WBJEE · Physics · Laws of Motion
Two blocks of \(2 \mathrm{~kg}\) and \(1 \mathrm{~kg}\) are in contact on a frictionless table. If a force of \(3 \mathrm{~N}\) is applied on \(2 \mathrm{~kg}\) block, then the force of contact between the two blocks will be :
- A \(0 \mathrm{~N}\)
- B \(1 \mathrm{~N}\)
- C \(2 \mathrm{~N}\)
- D \(3 \mathrm{~N}\)
Answer & Solution
Correct Answer
(B) \(1 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
Hints: Common acceleration \(=\frac{3}{3}=1 \mathrm{~m} / \mathrm{sec}^2\) \[ \mathrm{N}_1=1 \mathrm{~N} \]
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