Question

The maximum depth to which sinks before it returns to surface is:
A) $\left( {\dfrac{{h\sigma }}{{\rho - \sigma }}} \right)$
B) $h\dfrac{\sigma }{\rho }$
C) $h\left( {\dfrac{\rho }{{\rho - \sigma }}} \right)$
D) $h\left( {\dfrac{\rho }{\sigma } - 1} \right)$

Answer 1

In order to find the distance up to which a body goes we will apply the Newton’s law of distance ${v^2} = {u^2} - 2as$ where, v is the final velocity, u is the initial and a is the acceleration and s is the distance. For that we need acceleration which is $\dfrac{{upthrust - weight}}{{mass}}$ and velocity inside the water will be zero and initial velocity be $\sqrt {2gh}$

Step by step solution:
We need to find the maximum depth to which a body sinks before it returns to the surface.
We are supposing that we have a body of density $\sigma$ is dropped from the height h into the water of density $\rho$
So at the maximum depth the velocity of the body will be zero.
When that body collides to the surface its velocity (initial) will be $\sqrt {2gh}$ where, g is the gravity and h is the height from which it is thrown
That body will feel buoyancy force inside the water to upward or upthrust and weight of the body will be mg toward surface and getting a retardation in speed
So from here we can write acceleration of body =$\dfrac{{upthrust - weight}}{{mass}}$ ….. (1)
Upthrust will be the volume inside =V$\rho$ g
Weight of the body inside water will be product of volume and density =V$\rho$ g
And mass is volume of the body V$\sigma$
Putting these values in (1), we get $\dfrac{{V\rho g - V\sigma g}}{{V\sigma }}$ …… (2)
Cancelling V from the equation (2) we will get $\dfrac{{\rho - \sigma }}{\sigma }g$ …… (3)
Equation (3) is the acceleration due to which body will retard.
Final velocity is coming to zero; we will use Newton equation here ${v^2} = {u^2} - 2as$ where, v is the final velocity, u is the initial and a is the acceleration and s is the distance
Substituting the value we get, 0=2gh−2 $\dfrac{{\rho - \sigma }}{\sigma }g$ ${h_{MAX}}$
From here we will get ${h_{MAX}}$=$\dfrac{{\sigma h}}{{\rho - \sigma }}$
This will be the depth up to which a body will go inside the water.

Hence option A is correct.

Note: When a body is fully or partially submerged into the liquid, it feels an upward force which is called upthrust or buoyancy. We have taken the density of water as $\sigma$ and of water as $\rho$. The value may differ for different densities.