Question

A uniform glass U-tube is partically filled with a liquid. Horizontal length of the tube is 1 m. Tube is accelerating horizontally with constant acceleration of 6 m$s^{-2}$ . The difference in the heights of the liquid in the two arms of U-tube is-

• A. $\frac{6}{10}$ m
• B. $\frac{5}{3}$ m
• C. $\frac{2}{3}$ m
• D. $6$ m

Answer 1

The correct option is A $\frac{6}{10}$ m

given that, length of horitontal tube, $l=1m$ accelenation $a=6m/s^2$

let the two venticle side of the U tube is

${}^{\prime }{A}^{\prime }{\&}^{\prime }{B}^{\prime }$ and at $a=6m/s^2,$

liquid length of ${}^{\prime }{A}^{\prime }is=l_1,atm=P_0$

liquid length of ${}^{\prime }{B}^{\prime }$ is $=l_2$

now, let for the movement the length of liquid in

A is more than B

So, $l_1⟩l_2$

difference of liquid between ${}^{\prime }{A}^{\prime }an{d}^{\prime }{B}^{\prime }=(l_1-l_2)=h$

$\therefore P_0+\rho g{l}_1=P_0+\frac{ma}{A}\cdot l+\rho g{l}_2$

$\therefore P_0+\rho g{l}_1=P_0+\rho \cdot a\cdot l+\rho g{l}_2$

$\text{or}\rho g(l_1-l_2)=\rho \cdot 6\cdot 1$

$orh=\frac{6}{10}m$

Correct option is $\left(A\right)$