If the percentage error in measuring the surface area of a sphere is $α %,$ then the error in its volume, is

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$\frac{3}{2} α %$

$\frac{2}{3} α %$

$3 α %$

none of these

answer is A.

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Detailed Solution

Let $r$ be the radius, $S$ the surface area and $V$ the
volume of the sphere. Then,

$S = 4 π r^{2}$ and $V = \frac{4}{3} π r^{3}$

Let $∆ r$ , $∆ S$ and $∆ V$ be the errors in $r, S$ and $V$ respectively. Then,

$\begin{array}{l} Δ S = \frac{d S}{d r} Δ r and Δ V = \frac{d V}{d r} Δ r \\ \Rightarrow Δ S = 8 π r Δ r and Δ V = 4 π r^{2} Δ r \\ \Rightarrow \frac{Δ S}{S} = \frac{8 π r}{4 π r^{2}} Δ r and \frac{Δ V}{V} = \frac{4 π r^{2}}{4} \frac{Δ r}{3} π r^{3} \\ \Rightarrow \frac{Δ S}{S} \times 100 = 2 (\frac{Δ r}{r} \times 100) and \frac{Δ V}{V} \times 100 = 3 (\frac{Δ r}{r} \times 100) \\ \Rightarrow α = 2 (\frac{Δ r}{r} \times 100) and \frac{Δ V}{V} \times 100 = 3 (\frac{Δ r}{r} \times 100) \\ [∵ \frac{Δ S}{S} \times 100 = α (given)] \\ \Rightarrow \frac{Δ V}{V} \times 100 = 3 (\frac{α}{2}) = \frac{3}{2} α \end{array}$