Physics Solution: Time to Sink Calculation

The problem asks for the time taken for a body to sink through a specific depth in a lake. We need to determine the net force acting on the body and its resulting acceleration.

Understanding Forces on the Body

• The body has a density '$\rho$' and volume '$V$'. Its mass is $m = \rho V$.
• The downward gravitational force is $F_g = mg = (\rho V)g$.
• The upward buoyant force from the lake water (density '$\rho'$) is $F_b = (\rho' V)g$.
• We are given that the body's density is greater than the lake water's density ($\rho > \rho'$), so the body will sink.

Deriving the Acceleration

The net force ($F_{net}$) acting on the body is the difference between the gravitational force and the buoyant force:

$F_{net} = F_g - F_b = (\rho V)g - (\rho' V)g = (\rho - \rho')Vg$

According to Newton's second law, $F_{net} = ma$, where '$a$' is the acceleration of the body.

Substituting the mass $m = \rho V$:

$(\rho - \rho')Vg = (\rho V)a$

Solving for acceleration '$a$':

$a = \frac{(\rho - \rho')Vg}{\rho V} = \frac{\rho - \rho'}{\rho} g$

Calculating Sinking Time

Since the body is dropped slowly, we assume its initial velocity ($u$) is 0. We can use the kinematic equation for distance ($d$) under constant acceleration ($a$):

$d = ut + \frac{1}{2}at^2$

With $u=0$, this simplifies to:

$d = \frac{1}{2}at^2$

We need to find the time ($t$). Rearranging the equation:

$t^2 = \frac{2d}{a}$

Now, substitute the expression for acceleration '$a$' we found:

$t^2 = \frac{2d}{\left( \frac{\rho - \rho'}{\rho} g \right)} = \frac{2d\rho}{g(\rho - \rho')}$

Taking the square root to find the time '$t$':

$t = \left[ \frac{2d\rho}{g(\rho - \rho')} \right]^{\frac{1}{2}}$

Final Answer Comparison

Comparing this result with the given options, it matches Option 1.