The maximum percentage error in the measureme | vedclass

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(B) The density $\rho$ of a wire is given by the formula: $\rho = \frac{m}{V} = \frac{m}{\pi R^2 \ell}$.Taking the relative error,we have: $\frac{\Del

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$5$

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(B) The density $ho$ of a wire is given by the formula: $ho = \frac{m}{V} = \frac{m}{\pi R^2 \ell}$.Taking the relative error,we have: $\frac{\Delta ho}{ho} = \frac{\Delta m}{m} + 2 \frac{\Delta R}{R} + \frac{\Delta \ell}{\ell}$.Substituting the given values:$\frac{\Delta ho}{ho} = \left( \frac{0.003}{0.60} + 2 	imes \frac{0.01}{0.50} + \frac{0.05}{10.00} ight)$.Calculating each term:$\frac{0.003}{0.60} = 0.005$,$2 	imes \frac{0.01}{0.50} = 2 	imes 0.02 = 0.04$,$\frac{0.05}{10.00} = 0.005$.Summing these values: $0.005 + 0.04 + 0.005 = 0.05$.To find the percentage error,multiply by $100$:Percentage error $= 0.05 	imes 100 = 5 \%$.

The maximum percentage error in the measurement of the 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 $= (10.00 \pm 0.05) \ cm$]

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