JEE Mains · Physics · STD 11 - 13. oscillations
A cylindrical block of wood (density \(= 650\, kg\, m^{-3}\)), of base area \(30\,cm^2\) and height \(54\, cm\), floats in a liquid of density \(900\, kg\, m^{-3}\) . 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 ..... \(cm\) (nearly)
- A \(52\)
- B \(65\)
- C \(39\)
- D \(26\)
Answer & Solution
Correct Answer
(C) \(39\)
Step-by-step Solution
Detailed explanation
Required equivalent length \(=\frac{\rho_{\text {wood }}}{\rho_{\text {liquid }}} \times\) height of block \(=\frac{650}{900} \times 54 \times 10^{-2}\) \(\Rightarrow l=0.39 \mathrm{m}=39 \mathrm{cm}\)
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