Question

A wire has a mass (0.3+/- 0.003) g,radius (0.5 +/- 0.005) mm and length (6 +/- 0.06) cm.What is the maximum percentage error?

Answer 1

hey friend!!!!!

your answer is "4%"
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Explainatioan =>
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Since density is mass÷volume,

For a cylinder it becomes- mass÷(pi×radius^2×length)

Relative percent errors for-

Mass-0.003÷0.3×100=1%

Radius^2-2×0.005÷0.5×100=2%

Length-0.06÷0.6×100=1%

THE UNITS CANCEL THEMSELVES FROM THE NUMERATOR AND DENOMINATOR.

therefore,

maximum percentage error in density-1%+2%+1%=4%

so, the answer is 4%


I think my answer is capable to clear your confusion..

Answer 2

Answer:

The maximum percentage error in density is 4%.

Step-by-step explanation:

Given mass of wire, $m=0.3g$

          error in measurement of mass, $\Delta m =\pm 0.003 g$

          radius of wire, $r=0.5mm$

          error in measurement of radius, $\Delta r =\pm 0.005 mm$

          length of wire, $l=6cm$

          error in measurement of length, $\Delta l =\pm 0.06 cm$

Since mass, radius and length, assuming that the maximum percentage error in density is to be calculated,

Volume of the wire is given by, $V=\pi r^{2} l$

and density is the ratio of mass to the volume, $\rho =\frac{m}{V}$

The percentage error in density is

               $\frac{\Delta \rho}{\rho} \times 100=\frac{\Delta m}{m} \times 100+2\frac{\Delta r}{r} \times 100+\frac{\Delta l}{l} \times 100$

       $\implies \frac{\Delta \rho}{\rho} \times 100=\Big(\frac{\Delta m}{m} +2\frac{\Delta r}{r} +\frac{\Delta l}{l} \Big)\times 100$

                              $=\Big(\frac{0.003}{0.3} +2\frac{0.005}{0.5} +\frac{0.06}{6} \Big)\times 100$

                              $=(0.01 +2\times 0.01 +0.01\Big)\times 100$

                              $=(0.04 )\times 100$

                              $=4\%$

Therefore, the maximum percentage error in density is 4%.