A) a constant
B) proportional to the radius
C) inversely proportional to the radius
D) inversely proportional to the surface area
E) proportional to its surface area
Correct Answer: D
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.
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