Question

The outer and inner diameters of a hemispherical bowl are 17 cm and 15 cm 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 = $\frac{17}{2}$ cm , Inner radius = $\frac{15}{2}$ cm.

Area of outer surface = 2 $\pi R^2$ = $[2\pi \times \left(\frac{17}{2}\right)]c{m}^2=\left(\frac{289\pi }{2}\right)c{m}^2$ .

Area of inner surface = 2 $\pi r^2$ = $[2\pi \times \left(\frac{15}{2}\right)]c{m}^2=\left(\frac{225\pi }{2}\right)c{m}^2$ .

Area of the ring at the top = $\pi (R^2-r^2)=\pi \left[\right(8.5{)}^2-\left(7.5{)}^2\right]c{m}^2=\left(16\pi \right)c{m}^2$.

$\therefore$ Total area to be polished = $(\frac{289\pi }{2}+\frac{225\pi }{2}+16\pi )c{m}^2$

= (273 $\pi )c{m}^2=(273\times \frac{22}{7})c{m}^2=858c{m}^2.$

$\therefore$ Cost of polishing the bowl = Rs $\left(\frac{858\times 25}{100}\right)$ = Rs. 214.50