A stream-lined body falls through air from a height `h` on the surface of a liquid . Let `d` and `D` denote the densities of the materials of the body and the liquid respectively, if `D gt d`, then the time after which the body will be intantaneously at rest, is:

Options

A. `sqrt((2h)/(g))`
B. `sqrt((2h)/(g)(D)/(d))`
C. `sqrt((2h)/(g)(d)/(D))`
D. `sqrt((2h)/(g)) ((d)/(D-d))`

Explanation

To solve the problem, we need to determine the time after which a streamlined body falls through the air and comes to an instantaneous rest when it hits the surface of a liquid. The densities of the body and the liquid are denoted as \(d\) and \(D\) respectively, with the condition that \(D > d\).

### Step-by-Step Solution:

1. **Determine the velocity of the body just before it hits the liquid surface:**
- The body is dropped from a height \(h\) with an initial velocity \(u = 0\).
- Using the equation of motion:
\[ v^2 = u^2 + 2as \]
- Here, \(a = g\) (acceleration due to gravity), and \(s = h\).
- Thus, we have:
\[ v^2 = 0 + 2gh \implies v = \sqrt{2gh} \]

2. **Calculate the apparent weight when the body enters the liquid:**
- When the body enters the liquid, the true weight of the body is \(W = V \cdot d \cdot g\), where \(V\) is the volume of the body.
- The weight of the liquid displaced is \(W_{displaced} = V \cdot D \cdot g\).
- The apparent weight \(W_{apparent}\) is given by:
\[ W_{apparent} = W - W_{displaced} = Vdg - V Dg = Vg(d - D) \]

3. **Determine the acceleration (retardation) of the body in the liquid:**
- The mass of the body is \(m = V \cdot d\).
- The acceleration \(a\) (retardation in this case) can be calculated using Newton's second law:
\[ a = \frac{W_{apparent}}{m} = \frac{Vg(d - D)}{Vd} = \frac{g(d - D)}{d} \]

4. **Use the equation of motion to find the time taken to stop:**
- The initial velocity \(u = \sqrt{2gh}\) when the body enters the liquid.
- The final velocity \(v = 0\) when the body comes to rest.
- Using the equation of motion:
\[ v = u + at \]
- Substituting the values:
\[ 0 = \sqrt{2gh} + \left(-\frac{g(D - d)}{d}\right)t \]
- Rearranging gives:
\[ \frac{g(D - d)}{d}t = \sqrt{2gh} \]
- Solving for \(t\):
\[ t = \frac{d \sqrt{2h}}{g(D - d)} \]

### Final Result:
The time after which the body will be instantaneously at rest is given by:
\[ t = \frac{d}{D - d} \sqrt{\frac{2h}{g}} \]