## Volume and Surface Area Questions and Answers Part-1

1. 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

Explanation: 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}

2. The radius of a cylinder is 5 m more than its height. If the curved surface area of the cylinder is 792 m2, what is the volume of the cylinder ?
a) 5712 m3
b) 5244 m3
c) 5544 m3
d) 5306 m3

Explanation: Let the height of the cylinder be x cm
Then, radius = (x + 5) m
Curved surface area of the cylinder = $$2\pi rh$$
Now,
\eqalign{ & 2\pi \left( {x + 5} \right) \times x = 792 \cr & 2 \times \frac{{22}}{7} \times \left( {{x^2} + 5x} \right) = 792 \cr & {x^2} + 5x = \frac{{792 \times 7}}{{44}} = 126 \cr & {x^2} + 5x - 126 = 0 \cr & {x^2} + 14x - 9x - 126 = 0 \cr & x\left( {x + 14} \right) - 9\left( {x + 14} \right) = 0 \cr & \left( {x - 9} \right)\left( {x + 14} \right) = 0 \cr & x = 9, - 14{\text{(neglect negative value)}} \cr}
Height of cylinder = 9 m
Radius of cylinder = 9 + 5 = 14 m
Volume of cylinder :
\eqalign{ & = \pi {r^2}h \cr & = \frac{{22}}{7} \times 14 \times 14 \times 9 \cr & = 5544\,{m^3} \cr}

3. The radius of a sphere is equal to the radius of the base of a right circular cone, and the volume of the sphere is double the volume of the cone. The ratio of the height of the cone to the radius of its base is :
a) 2 : 1
b) 1 : 2
c) 2 : 3
d) 3 : 2

Explanation: Let the radius of cone and the sphere be R and the height of the cone be H
Volume of sphere $$= \frac{4}{3}\pi {r^3}$$
Volume of cone $$= \frac{1}{3}\pi {r^2}h$$
\eqalign{ & \Rightarrow \frac{4}{3}\pi {R^3} = 2 \times \frac{1}{3}\pi {R^2}H \cr & \Rightarrow 4R = 2H \cr & \Rightarrow \frac{H}{R} = \frac{4}{2}\,Or\,2:1 \cr}

4. The dimensions of an open box are 52 cm × 40 cm × 29 cm. It thickness is 2 cm. If 1 cu.cm of metal used in the box weight 0.5 gm, then the weight of the box is :
a) 6.832 kg
b) 37.576 kg
c) 7.76 kg
d) 8.56 kg

Explanation: Since the box is an open one, we have :
Internal length = (52 - 4) cm = 48 cm
Internal breadth = (40 - 4) cm = 36 cm
Internal depth = (29 - 2) cm = 27 cm
Volume of the metal used in the box :
= External volume - Internal volume
= [(52 × 40 × 29) - (48 × 36 × 27)] cm3
= (60320 - 46656) cm3
= 13664 cm3
Weight of the box :
\eqalign{ & = \left( {\frac{{13664 \times 0.5}}{{1000}}} \right){\text{kg}} \cr & = 6.832\,{\text{kg}} \cr}

5. Except for one face of a given cube, identical cubes are glued through their faces to all the other faces of the given cube. If each side of the given cube measures 3 cm, then what is the total surface area of the solid body thus formed ?
a) 225 cm2
b) 234 cm2
c) 270 cm2
d) 279 cm2

Explanation: Clearly, each of the 5 faces of the given cube are glued to a face of another cube
Total surface area of the solid :
= 5 × 5a2 + a2 = 26a2
= (26 × 32) cm2
= 234 cm2

6. The length of the longest rod that can be placed in a room of dimensions 10 m × 10 m × 5 m is :
a) 15$$\sqrt 3$$ m
b) 15 m
c) 10$$\sqrt 2$$ m
d) 5$$\sqrt 3$$ m

Explanation: Required length :
\eqalign{ & = \sqrt {{{\left( {10} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 5 \right)}^2}} \,m \cr & = \sqrt {225} \,m \cr & = 15\,m \cr}

7. 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

Explanation: Let the radius of the cylinder be r and height be h
Then, r + h = 19
Again, total surface area of cylinder = $$\left( {2\pi rh + 2\pi {r^2}} \right)$$
\eqalign{ & 2\pi r\left( {h + r} \right) = 1672 \cr & 2\pi r \times 19 = 1672 \cr & 38\pi r = 1672 \cr & \pi r = \frac{{1672}}{{38}} = 44\,m \cr & 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}

8. Given that 1 cu. cm of marble weights 25 gms, the weight of a marble block 28 cm in width and 5 cm thick is 112 kg. The length of the block is :
a) 26.5 cm
b) 32 cm
c) 36 cm
d) 37.5 cm

Explanation: Let length = x cm
\eqalign{ & x \times 28 \times 5 \times \frac{{25}}{{1000}} = 112 \cr & x = \left( {112 \times \frac{{1000}}{{25}} \times \frac{1}{{28}} \times \frac{1}{5}} \right) \cr & x = 32\,cm \cr}

9. An open box is made by cutting the congruent squares from the corners of a rectangular sheet of cardboard of dimension 20 cm × 15 cm. If the side of each square is 2 cm, the total outer surface area of the box is :
a) 148 cm2
b) 284 cm2
c) 316 cm2
d) 460 cm2

Explanation: Clearly,
$$l$$ = (20 - 4) cm = 16 cm
b = (15 - 4) cm = 11 cm and
h = 2 cm
Outer surface area of the box :
\eqalign{ & = \left[ {2\left( {l + b} \right) \times h} \right] + lb \cr & = \left[ {\left\{ {2\left( {16 + 11} \right) \times 2} \right\} + 16 \times 11} \right] \cr & = \left( {108 + 176} \right) \cr & = 284{\text{ c}}{{\text{m}}^2} \cr}

10.V1, V2, V3 and V4 are the volumes of four cubes of side lengths x cm, 2x cm, 3x cm and 4 cm respectively. Some statements regarding these volumes are given below :
a) 1 and 2
b) 2 and 3
c) 1 and 3
d) 1, 2 and 3

Explanation: Clearly, we have :
V1 = x3,
V2 = (2x)3 = 8x3
V3 = (3x)3 = 27x3
V4 = (4x)3 = 64x3
(i) V1 + V2 + 2V3
= x3 + 8x3 + 2 × 27x3
= 63x3 < V4
(ii) V1 + 4V2 + V3
= x3 + 4 × 8x3 + 27 x3
= 60x3 < V4
(iii) 2(V1 + V3) + V2
= 2 (x3 + 27x3) + 8x3
= 64x3 = V4