Question

A U-tube of base length 'l ' filled with same volume of two liquids of densities r and 2r is moving with an acceleration 'a' on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height h is given by –

• A. $\frac{a}{2g}\mathcal{l}$
• B. $\frac{3a}{2g}\mathcal{l}$
• C. $\frac{a}{g}\mathcal{l}$
• D. $\frac{2a}{3g}\mathcal{l}$

Answer 1

Answer: (B)

$\frac{3a}{2g}\mathcal{l}$

Answer 2

For the given situation, liquid of density 2r should be behind that of r from right limb :

P_A = P_atm + rgh

P_a = P_A + ra$\frac{\mathcal{l}}{2}$ = P_atm + rgh + ra$\frac{\mathcal{l}}{2}$

P_C = P_a + (2r) a$\frac{\mathcal{l}}{2}$ = P_atm + rgh + $\frac{3}{2}$ ral …(1)

But from left limb :

P_C = P_atm + (2r) gh …(2)

from (1) and (2)

P_atm + rgh + $\frac{3}{2}$ ral = P_atm + 2rgh

Ž h = $\frac{3a}{2g}$l