Question

A balloon which always remain spherical has a variable diameter $\frac{3}{2}(2x+3)$. The rate of change of its volume w.r.t $x$, is

• A. $\frac{27}{8}\pi {(2x+3)}^2$ cu.unit
• B. $\frac{7}{18}\pi {(2x+3)}^2$ cu.unit
• C. $\frac{27}{8}{(2x+3)}^2$ cu.unit
• D. None of the above

Answer 1

The correct option is A $\frac{27}{8}\pi {(2x+3)}^2$ cu.unit

Given that
Diameter $=\frac{3}{2}(2x+3)$
Radius $=\frac{3}{4}(2x+3)$
Therefore, volume $V=\frac{4}{3}\pi {\left[\frac{3}{4}\right(2x+3\left)\right]}^3=\frac{9}{16}\pi {(2x+3)}^3$
On differentiating w.r.t $x$, we get
$\frac{dV}{dx}=\frac{9}{16}\pi .3{(2x+3)}^2\cdot 2=\frac{27}{8}\pi {(2x+3)}^2$