A metal wire has mass <math> <mo>(</mo> <mn>0.4</mn> <mo>±</mo> <mn>0.002</mn> <mo>)</mo> <mi mathvariant="normal">g</mi> </math> <math></math> , radius <math> <mo>(</mo> <mn>0.3</mn> <mo>±</mo> <mn>0.001</mn> <mo>)</mo> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">m</mi> </math> <math><msub><mrow><mrow></mrow></mrow><mrow><mrow></mrow></mrow></msub></math> and length <math> <mo>(</mo> <mn>5</mn> <mo>±</mo> <mn>0.02</mn> <mo>)</mo> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">m</mi> </math> <math></math> . The maximum possible percentage error in the measurement of density will nearly be :

m = ρ π r 2 1 ρ = m π r 2 1 Δ ρ ρ = Δ m m + 2 Δ r r + Δ l 1 Δ ρ ρ × 100 = 0.002 0.4 × 100 + 2 × 0.001 0.3 × 100 + 0.02 5 × 100 = 0.2 0.4 + 0.2 0.3 + 2 5 = 0.5 + 0.67 + 0.4 = 1.57 = 1.6 %

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Physics Class 11th Units and Measurement

A metal wire has mass $(0.4 \pm 0.002) g$ , radius $(0.3 \pm 0.001) m m$ $_{}$ and length $(5 \pm 0.02) c m$ . The maximum possible percentage error in the measurement of density will nearly be :

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Raj Pandey | Contributor-Level 9

4 weeks ago

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Detailed Solution:

$m = ρ π r^{2} 1$

$\begin{array}{r} ρ = \frac{m}{π r^{2} 1} \\ \frac{Δ ρ}{ρ} = \frac{Δ m}{m} + \frac{2 Δ r}{r} + \frac{Δ l}{1} \\ \frac{Δ ρ}{ρ} \times 100 = \frac{0.002}{0.4} \times 100 + \frac{2 \times 0.001}{0.3} \times 100 + \frac{0.02}{5} \times 100 \\ = \frac{0.2}{0.4} + \frac{0.2}{0.3} + \frac{2}{5} \\ = 0.5 + 0.67 + 0.4 \\ = 1.57 \\ = 1.6 % \end{array}$

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A metal wire has mass $(0.4 \pm 0.002) g$ , radius $(0.3 \pm 0.001) m m$ $_{}$ and length $(5 \pm 0.02) c m$ . The maximum possible percentage error in the measurement of density will nearly be :

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A metal wire has mass ( 0.4 ± 0.002 ) g , radius ( 0.3 ± 0.001 ) m m and length ( 5 ± 0.02 ) c m . The maximum possible percentage error in the measurement of density will nearly be :

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A metal wire has mass $(0.4 \pm 0.002) g$ , radius $(0.3 \pm 0.001) m m$ $_{}$ and length $(5 \pm 0.02) c m$ . The maximum possible percentage error in the measurement of density will nearly be :