You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Liquid ammonia

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Oil

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Mercury

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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.

Suggest Corrections

Similar questions

Q. In the manometer as shown,

2

immiscible fluids - mercury

(ρ = 13600 {kg/m}^{3})

and water

(ρ = 1000 {kg/m}^{3})

are used as manometric fluids. The water end is exposed to the atmosphere

(100 kPa)

and the mercury end is exposed to a gas. At this position, the interface between the fluids is at the bottom most point of the manometer. The value of

h

is found to be

9.45 m

. The height of the mercury column is given to be

75 cm

. Find the gauge pressure of the gas.

(g = 9.8 {m/s}^{2})

Q. In the manometer as shown,

2

immiscible fluids - mercury

(ρ = 13600 {kg/m}^{3})

and water

(ρ = 1000 {kg/m}^{3})

are used as manometric fluids. The water end is exposed to the atmosphere

(100 kPa)

and the mercury end is exposed to a gas. At this position, the interface between the fluids is at the bottom most point of the manometer. The value of

h

is found to be

9.45 m

. The height of the mercury column is given to be

75 cm

. Find the gauge pressure of the gas.

(g = 9.8 {m/s}^{2})

The bulb of the open-end manometer shown above contains a gas. Find out the statement which is relevant to this situation.

I.	Only atmospheric pressure is exerted on the exposed mercury surface in the right side of the tube.
II.	The gas pressure is greater than atmospheric pressure.
III.	The difference in the height of mercury levels is equal to the pressure of the gas.

Q. A manometer consists of a

U -

shaped tube containing a liquid. One side is connected to the apparatus and other is open to the atmosphere. The pressure inside the apparatus is then determined from the difference in heights of the liquid. The external pressure is

760 t o r r

and density of liquid is

10 g c m^{- 3}

. Then open side is

10.0 c m

lower than the side connected to the apparatus. The pressure in the apparatus is (approx)

One arm of a glass U-tube is open to the atmosphere, whereas the other vertical arm is connected to a gas supply. The difference in levels of mercury (density = 13.6 g cm^-3) between the vertical arms of the U-tube manometer is h = 3 cm. If the atmospheric pressure is 760 mm Hg and the gravitational acceleration is 9.8 m s^-2, then the pressure of the gas is about:

Join BYJU'S Learning Program

Explore more

Liquid Inside a U-Tube

Standard XII Physics

Join BYJU'S Learning Program

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.