Question

A body of density $\rho$ is dropped from rest from a height h into a lake of density $\sigma \left( {\sigma > \rho } \right)$ the maximum depth the body sinks inside the liquid is (neglect viscous effect of liquid):
$\left( a \right)\,\,h\rho /\left( {\sigma - \rho } \right)$
$\left( b \right)\,\,h\sigma /\left( {\sigma - \rho } \right)$
$\left( c \right)\,\,h\rho /\sigma$
$\left( d \right)\,\,h\sigma /\rho$

Answer 1

A body is dropped into a lake so whenever something is dropped in water the buoyant force will come into picture. Buoyant force- it is a force which is exerted on objects when it is submerged in fluids. Now, we have to find out the velocity with which the body will enter into the lake for that we will use the formula $\sqrt {2gh}$ . After that we just have to make the work done equal to the change in the kinetic energy which is nothing but a concept of conservation of energy.

Complete Step by step solution:
According to the question there is a body with density $\rho$ which is dropped from rest from a height h into a lake of density $\sigma \left( {\sigma > \rho } \right)$ . We have to find the maximum depth the body sinks inside the liquid.
Okay, let’s suppose the body is dropped from the height $h$ . So, velocity with which it will touch the surface of the lake will be given by the equation $\sqrt {2gh}$ .
Now, with the help of energy conservation
The buoyant force will act in the upward direction and the gravity will act in the downward direction and besides this there will be no force which will be acting on the body.
Let buoyant force be $F'$ and force due to weight be $W$ and $x$ be the distance it moves inside the lake volume if the ball is $z$ .
Now, the work done will be equal to the change in the kinetic energy
$\Rightarrow \left( {\sigma zg} \right)\left( { - x} \right) + \rho zg\left( x \right) = 0 - 1/2z\rho \left( {2gh} \right)$ and we will name it equation $1$
Where $\sigma zg$ is the buoyant force and $- x$ , is written because it is acting in the opposite direction and
$\rho zg$ is weight
And the change is $- ve$ because at the deepest point body comes to rest
So by solving the equation $1$ we will get the value of which is $h\rho /(\sigma - \rho )$
Hence, the maximum depth to which the body can sink is $h\rho /(\sigma - \rho )$

Therefore, the option $\left( a \right)$ is correct.

Note: Remember the key formulas like $\sqrt {2gh}$ and the process we have followed which will be very helpful for future problem solving. Also, these kinds of questions can be very tricky and confusing. So, try to go step by step to get the desired answer without wasting much time and without getting confused in the mid-way of solving the question.