Question

An irregular shaped body of mass 250g is put in an Eureka can and water displaced is collected in a measuring cylinder in which water is already at mark 20$c{m^3}$. Now the water level rises to 60$c{m^3}$. Calculate
(a) Volume of body
(b) Density of body in CGS and SI units

Answer 1

Hint: Remember the concept of Archimedes principle and identify the given values to use the formula of Archimedes principle .i.e. water rises = the volume of the body to find the volume of the body and then using the density formula find the density of the body.

Complete Step-by-Step solution:
According to question body of mass 250g is dropped in eureka can and the water level in measuring cylinder rises from 20$c{m^3}$ to 60$c{m^3}$
Thus by the Archimedes principle the water rises = the volume of the bodyṣ
Therefore V = 60 – 20 = 40$c{m^3}$
Hence, Volume of body = 40$c{m^3}$
Now for density of the body \[\rho = \dfrac{m}{V}\]
$ \Rightarrow $$\dfrac{{250}}{{40}}$= 6.25
Also \[\rho = 6.25 \times 1000\]= 6250 \[kg/{m^3}\]

Note: In the above solution I have mentioned the basic principle named as Archimedes principle which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid that is displaced by that object. A resting body in a fluid is accomplished by a force pushing upwards called the buoyant force, which is equivalent to the mass of the fluid the body displaces.When the body is fully submerged, the volume of the displaced liquid becomes equivalent to the body volume. When the body is only partly submerged, the volume of the displaced fluid shall be equal to the volume of the submerged portion of the body. The theory of Archimedes is very useful in measuring an object's volume which does not have a normal shape. It can also be used to measure an object's density or specific gravity.