JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2026
- Medium
- MCQ
The rain drop of mass \(1\) g, starts with zero velocity from a height of \(1\) km. It hits the ground with a speed of \(5\) m/s. The work done by the unknown resistive force is _______ J. (take \(g = 10\) m/s\(^2\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(-9.98\)
Step-by-step Solution
Detailed explanation
Using the work-energy theorem, the total work done by all forces equals the change in kinetic energy:
\(W_{g} + W_{r} = \Delta K\)
\(mgh + W_{r} = \dfrac{1}{2}mv^2 - 0\)
Substituting the given values (\(m = 10^{-3}\) kg, \(g = 10\) m/s\(^2\), \(h = 1000\) m, \(v = 5\) m/s):
\((10^{-3})(10)(1000) + W_{r} = \dfrac{1}{2}(10^{-3})(5)^2\)
\(10 + W_{r} = 0.0125\)
\(W_{r} = 0.0125 - 10 = -9.9875\) J
\(W_{r} \approx -9.98\) J
Answer: \(-9.98\)
\(W_{g} + W_{r} = \Delta K\)
\(mgh + W_{r} = \dfrac{1}{2}mv^2 - 0\)
Substituting the given values (\(m = 10^{-3}\) kg, \(g = 10\) m/s\(^2\), \(h = 1000\) m, \(v = 5\) m/s):
\((10^{-3})(10)(1000) + W_{r} = \dfrac{1}{2}(10^{-3})(5)^2\)
\(10 + W_{r} = 0.0125\)
\(W_{r} = 0.0125 - 10 = -9.9875\) J
\(W_{r} \approx -9.98\) J
Answer: \(-9.98\)
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