Work Done by Unknown Resistive Force on Falling Raindrop

To find the work done by the unknown resistive force, we apply the Work-Energy Theorem, which states that the net work done by all forces acting on a body is equal to the change in its kinetic energy.

1. Given Data:

• Mass of the raindrop, $m = 1\text{ g} = 10^{-3}\text{ kg}$
• Height from which it falls, $h = 1\text{ km} = 1000\text{ m}$
• Initial velocity, $u = 0\text{ m/s}$
• Final speed on hitting the ground, $v = 5\text{ m/s}$
• Acceleration due to gravity, $g = 10\text{ m/s}^2$

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2. Work Done by Gravity ($W_g$):

The force of gravity acts vertically downwards in the direction of motion: $W_g = mgh$ $W_g = 10^{-3}\text{ kg} \times 10\text{ m/s}^2 \times 1000\text{ m} = 10\text{ J}$

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3. Change in Kinetic Energy ($\Delta K$):

$\Delta K = K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$ $\Delta K = \frac{1}{2} \times 10^{-3}\text{ kg} \times (5\text{ m/s})^2 - 0$ $\Delta K = \frac{1}{2} \times 10^{-3} \times 25 = 0.0125\text{ J}$

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4. Applying the Work-Energy Theorem:

Let $W_r$ be the work done by the resistive force. $W_{\text{net}} = \Delta K$ $W_g + W_r = \Delta K$

Substitute the known values into the equation: $10 + W_r = 0.0125$ $W_r = 0.0125 - 10 = -9.9875\text{ J} \approx -9.98\text{ J}$

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Conclusion:

The work done by the unknown resistive force is $-9.98\text{ J}$.

Correct Option: D