An open-ended U-tube of uniform cross-section | vedclass

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(C) Let $\rho_w = 1.0 \, g/cm^3$ be the density of water and $\rho_l = 4.0 \, g/cm^3$ be the density of the immiscible liquid.Initially,the water leve

Vedclass · Accepted Answer

$2/1$

Vedclass · Answer

(C) Let $\rho_w = 1.0 \, g/cm^3$ be the density of water and $\rho_l = 4.0 \, g/cm^3$ be the density of the immiscible liquid.Initially,the water level is $20 \, cm$ in both arms. The total volume of water is constant.When $5 \, cm$ of the immiscible liquid is added to the left arm,the water level in the left arm drops by $x$ and rises by $x$ in the right arm.The new height of the water column in the left arm is $h_w = 20 - x$. The total height in the left arm is $h_1 = h_w + 5 = 25 - x$.The height of the water column in the right arm is $h_2 = 20 + x$.Equating the pressure at the level of the interface (bottom of the immiscible liquid) in both arms:$P_{left} = P_{right}$$P_{atm} + \rho_l g (5) + \rho_w g (20 - x) = P_{atm} + \rho_w g (20 + x)$$4(5) + 1(20 - x) = 1(20 + x)$$20 + 20 - x = 20 + x$$20 = 2x \implies x = 10 \, cm$.Now,$h_1 = 25 - 10 = 15 \, cm$ and $h_2 = 20 + 10 = 30 \, cm$.The ratio $h_2/h_1 = 30/15 = 2/1$.

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