Question

A spherical balloon is being inflated so that its volume increases uniformly at the rate of $40c{m}^3/min.$ When $r=8$, then the increase in radius in the next $1/2min$ is

• A. $0.025$ cm
• B. $0.050$ cm
• C. $0.075$ cm
• D. $0.01$ cm

Answer 1

The correct option is A $0.025$ cm

Volume of sphere
$=\frac{4\pi r^3}{3}$
Hence
$\frac{dV}{dt}=4\pi .r^2.\frac{dr}{dt}$
Now
$\frac{dV}{dt}=4\pi .r^2{\frac{dr}{dt}}_{r=8}$
$40=4\pi .64.\frac{dr}{dt}$
$\frac{dr}{dt}=\frac{10}{64\pi }$
Or
$r=\frac{10}{64\pi }.t$. ($r=0$ at $t=0$).
Hence
$r_{1/2}=\frac{10}{64\pi .2}$
$=\frac{5}{64\pi }$.
Hence
$r_{1/2}-r_0=\frac{5}{64\pi }$
$=0.025cm$.