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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A block of mass \(m =1\, kg\) slides with velocity \(v=6\, m / s\) on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about \(O\) and swings as a result of the collision making angle \(\theta\) before momentarily coming to rest. If the rod has mass \(M =2 \,kg \) and length \(l=1\, m\) the value of \(\theta\) is approximately (Take \(\left.g=10 \,m / s ^{2}\right)\)
- A \(69\)
- B \(63\)
- C \(55\)
- D \(49\)
Answer & Solution
Correct Answer
(B) \(63\)
Step-by-step Solution
Detailed explanation
Angular momentum conservation \(mv l=\frac{ Ml ^{2}}{3} \omega+ m l^{2} \omega\) \(\Rightarrow \omega=\frac{1 \times 6 \times 1}{\frac{2}{3}+1}=\frac{18}{5}\) Now using energy consevation…
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