A U-tube has a left limb with a cross-section | vedclass

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(C) Let the cross-sectional area of the left limb be $A$ and the right limb be $4A$.Let the rise in the level of mercury in the right limb be $x \ cm$

Vedclass · Accepted Answer

$0.56$

Vedclass · Answer

(C) Let the cross-sectional area of the left limb be $A$ and the right limb be $4A$.Let the rise in the level of mercury in the right limb be $x \ cm$.Since the volume of mercury displaced must be conserved,the fall in the level of mercury in the left limb will be $4x \ cm$.The total height of the water column in the left limb becomes $h_w = 36 + 4x \ cm$.The new level of mercury in the right limb is $x \ cm$ above the initial level,and the mercury level in the left limb is $4x \ cm$ below the initial level. The difference in mercury levels between the two limbs is $x + 4x = 5x \ cm$.Equating the pressure at the interface of mercury and water (at the level of the lower mercury surface in the left limb):$P_{water} = P_{mercury}$$(36 + 4x) 	imes ho_{water} 	imes g = (5x) 	imes ho_{Hg} 	imes g$$(36 + 4x) 	imes 1 = 5x 	imes 13.6$$36 + 4x = 68x$$64x = 36$$x = \frac{36}{64} = 0.5625 \ cm \approx 0.56 \ cm$.

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