UPSC NDA Math Model Paper 7

At any instant t, let $r$ be the radius, $V$ be the volume and $S$ be the surface area of the balloon. Given: $\frac{dS}{dt} = 2 \,\mathrm{cm}^2/\mathrm{sec}$ As we know that, surface area of sphere is given by: $4\pi r^2$ where $r$ is the radius of the sphere. i.e $S = 4\pi r^2$ Now by differentiating S with respect to t we get, $\Rightarrow \frac{dS}{dt} = \frac{dS}{dr} \cdot \frac{dr}{dt}$------(By chain rule) $\Rightarrow \frac{dS}{dr} = \frac{d(4\pi r^2)}{dr} = 8\pi r$ By substituting the value of dS/dr in dS/dt we get $\Rightarrow \frac{dS}{dt} = 8\pi r \cdot \frac{dr}{dt}$ By substituting dS/dt = cm$^2$/sec in the above equation we get $\Rightarrow 2 = 8\pi r \cdot \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{4\pi r}$ As we know that volume of sphere is given by: $\frac{4}{3}\pi r^3$ where $r$ is the radius of the sphere. i.e $V = \frac{4}{3}\pi r^3$ Now differentiating V with respect to t we get $\Rightarrow \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$ -----(By chain Rule) As, $V = \frac{4}{3}\; \pi r^3$ $\Rightarrow \frac{dV}{dr} = \frac{d}{dr}\left( \frac{4}{3}\; \pi r^3 \right) = \frac{4}{3}\; \pi \cdot 3r^2 = 4\pi r^2$ By substituting dV/dr in dV/dt we get $\Rightarrow \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$ Now substitute dr/dt = 1/4πr in the above equation we get $\Rightarrow \frac{dV}{dt} = 4\pi r^2 \cdot \frac{1}{4\pi r} = r$ By substituting r = 6 cm in the above equation we get, $\Rightarrow \left[ \frac{dV}{dt} \right]_{r=6} = 6\,\mathrm{cm}^3/\mathrm{sec}$