A wire of length 'l' and resistance 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:
A wire of length 'l' and resistance 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
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6 months ago
Correct Option - 2
Detailed Solution:

Divided into 10 parts
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For uniformly charged spherical shell,
In first condition R1 = 36
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So
x = 4
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Physics Current Electricity 2025
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