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.
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