A U-tube contains water and methylated spirit | vedclass

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Question

(A) Initial state: The mercury levels are equal. Pressure at the mercury-liquid interface must be equal.$P_{atm} + h_{water} \rho_{water} g = P_{atm}

Vedclass · Accepted Answer

$0.22$

Vedclass · Answer

(A) Initial state: The mercury levels are equal. Pressure at the mercury-liquid interface must be equal.$P_{atm} + h_{water} ho_{water} g = P_{atm} + h_{spirit} ho_{spirit} g$$10 	imes 1 = 12.5 	imes 0.8 = 10$,which is consistent.Final state: Let $h$ be the difference in mercury levels. Let the mercury level in the water arm go down by $x$ and in the spirit arm go up by $x$,so $h = 2x$.The new height of the water column is $h_1 = 10 + 15 + x = 25 + x$.The new height of the spirit column is $h_2 = 12.5 + 15 - x = 27.5 - x$.Equating pressures at the lower mercury level (in the water arm):$P_{atm} + h_1 ho_{water} g = P_{atm} + h_2 ho_{spirit} g + h ho_{Hg} g$$(25 + x) 	imes 1 = (27.5 - x) 	imes 0.8 + (2x) 	imes 13.6$$25 + x = 22 - 0.8x + 27.2x$$25 + x = 22 + 26.4x$$3 = 25.4x$$x = 3 / 25.4 \approx 0.1181 \; cm$Difference in mercury levels $h = 2x = 2 	imes 0.1181 = 0.2362 \; cm$.Given the options,the closest value is $0.22 \; cm$.

Similar Questions

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 $\,cm$)?

Fill in the blanks:
$(i)$ $1 \text{ atm} = \dots \text{ Pascal}$
$(ii)$ $1 \text{ torr} = \dots \text{ Pascal}$
$(iii)$ $1 \text{ bar} = \dots \text{ Pa}$
$(iv)$ $1 \text{ atm} = \dots \text{ cm Hg}$

Select the wrong statement about pressure.

The pressure at the bottom of a tank of water is $3P$,where $P$ is the atmospheric pressure. If the water is drawn out until the level of water is lowered by one-fifth,the pressure at the bottom of the tank will now be:

If two liquids of same volume but different densities $\rho_1$ and $\rho_2$ are mixed,then the density of the mixture is given by:

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