A raindrop of mass `1g` falling from a height of `1km` hits is the ground with a speed of `50 ms^(-1)` If the resistence force is properition to the speed of the drop .then the work done by the resistence force is`("Taking" g: 10ms^(-2))`.

Options

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

Explanation

To solve the problem, we need to calculate the work done by the resistance force acting on the raindrop as it falls. We will follow these steps:

### Step 1: Convert the mass of the raindrop from grams to kilograms.
Given mass \( m = 1 \, \text{g} = 1 \times 10^{-3} \, \text{kg} \).

### Step 2: Calculate the kinetic energy of the raindrop just before it hits the ground.
The formula for kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times (1 \times 10^{-3} \, \text{kg}) \times (50 \, \text{m/s})^2 \]
Calculating this:
\[ KE = \frac{1}{2} \times (1 \times 10^{-3}) \times 2500 = \frac{2500 \times 10^{-3}}{2} = 1.25 \, \text{J} \]

### Step 3: Calculate the work done by the gravitational force.
The work done by the gravitational force (Wg) can be calculated using the formula:
\[ W_g = mgh \]
Where:
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 1 \, \text{km} = 1000 \, \text{m} \)

Substituting the values:
\[ W_g = (1 \times 10^{-3} \, \text{kg}) \times (10 \, \text{m/s}^2) \times (1000 \, \text{m}) = 10 \, \text{J} \]

### Step 4: Use the work-energy principle to find the work done by the resistance force.
According to the work-energy principle:
\[ KE = W_g + W_r \]
Where \( W_r \) is the work done by the resistance force. Rearranging gives:
\[ W_r = KE - W_g \]
Substituting the values we calculated:
\[ W_r = 1.25 \, \text{J} - 10 \, \text{J} = -8.75 \, \text{J} \]

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

---