A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless sprin - Physics MCQ JEE Main 2026 | ExamTest.live

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A uniform cylinder of length L and mass M having cross-sectional 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 σ at equilibrium position. The extension x_{0} of the spring when it is in equilibrium is:

Solution & Step-by-step Explanation

At equilibrium, the forces acting on the cylinder are:Downward: Gravity (M g)Upward: Spring force (k x_{0}) and Buoyant force (F_{B}) F_{B} = Weight of displaced liquid = (Volume submerged) \times σ \times g = (\frac{A L}{2}) σ g Equation of equilibrium: k x_{0} + \frac{A L σ g}{2} = M g k x_{0} = M g - \frac{A L σ g}{2} = M g (1 - \frac{L A σ}{2 M}) x_{0} = \frac{M g}{k} (1 - \frac{L A σ}{2 M})

Try it yourself before checking the explanation above.

A uniform cylinder of length L and mass M having cross-sectional 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 σ at equilibrium position. The extension x_{0} of the spring when it is in equilibrium is:

\frac{Mg}{k} \left( 1 - \frac{LA\sigma}{M} \right)

\frac{Mg}{k} \left( 1 - \frac{LA\sigma}{2M} \right)

\frac{Mg}{k} \left( 1 + \frac{LA\sigma}{M} \right)

\frac{Mg}{k}

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