Question

A liquid of density $\rho$ 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:

• A. $\frac{g(h_1-h_2)}{2l}towardsright$
• B. $\frac{g(h_1-h_2)}{2l}towardsleft$
• C. $\frac{g(h_1-h_2)}{l}towardsright$
• D. $\frac{g(h_1-h_2)}{l}towardsleft$

Answer 1

The correct option is C $\frac{g(h_1-h_2)}{l}towardsright$

Solution

REF.Image

Pressure at A

$P_A=fg{h}_1+P_0$

Similarly at B

$P_B=fg{h}_2+P_0$

We know that the pressure in

a fluid decreases in the direction

of its acceleration. So,

As $P_A>P_B$ acceleration will be

towards right (A to B)

Also, $P_A-P_B=falg$

$\Rightarrow fg{h}_1+P_0-pg{h}_2-P_0=fla$

$\Rightarrow fg\frac{[h_1-h_2]}{l}=a$

$\Rightarrow a=g\frac{[h_1-h_2]}{l}$

Option - C is correct.