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  4. If the volume of a sphere is increasing at a constant rate, then the rate at which its radius is increasing, is:

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Q. If the volume of a sphere is increasing at a constant rate, then the rate at which its radius is increasing, is:

KEAMKEAM 2004

Solution:

Given that, $ \frac{dV}{dt}=k $ (say) ...(i)
$ \because $ $ V=\frac{4}{3}\pi {{R}^{3}} $
On differentiating w.r.t. t, we get
$ \frac{dV}{dt}=4\pi {{R}^{2}}\frac{dR}{dt} $
$ \Rightarrow $ $ \frac{dR}{dt}=\frac{k}{4\pi {{R}^{2}}} $ [from(i)]
$ \Rightarrow $ Rate of increasing radius is inversely proportional to its surface area.