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JEE Main 2015 (11 Apr, Shift 1) · Other SHM Systems

Other SHM Systems question from JEE Main 2015 (11 Apr, Shift 1)

Q17JEE Main 2015 · 11 APR · SHIFT 1Medium

Other SHM Systems

A cylindrical block of wood (density = 650 kg m−3{}^{-3}), of base area 30 cm2 and height 54 cm, floats in a liquid of density 900 kg m−3{}^{-3}. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly):

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Answer: (c)

  1. T=2πmρlAgT = 2\pi\sqrt{\frac{m}{\rho_l A g}} with m=ρwAhm = \rho_w A h, so T=2πρwhρlgT = 2\pi\sqrt{\frac{\rho_w h}{\rho_l g}}.
  2. Equivalent pendulum length L=ρwhρl=650×54900=39 cmL = \frac{\rho_w h}{\rho_l} = \frac{650 \times 54}{900} = 39\text{ cm}
  3. So the answer is (c).
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