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 $m/s^2$.

• A. $0.0089cm$
• B. $0.0059cm$
• C. $0.89cm$
• D. $0.59cm$

Answer 1

The correct option is A $0.0089cm$

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
$=\frac{0.01kg}{9000kg/m^3}=\frac{1}{9}\times {10}^{-5}{m}^3$
This is also the volume of water displaced when the piece is immersed in water. The force of buoyancy
= weight of the liquid displaced
= $\frac{1}{9}\times {10}^{-5}{m}^3\times \left(1000kg/m^3\right)\times \left(10m/s^2\right)$
= 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), $x=\frac{0.089}{10}$ cm = 0.0089 cm.