{\text{m}}^3} \cr} $$

47.

The capacities of two hemispherical vessels are 6.4 litres and 21.6 litres. The areas of inner curved surfaces of the vessels will be in the ratio of :

• (A) $$\sqrt 2 $$ : $$\sqrt 3 $$
• (B) 2 : 3
• (C) 4 : 9
• (D) 16 : 81

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 Let their radii be R and r Then, $$\eqalign{ & \frac{{\frac{2}{3}\pi {R^3}}}{{\frac{2}{3}\pi {r^3}}} = \frac{{6.4}}{{21.6}} \cr & \Rightarrow {\left( {\frac{R}{r}} \right)^3} = \frac{8}{{27}} \cr & \Rightarrow {\left( {\frac{R}{r}} \right)^3} = {\left( {\frac{2}{3}} \right)^3} \cr & \Rightarrow \frac{R}{r} = \frac{2}{3} \cr} $$ ∴ Ratio of curved surface area : $$ = \frac{{2\pi {R^2}}}{{2\pi {r^2}}} = {\left( {\frac{R}{r}} \right)^2} = \frac{4}{9}\,or\,4:9$$

48.

A rectangular water tank is 8 m high, 6 m long and 2.5 m wide. How many litres of water can it hold ?

• (A) 120 litres
• (B) 1200 litres
• (C) 12000 litres
• (D) 120000 litres

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 Volume of the tank : $$\eqalign{ & = \left( {8 \times 100 \times 6 \times 100 \times 2.5 \times 100} \right)c{m^3} \cr & = 120000000\,c{m^3} \cr & = \left( {\frac{{120000000}}{{1000}}} \right){\text{litres}} \cr & = 120000{\text{ litres}} \cr} $$

49.

A closed box made of wood of uniform thickness has length, breadth and height 12 cm, 10 cm and 8 cm respectively. If the thickness of the wood is 1 cm, the inner surface area is :

• (A) 264 cm2
• (B) 376 cm2
• (C) 456 cm2
• (D) 696 cm2

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 Internal length = (12 - 2) cm = 10 cm Internal breadth = (10 - 2) cm = 8 cm Internal height = (8 - 2) cm = 6 cm Inner surface area : = 2 [10 × 8 + 8 × 6 + 10 × 6] cm2 = (2 × 188) cm2 = 376 cm2

50.

A rectangular tank measuring 5 m × 4.5 m × 2.1 m is dug in the centre of the field measuring 13.5 m by 2.5 m. The earth dug out is evenly spread over the remaining portion of the field. How much is the level of the field raised ?

• (A) 4 m
• (B) 4.1 m
• (C) 4.2 m
• (D) 4.3 m

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 Volume of earth dug out : $$\eqalign{ & = \left( {5 \times 4.5 \times 2.1} \right){{\text{m}}^{\text{3}}} \cr & = 47.25\,{{\text{m}}^{\text{3}}} \cr} $$ Area over which earth is spread : $$\eqalign{ & = \left( {13.5 \times 2.5 - 5 \times 4.5} \right){{\text{m}}^2} \cr & = \left( {33.75 - 22.5} \right){{\text{m}}^2} \cr & = 11.25\,{{\text{m}}^2} \cr} $$ $$\eqalign{ & \therefore {\text{Rise in level}} = \frac{{{\text{Volume}}}}{{{\text{Area}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{47.25}}{{11.25}}} \right)m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.2\,m \cr} $$