A cylindrical well of radius $2.5 \mathrm{~m}$ has water upto a height of $14 \mathrm{~m}$ from the bottom. If the water level is at a depth of $6 \mathrm{~m}$ from the top of the well, then the time taken (in minutes) to empty the well using a motor of $10 \mathrm{HP}$ is approximately, (Take,…

Question

A cylindrical well of radius $2.5 \mathrm{~m}$ has water upto a height of $14 \mathrm{~m}$ from the bottom. If the water level is at a depth of $6 \mathrm{~m}$ from the top of the well, then the time taken (in minutes) to empty the well using a motor of $10 \mathrm{HP}$ is approximately, (Take, $g=10 \mathrm{~ms}^{-2}$ )

Aicharya · Accepted Answer

A cylindrical well of radius $2.5 \mathrm{~m}$ is shown in the figure.

Let a small volume element $d x$ at a distance $x$ from the surface, then mass of the element, $ d m=\rho d V=\rho \pi r^2 d x $ So, the potential energy of the mass $d m$, $ d U=d m g x \Rightarrow d U=g \rho \pi r^2 x d x $ Integrating on the both sides, we get $ U=g \rho \pi r^2 \int_6^{20} x d x $ Here, $g=10 \mathrm{~m} / \mathrm{s}^2$, density of water, $\rho=10^3 \mathrm{~kg} / \mathrm{m}^3$ $ \begin{aligned} r=2.5 \mathrm{~m} & \\ U & =10 \times 10^3 \times 3.14 \times(2.5)^2\left[\frac{20^2}{2}-\frac{6^2}{2}\right] \\ & =10^4 \times 3.14 \times(2.5)^2 \times 182 \\ U & =35.71 \times 10^6 \mathrm{~J} \end{aligned} $ Now, the time taken by pump of power $10 \mathrm{HP}$ $ \begin{aligned} \text { time } & =\frac{\text { work done }}{\text { power }} \\ t & =\frac{35.71 \times 10^6}{10 \times 746 \times 60} \approx 80 \mathrm{~min} \quad[\because 1 \mathrm{HP}=746 \mathrm{~W}] \end{aligned} $ Hence, the correct option is (b)

A cylindrical well of radius $2.5 \mathrm{~m}$ has water upto a height of $14 \mathrm{~m}$ from the bottom.…

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