Practice MCQ Questions and Answer on Volume and Surface Area
61.
A solid cube just gets completely immersed in water when a 0.2 kg mass is placed on it. If the mass is removed, the cube is 2 cm above the water level. What is the length of each side of the cube ?
(A) 6 cm
(B) 8 cm
(C) 10 cm
(D) 12 cm
Solution:
Let the length of each side of the cube be a cm Then, Volume of the part of cube outside water = Volume of the mass placed on it ⇒ 2a2 = 0.2 × 1000 ⇒ 2a2 = 200 ⇒ a2 = 100 ⇒ a = 10
62.
A hemispherical bowl has 3.5 cm radius. It is to be painted inside as well as outside. The cost of painting it at the rate of Rs. 5 per 10 sq. cm will be :
(A) Rs. 77
(B) Rs. 100
(C) Rs. 175
(D) Rs. 50
Solution:
Radius of a hemispherical bowl = 3.5 cm Inner and outer surface area of the bowl : $$\eqalign{ & = 4\pi {r^2} \cr & = 4 \times \frac{{22}}{7} \times 3.5 \times 3.5 \cr & = 154{\text{ sq}}{\text{.cm}} \cr} $$ Total cost of painting at the rate of Rs. 5 per 10 sq.cm : $$\eqalign{ & = 154 \times \frac{5}{{10}} \cr & = {\text{Rs}}{\text{. 77}} \cr} $$
63.
A bucket is in the from of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket.
(A) 15 cm
(B) 20 cm
(C) 25 cm
(D) 30 cm
Solution:
Volume of bucket : = 28.490 litres = (28.490 ×1000) cm3 = 28490 cm3 Let the height of the bucket be h cm We have : r = 21 cm. and R = 28 cm $$\therefore \frac{\pi }{3}h\left[ {{{\left( {28} \right)}^2} + {{\left( {21} \right)}^2} + 28 \times 21} \right]$$ $$ = 28490$$ $$ \Rightarrow h\left( {784 + 441 + 588} \right) = $$ $$\frac{{28490 \times 21}}{{22}}$$ $$\eqalign{ & \Rightarrow 1813h = 27195 \cr & \Rightarrow h = \frac{{27195}}{{1813}} \cr & \Rightarrow h = 15\,cm \cr} $$
64.
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 = $$\frac{{\text{R}}}{2}$$ $$\eqalign{ & \therefore \frac{{{\text{Volume of reduced cylinder }}}}{{{\text{Volume of original cylinder}}}} \cr & = \frac{{\pi \times {{\left( {\frac{R}{2}} \right)}^2} \times h}}{{\pi \times {R^2} \times h}} \cr & = \frac{1}{4}\,Or\,1:4 \cr} $$
65.
A cylindrical tank of diameter 35 cm is full of water. If 11 litres of water is drawn off, the water level in the tank will drop by :
(A)
(B)
(C)
(D)
Solution:
Let the drop in the water level be h cm Then, $$\eqalign{ & \Rightarrow \frac{{22}}{7} \times \frac{{35}}{2} \times \frac{{35}}{2} \times h = 11000 \cr & \Rightarrow h = \left( {\frac{{11000 \times 7 \times 4}}{{22 \times 35 \times 35}}} \right){\text{ cm}} \cr & \Rightarrow h = \frac{{80}}{7}{\text{ cm}} \cr & \Rightarrow h = 11\frac{3}{7}{\text{ cm}} \cr} $$
66.
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
Solution:
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
67.
Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank, the radius of whose base is 60 cm. By how much will the level of water rise in 30 minutes ?
(A) 2 m
(B) 3 m
(C) 4 m
(D) 5 m
Solution:
Volume of water flown through the pipe in 30 min : $$\eqalign{ & = \left[ {\left( {\pi \times 0.01 \times 0.01 \times 6} \right) \times 30 \times 60} \right]{m^3} \cr & = \left( {1.08\pi } \right){m^3} \cr} $$ Let the rise in level of water be h metres Then, $$\eqalign{ & \pi \times 0.6 \times 0.6 \times h = 1.08\pi \cr & \Rightarrow h = \left( {\frac{{1.08}}{{0.6 \times 0.6}}} \right) \cr & \Rightarrow h = 3\,m \cr} $$
68.
A rectangular tank is 225 m by 162 m at the base. With what speed must water flow into it through an aperture 60 cm by 45 cm so that the level may be raised 20 cm in 5 hours ?