Question

A body of density '\( \rho \)' is dropped slowly on the surface of a lake of depth \( d \). If the density of the lake water be '\( \rho' \)' (\( \rho' < \rho \)) then the time taken by the body to reach the bottom of the lake is

• A. \( \left[\frac{2d\rho}{g(\rho - \rho')}\right]^{\frac{1}{2}} \)
• B. \( \left[\frac{2gd}{\rho(\rho - \rho')}\right]^{\frac{1}{2}} \)
• C. \( \left[\frac{2d\rho'}{\rho g(\rho - \rho')}\right]^{\frac{1}{2}} \)
• D. \( \left[\frac{g(\rho - \rho')}{2d\rho}\right]^{\frac{1}{2}} \)

Hint:

To quickly verify such algebraic formulas, use dimensional analysis. Since \( [t] = T \), the term inside the square root must have the dimensions of \( T^2 \) (which is \( \frac{L}{LT^{-2}} = T^2 \)). Only option (A) contains \( \frac{\text{length}}{\text{acceleration}} \) because the densities in the numerator and denominator cancel each other's units.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A body of density \( \rho \) is dropped at the surface of a lake of depth \( d \). Due to buoyancy from the water of density \( \rho' \), the net downward acceleration of the body is reduced. We need to find the time it takes to travel the distance \( d \) and reach the bottom of the lake starting from rest.

Step 2: Key Formula or Approach:
1. Net Force in a fluid:
\[ F_{\text{net}} = F_g - F_b \]
where \( F_g = m g = V \rho g \) is the weight of the body and \( F_b = V \rho' g \) is the buoyant force.
2. Net downward acceleration:
\[ a_{\text{net}} = \frac{F_{\text{net}}}{m} \]
3. Equation of Motion: Since the body is dropped slowly (initial velocity \( u = 0 \)) and accelerates uniformly:
\[ d = \frac{1}{2} a_{\text{net}} t^2 \implies t = \sqrt{\frac{2d}{a_{\text{net}}}} \]

Step 3: Detailed Explanation:
Let \( V \) be the volume of the body.
- Mass of the body: \( m = V \rho \)
- Net downward force:
\[ F_{\text{net}} = V \rho g - V \rho' g = V g (\rho - \rho') \]
- Net downward acceleration \( a_{\text{net}} \):
\[ a_{\text{net}} = \frac{F_{\text{net}}}{m} = \frac{V g (\rho - \rho')}{V \rho} = g \left(\frac{\rho - \rho'}{\rho}\right) \]
Now, calculate the time \( t \) to descend through depth \( d \):
\[ t = \sqrt{\frac{2d}{a_{\text{net}}}} \]
Substitute \( a_{\text{net}} \) into this equation:
\[ t = \sqrt{\frac{2d}{g \left(\frac{\rho - \rho'}{\rho}\right)}} = \left[ \frac{2d\rho}{g(\rho - \rho')} \right]^{\frac{1}{2}} \]

Step 4: Final Answer:
The time taken to reach the bottom is \( \left[\frac{2d\rho}{g(\rho - \rho')}\right]^{\frac{1}{2}} \).