Question

What is the density of a bowling ball with a mass of 3.0kg and a volume of $0.0050{{m}^{3}}$?

Answer 1

Hint: As it is known to us that the density of a substance or an object is its mass per unit volume. Here we have to calculate the density of the bowling ball using the given information regarding its mass and volume.
Formula used:
We will use the following formula in this solution:-
$\rho =\dfrac{M}{V}$
where,
$\rho$= density of the object
M = mass of the object
V = volume of the object

Complete answer:
Let us discuss about density followed by the calculations as follows:-
“Density of a substance or an object is its mass per unit volume”. Mathematically it can be written as follows:-
$\rho =\dfrac{M}{V}$
where,
$\rho$= density of the object
M = mass of the object
V = volume of the object
The values provided in the question are as follows:-
M = mass of the object = 3.0 kg
V = volume of the object = $0.0050{{m}^{3}}$
-Calculation of density of a bowling ball with a mass of 3.0kg and a volume of$0.0050{{m}^{3}}$:-
As we can see that values are provided in kg and${{m}^{3}}$, so the unit of density will be in $kg/{{m}^{3}}$.
On putting all the values in the formula of density, we get:-
$\begin{align}
  & \Rightarrow \rho =\dfrac{M}{V} \\
 & \Rightarrow \rho =\dfrac{3.0kg}{0.0050{{m}^{3}}} \\
 & \Rightarrow \rho =600kg/{{m}^{3}} \\
\end{align}$
-Therefore the density of the bowling ball is$600kg/{{m}^{3}}$.

Note:
-Remember that the density of a material or an object varies with temperature and pressure and this variation is very small for solids and liquids but much greater for gases. Since increase in the pressure on an object decreases its volume and thus increases its density and increase in the temperature of a substance decreases its density by increment in its volume.