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 froces as (1) Spring force \[{{K}_{x,}}\] where X = elongation of the spring. (2) Gravitational force \[{{F}_{b}}\] (or upward thrust) imparted. (3) Byoliant 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             \[{{K}_{x}}+{{F}_{b}}=mg\] \[\Rightarrow \]               \[X=\frac{mg-{{F}_{b}}}{R}=\frac{mg-Vdg}{K}\] Where,      \[V=(m/D)\]                 \[X=\frac{mg}{K}\,\,\left( 1-\frac{d}{D} \right)\]