Question

A hemispherical bowl is made from a metal sheet having thickness $0.3cm$. The inner radius of the bowl is $24.7cm$. Find the cost of polishing its outer surface at the rate of Rs.$4$ per $10{cm}^2$. (Take $\pi =3.14$)

• A. Rs.$1570$
• B. Rs.$137$
• C. Rs.$127$
• D. Rs.$107$

Answer 1

The correct option is A Rs.$1570$

Inner radius$=24.7cm$
Thickness$=0.3cm$
Outer radius$=(24.7+0.3)cm=25cm$
outer surface area of bowl$=2\pi r^2=2\times 3.14\times 25\times 25=3925c{m}^2$
cost of polishing $10c{m}^2$ surface area$=Rs.4$
$\therefore$ cost of polishing $3925c{m}^2$ surface area$=\frac{4\times 3925}{10}=Rs.1570$