A cylindrical vessel of radius 0.5 m is filled with oil at the rate of 0.25 $\pi$ m3/minute. The rate at which the surface of the oil is rising, is

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KollegeApply · Accepted Answer

1 m/minute \text { Le tr be the radius,h be the height and V be the volume of the cylindrical vessel at any time t.Then, } V = $\pi$ r^2 h \Rightarrow dV/dt = $\pi$ r^2 dh/dt \Rightarrow dh/dt = 1/$\pi$ r^2dV/dt \Rightarrow dh/dt = \frac{0 . 25$\pi$}{$\pi$ \left( 0 . 5 \right)^2} \Rightarrow dh/dt = 0 . $\frac{25}{0}$ . 25 \Rightarrow dh/dt = 1 m/\min

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