A raindrop of mass 1 g falling from a height of 1 km hits the ground with a speed of `50 m s^(-1)`. If the resistive force is proportional to the speed of the drop, then the work done by the resistive force is (Take `g = 10 m s^(-2)`)

Options

A. 10 J
B. -10 J
C. 8.75J
D. `-8.75 J`

Explanation

To solve the problem step by step, we will use the work-energy theorem, which states that the change in kinetic energy of an object is equal to the work done by all forces acting on it.

### Step 1: Identify the given data
- Mass of the raindrop, \( m = 1 \, \text{g} = 1 \times 10^{-3} \, \text{kg} \)
- Height from which the raindrop falls, \( h = 1 \, \text{km} = 1000 \, \text{m} \)
- Final speed of the raindrop when it hits the ground, \( v = 50 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the change in kinetic energy
The change in kinetic energy (\( \Delta KE \)) can be calculated using the formula:
\[ \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \]
where \( u \) is the initial velocity (which is 0 since the raindrop is falling from rest).

Substituting the values:
\[ \Delta KE = \frac{1}{2} \times (1 \times 10^{-3}) \times (50)^2 - \frac{1}{2} \times (1 \times 10^{-3}) \times (0)^2 \]
\[ = \frac{1}{2} \times (1 \times 10^{-3}) \times 2500 \]
\[ = \frac{1 \times 10^{-3} \times 2500}{2} = \frac{2.5}{2} = 1.25 \, \text{J} \]

### Step 3: Calculate the work done by gravity
The work done by gravity (\( W_g \)) is given by:
\[ W_g = mgh \]
Substituting the values:
\[ W_g = (1 \times 10^{-3}) \times (10) \times (1000) \]
\[ = 1 \times 10^{-3} \times 10000 = 10 \, \text{J} \]

### Step 4: Apply the work-energy theorem
According to the work-energy theorem:
\[ \Delta KE = W_g + W_r \]
where \( W_r \) is the work done by the resistive force. Rearranging gives:
\[ W_r = \Delta KE - W_g \]
Substituting the values we calculated:
\[ W_r = 1.25 - 10 \]
\[ = -8.75 \, \text{J} \]

### Conclusion
The work done by the resistive force is \( -8.75 \, \text{J} \).