# 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?

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

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

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

$=122.72m$

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

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

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

$\Rightarrow R'=2.5\Omega$

$\frac{R'}{R}=\frac{2.5}{10}=\frac{1}{4}$

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

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