#### provide solution for rd sharma maths class 12 chapter 12 derivatives as a rate measure exercise  fill in the blanks question 10

Answer: $\frac{2}{\pi }ft/minute$

Hint: Here we use the formula of volume of vertical cylindrical tank

Given: Radius=2 ft as the r of 8 cubic ft/min

Solution:    $V=\pi r^{2}h$

$\frac{dV}{dt}=4\pi \left ( \frac{dh}{dt} \right )$

$\left ( \frac{dh}{dt} \right )= \left ( \frac{1}{4\pi } \right )\left ( \frac{dV}{dt} \right )$

$\frac{dV}{dt}=\left ( \frac{1}{4\pi } \right )\times 8$

$=\frac{2}{\pi }ft/minute$

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