Question

radius of a spherical ballon is increasing with time as 2/pie then the rate of the change in the volume of balloons when radius is 0.5 is​

Answer 1

Volume of the spherical balloon is given by,

$\sf{\longrightarrow V=\dfrac{4}{3}\,\pi r^3}$

Then rate of change of volume will be,

$\sf{\longrightarrow \dfrac{dV}{dt}=\dfrac{d}{dt}\left(\dfrac{4}{3}\,\pi r^3\right)}$

$\sf{\longrightarrow \dfrac{dV}{dt}=\dfrac{4}{3}\,\pi \cdot\dfrac{d}{dt}\left(r^3\right)}$

$\sf{\longrightarrow \dfrac{dV}{dt}=\dfrac{4}{3}\,\pi \cdot\dfrac{d}{dr}\left(r^3\right)\cdot\dfrac{dr}{dt}}$

$\sf{\longrightarrow \dfrac{dV}{dt}=\dfrac{4}{3}\,\pi \cdot3r^2\cdot\dfrac{dr}{dt}}$

$\sf{\longrightarrow \dfrac{dV}{dt}=4\pi r^2\cdot\dfrac{dr}{dt}\quad\quad\dots(1)}$

Here rate of change of radius is given as,

$\sf{\longrightarrow \dfrac{dr}{dt}=\dfrac{2}{\pi}\ m\,s^{-1}}$

Then (1) becomes,

$\sf{\longrightarrow \dfrac{dV}{dt}=4\pi r^2\cdot\dfrac{2}{\pi}}$

$\sf{\longrightarrow \dfrac{dV}{dt}=8r^2}$

At $\sf{r=0.5\ m,}$

$\sf{\longrightarrow \dfrac{dV}{dt}=8(0.5)^2}$

$\sf{\longrightarrow\underline{\underline{\dfrac{dV}{dt}=2\ m^3\,s^{-1}}}}$

Hence (3) is the answer.