A copper piece of mass 10 g is suspended by a vertical spring. The spring elongates 1 cm over its natural length to keep the piece in equilibrium. A beaker containing water is now placed below the piece so as to immerse the piece completely in water. Find the elongation of the spring. Density of copper =9000 kgm^-3. `Take g=10ms^-2`

Explanation

Let the spring constant be k. When the piece is hanging in air the equilibrium condition gives
`k(1cm)=(0.01kg)(10ms^-2)`
`or k(1cm)=0.1N` …………..i
The volume of the copper piece
`=(0.01kg)/(9000kgm^-3)=1/9xx10^-5m^3`
This also the volume of water displaced when the piece is immersed in water. The force of buoyancy
`=weight of the liquid displaced
`=1/9xx10^-5m^3x(100kgm^-3)xx(10ms^-2)`
`=0.011N.
If the elongation of the spring is x when the piece is immersed in wter the equilibrium condition of the piece gives,
`=kx=0.1N-0.011N=0.089N.`........ii
By i and ii
`=x=0.089/0.1cm=0.89cm`