## Volume and Surface Area Questions and Answers Part-7

1. The height of a closed cylinder of given volume and the minimum surface area is :
a) Equal to its diameter
b) Half of its diameter
c) Double of its diameter
d) None of these

Explanation:
\eqalign{ & V = \pi {r^2}h{\text{ and }} \cr & S = 2\pi rh + 2\pi {r^2} \cr & \,\,\,\,\,\,\, = 2\pi r\left( {h + r} \right) \cr & {\text{Where, }}h = \frac{V}{{\pi {r^2}}} \cr & S = 2\pi r\left( {\frac{V}{{\pi {r^2}}} + r} \right) \cr & S = \frac{{2V}}{r} + 2\pi {r^2} \cr & \frac{{dS}}{{dr}} = \frac{{ - 2V}}{{{r^2}}} + 4\pi r{\text{ and}} \cr & \frac{{{d^2}S}}{{d{r^2}}} = \left( {\frac{{4V}}{{{r^3}}} + 4\pi } \right){\text{ > 0}} \cr}
S is minimum when :
\eqalign{ & \frac{{dS}}{{dr}} = 0 \cr & \frac{{ - 2V}}{{{r^2}}} + 4\pi r = 0 \cr & V = 2\pi {r^3} \cr & \pi {r^2}h = 2\pi {r^3} \cr & h = 2r \cr}

2. Water is poured into an empty cylindrical tank at a constant rate for 5 minutes. After the water has been poured into the tank. the depth of the water is 7 feet. The radius of the tank is 100 feet. Which of the following is the best approximation for the rate at which the water was poured into the tank ?
a) 140 cubic feet/sec
b) 440 cubic feet/sec
c) 700 cubic feet/sec
d) 2200 cubic feet/sec

Explanation: Volume of water flown into the tank in 5 min :
\eqalign{ & = \left( {\frac{{22}}{7} \times 100 \times 100 \times 7} \right){\text{cu}}{\text{.feet}} \cr & = 220000\,{\text{cu}}{\text{.feet}} \cr}
Rate of flow of water :
\eqalign{ & = \left( {\frac{{220000}}{{5 \times 60}}} \right){\text{cu}}{\text{.feet/sec}} \cr & = 733.3 \approx 700\,{\text{cu}}{\text{.feet/sec}} \cr}

3.The curved surface of a right circular cone of height 15 cm and base diameter 16 cm is :
a) 60π cm2
b) 68π cm2
c) 120π cm2
d) 136π cm2

Explanation: h = 15 cm, r = 8 cm
So,
\eqalign{ & l = \sqrt {{r^2} + {h^2}} \cr & \,\,\,\,\,\, = \sqrt {{8^2} + {{\left( {15} \right)}^2}} \cr & \,\,\,\,\,\, = 17\,cm \cr}
Curved surface area :
\eqalign{ & = \pi rl \cr & = \left( {\pi \times 8 \times 17} \right){\text{ c}}{{\text{m}}^2} \cr & = 136\pi {\text{ c}}{{\text{m}}^2} \cr}

4. A right circular cone and a right circular cylinder have equal base and equal height. If the radius of the base and the height are in the ratio 5 : 12, then the ratio of the total surface area of the cylinder to that of the cone is :
a) 3 : 1
b) 13 : 9
c) 17 : 9
d) 34 : 9

Explanation: Let their radius and height be 5x and 12x respectively
Slant height of the cone,
$$l = \sqrt {{{\left( {5x} \right)}^2} + {{\left( {12x} \right)}^2}} = 13x$$
\eqalign{ & \frac{{{\text{Total surface area of cylinder}}}}{{{\text{Total surface area of cone}}}} \cr & = \frac{{2\pi r\left( {h + r} \right)}}{{\pi r\left( {l + r} \right)}} \cr & = \frac{{2\left( {h + r} \right)}}{{\left( {l + r} \right)}} \cr & = \frac{{2 \times \left( {12x + 5x} \right)}}{{\left( {13x + 5x} \right)}} \cr & = \frac{{34x}}{{18x}} \cr & = \frac{{17}}{9}\,Or\,17:9 \cr}

5.The curved surface area of a sphere is 5544 sq.cm. Its volume is :
a) 22176 cm3
b) 33951 cm3
c) 38808 cm3
d) 42304 cm3

Explanation:
\eqalign{ & 4\pi {r^2} = 5544 \cr & {r^2} = \left( {5544 \times \frac{1}{4} \times \frac{7}{{22}}} \right) \cr & {r^2} = 441 \cr & r = 21 \cr}
Volume :
\eqalign{ & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times 21 \times 21 \times 21} \right){\text{ c}}{{\text{m}}^3} \cr & = 38808{\text{ c}}{{\text{m}}^3} \cr}

6. The diameter of a spare is 8 cm. It is melted and drawn into a wire of diameter 3 mm. The length of the wire is :
a) 36.9 m
b) 37.9 m
c) 38.9 m
d) 39.9 m

Explanation: Let the length of the wire be h
\eqalign{ & \pi \times \frac{3}{{20}} \times \frac{3}{{20}} \times h = \frac{4}{3}\pi \times 4 \times 4 \times 4 \cr & h = \left( {\frac{{4 \times 4 \times 4 \times 4 \times 20 \times 20}}{{3 \times 3 \times 3}}} \right)cm \cr & h = \left( {\frac{{102400}}{{27}}} \right)cm \cr & h = 3792.5\,cm \cr & h = 37.9\,m \cr}

7. The external and internal diameters of a hemispherical bowl are 10 cm and 8 cm respectively. What is the total surface area of the bowl ?
a) 257 cm2
b) 286 cm2
c) 292 cm2
d) 302 cm2

Explanation: Internal radius, r = 4 cm
External radius, R = 5 cm
Total surface area :
\eqalign{ & = 2\pi {R^2} + 2\pi {r^2} + \pi \left( {{R^2} - {r^2}} \right) \cr & = 3\pi {R^2} + \pi {r^2} \cr & = \left[ {\pi \left( {3 \times 25 + 16} \right)} \right]{\text{ c}}{{\text{m}}^2} \cr & = \left( {\frac{{22}}{7} \times 91} \right){\text{c}}{{\text{m}}^2} \cr & = 286{\text{ c}}{{\text{m}}^2} \cr}

8. A pyramid has an equilateral triangle as its base of which each side is 1 m. Its slant edge is 3 m. The whole surface are of the pyramid is equal to :
a) $$\frac{{\sqrt 3 + 2\sqrt {13} }}{4}sq.m$$
b) $$\frac{{\sqrt 3 + 3\sqrt {13} }}{4}sq.m$$
c) $$\frac{{\sqrt 3 + 3\sqrt {35} }}{4}sq.m$$
d) $$\frac{{\sqrt 3 + 2\sqrt {35} }}{4}sq.m$$

Explanation: Area of base :
\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2} \cr & = \frac{{\sqrt 3 }}{4}{m^2} \cr}
Clearly, the pyramid has 3 triangular faces each with sides 3m, 3m and 1 m
So, area of each lateral face :
\eqalign{ & = \sqrt {\frac{7}{2} \times \left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 1} \right)} {m^2} \cr & \left[ { s = \frac{{3 + 3 + 1}}{2} = \frac{7}{2}} \right] \cr & = \sqrt {\frac{7}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{5}{2}} {m^2} \cr & = \frac{{\sqrt {35} }}{4}{m^2} \cr}
Whole surface area of the pyramid :
\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} + 3 \times \frac{{\sqrt {35} }}{4}} \right){m^2} \cr & = \frac{{\sqrt 3 + 3\sqrt {35} }}{4}{m^2} \cr}

9. A hemisphere and a cone have equal bases. If their heights are also equal, then the ratio of their curved surface will be :
a) $$\sqrt 2 :1$$
b) $$1:\sqrt 2$$
c) 2 : 1
d) 1 : 2

Explanation: Let,
\eqalign{ & OP = OQ = OR = r \cr & OR = h = r \cr}
Curved surface area of the hemisphere = $$2\pi {r^2}$$
Curved surface area of a cone = $$\pi rl$$
Where,
\eqalign{ & l = \sqrt {{h^2} + {r^2}} \cr & \,\,\,\,\, = \sqrt {{r^2} + {r^2}} \cr & \,\,\,\,\, = r\sqrt 2 \cr}
Required ratio :
\eqalign{ & = \frac{{2\pi {r^2}}}{{\pi rl}} \cr & = \frac{{2\pi {r^2}}}{{\pi r \times r\sqrt 2 }} \cr & = \frac{2}{{\sqrt 2 }} \cr & = \frac{{2 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} \cr & = \frac{{2\sqrt 2 }}{2} \cr & = \frac{{\sqrt 2 }}{1}\,Or\,\sqrt 2 :1 \cr}

10. A swimming pool 9 m wide and 12 m long and 1 m deep on the shallow side and 4 m deep on the deeper side. Its volume is :
a) 360 m3
b) 270 m3
c) 420 m3
d) None of these

\eqalign{ & = 9 \times 12 \times \left( {\frac{{1 + 4}}{2}} \right) \cr & = 9 \times 12 \times \frac{5}{2} \cr & = 270\,\text{cu. metre} \cr}