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In a sphere the rate of change of volume with respect to(w.r.t) time is

Solution

✅ Correct Option: 2

With V=43πr3V=\frac{4}{3}\pi r^3, the chain rule gives dVdt=4πr2drdt\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.

Since 4πr24\pi r^2 is the surface area, the rate of change of volume equals the surface area times drdt\frac{dr}{dt}.

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