A wire has a mass 0.3 pm 0.003,g, radius 0.5 | vedclass

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Question

(d) Density, $\rho = \frac{M}{V} = \frac{M}{{\pi {r^2}L}}$

$ \Rightarrow \frac{{\Delta \rho }}{\rho } = \frac{{\Delta M}}{M} + 2\frac{{\Delta r}}{r

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(d) Density, $ho = \frac{M}{V} = \frac{M}{{\pi {r^2}L}}$

$ \Rightarrow \frac{{\Delta ho }}{ho } = \frac{{\Delta M}}{M} + 2\frac{{\Delta r}}{r} + \frac{{\Delta L}}{L}$

$ = \frac{{0.003}}{{0.3}} + 2 	imes \frac{{0.005}}{{0.5}} + \frac{{0.06}}{6}$

$ = 0.01 + 0.02 + 0.01 = 0.04$

$	herefore $ Percentage error $ = \frac{{\Delta ho }}{ho } 	imes 100 = 0.04 	imes 100 = 4\% $

A wire has a mass $0.3 \pm 0.003\,g$, radius $0.5 \pm 0.005\,mm$ and length $6 \pm 0.06\,cm$. The maximum percentage error in the measurement of its density is .......... $\%$

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