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

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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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