MCQ
The relative error in the determination of the surface area of a sphere is $\alpha $. Then the relative error in the determination of its volume is
- A$\frac{2}{3}\alpha $
- B$\frac{5}{2}\alpha $
- ✓$\frac{3}{2}\alpha $
- D$\alpha $
✓
Answer
Correct option: C.
$\frac{3}{2}\alpha $
c
$\begin{array}{l}
{\mathop{\rm Re}\nolimits} lative\,\,eroor\,\,in\,\,surface\,\,area,\\
\frac{{\Delta s}}{s} = 2 \times \frac{{\Delta r}}{r} = \alpha \,and\,relative\,error\,in\\
volume,\,\frac{{\Delta v}}{v} = 3 \times \frac{{\Delta r}}{r}\\
\therefore \,{\mathop{\rm Re}\nolimits} lative\,\,error\,in\,volume\,w.r.t\,relative\,error\,in\,area,\\
\frac{{\Delta v}}{v} = \frac{3}{2}\alpha
\end{array}$
$\begin{array}{l}
{\mathop{\rm Re}\nolimits} lative\,\,eroor\,\,in\,\,surface\,\,area,\\
\frac{{\Delta s}}{s} = 2 \times \frac{{\Delta r}}{r} = \alpha \,and\,relative\,error\,in\\
volume,\,\frac{{\Delta v}}{v} = 3 \times \frac{{\Delta r}}{r}\\
\therefore \,{\mathop{\rm Re}\nolimits} lative\,\,error\,in\,volume\,w.r.t\,relative\,error\,in\,area,\\
\frac{{\Delta v}}{v} = \frac{3}{2}\alpha
\end{array}$
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