A tent is of the shape of a right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs 2 per square metre, if the radius of the base is 14 metres.
Let r m be the radius of cylindrical base of cylinder of height by m
r = 14 m and h1 = 3m
Curved surface area of cylinder
$=2 \pi r h_{1} m^{2}$
$=25 \frac{22}{7} \times 14 \times 3 \mathrm{~m}^{2}$
$=264 \mathrm{~m}^{2}$
The radius of cylindrical box of cylinder is also equal to the radius of right circular cons.
Let h2 be the height of cone and l be the slant height of cone
$r=14 m$ and $h_{2}=(13.5-3)$
$=10.5 \mathrm{~m}$
$l^{2}=r^{2}+h_{2}^{2}$
$l^{2}=(14)^{2}+(10.5)^{2}$
$l^{2}=(14)^{2}+(10.5)^{2}$
$I=\sqrt{196+110.25}$
$\sqrt{306.25}=17.5 \mathrm{~m}$
Curved surface area of the cone
$=\pi r l$
$=\frac{22}{7} \times 14 \times 17.5$
$=770 \mathrm{~m}^{2}$
Therefore,
Total area of tent which is to be painted
= curved surface area of cylinder + curved surface area of cone
$=(264+770) \mathrm{m}^{2}$
$=1034 \mathrm{~m}^{2}$
Now cost of painting 1 m2 of inner side of tent = Rs. 2
Cost of painting 1034 m2 inner side of tent
$=2 \times 1034$
$=$ Rs. 2068