Verified Solution • 2025–26 Session moderate

QJEEM19Ph9ced: JEE Main 2019 Physics - Class 12 JEEM19Ph9ced Solution: A metal wire of resistance 3 is elongated to make a uniform…

A metal wire of resistance 3 Ω\Omega is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at the centre, the equivalent resistance between these two points will be :-

Multiple Choice Options

  • A 5 /2 Ω\Omega
  • B 12/5 Ω\Omega
  • C 7/2 Ω\Omega
  • D 5 / 3 Ω\Omega

Step-by-Step Verified Solution

R=ρlA=ρl(V/l)=ρl2V(V→Volume of wire)R = {{\rho l} \over A} = {{\rho l} \over {\left( {V/l} \right)}} = {{\rho {l^2}} \over V}\left( {V{\rm{ }} \to {\rm{ }}Volume{\rm{ }}\,of{\rm{ }}\,wire} \right)

⇒\Rightarrow Final resistance = 3 × (B)2 = 12 Ω\Omega

Req=2Ω∥10Ω=53Ω{R_{eq}} = 2\Omega \parallel 10\Omega = {5 \over 3}\Omega

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