Correct answer: 3. 2 cm u sin2 R 5 3 g    m Let M be the mass of each fragment  M 3 (n 1)M 7 3 5 3 nM      n = 3

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AITS & Test SerieshardNUMERICAL

Question

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Answer: 3

💡 Solution & Explanation

2 cm u sin2 R 5 3 g    m Let M be the mass of each fragment  M 3 (n 1)M 7 3 5 3 nM      n = 3

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