Last Activity: 10 Years ago
Sol. ℓ ‘ = 2 ℓ volume of the wire remains constant. A ℓ = A’ ℓ ‘ ⇒ A ℓ = A’ * 2 ℓ ⇒ A’ = A/2 f = Specific resistance R = fℓ/A : R’ = fℓ’/A’ 100 Ω = f2ℓ/A / 2 = 4fℓ/A = 4R ⇒ 4 * 100 Ω = 400 Ω
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Get your questions answered by the expert for free
Last Activity: 2 Year ago(s)