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 ⁻¹. 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: (c). (i) Wg = mgh = 10⁻³ × 10 × 1000 = 10 J Final KE = 1/2(10⁻³)(50)² = 1.25 J (ii) Wg + Wair = Δ K ⇒ Wair = 1.25 - 10 = -8.75 J So the answer is (c).

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Q21NEET 2017Easy

Work-Energy Theorem

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 ${}^{-1}$ . 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

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Answer: (c)

(i) $W_g = mgh = 10^{-3} \times 10 \times 1000 = 10\text{ J}$
Final KE $= \frac{1}{2}(10^{-3})(50)^2 = 1.25\text{ J}$
(ii) $W_g + W_{air} = \Delta K \Rightarrow W_{air} = 1.25 - 10 = -8.75\text{ J}$
So the answer is (c).

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