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 Concept:
The problem involves a body moving through a fluid medium under the influence of gravity and buoyancy.
When an object is immersed in a fluid, it experiences an upward force called the buoyant force, as described by Archimedes' Principle.
Since the density of the body \(\rho\) is greater than the density of the water \(\rho'\), the downward weight exceeds the upward buoyancy, causing the body to sink.
The motion of the body is characterized by a net downward force, which produces a constant acceleration.
Because the body is dropped "slowly," we can assume its initial velocity at the surface is zero.
We will use Newton's Second Law to determine the acceleration and then apply the equations of motion to find the time taken to travel the distance \(d\).
Step 2: Key Formula or Approach:
The net downward force \(F_{net}\) is given by:
\[ F_{net} = \text{Weight} - \text{Buoyant Force} \]
Mass of the body \(m = V\rho\), where \(V\) is the volume.
Weight \(W = mg = V\rho g\).
Buoyant force \(F_b = V\rho' g\).
Equation of motion: \(s = ut + \frac{1}{2}at^2\).
Step 3: Detailed Explanation:
First, we calculate the net force acting on the body while it is submerged.
The weight acts downwards, and the upthrust (buoyancy) acts upwards.
\[ F_{net} = V\rho g - V\rho' g = Vg(\rho - \rho') \]
According to Newton's second law, \(F_{net} = ma\), where \(a\) is the acceleration.
Substituting the mass \(m = V\rho\):
\[ V\rho a = Vg(\rho - \rho') \]
Dividing both sides by \(V\rho\), we get:
\[ a = \frac{g(\rho - \rho')}{\rho} \]
This acceleration is constant throughout the motion.
Now, we consider the kinematics of the fall. The distance traveled is the depth of the lake, \(d\).
The initial velocity \(u = 0\) as it is dropped slowly from the surface.
Using the second equation of motion:
\[ d = (0)t + \frac{1}{2}at^2 \]
\[ d = \frac{1}{2} \left[ \frac{g(\rho - \rho')}{\rho} \right] t^2 \]
To find the time \(t\), we rearrange the equation:
\[ t^2 = \frac{2d\rho}{g(\rho - \rho')} \]
\[ t = \sqrt{\frac{2d\rho}{g(\rho - \rho')}} \]
\[ t = \left[ \frac{2d\rho}{g(\rho - \rho')} \right]^{\frac{1}{2}} \]
This result matches option (A).
Step 4: Final Answer:
The acceleration of the body in water is reduced due to buoyancy, leading to a modified gravitational effect.
Applying the standard kinematic relation for distance and time yields the final expression for \(t\).