A boy of $45 k g$ weight is standing on the piston of cross sectional area $900 c m^{2}$ . Find the difference in the levels of water in the two tubes.

$5 c m$

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$0.50 c m$

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$50 m$

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$50 c m$

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Solution

The correct option is D
$50 c m$

Given area, $A = 900 {c m}^{2} = \frac{900}{10^{4}} m^{2}$
Density of water, $ρ = \frac{1000 k g}{m^{3}}$

Let the difference in the height of two columns be $H$ .

Using Pascal's Law, $P_{A} = P_{B}$
$\frac{M g}{a} + P_{a t m} = P_{a t m} + ρ g H$
$\frac{45}{\frac{900}{10^{4}}} = 1000 \times H$
$H = 0.50 m = 50 c m$

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Q. A boy of

45 k g

weight is standing on the piston of cross sectional area

900 c m^{2}

. Find the difference in the levels of water in the two tubes.

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A boy of 45 kg weight is standing on the piston of cross sectional area 900 cm2. Find the difference in the levels of water in the two tubes.

The correct option is D 50 cm Given area, A=900 cm2=900104 m2 Density of water, ρ=1000 kgm3 Let the difference in the height of two columns be H. Using Pascal's Law, PA=PB Mga+Patm=Patm+ρgH 45900104=1000×H H=0.50 m=50 cm

A boy of $45 k g$ weight is standing on the piston of cross sectional area $900 c m^{2}$ . Find the difference in the levels of water in the two tubes.