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Q.

A wire has a mass 0.4 $\pm$ 0.004g and length 8 $\pm$ 0.08 (cm) . The maximum percentage error in the measurement of its density is 4%. The radius of the wire is $r \pm Δr, find Δr$

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a

0.03r

b

0.01r

c

0.02r

d

0.1r

answer is B.

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Detailed Solution

$ρ = \frac{m}{{πr}^{2} L}$

$\Rightarrow \frac{Δρ}{ρ} = [\frac{Δm}{m} + 2 \frac{Δr}{r} + \frac{ΔL}{L}]$

$\frac{Δρ}{ρ} \times 100 = (0 .01 + 2 \frac{Δr}{r} + 0 .01) \times 100 = 4 (given)$
or $100 \frac{Δr}{r} = 1 \Rightarrow Δr = 0 .01 r$

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A wire has a mass 0.4 ± 0.004g and length 8± 0.08 (cm) . The maximum percentage error in the measurement of its density is 4%. The radius of the wire is r±Δr, find Δr