A cube of mass m and densituy 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, where D gt d. What is the elongation in the spring ?

Answer

`(mg)/(k)( 1 -(d)/(D))`

Explanation

Refer to

the cube is in equilibrium undeer the effect of following three forces (i) Spring force = kx, acting upwards, where x is the extesnion of the spring. (ii) Gravitational force = mg, acting downwards. (iii) Bouyent force `F_B` = upward thrust on cube due to liquid = Vdg Where V is the volume of the cube completely immersed in liquid. Therefore, for equilibrium of the cube `kx + F_B = mg`
`or x = (mg - F_B)/(k) =(mg - V dg)/(k)`
` =(mg - (m//D)dg)/(k) = (mg)/(k) [(1 - (d)/(D))]`