Practice MCQ Questions and Answer on Volume and Surface Area

41.

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

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

Show Answer

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

42.

A rectangular paper of 44 cm long and 6 cm wide is rolled to form a cylinder of height equal to width of the paper. The radius of the base of the cylinder so rolled is :

• (A) 3.5 cm
• (B) 5 cm
• (C) 7 cm
• (D) 14 cm

Show Answer

 Length of rectangle paper = Circumference of the base of cylinder If r is the radius of the cylinder : $$\eqalign{ & \Rightarrow 44 = 2\pi r \cr & \Rightarrow r = \frac{{44 \times 7}}{{2 \times 22}} \cr & \Rightarrow r = 7\,cm \cr} $$

43.

A hemispherical bowl is 176 cm round the brim. Supposing it to be half full, how many persons may be served from it in hemispherical glasses 4 cm in diameter at the top ?

• (A) 1172
• (B) 1272
• (C) 1372
• (D) 1472

Show Answer

 Let the radius of the hemispherical bowl be r cm Then, $$\eqalign{ & 2\pi r = 176 \cr & \Rightarrow r = \frac{{176 \times 7}}{{2 \times 22}} \cr & \Rightarrow r = 28 \cr} $$ Volume of liquid in the bowl : $$\eqalign{ & = \frac{1}{2} \times \left( {\frac{2}{3} \times \pi \times 28 \times 28 \times 28} \right){\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{2}{3} \times \pi \times 14 \times 28 \times 28} \right){\text{ c}}{{\text{m}}^3} \cr} $$ Volume of 1 glass :$$ = \left( {\frac{2}{3} \times \pi \times 2 \times 2 \times 2} \right){\text{ c}}{{\text{m}}^3}$$ ∴ Required number of person : $$\eqalign{ & = \frac{{{\text{Volume of liquid in the bowl}}}}{{{\text{Volume of 1 glass}}}} \cr & = \left( {\frac{{14 \times 28 \times 28}}{{2 \times 2 \times 2}}} \right) \cr & = 1372 \cr} $$

44.

66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be:

• (A) 84 meters
• (B) 90 meters
• (C) 168 meters
• (D) 336 meters

Show Answer

 $$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{wire}}\,{\text{be h}} \cr & {\text{Radius}} = \frac{1}{2}mm = \frac{1}{{20}}cm.\,{\text{Then}}, \cr & \Rightarrow \frac{{22}}{7} \times \frac{1}{{20}} \times \frac{1}{{20}} \times h = 66 \cr & \Rightarrow h = {\frac{{66 \times 20 \times 20 \times 7}}{{22}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8400\,cm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 84\,meters \cr} $$

45.

The sum of the radius and the height of a cylinder is 19 m. The total surface area of the cylinder is 1672 m², what is the volume of the cylinder ?

• (A) 3080 m3
• (B) 2940 m3
• (C) 3220 m3
• (D) 2660 m3

Show Answer

 Let the radius of the cylinder be r and height be h Then, r + h = 19 Again, total surface area of cylinder = $$\left( {2\pi rh + 2\pi {r^2}} \right)$$ Now, $$\eqalign{ & 2\pi r\left( {h + r} \right) = 1672 \cr & \Rightarrow 2\pi r \times 19 = 1672 \cr & \Rightarrow 38\pi r = 1672 \cr & \therefore \pi r = \frac{{1672}}{{38}} = 44\,m \cr & \therefore r = \frac{{44 \times 7}}{{22}} = 14\,m \cr} $$ ∴ Height = 19 - 14 = 5 m Volume of cylinder : $$\eqalign{ & = \pi {r^2}h \cr & = \frac{{22}}{7} \times 14 \times 14 \times 5 \cr & = 22 \times 2 \times 14 \times 5 \cr & = 3080\,{m^3} \cr} $$

46.

A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side of 3 cm to form a cone. The volume of the cone so formed is :

• (A) 12π cm3
• (B) 15π cm3
• (C) 16π cm3
• (D) 20π cm3

Show Answer

 Clearly, we have r = 3 cm and h = 4 cm ∴ Volume : $$\eqalign{ & = \frac{1}{3}\pi {r^2}h \cr & = \left( {\frac{1}{3} \times \pi \times {3^2} \times 4} \right)\pi {r^3} \cr & = 12\pi {\text{ c}}{{\text{m}}^3} \cr} $$

47.

A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients ?

• (A) 38L
• (B) 40L
• (C) 39.5L
• (D) 38.5L

Show Answer

 Diameter of bowl = 7 cm ∴ Radius of bowl = $$\frac{2}{7}$$ cm Height = 4 cm ∴ Volume of cylindrical bowl : $$\eqalign{ & = \pi {r^2}h \cr & = \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2} \times 4 \cr & = 154\,cu.cm \cr} $$ Hence, volume of soup for 250 patients : $$\eqalign{ & = 154 \times 250 \cr & = 38500{\text{ c}}{{\text{m}}^3} \cr & = 38.5{\text{L}} \cr} $$

48.

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

Show Answer

 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, $$\eqalign{ & \frac{{22}}{7} \times 1 \times 1 \times h = \left( {\frac{{11 \times 705}}{{56}}} \right) \cr & \Rightarrow h = \left( {\frac{{11 \times 705}}{{56}} \times \frac{7}{{22}}} \right) \cr & \Rightarrow h = 44.0625\,{\text{cm}} \cr} $$

49.

The capacity of a cylindrical tank is 246.4 litres. If the height is 4 metres, what is the diameter of the base ?

• (A) 1.4 cm
• (B) 2.8 cm
• (C) 14 cm
• (D) 28 cm

Show Answer

 Volume of the tank = 246.4 litres = 246400 cm3 Let the radius of the base be r cm Then, $$\eqalign{ & \left( {\frac{{22}}{7} \times {r^2} \times 400} \right) = 246400 \cr & \Rightarrow {r^2} = \left( {\frac{{246400 \times 7}}{{22 \times 400}}} \right) \cr & \Rightarrow {r^2} = 196 \cr & \Rightarrow r = 14 \cr} $$ ∴ Diameter of the base = 2r = 28 cm

50.

Consider the volumes of the following
1. A parallelepiped of length 5 cm, breadth 3 cm and height 4 cm
2. A cube of each side 4 cm
3. A cylinder of radius 3 cm and length 3 cm
4. A sphere of radius 3 cm
The volumes of these in the decreasing order is :

• (A) 1, 2, 3, 4
• (B) 1, 3, 2, 4
• (C) 4, 2, 3, 1
• (D) 4, 3, 2, 1

Show Answer

 Volume of parallelepiped : $$\eqalign{ & = \left( {5 \times 3 \times 4} \right)c{m^3} \cr & = 60\,c{m^3} \cr} $$ Volume of cube : $$\eqalign{ & = {\left( 4 \right)^3}\,c{m^3} \cr & = 64\,c{m^3} \cr} $$ Volume of cylinder : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 3 \times 3 \times 3} \right)c{m^3} \cr & = 84.86\,c{m^3} \cr} $$ Volume of spare : $$\eqalign{ & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times 3 \times 3 \times 3} \right) \cr & = 113.14\,c{m^3} \cr} $$ So, Option D is correct decreasing order