Practice MCQ Questions and Answer on Volume and Surface Area

51.

A metallic sphere of radius 5 cm is melted to make a cone with base of the same radius. What is the height of the cone ?

• (A) 5 cm
• (B) 10 cm
• (C) 15 cm
• (D) 20 cm

Show Answer

 Let the height of the cone be h cm Then, $$\eqalign{ & \frac{4}{3}\pi \times {\left( 5 \right)^3} = \frac{1}{3}\pi \times {\left( 5 \right)^2} \times h \cr & \Rightarrow h = 20\,cm \cr} $$

52.

The diameter of the base of a cylindrical drum is 35 dm and the height is 24 dm. It is full of kerosene. How many tins each of size 25 cm × 22 cm × 35 cm can be filled with kerosene from the drum ?

• (A) 120
• (B) 600
• (C) 1020
• (D) 1200

Show Answer

 $$\eqalign{ & {\text{Number of tins}} \cr & = \frac{{{\text{Voulme of the drum}}}}{{{\text{Volume of each tin}}}} \cr & = \frac{{\left( {\frac{{22}}{7} \times \frac{{35}}{2} \times \frac{{35}}{2} \times 24} \right)}}{{\left( {\frac{{25}}{{10}} \times \frac{{22}}{{10}} \times \frac{{35}}{{10}}} \right)}} \cr & = 1200 \cr} $$

53.

A cube of length 1 cm is taken out from a cube of length 8 cm. What is the weight of the remaining portion ?

• (A) $$\frac{7}{8}$$ of the weight of the original cube
• (B) $$\frac{8}{9}$$ of the weight of the original cube
• (C) $$\frac{63}{64}$$ of the weight of the original cube
• (D) $$\frac{511}{512}$$ of the weight of the original cube

Show Answer

 Volume of the bigger cube = (83) cm3 = 512 cm3 Volume of the cut-out cube = (13) cm3 = 1 cm3 Volume of the remaining portion = (512 - 1) cm3 = 511 cm3 $$\frac{{{\text{Weight of the remaining portion}}}}{{{\text{Weight of the original cube}}}}$$      $$ = \frac{{511}}{{512}}$$

54.

By what percent the volume of a cube increases if the length of each edge was increased by 50%

• (A) 50%
• (B) 125%
• (C) 237.5%
• (D) 273.5%

Show Answer

 Let original edge = a Then, original volume = a3 New edge : $$\eqalign{ & = \frac{{150}}{{100}}a \cr & = \frac{{3a}}{2} \cr} $$ New volume : $$\eqalign{ & = {\left( {\frac{{3a}}{2}} \right)^3} \cr & = \frac{{27{a^3}}}{8} \cr} $$ Increase in volume : $$\eqalign{ & = \left( {\frac{{27{a^3}}}{8} - {a^3}} \right) \cr & = \frac{{19{a^3}}}{8} \cr} $$ ∴ Increase % : $$\eqalign{ & = \left( {\frac{{19{a^3}}}{8} \times \frac{1}{{{a^3}}} \times 100} \right)\% \cr & = 237.5\% \cr} $$

55.

Rita and Meeta both are having lunch boxes of a cuboid shape. Length and breadth of Rita's lunch box are 10% more than that of Meeta's lunch box, but the depth of Rita's lunch box is 20% less than that of Meeta's lunch box. The ratio of the capacity of Rita's lunch box to that of Meeta's lunch box is :

• (A) 11 : 15
• (B) 15 : 11
• (C) 121 : 125
• (D) 125 : 121

Show Answer

 let l, b and h denote the length, breadth and depth of Meeta's lunch box Then, length of Rita's lunch box : $$ = 110\% {\text{ of }}l = \frac{{11l}}{{10}}$$ Breadth of Rita's lunch box : $$ = 110\% {\text{ of }}b = \frac{{11b}}{{10}}$$ Depth of Rita's lunch box : $$ = 80\% {\text{ of }}h = \frac{{4h}}{5}$$ ∴ Ratio of the capacities of Rita's and Meeta's lunch boxes : $$\eqalign{ & = \frac{{11l}}{{10}} \times \frac{{11b}}{{10}} \times \frac{{4h}}{5}:lbh \cr & = \frac{{121}}{{125}}:1 \cr & = 121:125 \cr} $$

56.

The sum of perimeters of the six faces of a cuboid is 72 cm and the total surface area of the cuboid is 16 cm². Find the longest possible length that can be kept inside the cuboid :

• (A) 5.2 cm
• (B) 7.8 cm
• (C) 8.05 cm
• (D) 8.36 cm

Show Answer

 Sum of perimeters of the six faces : $$\eqalign{ & = 2\left[ {2\left( {l + b} \right) + 2\left( {b + h} \right) + 2\left( {l + h} \right)} \right] \cr & = 4\left( {2l + 2b + 2h} \right) \cr & = 8\left( {l + b + h} \right) \cr} $$ Total surface area $$ = 2\left( {lb + bh + lh} \right)$$ $$\eqalign{ & \therefore 8\left( {l + b + h} \right) = 72 \cr & \Rightarrow l + b + h = 9 \cr & 2\left( {lb + bh + lh} \right) = 16 \cr & \Rightarrow lb + bh + lh = 8 \cr} $$ Now, $${\left( {l + b + h} \right)^2} = {l^2} + {b^2} + {h^2} + 2$$     $$\left( {lb + bh + lh} \right)$$ $$\eqalign{ & \Rightarrow {\left( 9 \right)^2} = {l^2} + {b^2} + {h^2} + 16 \cr & \Rightarrow {l^2} + {b^2} + {h^2} = 81 - 16 \cr & \Rightarrow {l^2} + {b^2} + {h^2} = 65 \cr} $$ Required length : $$\eqalign{ & = \sqrt {{l^2} + {b^2} + {h^2}} \cr & = \sqrt {65} \cr & = 8.05\,cm \cr} $$

57.

The volume of a sphere is $$2145\frac{{11}}{{21}}{\text{c}}{{\text{m}}^3}.$$   Its radius is equal to :

• (A) 7 cm
• (B) 8 cm
• (C) 9 cm
• (D) None of these

Show Answer

 $$\eqalign{ & \frac{4}{3} \times \frac{{22}}{7} \times {r^3} = \frac{{45056}}{{21}} \cr & \Rightarrow {r^3} = \left( {\frac{{45056}}{{21}} \times \frac{3}{4} \times \frac{7}{{22}}} \right) \cr & \Rightarrow {r^3} = 512 \cr & \Rightarrow r = \root 3 \of {512} \cr & \Rightarrow r = 8\,cm \cr} $$

58.

If three equal cubes are placed adjacently in a row, then the ratio of the total surface area of the new cuboid to the sum of the surface areas of the three cubes will be ?

• (A) 1 : 3
• (B) 2 : 3
• (C) 5 : 9
• (D) 7 : 9

Show Answer

 Let the length of each edge of each cube be a Then, the cuboid formed by placing 3 cubes adjacently has the dimensions 3a , a and a Surface area of the cuboid : $$\eqalign{ & = 2\left[ {3a \times a + a \times a + 3a \times a} \right] \cr & = 2\left[ {3{a^2} + {a^2} + 3{a^2}} \right] \cr & = 14{a^2} \cr} $$ Sum of surface area of 3 cubes : $$\eqalign{ & = \left( {3 \times 6{a^2}} \right) \cr & = 18{a^2} \cr} $$ ∴ Required ratio : $$\eqalign{ & = 14{a^2}:18{a^2} \cr & = 7:9 \cr} $$

59.

What is the volume in cubic cm of a pyramid whose area of the base is 25 sq cm and height 9 cm ?

• (A) 60 cm3
• (B) 75 cm3
• (C) 90 cm3
• (D) 105 cm3

Show Answer

 Volume of pyramid :$$\eqalign{ & = \frac{1}{3} \times {\text{Area of base}} \times {\text{Height}} \cr & = \left( {\frac{1}{3} \times 25 \times 9} \right){\text{c}}{{\text{m}}^3} \cr & = 75\,{\text{c}}{{\text{m}}^3} \cr} $$

60.

If the height of a right circular cone is increased by 200% and the radius of the base is reduced by 50%, then the volume of the cone :

• (A) Remains unaltered
• (B) Decrease by 25%
• (C) Increase by 25%
• (D) Increase by 50%

Show Answer

 Let the radius of a right circular cine be R cm and height be H cm Volume of right circular cone $$ = \frac{1}{3}\pi {R^2}H{\text{ cu}}{\text{.cm}}$$ When height of right circular cone is increased by 200% and radius of the base is reduce by 50% New volume :$$\eqalign{ & {\text{ = }}\frac{1}{3}\pi {\left( {\frac{R}{2}} \right)^2}.3H \cr & = \frac{1}{3}\pi \frac{{{R^2}4}}{4}.3H \cr & = \frac{{\pi {R^2}H}}{4} \cr} $$ Difference : $$\eqalign{ & = \pi {R^2}H\left( {\frac{1}{3} - \frac{1}{4}} \right) \cr & = \frac{1}{{12}}\pi {R^2}H \cr} $$ Decrease percentage : $$\eqalign{ & = \frac{{\frac{1}{{12}}\pi {R^2}H}}{{\frac{1}{3}\pi {R^2}H}} \times 100 \cr & = 25\% \cr} $$