Question

The volume of water in a measuring cylinder is 55 mL When a stone tied to a string is immersed in the water the water level rises to 83 mL. Find the volume of the stone.

A. 28m$l^2$
B. 28ml
C. 38ml
D. 28$m^2$l

Answer 1

Hint: As we know that, total matter enclosed by the closed surface is called volume. It is a three-dimensional quantity. A volume is obtained from sweeping an area in the third dimension of just integrating a differential area over a linear dimension.

 Complete step by step answer:
We were given the data as,

We know that the Initial Volume of water is ${V_i} = 55\;{\rm{mL}}$.
We know that the Final Volume of water is ${V_f} = 83\;{\rm{mL}}$

We can conclude that therefore, the volume V of the displaced water is,

$V = {V_f} - {V_i}$

We can substitute ${V_f} = 83\;{\rm{mL}}$, ${V_i} = 55\;{\rm{mL}}$ to find the value of V.

$\begin{array}{l}
V = 83\;{\rm{mL}} - 55\;{\rm{mL}}\\
V = 28\;{\rm{mL}}
\end{array}$

Hence the correct option is B that is the displaced volume which is the volume of the stone that is equal to 28mL.


Additional Information:
When a stone is dipped in water, the stone experiences weightlessness due to the force which is buoyant in nature acts on it. The force acts in an upward direction which lowers the net weight force of the dropped stone. Firstly, while traversing downwards, the velocity of the stone increases then it becomes steady and constant. We call this velocity as the terminal velocity which is always constant because the net acceleration acting on it becomes zero. And we all know that zero acceleration equals constant velocity.

Note:
When an object is dipped inside the volume of water in a bucket, the object is displaced by the object. So to compensate for that displaced volume of the object the water level rises in the bucket and the water level increases in the third dimension.