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Match List-I (Lengths of two simple pendulums) with List-II (Minimum number of oscillations of the shorter pendulum after which they are again in phase at the mean position) and select the correct option. List-I List-II (A) 144 cm and 100 cm (I) 8 (B) 64 cm and 49 cm (II) 9 (C) 81 cm and 64 cm (III) 13 (D) 169 cm and 144 cm (IV) 6

Options

  1. A(A) (IV), (B) (II), (C) (I), (D) (III)
  2. B(A) (IV), (B) (I), (C) (II), (D) (III)
  3. C(A) (IV), (B) (I), (C) (III), (D) (II)
  4. D(A) (I), (B) (IV), (C) (II), (D) (III)

Correct answer

B. (A) (IV), (B) (I), (C) (II), (D) (III)

Step-by-step solution

The time period of a simple pendulum is T l . Let the shorter pendulum complete n oscillations and the longer pendulum complete (n-1) oscillations when they are again in phase. nT_ short = (n-1)T_ long n n-1 = T_ long T_ short = l_ long l_ short For (A): 144 100 = 12 10 = 6 5 n = 6 . Matches with (IV). For (B): 64 49 = 8 7 n = 8 . Matches with (I). For (C): 81 64 = 9 8 n = 9 . Matches with (II). For (D): 169 144 = 13 12 n = 13 . Matches with (III). Therefore, the correct matching is (A) (IV), (B) (I), (C) (II), (D)

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