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PhysicsCurrent ElectricityEquivalent ResistanceMedium2 minPYQ_2018

PhysicsMediumsingle choice

A conducting wire has length $L_{1}$ and diameter $d_{1}$ . After stretching the same wire length becomes $L_{2}$ and diameter $d_{2}$ . The ratio of resistance before and after strectching is

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Answer:

Solution:

$\frac{R_{1}}{R_{2}} = \frac{L_{1}}{L_{2}} \times \frac{A_{2}}{A_{1}} = \frac{L_{1}}{L_{2}} \times \frac{\frac{π d^{22}}{4}}{\frac{π d_{1}^{2}}{4}}$ ...(1)
$L_{1} d_{1}^{2} = L_{2} d_{2}^{2}$ ...(2)
$\frac{L_{1}}{L_{2}} = {(\frac{d_{2}}{d_{1}})}^{2}$
$\frac{R_{1}}{R_{2}} = (\frac{L_{1}}{L_{2}}) \times {(\frac{d_{2}}{d_{1}})}^{2}$
$= \frac{d_{2}^{4}}{d_{1}^{4}}$

Stream:MHT_CETSubject:PhysicsTopic:Current ElectricitySubtopic:Equivalent Resistance

⏱ 2mℹ️ Source: PYQ_2018

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