A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length…

1. $\frac{\text{Mg}}{\text{k}}\left(1-\frac{\text{LA}\sigma }{2\text{M}}\right)$
2. $\frac{\text{Mg}}{\text{k}}\left(1+\frac{\text{LA}\sigma }{ \text{M}}\right)$
3. $\frac{\text{Mg}}{\text{k}}$
4. $\frac{\text{Mg}}{\text{k}}\left(1-\frac{\text{LA}\sigma }{\text{M}}\right)$

Solution

${\text{kx}}_0+{\text{F}}_{\text{B}}=\text{mg}$

${\text{kx}}_0+\sigma \frac{\text{L}}{2}\text{Ag}=\text{Mg}$


${\text{x}}_0=\frac{\text{Mg}-\frac{\sigma \text{LAg}}{2}}{\text{k}}$

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

Asked in: JEE Main 2013 (07 Apr)

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