JEE Main 20137 April
- Exam
- JEE Main
- Year
- 2013
- Date
- 7 April
- Subject
- Physics
- Chapter
- Laws of Motion
JEE Main 2013 · 7 April · Physics · Laws of Motion
A uniform cylinder of length L and mass M having cross-sectional...
A uniform cylinder of length and mass having cross-sectional area 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 at equilibrium position. The extension of the spring when it is in equilibrium is :
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Options
- A\dfrac{\text{Mg}}{\text{k}} \left(1 - \dfrac{\text{LA} σ}{2 \text{M}}\right)
- B\dfrac{\text{Mg}}{\text{k}} \left(1 + \dfrac{\text{LA} σ}{ \text{M}}\right)
- C
- D\dfrac{\text{Mg}}{\text{k}} \left(1 - \dfrac{\text{LA} σ}{\text{M}}\right)
Correct answer
A. \dfrac{\text{Mg}}{\text{k}} \left(1 - \dfrac{\text{LA} σ}{2 \text{M}}\right)
Step-by-step solution
{\text{kx}}_{0} + {\text{F}}_{\text{B}} = \text{mg}{\text{kx}}_{0} + \sigma \dfrac{\text{L}}{2} \text{Ag} = \text{Mg}{\text{x}}_{0} = \dfrac{\text{Mg} - \dfrac{\sigma \text{LAg}}{2}}{\text{k}}= \dfrac{\text{Mg}}{\text{k}} \left(1 - \dfrac{\sigma \text{LA}}{2 \text{M}}\right)