Question

A U-tube of base length $L$ filled with the same volume of two liquids of densities $\rho $ and $2\rho $ is moving with 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. $\frac{aL}{2 g}$
• B. $\frac{3 aL}{2 g}$
• C. $\frac{aL}{g}$
• D. $\frac{2 aL}{3 g}$

Answer 1

Answer: (B)

The pressure at the left bottom due to left vertical pipe $\left(= P\right)_{L}=P_{a t m}+h\left(2 \rho \right)g$
The pressure at the right bottom due to right vertical pipe $=P_{R}=P_{a t m}+h\left(\rho \right)g$
and Pressure difference in the horizontal tube due to acceleration
$P_{L R}=\left(\frac{L}{2}\right)\rho a+\left(\frac{L}{2}\right)2\rho a$
$P_{L R}=\frac{3 Lρa}{2}$
The pressure at the left bottom due to right vertical pipe and horizontal pipe
$P_{L}=P_{R}+P_{L R}$
$P_{L}=P_{a t m}+h\rho g+\frac{3 Lρa}{2}$
The pressure at the left bottom due to left vertical pipe=Pressure at the left bottom due to right vertical pipe and horizontal pipe
$P_{a t m}+h\left(2 \rho \right)g=P_{a t m}+h\rho g+\frac{3 Lρa}{2}$
$\Rightarrow h=\frac{3 aL}{2 g}$