Question

A beaker containing water is placed on the platform of a spring balance. The balance reads $1.5\text{kg}$. A stone of mass $0.5\text{kg}$ and density $500{\text{kg/m}}^3$ is immersed in water without touching the walls of the beaker. What will be the reading of spring balance now?

• A. $2\text{kg}$
• B. $2.5\text{kg}$
• C. $1\text{kg}$
• D. $3\text{kg}$

Answer 1

The correct option is B $2.5\text{kg}$

Volume of stone $=\frac{m}{\rho }=\frac{0.5}{500}={10}^{-3}{\text{m}}^3$
Force of Buoyancy $=\rho Vg=1000\times {10}^{-3}\times 10=10\text{N}$

There will be a reaction force to the buoyant force, by the stone on the water.
Therefore, net force on spring balance is sum of force of buoyancy and weight of water and beaker $=F_B+mg=10+1.5g=10+15=25N$
Net reading of spring balance$=2.5\text{kg}$