Question

The product and difference of inner and outer radius of a hemi-spherical shell are $12c{m}^2$ and 1$cm$ respectively. The sum of inner and outer curved surface areas of the shell is equal to

• A. $90\pi c{m}^2$
• B. $200\pi c{m}^2$
• C. $50\pi c{m}^2$
• D. $150\pi c{m}^2$

Answer 1

The correct option is C $50\pi c{m}^2$

Let $r_1$ and $r_2$ be the inner and the outer radii of a hemi-spherical shell, respectively.
Now, $r_1\times r_2=12c{m}^2$ and $r_2–r_1=1cm$
$\therefore (r_2-r_1{)}^2=r_2^2+r_1^2-2r_1{r}_2$
$\Rightarrow 1^2=r_1^2+r_2^2-2\times 12$
$\Rightarrow r_1^2+r_2^2=25$
Thus, the sum of inner and outer CSA of the hemi-spherical shell
$=2\pi r_1^2+2\pi r_2^2$
$=2\pi (r_1^2+r_2^2)$
$=2\pi \times 25$
$=50\pi c{m}^2$
Hence, the correct answer is option (c).