A cylindrical well of radius 2.5 m has water | vedclass

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(B) The total depth of the well is $6 \ m + 14 \ m = 20 \ m$. Let a small volume element $dx$ be at a distance $x$ from the top surface. The mass of t

Vedclass · Accepted Answer

$80$

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(B) The total depth of the well is $6 \ m + 14 \ m = 20 \ m$. Let a small volume element $dx$ be at a distance $x$ from the top surface. The mass of this element is $dm = ho dV = ho \pi r^2 dx$.The potential energy required to lift this element to the top is $dU = dm \cdot g \cdot x = ho \pi r^2 g x dx$.Integrating from $x = 6 \ m$ to $x = 20 \ m$ to find the total work done $W$:$W = \int_{6}^{20} ho \pi r^2 g x dx = ho \pi r^2 g \left[ \frac{x^2}{2} ight]_{6}^{20} = ho \pi r^2 g \left( \frac{400 - 36}{2} ight) = ho \pi r^2 g (182)$.Given $ho = 10^3 \ kg/m^3$,$r = 2.5 \ m$,$g = 10 \ m/s^2$:$W = 10^3 	imes 3.14 	imes (2.5)^2 	imes 10 	imes 182 = 35.71 	imes 10^6 \ J$.Power $P = 10 \ HP = 10 	imes 746 \ W = 7460 \ W$.Time $t = \frac{W}{P} = \frac{35.71 	imes 10^6}{7460} \approx 4787 \ s$.Converting to minutes: $t = \frac{4787}{60} \approx 79.78 \ min \approx 80 \ min$.Thus,the correct option is $B$.

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

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