question_answer

A cube of mass m and density D is suspended from the point P by a spring of stiffness k. The system is kept inside a beaker filled with a liquid of density d. The elongation in the spring, assuming D > d, is

A)  \[\frac{mg}{k}\left( 1-\frac{d}{D} \right)\]               

B)  \[\frac{mg}{k}\left( 1-\frac{D}{d} \right)\]

C)  \[\frac{mg}{k}\left( 1+\frac{d}{D} \right)\]             

D)  None of these

Correct Answer: A

Solution :

 The cube is in equilibrium under three forces as, Spring force \[kx,\] where \[x=\] elongation of the spring, Gravitational force \[w,\] weight of the cube = \[=mg\] Buoyant force, \[{{F}_{b}}\] (or upward thrust) imparted by the liquid on the cube given as \[{{F}_{b}}=Vdg,\] where V = volume of the immersed portion of the cube. For complete immersion, V = volume of the cube. For equilibrium of the cube, \[kx+{{F}_{b}}=mg\] \[\Rightarrow \]            \[x=\frac{mg-{{F}_{b}}}{k}\] \[=\frac{mg-Vdg}{k}\]           where, \[V=\left( \frac{m}{D} \right)\] \[\Rightarrow \]            \[x=\frac{mg}{k}\left( 1-\frac{d}{D} \right)\]