Question

A beaker containing full of 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 ${10}^4{\text{kg/m}}^3$ is immersed in water without touching the walls of the beaker. What will be the final balance reading ?

• A. $2\text{kg}$
• B. $2.5\text{kg}$
• C. $1\text{kg}$
• D. $1.95\text{kg}$

Answer 1

The correct option is D $1.95\text{kg}$

Since, beaker is completely filled with water, so volume of spilled water after immersing of stone will be

$V=\frac{m_{stone}}{{\rho }_{stone}}=\frac{0.5}{{10}^4}=5\times {10}^{-5}{\text{m}}^3$

Mass of spilled water,

$m_{spilled}={\rho }_{w}V={10}^3\times 5\times {10}^{-5}=0.05\text{kg}$

The reading of the spring balance = mass of remaining water + mass of stone

$=(1.5-0.05)+0.5=1.95\text{kg}$

$\therefore W_{reading}=1.95\text{kg}$

Hence, $\left(\text{D}\right)$ is the correct answer.