Math, asked by khushighosh1910, 8 months ago

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?

Answers

Answered by Anonymous
117

 {\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\bigstarANSWER:-

✦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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