The vertical limbs of a U shaped tube are filled with a liquid of density - upto a height h on each side- The horizontal portion of the U tube of length 2h contains a liquid of density 2- The U tube is moved horizontally with an acceleration g-2 parallel to the horizontal arm- What is the difference in heights in the liquid levels in the two vertical limbs- at the current steady state-

At steady state, liquid profile be like → Let water displace by distance x So, we can write P _ 0 ρ( h x ) g+ρ g 2 x+2ρ( 2h x ) g 2 +2ρ xg+ρ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Mechanism of Formation of P-N Junction

The vertical ...

Question

The vertical limbs of a U shaped tube are filled with a liquid of density $ρ$ upto a height $h$ on each side. The horizontal portion of the U tube of length $2 h$ contains a liquid of density $2 ρ$ . The U tube is moved horizontally with an acceleration $g / 2$ parallel to the horizontal arm. What is the difference in heights in the liquid levels in the two vertical limbs, at the current steady state.

Open in App

Solution

At steady state, liquid profile be like
$\to$ Let water displace by distance $x$
So, we can write
$P_{0} - ρ (h - x) g + ρ \frac{g}{2} x + 2 ρ (2 h - x) \frac{g}{2} + 2 ρ x g + ρ g h = P_{0} - h + x + \frac{x}{2} - x + 4 h + 2 x + h = 0 \frac{3 x}{2} = 4 h x = \frac{8 h}{3}$

Suggest Corrections

Similar questions

Q. The vertical limbs of a

U

shaped tube are filled with a liquid of density

ρ

upto height

h

on each side. The horizontal portion of

U

tube having

2 h

contains a liquid of density

2 ρ

. the

U

tube is moved horizontally with an accelerator

g / 2

parallel to the horizontal arm. Find the difference in liquid levels in the two vertical limbs, at steady state.

Q. The vertical limbs of a U shaped tube are filled with a liquid of density

ρ

upto a height

h

on each side. The horizontal portion of the U tube having length

2 h

contains a liquid of density 2

ρ

. The U tube is moved horizontally with an accelerator

g / 2

parallel to the horizontal arm. Find the difference in heights in liquid levels in the two vertical limbs, at steady state.

Q. An

U -

tube of base length

^{'} l^{'}

filled with the same volume of two liquids of densities

ρ

and

2 ρ

is moving with an acceleration

^{'} a^{'}

on the horizontal plane, as shown in the figure. If the difference in heights of the liquid columns in the two vertical limbs (open to the atmosphere) is zero, then the height

h

is given by

Q. An

U -

tube of base length

^{'} l^{'}

filled with the same volume of two liquids of densities

ρ

and

2 ρ

is moving with an acceleration

^{'} a^{'}

on the horizontal plane, as shown in the figure. If the difference in heights of the liquid columns in the two vertical limbs (open to the atmosphere) is zero, then the height

h

is given by

Q. A liquid of density

ρ

is filled in a U-tube. The tube is accelerated with an acceleration a so that the height of liquid in its two vertical arms are

h_{1}

and

h_{2}

as shown in the fig, If l is the length of horizontal arm of the tube, the acceleration a is:

Join BYJU'S Learning Program

At steady state, liquid profile be like→ Let water displace by distance xSo, we can writeP0−ρ(h−x)g+ρg2x+2ρ(2h−x)g2+2ρxg+ρgh=P0−h+x+x2−x+4h+2x+h=03x2=4hx=8h3

At steady state, liquid profile be like
$\to$ Let water displace by distance $x$
So, we can write
$P_{0} - ρ (h - x) g + ρ \frac{g}{2} x + 2 ρ (2 h - x) \frac{g}{2} + 2 ρ x g + ρ g h = P_{0} - h + x + \frac{x}{2} - x + 4 h + 2 x + h = 0 \frac{3 x}{2} = 4 h x = \frac{8 h}{3}$