The percentage error in measuring the radius of a sphere is $1 % .$ If $S_{1} =$ percentage error in volume of the sphere, $S_{2} =$ percentage error
in surface area of the sphere, $S_{3} =$ relative error in the diameter of the sphere then arrange $S_{1} S_{2} S_{3}$ according to the values in ascending order

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$S_{3}, S_{2}, S_{1}$

$S_{1}, S_{2}, S_{3}$

$S_{2}, S_{1}, S_{3}$

$S_{3}, S_{1}, S_{2}$

answer is A.

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Detailed Solution

$\begin{array}{rcl} \frac{δ r}{r} \times 100 & = & 1 \\ \begin{matrix} v \frac{4}{3} {πr}^{3} \\ \log v = \log \frac{4}{3} π + 3 \log r \\ \frac{1}{v} δ v \times 100 = \frac{3}{4} δ r \times 100 \\ S_{1} = 3 \end{matrix} \begin{matrix} S = 4 {πr}^{2} \\ \log S = \log 4 π + 2 \log r \\ \frac{1}{S} δ s \times 100 = \frac{2}{r} δ r \times 100 \\ S_{2} = 2 \\ S_{3} = \frac{δ r}{r} = \frac{1}{100} \end{matrix} = \frac{1}{100} \\ S_{3} & < & S_{2} < S_{1} \end{array}$