
For ADER′1=2x1+101 or R′=10+2x20x RBC=10+2x20x+20−x+20−x or 10+2x20x+40=2x Solving we get x=10Ω Putting the value of x=10Ω in equation (i) We get RBC=10+2×1020×10+20−10+20−10=380=26.7Ω
JEE Main 2013 — Physics Electromagnetism
A letter ′A′ is constructed of a uniform wire with resistance 1.0Ω per cm, The sides of the letter are 20 cm and the cross piece in the middle is 10 cm long. The apex angle is 60 . The resistance between the ends of the legs is close to:
Held on 9 Apr 2013 · Verified 6 Jul 2026.
50.0Ω
10Ω
36.7Ω
26.7Ω
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