Question

If the length of the cylinder is measured as $25mm$, the diameter is $3.09cm$ and the mass of the cylinder is measured as $50.0gm$. Find the density of the cylinder in proper significant figures.
A) $2.700gm/c{m^3}$
B) $2.7gm/c{m^3}$
C) $0.27gm/c{m^3}$
D) $2.70gm/c{m^3}$

Answer 1

Density is defined as mass per unit of the volume. In this question, mass is directly given to us but volume is not given directly so we need to calculate it first. Significant figures of a mathematical operation are determined by the least precise quantity involved.

Formula used:
$\rho = \dfrac{M}{V}$

Complete step by step solution:
The density of an object is defined as
$\rho = \dfrac{M}{V}$, where $\rho$=density of the object, M=mass of an object, and V= volume of the object
First, let’s calculate the volume of the cylinder which we know is given by
$V = \pi {R^2}L$, where V=volume of the cylinder, R=radius of the cylinder, and L=length of the cylinder.

So, Length $\left( L \right) = 25mm$ but we have to convert it to centimeters. As we know that
$1mm = {10^{ - 1}}cm$
Hence we will get
Length $\left( L \right) = 25 \times {10^{ - 1}}cm$
Now we will find the value of the radius of the cylinder.
As we know that
$Radius(R) = \dfrac{{Diameter}}{2}$
And it is given that
$Diameter = 3.09cm$
$\Rightarrow Radius = \dfrac{{3.09}}{2}$
$\Rightarrow Radius(R) = 1.54cm$
Now we will substitute these values in the above the mentioned formula. Hence
$V = \pi \times {\left( {1.54} \right)^2} \times 2.5c{m^3}$
$\Rightarrow V = 18c{m^3}$

It has only two significant digits because the length of the cylinder was least precise and had only 2 significant digits. The mass given is 50 gm so now we will substitute the values of mass and volume in the formula of density. Hence,
$\rho = \dfrac{{50}}{{18}} gm/c{m^3}$

$\Rightarrow \rho = 2.7gm/c{m^3}$

Note:
1. Source of error: Always convert all physical quantities to the desired units then proceed with calculations.
2. Significant figures provide a proper solution to deal with the problem that the resulting quantity cannot be more precise than the ones from which it is calculated.