The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is x100%. If the percentage errors in measuring the mass and the diameter are 6.0% and 1.5% respectively, the value of ‘x’ is………….

ρ=m43πr3=6mπd3⇒Δρρ×100=Δmm×100+3Δdd×100=6+3×1.5=10.5%=1050100%

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The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is $(\frac{x}{100}) %$ . If the percentage errors in measuring the mass and the diameter are 6.0% and 1.5% respectively, the value of ‘x’ is………….

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answer is 1050.

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Detailed Solution

$\begin{array}{l} ρ = \frac{m}{\frac{4}{3} π r^{3}} = \frac{6 m}{π d^{3}} \\ \Rightarrow \frac{Δ ρ}{ρ} \times 100 = \frac{Δ m}{m} \times 100 + 3 \frac{Δ d}{d} \times 100 = 6 + 3 \times 1.5 = 10.5 % = \frac{1050}{100} % \end{array}$

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