Home/
Competitive Exams/
JEE Main & Advanced/
Physics/
Laws of Motion and Friction/
Question 30247
JEE Physics · Medium

Laws of Motion and Friction: Force — Impulse Application MCQ

Solve this quality-checked JEE multiple-choice question, then review the correct answer and explanation.

Timer-based practiceDetailed answer reviewMobile-friendly flow
Force — Impulse ApplicationMediumQuestion 30247

Question

A cricket ball (0.15 kg) is stopped from 30 m/s to 0 in 0.05 s by a fielder. Average force exerted is:
  1. A
    9 N
  2. B
    45 N
  3. C
    90 N
    Correct
  4. D
    450 N

Correct answer

90 N

Explanation

F = Δp/Δt = (0.15×30)/0.05 = 4.5/0.05 = 90 N.

Review Laws of Motion and Friction notesStart timed chapter practice