Question

In the figure, the cross-sectional area of the smaller tube is $a$ and that of the larger tube is $2a$. A block of mass $m$ is kept in the smaller tube having the same base area $a$, as that of the tube. The difference between water levels of the two tubes is

• A. $\frac{P_0}{\rho g}+\frac{m}{a\rho }$
• B. $\frac{P_0}{\rho g}+\frac{m}{2a\rho }$
• C. $\frac{m}{a\rho }$
• D. $\frac{m}{2\rho a}$

Answer 1

The correct option is C $\frac{m}{a\rho }$

By Pascal's Law, pressure is same at the same horizontal level in a liquid.
Let the heights of fluid above this level be $h$ and $h_1$ as shown in the figure.
Pressure from part 1 is $P_0+h_1\rho g+\frac{mg}{a}$
Pressure from part 3 is $P_0+h\rho g$
Equating these two we have
$P_0+h_1\rho g+\frac{mg}{a}=P_0+h\rho g$
$\therefore h-h_1=\frac{m}{a\rho }$
Therefore difference in height of liquid in the two arms is $\frac{m}{a\rho }$