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 ${10^4}\,{\text{kg/}}{{\text{m}}^3}$ is immersed in water without touching the walls of the beaker. What will be the balance reading now?
A. $2\,{\text{kg}}$
B. $2.5\,{\text{kg}}$
C. $1\,{\text{kg}}$
D. $3\,{\text{kg}}$

Answer 1

Use the formula for the upward thrust acting on an object immersed in the liquid. This formula gives the relation between the density of liquid, volume of the object and acceleration due to gravity. The new reading of the balance is the sum of the mass of the beaker with water and the upward thrust force acting on the stone.

Formulae used:
The upward thrust force $U$ acting on an object immersed in the liquid is given by
$U = \rho Vg$ …… (1)
Here, $\rho$ is density of the liquid, $V$ is volume of the object immersed in the liquid and $g$ is acceleration due to gravity.
The density $\rho$ of an object is given by
$\rho = \dfrac{M}{V}$ …… (2)
Here, $M$ is the mass of the object and $V$ is the volume of the object.

Complete step by step answer:
We have given that the mass of the water with the beaker is $1.5\,{\text{kg}}$.
${m_{Water}} = 1.5\,{\text{kg}}$
We have also given that the mass of the stone inserted in the beaker is $0.5\,{\text{kg}}$ and its density is ${10^4}\,{\text{kg/}}{{\text{m}}^3}$.
${m_{Stone}} = 0.5\,{\text{kg}}$
$\Rightarrow{\rho _{Stone}} = {10^4}\,{\text{kg/}}{{\text{m}}^3}$
We are asked to calculate the new weight of the beaker with the water and the stone.

The new mass $m$ of the beaker is the sum of the weight ${m_{Water}}$ of the beaker with the water and the upward thrust $U$ acting on the beaker due to the stone.
$m = {m_{Water}} + U$ …… (3)
The volume of the stone according to equation (2) is given by
$V = \dfrac{{{\rho _{Stone}}}}{{{m_{Stone}}}}$
Substitute ${10^4}\,{\text{kg/}}{{\text{m}}^3}$ for ${\rho _{Stone}}$ and $0.5\,{\text{kg}}$ for ${m_{Stone}}$ in the above equation.
$V = \dfrac{{0.5\,{\text{kg}}}}{{{{10}^4}\,{\text{kg/}}{{\text{m}}^3}}}$
$\Rightarrow V = 5 \times {10^{ - 5}}\,{{\text{m}}^3}$
Hence, the volume of the stone is $5 \times {10^{ - 5}}\,{{\text{m}}^3}$.

The density of the water is ${10^3}\,{\text{kg/}}{{\text{m}}^3}$.
${\rho _{Water}} = {10^3}\,{\text{kg/}}{{\text{m}}^3}$
Rewrite equation (1) for the upward thrust.
$U = {\rho _{Water}}{V_{Stone}}g$
Substitute ${10^3}\,{\text{kg/}}{{\text{m}}^3}$ for ${\rho _{Water}}$, $5 \times {10^{ - 5}}\,{{\text{m}}^3}$ for ${V_{Stone}}$ and $10\,{\text{m/}}{{\text{s}}^2}$ for $g$ in the above equation.
$U = \left( {{{10}^3}\,{\text{kg/}}{{\text{m}}^3}} \right)\left( {5 \times {{10}^{ - 5}}\,{{\text{m}}^3}} \right)\left( {10\,{\text{m/}}{{\text{s}}^2}} \right)$
$\Rightarrow U = 0.5\,{\text{kg}}$
Substitute $1.5\,{\text{kg}}$ for ${m_{Water}}$ and $0.5\,{\text{kg}}$ for $U$ in equation (3).
$m = \left( {1.5\,{\text{kg}}} \right) + \left( {0.5\,{\text{kg}}} \right)$
$\therefore m = 2\,{\text{kg}}$
Therefore, the new reading of the balance is $2\,{\text{kg}}$.

Hence, the correct option is A.

Note: The students may think that the upward thrust force on the stone is acting on the upward direction then how can we add it in the weight of the system. But the students should keep in mind that the upward thrust acting on the stone in the water exerts an equal downward force on the balance. Hence, this upward force adds in the mass of the system measured by the balance.