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Work Energy and Power question

2017 · Q149
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Work Energy and Power question

2017 · Q149

NEETPhysicsWork Energy and PowerMCQ+4 / −1
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
  1. A
    (i) 1.25 j   (ii) −-−8.25 J
  2. B
    (i) 100 j   (ii) 8.75 J
  3. C
    (i) 10 j   (ii) −-−8.75 J
  4. D
    (i) −-−10 j   (ii) −-−8.25 J
View written solutionFree

Correct answer: C

From work-energy theorem,
Wg + Wa = Δ\Delta ΔK.E

or mgh + Wa = 12mv2−0{1 \over 2}m{v^2} - 021​mv2−0

10−3×10×103+Wa=12×10−3×(50)2{10^{ - 3}} \times 10 \times {10^3} + {W_a} = {1 \over 2} \times {10^{ - 3}} \times {\left( {50} \right)^2}10−3×10×103+Wa​=21​×10−3×(50)2

⇒Wa=−8.75 J \Rightarrow {W_a} = - 8.75\,J⇒Wa​=−8.75J

which is the work done due to air resistance

Work done due to gravity = mgh

= 10–3 × 10 × 103 = 10 J

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