The limbs of a U-tube are lowered into vessel | vedclass

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(B) When air is pumped out from the top of the $U$-tube,the pressure at the top of both liquid columns becomes equal to the reduced pressure $P_{top}$

Vedclass · Accepted Answer

$0.83$

Vedclass · Answer

(B) When air is pumped out from the top of the $U$-tube,the pressure at the top of both liquid columns becomes equal to the reduced pressure $P_{top}$.Let $P_{atm}$ be the atmospheric pressure and $h_A = 10\, cm$ and $h_B = 12\, cm$ be the heights of the liquid columns in the limbs.The pressure at the free surface of the liquid in vessel $A$ is $P_{atm}$. The pressure at the top of the liquid column in limb $A$ is $P_{top} = P_{atm} - ho_w g h_A$.Similarly,for vessel $B$,the pressure at the top of the liquid column is $P_{top} = P_{atm} - ho_B g h_B$.Equating the two expressions for $P_{top}$:$P_{atm} - ho_w g h_A = P_{atm} - ho_B g h_B$This simplifies to:$ho_w g h_A = ho_B g h_B$Given that $ho_w = 1\, g/cm^3$,$h_A = 10\, cm$,and $h_B = 12\, cm$:$1 	imes 10 = ho_B 	imes 12$$ho_B = \frac{10}{12} = \frac{5}{6} \approx 0.833\, g/cm^3$.Thus,the density of liquid $B$ is approximately $0.83\, g/cm^3$.

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