On opposite sides of a wide vertical vessel f | vedclass

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(B) Let the height of the upper hole from the surface be $h_1$ and the height of the lower hole be $h_2$. The difference in height is $h_2 - h_1 = x$.

Vedclass · Accepted Answer

$2\rho Agx$

Vedclass · Answer

(B) Let the height of the upper hole from the surface be $h_1$ and the height of the lower hole be $h_2$. The difference in height is $h_2 - h_1 = x$.The velocity of efflux from the upper hole is $v_1 = \sqrt{2gh_1}$ and from the lower hole is $v_2 = \sqrt{2gh_2}$.The force of reaction (thrust) exerted by the water flowing out of a hole of area $A$ is given by $F = \rho A v^2$.For the upper hole, $F_1 = \rho A (2gh_1) = 2\rho Agh_1$.For the lower hole, $F_2 = \rho A (2gh_2) = 2\rho Agh_2$.Since the holes are on opposite sides, the forces act in opposite directions.The resultant force $F_{net} = |F_2 - F_1| = |2\rho Agh_2 - 2\rho Agh_1| = 2\rho Ag(h_2 - h_1)$.Substituting $h_2 - h_1 = x$, we get $F_{net} = 2\rho Agx$.

On opposite sides of a wide vertical vessel filled with water (density $\rho$), two identical holes are drilled, each having cross-section area $A$. The height difference between holes is $x$. The resultant force of reaction of water flowing out of the vessel is ($g$ = acceleration due to gravity).

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