AP EAMCET · PHYSICS · Center of Mass Momentum and Collision
Three identical uniform thin rods each of mass ' \(\mathrm{m}\) ' and length ' \(\mathrm{L}\) ' are arranged in the \(\mathrm{XY}\) plane as shown in the figure. A fourth uniform thin rod of mass ' \(3 \mathrm{~m}\) ' is placed as shown in the figure in the XY plane. The value of length of the fourth rod such that the centre of mass of all the four rods lies at the origin is
- A \(3 \mathrm{~L}\)
- B \(2 \mathrm{~L}\)
- C \(\frac{\mathrm{L}(\sqrt{2}+1)}{3}\)
- D \(\frac{\mathrm{L}(2 \sqrt{2}+1)}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{L}(\sqrt{2}+1)}{3}\)
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Detailed explanation
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