kx0+FB=mg
kx0+σ2LAg=Mg

x0=kMg−2σLAg
=kMg(1−2MσLA)
JEE Main 2013 — Physics Mechanics
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 x0 of the spring when it is in equilibrium is :
Held on 7 Apr 2013 · Verified 6 Jul 2026.
kMg(1−2MLAσ)
kMg(1+MLAσ)
kMg
kMg(1−MLAσ)
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