Practice MCQ Questions and Answer on Volume and Surface Area
1.
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) decreases by 25%
(C) increases by 25%
(D) increases by 50%
Solution:
Let the original radius and height of the cone be r and h respectively Then, Original volume = $$\frac{1}{3}\pi {r^2}h$$ New radius = $$\frac{r}{2}$$ and new hight = 2h New volume : $$\eqalign{ & = \frac{1}{3} \times \pi \times {\left( {\frac{r}{2}} \right)^2} \times 3h \cr & = \frac{3}{4} \times \frac{1}{3}\pi {r^2}h \cr} $$ ∴ Decrease % : $$\eqalign{ & = \left( {\frac{{\frac{1}{4} \times \frac{1}{3}\pi {r^2}h}}{{\frac{1}{3}\pi {r^2}h}} \times 100} \right)\% \cr & = 25\% \cr} $$
2.
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
Solution:
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 : $$\eqalign{ & = 14{a^2}:18{a^2} \cr & = 7:9 \cr} $$
3.
The radius of the base and height of a metallic solid cylinder are r cm and 6 cm respectively. It is melted and recast into a solid cone of the same radius of base. The height of the cone is :
(A) 9 cm
(B) 18 cm
(C) 27 cm
(D) 54 cm
Solution:
Let the height of the cone be h cm Then, $$\eqalign{ & \pi \times {r^2} \times 6 = \frac{1}{3} \times \pi \times {r^2} \times h \cr & \Rightarrow h = 18\,cm \cr} $$
4.
Diameter of a jar cylindrical in shape is increased by 25%. By what percent must the height be decreased so that there is no change in its volume :
(A) 10%
(B) 25%
(C) 36%
(D) 50%
Solution:
Let original radius = r and original height = h Original volume = $$\pi {r^2}h$$ New radius = 125% of r = $$\frac{5r}{4}$$ Let new height = H Then, $$\eqalign{ & \pi {r^2}h = \pi {\left( {\frac{{5r}}{4}} \right)^2} \times H \cr & Or,\,H = \frac{{16}}{{25}}h \cr} $$ Decrease in height : $$\eqalign{ & = \left( {h - \frac{{16h}}{{25}}} \right) \cr & = \frac{{9h}}{{25}} \cr} $$ ∴ Decrease % : $$\eqalign{ & = \left( {\frac{{9h}}{{25}} \times \frac{1}{h} \times 100} \right)\% \cr & = 36\% \cr} $$
5.
50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m3, then the rise in the water level in the tank will be:
A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m3) is:
(A) 4830
(B) 5120
(C) 6420
(D) 8960
Solution:
Clearly, l = (48 - 16)m = 32 m, b = (36 -16)m = 20 m, h = 8 m. ∴ Volume of the box = (32 x 20 x 8) m3 = 5120 m3
7.
If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the reduced cylinder to that of the original one ?
(A) 1 : 2
(B) 1 : 4
(C) 1 : 8
(D) 8 : 1
Solution:
Let original radius = R Then, new radius = $$\frac{{\text{R}}}{2}$$ $$\eqalign{ & \therefore \frac{{{\text{Volume of reduced cylinder }}}}{{{\text{Volume of original cylinder}}}} \cr & = \frac{{\pi \times {{\left( {\frac{R}{2}} \right)}^2} \times h}}{{\pi \times {R^2} \times h}} \cr & = \frac{1}{4}\,Or\,1:4 \cr} $$
8.
The breadth of a room is twice its height and half its length. The volume of the room is 512 cu.m. The length of the room is :
(A) 16 m
(B) 18 m
(C) 20 m
(D) 32 m
Solution:
Let the height of the room be x metres Then, breadth = 2x metres and length = 4x metres ∴ Volume of the room : = (4x × 2x × x) m3 = (8x3) m3 8x3 = 512 ⇒ x3 = 64 ⇒ x = 4 ∴ Length of the room is : = 4x = (4 × 4) = 16 m
9.
A larger cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. The ratio of the total surface areas of the smaller cubes and the larger cube is :
(A) 2 : 1
(B) 3 : 2
(C) 25 : 18
(D) 27 : 20
Solution:
Volume of the larger cube : $$\eqalign{ & = \left( {{3^3} + {4^3} + {5^3}} \right){\text{ c}}{{\text{m}}^3} \cr & = 216{\text{ c}}{{\text{m}}^3} \cr} $$ Let the edge of the larger cube be a cm $$\eqalign{ & \therefore {a^3} = 216 \cr & \Rightarrow a = 6 \cr} $$ Required ratio : $$\eqalign{ & = \frac{{6\left( {{3^2} + {4^2} + {5^2}} \right)}}{{6 \times {6^2}}} \cr & = \frac{{6 \times 50}}{{6 \times 36}} \cr & = \frac{{25}}{{18}}\,Or\,25:18 \cr} $$
10.
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
Solution:
Let the height of the cone be h cm Then, $$\eqalign{ & \frac{4}{3}\pi \times {\left( 5 \right)^3} = \frac{1}{3}\pi \times {\left( 5 \right)^2} \times h \cr & \Rightarrow h = 20\,cm \cr} $$