## Volume and Surface Area Questions and Answers Part-10

1.A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. The volume of the cone so formed is:
a) 12$$\pi$$ cm3
b) 15$$\pi$$ cm3
c) 16$$\pi$$ cm3
d) 20$$\pi$$ cm3

Explanation:

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

2. In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:
a) 75 cu. m
b) 750 cu. m
c) 7500 cu. m
d) 75000 cu. m

Explanation:
\eqalign{ & {\text{1}}\,{\text{hectare}} = 10000\,{m^2} \cr & {\text{So,}}\,{\text{Area}} = \left( {1.5 \times 10000} \right){m^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 15000\,{m^2} \cr & {\text{Depth}} = \frac{5}{{100}}m = \frac{1}{{20}}m \cr & {\text{Volume}} = \left( {{\text{Area}}\, \times \,{\text{Depth}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {15000 \times \frac{1}{{20}}} \right){m^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 750\,{m^3} \cr}

3. A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:
a) 720 m3
b) 900 m3
c) 1200 m3
d) 1800 m3

Explanation:
\eqalign{ & 2\left( {15 + 12} \right) \times h = 2\left( {15 \times 12} \right) \cr & \Rightarrow h = \frac{{180}}{{27}}m = \frac{{20}}{3}m \cr & {\text{Volume}} = \left( {15 \times 12 \times \frac{{20}}{3}} \right){m^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1200\,{m^3} \cr}

4. 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

Explanation:
\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 & \frac{{22}}{7} \times \frac{1}{{20}} \times \frac{1}{{20}} \times h = 66 \cr & h = {\frac{{66 \times 20 \times 20 \times 7}}{{22}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8400\,cm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 84\,meters \cr}

5. A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weights 8 g/cm3, then the weight of the pipe is:
a) 3.6 kg
b) 3.696 kg
c) 36 kg
d) 36.9 kg

Explanation:
\eqalign{ & {\text{External}}\,{\text{radius}} = 4\,cm \cr & {\text{Internal}}\,{\text{radius}} = 3\,cm \cr & {\text{Volume}}\,{\text{of}}\,{\text{iron}} \cr & = \left( {\frac{{22}}{7} \times \left[ {{{\left( 4 \right)}^2} - {{\left( 3 \right)}^2}} \right] \times 21} \right)c{m^3} \cr & = \left( {\frac{{22}}{7} \times 7 \times 1 \times 21} \right)c{m^3} \cr & = 462\,c{m^3} \cr & {\text{Weight}}\,{\text{of}}\,{\text{iron}} = \left( {462 \times 8} \right)gm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3696\,gm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3.696\,kg \cr}

6. 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

Explanation:
\eqalign{ & {\text{Number of tins}} = \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}

7. The radius of a cylindrical cistern is 10 metres and its height is 15 metres. Initially the cistern is empty. We start filling the cistern with water through a pipe whose diameter is 50 cm. Water is coming out of the pipe with a velocity of 5 m/sec. How many minutes will it take in filling the cistern with water ?
a) 20 min
b) 40 min
c) 60 min
d) 80 min

Explanation: Volume of cistern :
\eqalign{ & = \left( {\pi \times {{10}^2} \times 15} \right){m^3} \cr & = 1500\pi {m^3} \cr}
Volume of water flowing through the pipe in 1 sec :
\eqalign{ & = \left( {\pi \times 0.25 \times 0.25 \times 5} \right){m^3} \cr & = 0.3125\pi {m^3} \cr}
Time taken to fill the cistern :
\eqalign{ & = \left( {\frac{{1500\pi }}{{0.3125\pi }}} \right) \cr & = \left( {\frac{{1500 \times 10000}}{{3125}}} \right) \cr & = 4800\,\sec \cr & = \left( {\frac{{4800}}{{60}}} \right)\min \cr & = 80\,\min \cr}

8. 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

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

9. 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%

Explanation: 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}

10. 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

$$\frac{\pi }{3}h\left[ {{{\left( {28} \right)}^2} + {{\left( {21} \right)}^2} + 28 \times 21} \right]$$     $$= 28490$$
$$h\left( {784 + 441 + 588} \right) =$$     $$\frac{{28490 \times 21}}{{22}}$$
\eqalign{ & 1813h = 27195 \cr & h = \frac{{27195}}{{1813}} \cr & h = 15\,cm \cr}