Question

The area of the 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.

Answer 1

Applying Pascal's law,

F₁/A₁ = F₂/A₂

∴ F₁ is the weight of the person,

∴ F₁ = mg = 45 × 10 = 450 N.

A₁ (wider section) = 900 cm² = 900 × 10⁻⁴ m² = 9 × 10⁻² m².

Now, Pressure = F₁/A₁ = 450/(9 × 10⁻²)

= 5000 Pa.

This pressure will be equal both the sides, by Pascal's law.

Also, Pressure = hρg, where h is the difference in height and row is the density of water.

∴  5000  = h × 1000 × 10

∴ h = 0.5 m. = 50 cm.

Hope it helps .

Answer 2

$\huge\mathfrak\red{Answer :}$


According to question we have given;

Area of cross section (a) = 900 cm² = 0.09 m²

F = mg

F = 45 × 10 = 450 N

Now;

P = $\dfrac{F}{A}$

P = $\dfrac{450}{0.09}$

P = 5000 Pa (pascal)

Also,

P = h$\rho$g

5000 = h × 1000 × 10

h = $\dfrac{5000}{10000}$

h = 0.5 m

and in cm

h = 0.5 × 10 = 50 cm

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