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

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(C) Let the initial height of water in each arm be $H = 20 	ext{ cm}$. The total volume of water remains constant. When the immiscible liquid is adde

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$2/1$

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(C) Let the initial height of water in each arm be $H = 20 	ext{ cm}$. The total volume of water remains constant. When the immiscible liquid is added,the water level in one arm drops by $x$ and rises in the other arm by $x$.Thus,$h_1 = H - x + 5 = 20 - x + 5 = 25 - x$ and $h_2 = H + x = 20 + x$.Equating the pressure at the bottom of the $U$-tube:$P_{	ext{left}} = P_{	ext{right}}$$5 	imes 4 	imes g + (h_1 - 5) 	imes 1 	imes g = h_2 	imes 1 	imes g$$20 + (25 - x - 5) = 20 + x$$20 + 20 - x = 20 + x$$2x = 20 \Rightarrow x = 10 	ext{ cm}$.Now,$h_1 = 25 - 10 = 15 	ext{ cm}$ and $h_2 = 20 + 10 = 30 	ext{ cm}$.The ratio $h_2/h_1 = 30/15 = 2/1$.

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