Question

One end of a U-tube of uniform bore (area $A$) containing mercury is connected to a suction pump. Because of its level of liquid of density $\rho $ falls in one limb. When the pump is removed, the restoring force in the other limb is:

A. $2x\rho Ag$
B. $x\rho g$
C. $A\rho g$
D. $x\rho Ag$

Answer 1

Hint: Restoring force is the total force required to bring back the fluid to the original arrangement. Since there are two limbs the total force is equal to two times the weight of fluid displaced in one limb. Substitute the density and area of the tube to calculate the weight of the fluid.

Complete answer:
In the above diagram, we have been provided the final arrangement of the fluid when the suction pressure is applied at one end of the tube. The dotted line shows the original level of fluid, before the suction was applied. Therefore total height of column of fluid on the right limb with respect to the level of fluid in left limb in new arrangement = $2x$ (Note that if you displace unit length fluid from one level in one limb, then equal unit length of fluid will rise above the same level in other limb, considering the cross sectional area of both the limbs are constant). Weight of this column of fluid, given that the density of the fluid is, $\rho $ and area of the tube, $A$ = density $\times$ volume $\times$ acceleration due to gravity, $g$ = $\rho \times 2xA \times g = 2x\rho Ag$. (Since mass = density $\times$ volume)
Therefore to bring the arrangement back to original, the restoring force should be equal to the weight of the fluid and thus comparing the choices given in the problem above with the last result, we can say that option A is correct.

Note: Restoring force is the difference in weights of fluid before and after the suction is applied. Before the suction pressure, the level of fluid was the same and thus weight of the fluid is zero. When the suction pressure is applied, the weight of fluid causing the level difference is twice the fluid dropped in one limb.