Question

Consider a drop of rain water having mass $1\,g$ falling from a height of $1\,km$. It hits the ground with a speed of $50\,m/s$. Take 'g' constant with a value ${10 \, m/s^2}$. The work done by the (i) gravitational force and the (ii) resistive force of air is :-

• A. (i) 1.25 J (ii) - 8.25 J
• B. (i) 100 J (ii) 8.75 J
• C. (i) 10 J (ii) - 8.75 J
• D. (i) - 10 J (ii) - 8.25 J

Answer 1

The Correct Option is C

 To solve this problem, we need to calculate the work done by gravitational force and resistive force of air on the rain drop.

Step 1: Work Done by Gravitational Force

The work done by gravitational force can be calculated using the formula:

\(W_{\text{grav}} = m \cdot g \cdot h\)

• \(m = 1\,g = 0.001\,kg\) (conversion from grams to kilograms)
• \(g = 10\,m/s^2\) (acceleration due to gravity)
• \(h = 1000\,m\) (conversion from kilometers to meters)

Substituting the values, we get:

\(W_{\text{grav}} = 0.001 \times 10 \times 1000 = 10\,J\)

So, the work done by the gravitational force is 10 J.

Step 2: Work Done by Resistive Force

The total mechanical energy change involves kinetic energy (KE) and the work done by resistive force.

Initial potential energy (PE) is fully converted to kinetic energy and work done by resistive force:

\(PE = KE + W_{\text{resistive}}\)

Kinetic energy at the ground can be calculated using the formula:

\(KE = \frac{1}{2} m v^2\)

• \(v = 50\,m/s\) (given)

Substituting the values, we get:

\(KE = \frac{1}{2} \times 0.001 \times 50^2 = \frac{1}{2} \times 0.001 \times 2500 = 1.25\,J\)

Therefore, the potential energy is converted into kinetic energy and work done against air resistance:

\(10 = 1.25 + W_{\text{resistive}}\)

Solving for \(W_{\text{resistive}}\):

\(W_{\text{resistive}} = 10 - 1.25 = - 8.75\,J\)

The work done by the resistive force is -8.75 J.

Conclusion

Thus, the correct answers are:

• (i) Work done by gravitational force is 10 J.
• (ii) Work done by the resistive force of air is -8.75 J.

The correct option is:

(i) 10 J (ii) - 8.75 J