Question

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A wire of given material having length l and area of cross section A has a resistance of 4ohm. What would be the resistance of another wire of the same material having length l/2 and area of cross section 2A???
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Answer 1

Given :

$\sf\:For~1^{st}~wire~-$

• Length = l

• Area of cross section = A

• Resistance = $\tt\:4~Ω$

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$\sf\:For~2^{nd}~wire~-$

• Length = $\sf\:\frac{l}{2}$

• Area of cross section = $\sf\:2~A$

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To find :

• Resistance of second wire.

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Formula to be used :

$\bf\:\dag$ $\bf\color{aqua}{Resistance~of~a~wire~=~\rho \times \frac{l}{A}}$

where,

• $\sf\:\rho$ = resistivity of the material

• l = length of the wire

• A = Area of cross section of wire

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Solution :

Resistance of first wire -

$\sf\:\rho \times \frac{l}{A}$

$\sf\implies\:4~Ω$ -- [ given ]

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Let $\sf\:\rho \times \frac{l}{A}=4~Ω$ -- $\small\fbox{1}$

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Resistance of second wire -

$\sf\:\rho \times \frac{l}{A}$

Substituting the given values for second wire

$\sf\implies\:\rho \times \frac{\frac{l}{2}}{2A}$

Converting division into multiplication

$\sf\implies\:\rho \times \frac{l}{2} \times \frac{1}{2A}$

Proceeding with simple calculation

$\sf\implies\:\frac{\rho \times l}{A}(\frac{1}{2} \times \frac{1}{2})$

$\sf\implies\:\frac{\rho \times l}{A}(\frac{1}{4})$

Substituting the value of ρl/A from 1

$\sf\implies\:4(\frac{1}{4})$

$\sf\implies\:4 \times \frac{1}{4}$

$\sf\implies\:\frac{4}{4}$

$\sf\implies\:\cancel{\frac{4}{4}}$

$\small{\underline{\boxed{\mathrm{\implies\:1~Ω}}}}$ $\tt\color{teal}{\bigstar}$

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Henceforth,

❒ The resistance of the second wire is $\large{\mathfrak\red{1~Ω}}$

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