Question

A copper wire has diameter 0.5mm and
resistivity of 1.6 x 10-6ohm m what will be the
length of
this wire to make its resistance 10 ohm?
How much does the resistance change if the
diameter is doubled?
​

Answer 1

${\pink\bigstar}$QUESTION:-

A copper wire has diameter 0.5mm and resistivity of 1.6 x 10-6Ω m what will be the length of this wire to make its resistance 10 ohm?

How much does the resistance change if the diameter is doubled?

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$\blue\bigstar$ANSWER:-

✦Let diameter=d,

✦radius=r

✦Area of cross section=A,

✦Resistance=R

✦Resistivity=$\sf\rho$ and

✦ Length = l

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i)Diameter ,d=0.5mm

Radius,

r$\sf=\dfrac{0.5}{2}mm=0.25mm=0.25\times10^{-3}m$

•Area of cross-section,

A$\sf=\pi r^{2}=\dfrac{22}{7}\times(0.25\times 10^{-3})^{2}=0.1964\times10^{-6}m^{2}$

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•Resistance, R=10Ω

Resistivity,$\sf\rho=1.6\times 10^{-8}$Ωm,

Length,l=?

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We know that,

$\sf\red{ R=\dfrac{\rho l}{A}}$

$\sf\therefore \red{l=\dfrac{RA}{\rho}}$

$\sf =\dfrac{10\times 0.1964\times 10^{-6}}{1.6\times 10^{-8}}=\dfrac{1964}{16}=122.75m$

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ii)The area of cross section,

$\sf{ A_{1}=\pi r^{2}=\pi \bigg(\dfrac{d}{2}\bigg)=\dfrac{\pi r^{2}}{4}}$

Its resistance,

$\sf R_{1}=\dfrac{\rho l}{A_{1}}=\dfrac{4\rho l}{\pi d^{2}}.......(1)$

★When the diameter is doubled, area of cross-section,

$\sf A_{2}=\pi \bigg(\dfrac{2d}{2}\bigg)^{2}=\pi d^{2}$

Its resistance,

$\sf R_{2}=\dfrac{\rho l}{A_{2}}=\dfrac{\rho l}{\pi d^{2}}........(2)$

•From (1) and (2) ,we have

$\sf{\dfrac{R_{2}}{R_{1}}=\dfrac{\rho l}{\pi d^{2}}\times \dfrac{\pi d^{2}}{4\rho l}=\dfrac{1}{4}}$

or $\sf\boxed{\pink{ R_{2}=\dfrac{1}{4}R_{1}}}$

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