An open U-tube contains mercury. When 11.2 ,c | vedclass

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(C) Let the mercury level in the $U$-tube be initially at the same height in both arms. When $11.2 \,cm$ of water is poured into the left arm, the mer

Vedclass · Accepted Answer

$0.41$

Vedclass · Answer

(C) Let the mercury level in the $U$-tube be initially at the same height in both arms. When $11.2 \,cm$ of water is poured into the left arm, the mercury level in the left arm drops by $x \,cm$ and rises by $x \,cm$ in the right arm. The total difference in the mercury levels between the two arms becomes $2x \,cm$.Equating the pressure at the same horizontal level (the interface of water and mercury in the left arm, point $A$, and the corresponding level in the right arm, point $B$):$p_A = p_B$$h_{water} 	imes ho_{water} 	imes g = h_{Hg} 	imes ho_{Hg} 	imes g$Given $h_{water} = 11.2 \,cm = 0.112 \,m$, $ho_{water} = 1000 \,kg/m^3$, and $ho_{Hg} = 13600 \,kg/m^3$.$0.112 	imes 1000 = 2x 	imes 13600$$112 = 27200x$$x = \frac{112}{27200} \,m \approx 0.004117 \,m = 0.41 \,cm$.Thus, the mercury rises by $0.41 \,cm$ in the other arm.

An open $U$-tube contains mercury. When $11.2 \,cm$ of water is poured into one of the arms of the tube, how high does the mercury rise in the other arm from its initial level (in $\,cm$)?

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