Question

do the problem......​

Answer 1

• ✯ Length = 122.7 m ✯
• ✯ Resistance = 2.5 Ω ✯

Given:-

• Diameter of the copper wire = 0.5 mm
• Resistivity of the copper wire $(\rho)$= 1.6 × $\sf{10^{-8}}$ Ω m
• Resistance (R) = 10 Ω

To find:-

• Length of the wire to make its resistance 10 Ω.
• Change of resistance if the diameter is doubled.

Solution:-

• Diameter = 0.5 mm

Then,

Radius of the wire (r) = 0.5/2 mm

$\implies$ Radius (r) = 0.25 mm

$\implies$ Radius (r) = 2.5 × $\sf{10^{-4}}$ m

We know that,

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

To find l , we need to find A ( area of cross section).

$\implies\sf{A=\pi\:r^2}$

$\implies\sf{A=3.14\times\:(2.5\times\:10^{-4})^2}$

$\implies\sf{A=3.14\times\:(2.5)^2\times\:(10^{-4})^2}$

$\implies\sf{A=3.14\times\:6.25\times\:10^{-8}}$

$\implies\sf{A=19.625\times\:10^{-8}}$

Therefore, area of the cross section is $\sf{19.625\times\:10^{-8}}$ m².

Now find the length (l) of the wire.

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

$\implies\sf{l=R(\dfrac{A}{\rho})}$

Put values.

$\implies\sf{l=\dfrac{10\times\:19.625\times\:10^{-8}}{1.6\times\:10^{-8}}}$

$\implies\sf{l=\dfrac{10\times\:19.625}{1.6}}$

$\implies\sf{l=122.65}$

$\implies\sf{l=122.7}$

Therefore, the length of the wire is 122.7 m.

_________________

• If the diameter of the wire is doubled,
• New diameter = 2× old diameter
• New radius = 2× old radius = 2r m

Then,

New area(A') = πr²

• → New area(A') = π(2r)²
• → New area(A')= 4πr²
• → New area(A')= 4 × old area
• → New area = 4A

Let new resistance be Rʹ

$\implies\sf{R'=\rho(\dfrac{l}{A'})}$

$\implies$R’ = ρ ( l ) / ( 4A )

$\implies$R’ = ρ (l) X 1/( 4A )

• Hence if diameter doubles, resistance becomes 1/4 times.

•Therefore new resistance= 2.5 Ω.

Answer 2

$\sf Given \begin{cases} & \sf{diameter,\:d = \bf{0.5\;mm}} \\ & \sf{resistivity, \: ρ= \bf{1.6\times{10}^{ - 8}} \: Ωm} \\ & \sf{Resistance, \: R= \bf{10 \: Ω}}\end{cases}$

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☯ We need to find the length of the wire.

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Let the length of the wire be l.

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$\sf{A=\pi d^2/4}$

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$\sf{R=ρl/A}$

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$\sf{\implies{\dfrac{ρl}{\pi d^2/4}}}$

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$\sf{\implies{l=\dfrac{R\pi d^2}{ρ}}}$

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$\sf{\implies{l=\dfrac{10×3.14×(0.5×{10}^{-3}){}^{2}}{4×1.6×{10}^{-8}}}}$

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$\implies{\underline{\boxed{\purple{\sf{l ={\frak{ 122\;{\sf{m}}}}}}}}}\;\bigstar$

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$\sf{R∝1/d^2}$

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$\therefore\;{\sf{The\:length\:of\:the\:wire\:is\:{\bf{122\:m}}.}}$

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☯ We also need to find change of resistance if the diameter is doubled.

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If the diameter is doubled, resistance will be one-fourth.

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$\therefore\;{\sf{New\: resistance={\bf{2.5\:Ω}}}}$