Question

A beaker containing water with a total mass of $10kg$ is placed on the pan of a balance A . A solid body of mass $5kg$ and density $5g/cc$ suspended from a spring balance B is gently lowered in the water contained in the beaker. So that it gets fully immersed without any contact with the beaker. Find the ratio of readings shown by the balance A and B.
A. $\dfrac{{11}}{4}$
B. $\dfrac{7}{4}$
C. $\dfrac{9}{4}$
D. None of these

Answer 1

Hint:We know that the balance readings are ultimately the value of force acting on that balances .In these forces, we also need to include the weight of the solid body and beaker itself. Thus, we need to find the total force on both the balances first and then take their ratio.

Complete step by step answer:
Here we will first find the reading shown by balance A.We know that balance A shows the total of the weight of the water in the beaker and the upthrust force.We know that there is $10kg$ of water in the beaker. Thus the weight of water is $10g = 10 \times 10 = 100N$, where $g$ is the gravitational acceleration.

Now, we will find the upthrust force which is given by $U = V{\rho _w}g$,
Where, $V$ is the volume of the water replaced by the solid body, ${\rho _w}$ is the density of water and $g$ is the gravitational acceleration
In the question we are given that solid body is of mass $5kg$ and density $5g/cc$
$5g/cc = 5 \times {10^{ - 3}}kg/{m^3}$
$V = \dfrac{{5 \times {{10}^{ - 3}}}}{5} = {10^{ - 3}}{m^3}$
We know that density of water ${\rho _w} = {10^3}kg/{m^3}$ and $g = 10m/{s^2}$
$ \Rightarrow U = V{\rho _w}g = {10^{ - 3}} \times {10^3} \times 10 = 10N$
Thus, the reading shown by balance A is: $100 + 10 = 110N$
Now, If we consider the solid body, it experiences two forces in upward direction: the tensile force and upthrust force and its weight in the downward direction.
Therefore,
$
T + U = 5g \\
\Rightarrow T = \left( {5 \times 10} \right) - 10 \\
\Rightarrow T = 50 - 10 \\
\Rightarrow T = 40N \\ $
Thus, the reading shown by balance B is same as the tension which is $40N$.The ratio of readings shown by balance A and B is :
$\dfrac{{110}}{{40}} = \dfrac{{11}}{4}$
$\therefore\dfrac{{110}}{{40}} = \dfrac{{11}}{4}$

Hence, option A is the right answer.

Note:In this problem, one important thing we have considered is the upthrust force. This force can be defined as the upward force exerted by water when a body is immersed in it. The body will also exert an equal and opposite force downward on the water because every action has an equal and opposite reaction according to the third law of Newton.