The radius of a sphere is measured to be (7.50±0.85)cm . Suppose the percentage error in its volume is x. The value of x, to the nearest x, is_____

Volume of sphere, V=43πR3 ΔVV=3Δrr ΔVV×100=3Δrr×100 ΔVV×100=34

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The radius of a sphere is measured to be $(7.50 \pm 0.85) c m$ . Suppose the percentage error in its volume is x. The value of x, to the nearest x, is_____

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answer is 34.

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Detailed Solution

Volume of sphere, $V = \frac{4}{3} π R^{3}$
$\frac{Δ V}{V} = \frac{3 Δ r}{r} \frac{Δ V}{V} \times 100 = \frac{3 Δ r}{r} \times 100 \frac{Δ V}{V} \times 100 = 34$

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