Question

The U tube having identical limbs contains mercury (density ${{\rho }_{m}}$) to a level as shown in the figure. If the left limbs is filled to the top with water (${{\rho }_{w}}$), then the rise of mercury level in the right limb will be

A.$12\rho m12\rho m$
B.$\dfrac{\rho mL}{{{\rho }_{w}}}$
C.$\dfrac{L{{\rho }_{w}}}{2{{\rho }_{m}}+{{\rho }_{w}}}$
D.$\dfrac{L{{\rho }_{w}}}{2{{\rho }_{m}}-{{\rho }_{w}}}$

Answer 1

Since the level of mercury is the same in the two arms of the U tube, therefore Pressure of water column on the surface of mercury in one arm = Pressure of oil column on the surface of mercury in the other arm. Because pressure depends on the height and here mercury is filled up to the same level.
Formula for Pressure is given as $P=\rho g(L)$, We will use this formula to solve this question.

Complete answer:

Since the level of mercury is same so the pressure exerted at the center of the bottom will be the same, but according to the question we are adding water on the left limb so the water level will change on the right limb too, But the bottom center will act as a equilibrium point. So, we will equate the pressure from both the sides with respect to the center of the U tube
The pressure at the center of the bottom is given by
${{P}_{left}}={{\rho }_{w}}g(L+X)$
And ${{P}_{right}}={{\rho }_{m}}g(2X)$
Now Put ${{P}_{left}}={{P}_{right}}$
${{\rho }_{w}}g(L+X)={{\rho }_{m}}g(2X)$
${{\rho }_{w}}gL+{{\rho }_{w}}gX={{\rho }_{m}}g(2X)$
On solving the above equation, we get,
$X=\dfrac{L{{\rho }_{w}}}{2{{\rho }_{m}}-{{\rho }_{w}}}$

Hence the correct option is option (D) That is $X=\dfrac{L{{\rho }_{w}}}{2{{\rho }_{m}}-{{\rho }_{w}}}$

Note:
We have taken the centre of the bottom as the equilibrium, because that is the only point where the pressure exerted will be the same before and after adding water to any of the Limbs of U tube. The bottom centre point is the only place which will be common for both the conditions, the mercury will never leave from that position.