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Answer»

Explanation:

Answer is 2sec

T = 2π√(I/MB)

For a rod I = (Mℓ2/I2)

For half length I' = [{(m/2)(ℓ/2)2}/12] = (mℓ2/12) × (1/8) = (I/8) and magnetic moment = M' = (M/2)

HENCE new time period = T' = 2π√(I'/MB) T' = 2π√[(I/8)/{(M/2)B}] = 2π√(I/MB) ∙ √(2/8) = (2π/2) √(I/MB) = (1/2)[2π√(I/MB)] T' = (1/2)T T' = (T/2) Given T = 4 SEC

T' = (4/2) = 2sec

done


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