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Question: The radius of a sphere is measured to be \(5.3 \pm 0.1cm\). Calculate the percentage error in the me...

The radius of a sphere is measured to be 5.3±0.1cm5.3 \pm 0.1cm. Calculate the percentage error in the measure of its volume.
(A) 3.45%3.45\%
(B) 5.66%5.66\%
(C) 2.4%2.4\%
(D) 1.05%1.05\%

Explanation

Solution

The volume of a sphere can be found out from the formula, V=43πr3V = \dfrac{4}{3}\pi {r^3}. We can calculate the error in the calculation of VV from the small change in the value of rr 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
V=43πr3\Rightarrow V = \dfrac{4}{3}\pi {r^3}3
where VV is the volume of the sphere, rr is the radius of the sphere.
And ΔVV=3(Δrr)\dfrac{{\Delta V}}{V} = 3\left( {\dfrac{{\Delta r}}{r}} \right) for the calculation of error, where Δr\Delta r is the smallest change in radius and ΔV\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=43πr3V = \dfrac{4}{3}\pi {r^3}.
In the question, we are given the radius of the sphere as, r=5.3cmr = 5.3cm. So by substituting that value in the formula, we get the volume as,
V=43×227×(5.3)3\Rightarrow V = \dfrac{4}{3} \times \dfrac{{22}}{7} \times {\left( {5.3} \right)^3}
where the approximate value of π=227\pi = \dfrac{{22}}{7}.
On doing the above calculation, we get
V=623.86cm3\Rightarrow V = 623.86c{m^3}
Now, for the error in the calculation of the volume is given by ΔVV\dfrac{{\Delta V}}{V} and from the formula of volume we have,
ΔVV=3(Δrr)\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,
ΔVV×100%=3(Δrr)×100%\Rightarrow \dfrac{{\Delta V}}{V} \times 100\% = 3\left( {\dfrac{{\Delta r}}{r}} \right) \times 100\%
Now from the question we have, Δr=0.1cm\Delta r = 0.1cm
So substituting the value we get,
ΔVV×100%=3(0.15.3)×100%\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,
ΔVV×100%=3×0.0188×100%\Rightarrow \dfrac{{\Delta V}}{V} \times 100\% = 3 \times 0.0188 \times 100\%
ΔVV×100%=0.0566×100%\therefore \dfrac{{\Delta V}}{V} \times 100\% = 0.0566 \times 100\%
Therefore the value of the percent error in volume is ΔVV×100%=5.66%\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.