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 at equilibrium position. The extension of the s

Question

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: [2012]

Smart Achievers · Accepted Answer

(3)

$From figure, k x_{0} + F_{B} = M g$

$k x_{0} + σ \frac{L}{2} A g = M g$

$[∵ mass = density \times volume]$

$\Rightarrow k x_{0} = M g - σ \frac{L}{2} A g$

$\Rightarrow x_{0} = \frac{M g - \frac{σ L A g}{2}}{k} = \frac{M g}{k} (1 - \frac{L A σ}{2 M})$

$Hence, extension of the spring when it is in equilibrium is,$

$x_{0} = \frac{M g}{k} (1 - \frac{L A σ}{2 M})$

Solution

1 $\frac{M g}{k}$

2 $\frac{M g}{k} (1 - \frac{L A σ}{M})$

3 $\frac{M g}{k} (1 - \frac{L A σ}{2 M})$

4 $\frac{M g}{k} (1 + \frac{L A σ}{M})$

Want Us to Email you About Special Offers & Updates?

Solution

1 Mgk

2 Mgk(1-LAσM)

3 Mgk(1-LAσ2M)

4 Mgk(1+LAσM)

Ans. (3) From figure, kx0+FB=Mg kx0+σL2Ag=Mg [∵ mass = density × volume] ⇒kx0=Mg-σL2Ag ⇒x0=Mg-σLAg2k=Mgk(1-LAσ2M) Hence, extension of the spring when it is in equilibrium is, x0=Mgk(1-LAσ2M)

Want Us to Email you About Special Offers & Updates?

1 $\frac{M g}{k}$

2 $\frac{M g}{k} (1 - \frac{L A σ}{M})$

3 $\frac{M g}{k} (1 - \frac{L A σ}{2 M})$

4 $\frac{M g}{k} (1 + \frac{L A σ}{M})$