
The radius of a sphere is measured to be $5.3 \pm 0.1cm$. Calculate the percentage error in the measure of its volume.
(A) $3.45\% $
(B) $5.66\% $
(C) $2.4\% $
(D) $1.05\% $
Answer
565.5k+ views
Hint:
The volume of a sphere can be found out from the formula, $V = \dfrac{4}{3}\pi {r^3}$. We can calculate the error in the calculation of $V$ from the small change in the value of $r$ as given in the question. Then by multiplying 100% with the answer we can get the percentage error in our answers.
In the given question, we use the formula
$\Rightarrow V = \dfrac{4}{3}\pi {r^3}$3
where $V$ is the volume of the sphere, $r$ is the radius of the sphere.
And $\dfrac{{\Delta V}}{V} = 3\left( {\dfrac{{\Delta r}}{r}} \right)$ for the calculation of error, where $\Delta r$ is the smallest change in radius and $\Delta V$ gives us the volume error.
Complete step by step answer:
For the solution, we need to find the volume of a sphere. The volume of a sphere is given by the formula $V = \dfrac{4}{3}\pi {r^3}$.
In the question, we are given the radius of the sphere as, $r = 5.3cm$. So by substituting that value in the formula, we get the volume as,
$\Rightarrow V = \dfrac{4}{3} \times \dfrac{{22}}{7} \times {\left( {5.3} \right)^3}$
where the approximate value of $\pi = \dfrac{{22}}{7}$.
On doing the above calculation, we get
$\Rightarrow V = 623.86c{m^3}$
Now, for the error in the calculation of the volume is given by $\dfrac{{\Delta V}}{V}$ and from the formula of volume we have,
$\Rightarrow \dfrac{{\Delta V}}{V} = 3\left( {\dfrac{{\Delta r}}{r}} \right)$
So for the percentage of error, we multiply 100% on both sides of the equation.
Therefore we have,
$\Rightarrow \dfrac{{\Delta V}}{V} \times 100\% = 3\left( {\dfrac{{\Delta r}}{r}} \right) \times 100\% $
Now from the question we have, $\Delta r = 0.1cm$
So substituting the value we get,
$\Rightarrow \dfrac{{\Delta V}}{V} \times 100\% = 3\left( {\dfrac{{0.1}}{{5.3}}} \right) \times 100\% $
On doing the calculation in the R.H.S we get,
$\Rightarrow \dfrac{{\Delta V}}{V} \times 100\% = 3 \times 0.0188 \times 100\% $
$\therefore \dfrac{{\Delta V}}{V} \times 100\% = 0.0566 \times 100\% $
Therefore the value of the percent error in volume is $\dfrac{{\Delta V}}{V} \times 100\% = 5.66\% $
So the correct option will be (B).
Note:
In any formula, the most probable error is given by the sum of the errors in all the variables. These errors may be caused due to a lot of factors like errors in measurement of the individual variables due to problems in an instrument or taking a reading incorrectly, physical conditions like temperature, and various other factors.
The volume of a sphere can be found out from the formula, $V = \dfrac{4}{3}\pi {r^3}$. We can calculate the error in the calculation of $V$ from the small change in the value of $r$ as given in the question. Then by multiplying 100% with the answer we can get the percentage error in our answers.
In the given question, we use the formula
$\Rightarrow V = \dfrac{4}{3}\pi {r^3}$3
where $V$ is the volume of the sphere, $r$ is the radius of the sphere.
And $\dfrac{{\Delta V}}{V} = 3\left( {\dfrac{{\Delta r}}{r}} \right)$ for the calculation of error, where $\Delta r$ is the smallest change in radius and $\Delta V$ gives us the volume error.
Complete step by step answer:
For the solution, we need to find the volume of a sphere. The volume of a sphere is given by the formula $V = \dfrac{4}{3}\pi {r^3}$.
In the question, we are given the radius of the sphere as, $r = 5.3cm$. So by substituting that value in the formula, we get the volume as,
$\Rightarrow V = \dfrac{4}{3} \times \dfrac{{22}}{7} \times {\left( {5.3} \right)^3}$
where the approximate value of $\pi = \dfrac{{22}}{7}$.
On doing the above calculation, we get
$\Rightarrow V = 623.86c{m^3}$
Now, for the error in the calculation of the volume is given by $\dfrac{{\Delta V}}{V}$ and from the formula of volume we have,
$\Rightarrow \dfrac{{\Delta V}}{V} = 3\left( {\dfrac{{\Delta r}}{r}} \right)$
So for the percentage of error, we multiply 100% on both sides of the equation.
Therefore we have,
$\Rightarrow \dfrac{{\Delta V}}{V} \times 100\% = 3\left( {\dfrac{{\Delta r}}{r}} \right) \times 100\% $
Now from the question we have, $\Delta r = 0.1cm$
So substituting the value we get,
$\Rightarrow \dfrac{{\Delta V}}{V} \times 100\% = 3\left( {\dfrac{{0.1}}{{5.3}}} \right) \times 100\% $
On doing the calculation in the R.H.S we get,
$\Rightarrow \dfrac{{\Delta V}}{V} \times 100\% = 3 \times 0.0188 \times 100\% $
$\therefore \dfrac{{\Delta V}}{V} \times 100\% = 0.0566 \times 100\% $
Therefore the value of the percent error in volume is $\dfrac{{\Delta V}}{V} \times 100\% = 5.66\% $
So the correct option will be (B).
Note:
In any formula, the most probable error is given by the sum of the errors in all the variables. These errors may be caused due to a lot of factors like errors in measurement of the individual variables due to problems in an instrument or taking a reading incorrectly, physical conditions like temperature, and various other factors.
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