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Question: The radius of the sphere is \[\left( {4.6 \pm 0.1} \right)cm\]. The percentage error in its volume i...

The radius of the sphere is (4.6±0.1)cm\left( {4.6 \pm 0.1} \right)cm. The percentage error in its volume is:
A. 0.14.6×100%\dfrac{{0.1}}{{4.6}} \times 100\%
B. 3×0.14.6×100%3 \times \dfrac{{0.1}}{{4.6}} \times 100\%
C. (0.14.6×100)3%{\left( {\dfrac{{0.1}}{{4.6}} \times 100} \right)^3}\%
D. 13×0.14.6×100%\dfrac{1}{3} \times \dfrac{{0.1}}{{4.6}} \times 100\%

Explanation

Solution

Complete step by step answer:
Error is the difference that we find when the difference between the actual(true) value and measured(observed) value of a given physical quantity is called an error.
Error= actual(true) value- measured(observed) value.
Under the errors there are 2 types of errors:

  1. systematic error and
  2. random errors
    Percentage error is given by
    Percentage error= Relative error ×100%\times 100\%
    δa%=Δaˉaˉ×100%\delta a\% = \dfrac{{\Delta \bar a}}{{\bar a}} \times 100\%
    Where Δaˉ=\Delta \bar a = mean absolute error and aˉ=\bar a = arithmetic mean.
    The percentage error again has many types :
    -Error in addition
    -Error is subtraction
    -Error in division
    -And Error in multiplication.

In the above mentioned question radius has been given and volume is to be found, hence we use error in multiplication. Given: Radius(r) =(4.6±0.1)cm\left( {4.6 \pm 0.1} \right)cm
V=43πr3V = \dfrac{4}{3}\pi {r^3}
V=43×π×(4.6)3=406.4cm3\Rightarrow V = \dfrac{4}{3} \times \pi \times {\left( {4.6} \right)^3} = 406.4c{m^3}
Now in error in multiplication :if X and Y are 2 individual quantities to be multiplied and is the resultant product like Z=XY, then the maximum possible error is given by
ΔZZ=ΔXX+ΔYY \dfrac{{\Delta Z}}{Z} = \dfrac{{\Delta X}}{X} + \dfrac{{\Delta Y}}{Y} \\\
And percentage error is given by:
ΔZZ×100%=ΔXX×100%+ΔYY×100% \dfrac{{\Delta Z}}{Z} \times 100\% = \dfrac{{\Delta X}}{X} \times 100\% + \dfrac{{\Delta Y}}{Y} \times 100\% \\\
Where ΔZZ,ΔXXandΔYY\dfrac{{\Delta Z}}{Z},\dfrac{{\Delta X}}{X}and\dfrac{{\Delta Y}}{Y} are relative errors in X,Y and Z.
Hence applying the above way we get:
ΔVV=3(Δrr)×100% \dfrac{{\Delta V}}{V} = 3\left( {\dfrac{{\Delta r}}{r}} \right) \times 100\% \\\
ΔVV=3×(0.14.6)×100%\therefore \dfrac{{\Delta V}}{V} = 3 \times \left( {\dfrac{{0.1}}{{4.6}}} \right) \times 100\%

Hence the correct option is option B.

Note: Percentage error should always be mentioned in %. We should not leave it, otherwise it is relative error.it can also be written as ΔV=V×3×(0.14.6)×100%\Delta V = V \times 3 \times \left( {\dfrac{{0.1}}{{4.6}}} \right) \times 100\% but we have written as required in the options. Also remember that when physical quantities are multiplied the maximum error in the resultant physical quantity is equal to the sum of percentage errors of individual quantities.