The maximum percentage error in the measurement of density of a wire is [Given, mass of wire = (0.60 0.003) g radius of wire = (0.50 0.01) g length of wire = (10.00 0.05) cm] [2025]

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Solution

Q.
The maximum percentage error in the measurement of density of a wire is

[Given, mass of wire = (0.60 $\pm$ 0.003) g

radius of wire = (0.50 $\pm$ 0.01) cm

length of wire = (10.00 $\pm$ 0.05) cm] [2025]

1 4

2 5

3 8

4 7

Ans.
(2)

$ρ = \frac{m}{v o l .} = \frac{m}{π R^{2} l} \Rightarrow \frac{d ρ}{ρ} = \frac{d m}{m} + \frac{2 d R}{R} + \frac{d l}{l}$

$∴ \frac{d ρ}{ρ} = (\frac{0.003}{0.6} + \frac{2 \times 0.01}{0.5} + \frac{0.05}{10}) 100 = 5 %$

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