Question

A uniform cylinder of length L and M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in liquid of density $\sigma$ at equilibrium position. The extension ${{x}_{0}}$ of the spring when it is in equilibrium, is
A. $\dfrac{Mg}{k}\left( 1-\dfrac{LA\sigma }{M} \right)$
B. $\dfrac{Mg}{k}\left( 1-\dfrac{LA\sigma }{2M} \right)$
C. $\dfrac{Mg}{k}\left( 1+\dfrac{LA\sigma }{M} \right)$
D. $\dfrac{Mg}{k}$

Answer 1

Hint:First, identify the forces acting on the given cylinder. Then find the magnitude and direction of each force and find the net force acting on the cylinder. Since the cylinder is in equilibrium, equate the net force to zero.
Formula used:
${{F}_{g}}=Mg$
${{F}_{B}}={{V}_{s}}g\sigma$
${{F}_{s}}=k{{x}_{0}}$

Complete step by step answer:
At the equilibrium position, the uniform cylinder will be standing straight up.
There will be three forces acting on the cylinder and that are the gravitational force, the spring force and the force of buoyancy exerted by the liquid.

The earth exerts the gravitational force. The gravitational force on a body of mass m is given as ${{F}_{g}}=mg$, where g is the acceleration due to gravity. The gravitational force on the body is always direction downwards.
This means that in the case, the gravitational force on the cylinder is equal to ${{F}_{g}}=Mg$ and is
directed in the downward direction.
When a body is submerged in a liquid, the liquid exerts an upward force on the body called buoyant force. The magnitude of the buoyant force is given as ${{F}_{B}}={{V}_{s}}g\sigma$,
where ${{V}_{s}}$ is the volume of the body submerged in the liquid and $\sigma$ is the density of the liquid.
It is given that half of the cylinder is submerged in the liquid and the volume of a cylinder is equal to $V=AL$, where A is the cross sectional area and L is its length.
This means that ${{F}_{B}}=ALg\sigma$ and is directed upwards.
The cylinder is given to be in equilibrium. Therefore, the cylinder will be at rest and the net force on it will be zero.
Let the extension in the spring be ${{x}_{0}}$. Therefore, the spring force exerted on the cylinder is equal to ${{F}_{s}}=k{{x}_{0}}$, where k is the spring constant of the spring. The spring will be in the upward direction.
Therefore, the net force on the cylinder is ${{F}_{net}}={{F}_{s}}+{{F}_{B}}-{{F}_{g}}$.
$\Rightarrow {{F}_{net}}=k{{x}_{0}}+ALg\sigma -Mg$.
But we know that ${{F}_{net}}=0$.
$\Rightarrow k{{x}_{0}}+ALg\sigma -Mg=0$
$\Rightarrow {{x}_{0}}=\dfrac{Mg-ALg\sigma }{k}$
Take $\dfrac{Mg}{k}$ as a common term.
$\Rightarrow {{x}_{0}}=\dfrac{Mg}{k}\left( 1-\dfrac{AL\sigma }{M} \right)$.
Therefore, the correct option is A.

Note: Note that the direction of the gravitational force on a body is always in the downward direction if the body is on earth as the earth pulls down the body with the gravitational force. The direction of the buoyant force exerted by a liquid is always in the upward direction as it tries to push the body outwards.