Question

If the percentage error in measuring the surface area of a sphere is \[\alpha \% \] then the percentage error in volume is
A) \[\dfrac{3}{2}\alpha \% \]
B) \[\dfrac{2}{3}\alpha \% \]
C) \[3\alpha \% \]
D) None of these

Answer 1

Hint: We will use the formula of the surface area of a sphere and then differentiate it in order to get the percentage error in the radius of the sphere and then put that value in the formula of volume of the sphere.

Complete step by step answer:
Let the surface area of the sphere be S
Then, it is given that :
\[\dfrac{{\Delta S}}{S} \times 100 = \alpha \]
where \[\Delta S\]is the change in surface area and S is the original surface area of the sphere.
Also, we know that the surface area of a sphere is given by
\[S = 4\pi {r^2}\] where r is the radius of the sphere
Now differentiating both sides and dividing by S we get:
\[
\Rightarrow \dfrac{{\Delta S}}{S} = \dfrac{{4\pi \left( {2r} \right)\Delta r}}{S} \\
\Rightarrow \dfrac{{\Delta S}}{S} = \dfrac{{8\pi r\Delta r}}{S} \\
 \]
where \[\Delta r\]is the change in radius of the sphere.
Now putting the value of S on the right-hand side we get
\[
\Rightarrow \dfrac{{\Delta S}}{S} = \dfrac{{8\pi r\Delta r}}{{4\pi {r^2}}} \\
\Rightarrow \dfrac{{\Delta S}}{S} = \dfrac{{2\Delta r}}{r} \\
 \]
Now multiplying both the sides with 100 to get the error percentage:
\[
\Rightarrow \dfrac{{\Delta S}}{S} \times 100 = \dfrac{{2\Delta r}}{r} \times 100 \\
  \Rightarrow \alpha = \dfrac{{2\Delta r}}{r} \times 100 \\
 \Rightarrow \dfrac{{\Delta r}}{r} = \dfrac{\alpha }{2} \times 100 \\
 \]
Hence error percentage of radius is \[\dfrac{\alpha }{2}\]
Now we know that the volume of the sphere is given by:
\[V = \dfrac{4}{3}\pi {r^3}\]
Differentiating both the sides of the above equation and then dividing it by V we get:
\[
\Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{{\dfrac{4}{3}\pi \left( {3{r^2}} \right)\Delta r}}{V} \\
\Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{{4\pi {r^2}\Delta r}}{V} \\
 \]
Where \[\Delta V\] is the change in volume and V is the original volume of the sphere.
Now putting the value of V on the right-hand side of the above equation we get:
\[
\Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{{4\pi {r^2}\Delta r}}{{\dfrac{4}{3}\pi {r^3}}} \\
 \Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{{3\Delta r}}{r} \\
 \]
Multiplying both the sides with 100 to get the percentage error:
\[
\Rightarrow \dfrac{{\Delta V}}{V} \times 100 = \dfrac{{3\Delta r}}{r} \times 100 \\
\Rightarrow \dfrac{{\Delta V}}{V} \times 100 = \dfrac{{3\alpha }}{2} \\
 \]

Therefore the percentage error of volume of sphere is \[\dfrac{{3\alpha }}{2}\]. Hence, option A is correct.

Note:
The surface area is the area that describes the material that will be used to cover a solid geometrical figure such as a sphere. It is measured in square units.
Volume is the measure of quantity a geometrical figure can hold inside it and it is measured in cubic units.