Question

Find the value of volume in appropriate significant figures if the diameter and height of a cylinder are measured by meter scale to be $12.6 \pm 0.1cm$ and $34.2 \pm 0.1cm$ , respectively.
A. $4260 \pm 80c{m^3}$
B. $4300 \pm 80c{m^3}$
C. $4264 \pm 80c{m^3}$
D. $4264 \pm 81c{m^3}$

Answer 1

Hint:Find the volume of cylinder by using the physical quantity of diameter and height of cylinder. Then, calculate the relative error by using the expression:
$\dfrac{{\Delta V}}{V} = \dfrac{{2\Delta d}}{d} + \dfrac{{\Delta h}}{h}$
Put the relative values of diameter and height in the places of $\Delta d$ and $\Delta h$ respectively.

Complete step by step answer:
The relative error can be defined as the ratio of mean absolute error in the measurement of physical quantity to the most probable value. Let the diameter of the cylinder be $d$ and height of cylinder be $h$. Therefore, $\Delta d$ will be the relative value of diameter of cylinder and $\Delta h$ will be the relative height of cylinder.
So, according to the question, it is given that –
$d = 12.6cm$
$\Rightarrow\Delta d = 0.1cm$
$\Rightarrow h = 34.2cm$
$\Rightarrow \Delta h = 0.1cm$
First of we have to calculate the physical quantity of volume of cylinder which is calculated by the formula –
$V = \pi {r^2}h$
where, $r$ is the radius of cylinder and is the two times of the diameter of cylinder
$
d = 2r \\
\Rightarrow r = \dfrac{d}{2} \\ $
Putting the value of height of cylinder in the formula of volume of cylinder –
$
V = \pi {\left( {\dfrac{d}{2}} \right)^2} \times 34.2 \\
\Rightarrow V = \pi \dfrac{{{d^2}}}{4} \times 34.2 \\ $
Putting the value of diameter of cylinder in the formula of volume of cylinder –
$
V = \pi {\left( {12.6} \right)^2} \times 34.2 \\
\Rightarrow V = 4260c{m^3} \\ $
Now, the relative error of the volume can be calculated as –
$\dfrac{{\Delta V}}{V} = \dfrac{{2\Delta d}}{d} + \dfrac{{\Delta h}}{h}$
Putting the values in their respective places –
$
\Delta V = 2 \times \dfrac{{0.1V}}{{12.6}} + \dfrac{{0.1V}}{{34.2}} \\
\Rightarrow \Delta V = \dfrac{{0.2}}{{12.6}} \times 4260 + \dfrac{{0.1 \times 4260}}{{34.2}} \\
\Rightarrow \Delta V = 67.61 + 12.45 \\
\therefore \Delta V \simeq 80c{m^3} $
Therefore, the volume of cylinders in appropriate significant figures can be expressed as $4260 \pm 80c{m^3}$.

Hence, the correct option is A.

Note: Tolerance refers to the greatest range of variation that can be allowed or is acceptable and it is one half of the precision of an instrument. So, to obtain the range, we add and subtract one half of the precision of measuring instruments.