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:

Option 1 - <p><span contenteditable="false"> <math> <mrow> <mn>2</mn> <mn>6</mn> <mi>Ω</mi> </mrow> </math> </span></p>
Option 2 - <p><span contenteditable="false"> <math> <mrow> <mn>5</mn> <mn>2</mn> <mi>Ω</mi> </mrow> </math> </span></p>
Option 3 - <p><span contenteditable="false"> <math> <mrow> <mn>5</mn> <mn>5</mn> <mi>Ω</mi> </mrow> </math> </span></p>
Option 4 - <p><span contenteditable="false"> <math> <mrow> <mn>6</mn> <mn>0</mn> <mi>Ω</mi> </mrow> </math> </span></p>
2 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
A
6 months ago
Correct Option - 2
Detailed Solution:

Divided into 10 parts

R = ρ l A

R = ρ l 1 0 A = R 1 0

RS=5*R10 [series] 

R S = 5 0

R P = R 5 0 [  parallel  ]

R eq  = R S + R P

= 5 2 Ω

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Physics Current Electricity 2025

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