JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A cricket player catches a ball of mass \(120 \mathrm{~g}\) moving with \(25 \mathrm{~m} / \mathrm{s}\) speed. If the catching process is completed in \(0.1 \mathrm{~s}\) then the magnitude of force exerted by the ball on the hand of player will be _______ (in SI unit).
- A \(24\)
- B \(12\)
- C \(25\)
- D \(30\)
Answer & Solution
Correct Answer
(D) \(30\)
Step-by-step Solution
Detailed explanation
\(\mathrm{F}_{\mathrm{ar}}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}\) \(=\frac{0.12 \times 25}{0.1}=30 \mathrm{~N}\)
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