The area of cross section of the wider tube shown in figure is 900 cm². If the boy standing on the piston weighs 45 kg, find the difference in the levels of water in the two tubes.

5 cm

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0.50 cm

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50 m

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50 cm

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Solution

The correct option is D
50 cm

consider the following diagram,

Given area $a = 900 {c m}^{2} = \frac{900}{100 \times 100} m^{2}$

Density of water $ρ = 1000 \frac{k g}{m^{3}}$

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

Equating, $P_{A} = P_{B}$ {Pascal's Law}

$\frac{M g}{a} + P_{a t m} = P_{a t m} + ρ g H$
$\frac{45 \times 10 \times 100 \times 100}{900} = 1000 \times 10 \times H$
$\Rightarrow H = 0.50 m = 50 c m$

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weight is standing on the piston of cross sectional area

900 c m^{2}

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. If the boy standing on the piston weighs 45 kg, the difference in the levels of water in the two tubes.

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The area of cross section of the wider tube shown in figure is 900 cm2. If the boy standing on the piston weighs 45 kg, find the difference in the levels of water in the two tubes.

The correct option is D 50 cm consider the following diagram, Given area a=900cm2=900100×100m2 Density of water ρ=1000kgm3 Let the difference in the height of two columns be H. Equating,PA=PB {Pascal's Law} Mga+Patm=Patm+ρgH 45×10×100×100900=1000×10×H ⇒H=0.50m=50 cm

The area of cross section of the wider tube shown in figure is 900 cm². If the boy standing on the piston weighs 45 kg, find the difference in the levels of water in the two tubes.