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Q. 6.     A copper wire has diameter 0.5 mm and resistivity of 1.6\times 10^{-8}\; \Omega m. What will be the length of this wire to make its resistance 10\; \Omega ? How much does the resistance change if the diameter is doubled?

Answers (1)


Given : diameter=d= 0.5 mm and resistivity = \rho=1.6\times 10^{-8}\; \Omega m.,resistance =R=10

Area =A

A=\frac{\pi d^2}{4}=\frac{3.14\times 0.5\times 0.5}{4}

\Rightarrow A=0.000000019625 m^2

We know 

                  R=\frac{\rho l}{A}

            \Rightarrow l=\frac{RA}{\rho }=\frac{10\times 0.000000019625}{1.6\times 10^-^8}


If the diameter is doubled.

d=1 mm

Area =A'

A'=\frac{\pi d^2}{4}=\frac{3.14\times 1\times 1}{4}

\Rightarrow A=0.000000785 m^2

We know 

                  R'=\frac{\rho l}{A'}

            \Rightarrow R'=\frac{1.6\times 10^{-8}\times 122.72}{0.000000785}

            \Rightarrow R'=2.5\Omega


Hence, new resistance is \frac{1}{4}  of original resistance.



Posted by

seema garhwal

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