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Standard XII

Physics

Liquid Inside a U-Tube

In a U tube m...

Question

In a $U$ tube manometer, one end is open to the atmosphere, the other end attached to a pressurized gas of gauge pressure $40 kPa$ . The height of the fluid column in the side open to the atmosphere is $60 cm$ and that on the gas side is $30 cm$ . The manometric fluid used is (Take $g = 9.8 {m/s}^{2})$

Water

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Liquid ammonia

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Oil

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Mercury

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Solution

The correct option is D Mercury
Given.
Difference in pressure = Gauge pressure of gas $Δ P = 40 kPa = 40, 000 Pa$
Height difference $Δ h = h_{2} - h_{1} = 60 - 30 = 30 cm$
As we know,
$Δ P = ρ g (h_{2} - h_{1})$
$40, 000 = ρ \times 9.8 \times (30 \times 10^{- 2}$ )
$\Rightarrow ρ = 13606 {kg/m}^{3}$
$13606 {kg/m}^{3} \approx$ (density of mescury)
So, the fluid used is mercury.

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Liquid Inside a U-Tube

Standard XII Physics

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In a U tube manometer, one end is open to the atmosphere, the other end attached to a pressurized gas of gauge pressure 40 kPa. The height of the fluid column in the side open to the atmosphere is 60 cm and that on the gas side is 30 cm. The manometric fluid used is (Take g=9.8 m/s2)

The correct option is D Mercury Given. Difference in pressure = Gauge pressure of gas ΔP=40 kPa=40,000 Pa Height difference Δh=h2−h1=60−30=30 cm As we know, ΔP=ρg (h2−h1) 40,000=ρ×9.8×(30×10−2) ⇒ρ=13606 kg/m3 13606 kg/m3≈ (density of mescury) So, the fluid used is mercury.