Question

The outer and inner diameters of a hemispherical bowl are $17cm$ and $15cm$ respectively. Find the cost of polishing it all over at $25$ paise per $c{m}^2$. ( Take $\pi$ $=\frac{22}{7}$).

Answer 1

Outer radius $R=\frac{17}{2}=7.5cm$ , Inner radius, $r=\frac{15}{2}=7.5cm$.

Area of outer surface $=2\pi R^2$ $=[2\pi \times {\left(\frac{17}{2}\right)}^2]c{m}^2$

$=2\pi \times \frac{17\times 17}{2\times 2}=\left(\frac{289\pi }{2}\right)c{m}^2$ .

Area of inner surface $=2\pi r^2$ $=[2\pi \times {\left(\frac{15}{2}\right)}^2]c{m}^2$

$=2\pi \times \frac{15\times 15}{2\times 2}=\left(\frac{225\pi }{2}\right)c{m}^2$ .

Area of the ring at the top $=\pi (R^2-r^2)$

$=\pi \left[\right(8.5{)}^2-\left(7.5{)}^2\right]c{m}^2=\pi [8.5+7.5][8.5-7.5]$ $[\because (a^2-b^2)=(a+b\left)\right(a-b\left)\right]$

$=\left(16\pi \right)c{m}^2$ .

$\therefore$ Total area to be polished $=$ Outer Surface Area $+$ Inner Surface Area $+$ Top Area

$=(\frac{289\pi }{2}+\frac{225\pi }{2}+16\pi )c{m}^2$

$=\frac{289\pi +225\pi }{2}+16\pi$

$=257\pi +16\pi$

$=\left(273\pi \right)c{m}^2$

$=(273\times \frac{22}{7})c{m}^2$

$=858c{m}^2.$

$\therefore$ Cost of polishing the bowl @ $25$ paise per $c{m}^2$$=Rs.\left(\frac{858\times 25}{100}\right)$ $=Rs.214.50$