Question

A metal piece of mass `10 g` is suspended by a vertical spring. The spring elongates `10 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 metal `= 9000 kg//m^(3)`. Take `g = 10 m//s^(2)`.

Answer 1

Let the spring constant be `k`. When the piece is hanging in air, the equilibrium condition gives
`k(10 cm) = (0.01 kg) (10 m//s^(2))`
or `k (10 cm) = 0.1 N`. ………..(i)
The volume of the metal piece
`= (0.01kg)/(9000kg//m^(3)) = (1)/(9) xx 10^(-5)m^(3)`.
This is also the volume of water displaced when the piece is immersed in water. The force of buoyancy
`=` weight of the lilquid displace
`=(1)/(9) xx 10^(-5)m^(3) xx (1000 kg//m^(3)) xx (10 m//s^(2))`
`= 0.011 N`.
If the elongation of the spring is `x` when the piece is immersed in water, the equilibrium condition of the piece gives,
`kx = 0.1 N - 0.011 N = 0.089 N`. ..........(ii)
By `(i)` and `(ii), x = (0.089)/(10) cm = 0.0089 cm`.