UPSC NDA Math Model Paper 6

At any instant t, let r be the radius, V be the volume and S be the surface area of the balloon. Given: dV/dt = 20cm$^3$/sec 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 $\Rightarrow \frac{d V}{d t} = \frac{d V}{d r} \cdot \frac{d r}{d t}$ ......(By chain Rule) As, $V=\frac{4}{3}\pi r^{3}$ $\Rightarrow \frac{d V}{d r} = \frac{d}{dr}\left(\frac{4}{3}\pi r^{3}\right) = \frac{4}{3}\pi \cdot 3 r^{2}$ $=4\pi r^{2}$ By substituting dV/dr in dV/dt we get $\Rightarrow \frac{d V}{d t} = 4\pi r^{2} \cdot \frac{d r}{d t}$ Now substitute dV/dt = 20 cm$^3$/sec in the above equation we get $\Rightarrow 20 = 4\pi r^{2} \cdot \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{5}{\pi r^{2}}$ As we know that, $S = 4\pi r^{2}$ By differentiating S with respect to t we get $\Rightarrow \frac{d S}{d t} = \frac{d S}{d r} \cdot \frac{d r}{d t}$ .....(By chain Rule) $\Rightarrow \frac{d S}{d r} = \frac{d(4\pi r^{2})}{d r} = 8\pi r$ By substituting the value of dS/dr in dS/dt we get $\Rightarrow \frac{d S}{d t} = 8\pi r \cdot \frac{d r}{d t}$ By substituting $\frac{dr}{dt} = \frac{5}{\pi r^{2}}$ in the above equation we get $\Rightarrow \frac{dS}{dt} = 8\pi r \cdot \frac{5}{\pi r^{2}} = \frac{40}{r}$ By substituting r = 8 cm in the above equation we get $\Rightarrow \left[ \frac{dS}{dt} \right]_{r=8} = \frac{40}{8} = 5\,\text{cm}^2/\text{sec}$