Question

A beaker containing water is placed on the platform of a spring balance. The platform-balance reads $1.5\text{kg}$. A stone of mass $0.5\text{kg}$ and density ${10}^4{\text{kg/m}}^3$ is immersed in water without touching the walls of the beaker. What will the platform-balance read now?

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

Answer 1

The correct option is A $2\text{kg}$

Solution: Buoyant force acts on the body and the body will exert an equal force on the floor of the beaker.

Hence, the reading of the platform balance will be equal to the sum of the weights of the stone and the liquid.

i.e. $W=W_s+W_l=1.5+0.5=2.0\text{kg}$

Also, the weight is exerted on the beaker. Therefore, the reading will be $1.5+5=2\text{kg}$

Hence, $\left(A\right)$ is the correct answer.

Why this Q ? Buoyant force is equal to the weight of the fluid displaced by the body. Reaction of this force acts on the fluid, displaced by the body and in the opposite direction.