A wire of length 'l' and resistance <math> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mi>Ω</mi> </mrow> </math> &nbsp;is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

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Physics Class 11th Units and Measurement

A wire of length 'l' and resistance $100 Ω$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

Option 1 -

$26 Ω$

Option 2 -

$52 Ω$

Option 3 -

$55 Ω$

Option 4 -

$60 Ω$

6 Views | Posted a month ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

Divided into 10 parts

$R = \frac{ρ l}{A}$

$R^{'} = \frac{ρ l}{10 A} = \frac{R}{10}$

$R_{S} = 5 \times \frac{R}{10} [series]$

$R_{S} = 50$

$R_{P} = \frac{R}{50} [parallel]$

$R_{eq} = R_{S} + R_{P}$

$= 52 Ω$

<div><div><picture><source srcset="" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1754991777phpfJsjIb.jpeg" alt="" width="280" height="48" loading="lazy"></picture></div></div>Divided into 10 parts <math> <mrow> <mi>R</mi> <mo>=</mo> <mfrac> <mrow> <mi>ρ</mi> <mi>l</mi> </mrow> <mrow> <mi>A</mi> </mrow> </mfrac> </mrow> </math>  <math> <mrow> <msup> <mi>R</mi> <mo>′</mo> </msup> <mo>=</mo> <mfrac> <mrow> <mi>ρ</mi> <mi>l</mi> </mrow> <mrow> <mn>1</mn> <mn>0</mn> <mi>A</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>R</mi> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </mfrac> </mrow> </math>  <math> <mrow><msub><mrow><mi>R</mi></mrow><mrow><mi>S</mi></mrow></msub><mo>=</mo><mn>5</mn><mo>×</mo><mfrac><mrow><mi>R</mi></mrow><mrow><mn>1</mn><mn>0</mn></mrow></mfrac> [series] </mrow> </math>  <math> <mrow> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mi>S</mi> </mrow> </msub> <mo>=</mo> <mn>5</mn> <mn>0</mn> </mrow> </math>  <math> <mrow> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mi>P</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mi>R</mi> </mrow> <mrow> <mn>5</mn> <mn>0</mn> </mrow> </mfrac> <mrow> <mo>[</mo> <mrow> parallel </mrow> <mo>]</mo> </mrow> </mrow> </math>  <math> <mrow> <msub> <mrow> <mi>R</mi> </mrow> <mrow>eq </mrow> </msub> <mo>=</mo> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mi>S</mi> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mi>P</mi> </mrow> </msub> </mrow> </math>  <math> <mrow> <mo>=</mo> <mn>5</mn> <mn>2</mn> <mi>Ω</mi> </mrow> </math>

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A wire of length 'l' and resistance $100 Ω$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

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A wire of length 'l' and resistance 1 0 0 Ω is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

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A wire of length 'l' and resistance $100 Ω$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is: