JEE Main 2020 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 3 September 2020, Shift 1, Physics Q21

Question 21 of 75 in this shift, Physics question 21 of 25.

A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 m after leaving the machine. If the part pushing the ball applies a constant force F on the ball and moves horizontally a distance of 0.2 m while launching the ball, the value of F (in N) is (g = 10 ms

^{-2}

) __________.

Official answer

150.00

NTA final key (Sep 2020).

Chapter: Work, Energy and Power
Topic: Work and Kinetic Energy
Idea tested: Work-Energy Theorem

Same idea in other shifts

Asked 10× in all

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.