JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
A uniform cylinder of length \(L\) and mass \(M\) having crosssectional area \(A\) is suspended, with its length vertical, from a fixed point by a massless spring such that it is half submerged in a liquid of density \(\sigma\) at equilibrium position. The extension \(x_0\) of the spring when it is in equilibrium is
- A \(\frac{{Mg}}{k}\)
- B \(\;\frac{{Mg}}{k}\left( {1 - \frac{{LA\sigma }}{M}} \right)\)
- C \(\;\frac{{Mg}}{k}\left( {1 - \frac{{LA\sigma }}{{2M}}} \right)\)
- D \(\;\frac{{Mg}}{k}\left( {1 + \frac{{LA\sigma }}{M}} \right)\)
Answer & Solution
Correct Answer
(C) \(\;\frac{{Mg}}{k}\left( {1 - \frac{{LA\sigma }}{{2M}}} \right)\)
Step-by-step Solution
Detailed explanation
From figure, \(k{x_0} + {F_B} = Mg\) \(k{x_0} + \sigma \frac{L}{2}Ag = Mg\) \(\left[ {mass = density \times volume} \right]\) \( \Rightarrow k{x_0} = Mg - \sigma \frac{L}{2}Ag\)…
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