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

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(B) Let the initial height of water in each arm be $H = 0.29 \ m$. The total volume of water is constant. When kerosene of height $h_k = 0.1 \ m$ is a

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$\frac{35}{33}$

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(B) Let the initial height of water in each arm be $H = 0.29 \ m$. The total volume of water is constant. When kerosene of height $h_k = 0.1 \ m$ is added to the left arm,the water level in the left arm drops by $x$ and rises by $x$ in the right arm.Total height of liquid in left arm: $h_1 = (H - x) + h_k = 0.29 - x + 0.1 = 0.39 - x$.Height of liquid in right arm: $h_2 = H + x = 0.29 + x$.Equating the pressure at the bottom of the $U$-tube:$P_{left} = P_{right}$$P_0 + ho_k g h_k + ho_w g (h_1 - h_k) = P_0 + ho_w g h_2$$ho_k h_k + ho_w (h_1 - h_k) = ho_w h_2$$800 	imes 0.1 + 1000 	imes (h_1 - 0.1) = 1000 	imes h_2$$80 + 1000 h_1 - 100 = 1000 h_2$$1000 (h_1 - h_2) = 20 \implies h_1 - h_2 = 0.02 \ m$.We have the system of equations:$1$) $h_1 + h_2 = (0.29 - x + 0.1) + (0.29 + x) = 0.68 \ m$.$2$) $h_1 - h_2 = 0.02 \ m$.Adding the equations: $2h_1 = 0.70 \implies h_1 = 0.35 \ m$.Subtracting the equations: $2h_2 = 0.66 \implies h_2 = 0.33 \ m$.Therefore,the ratio $\frac{h_1}{h_2} = \frac{0.35}{0.33} = \frac{35}{33}$.

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