JEE MAIN - Physics (2013 (Offline) - No. 23)
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 mass-less 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:
$${{Mg} \over k}$$
$${{Mg} \over k}\left( {1 - {{LA\sigma } \over M}} \right)$$
$${{Mg} \over k}\left( {1 - {{LA\sigma } \over {2M}}} \right)$$
$${{Mg} \over k}\left( {1 + {{LA\sigma } \over M}} \right)$$
Explanation
_en_23_1.png)
From figure, $$k{x_0} + {F_B} = Mg$$
$$k{x_0} + \sigma {L \over 2}Ag = Mg$$
[ as mass $$=$$ density $$ \times $$ volume ]
$$ \Rightarrow k{x_0} = Mg - \sigma {L \over 2}Ag$$
$$ \Rightarrow {x_0} = {{Mg - {{\sigma LAg} \over 2}} \over k}$$
$$ = {{Mg} \over k}\left( {1 - {{LA\sigma } \over {2M}}} \right)$$
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