A letter 'A' is constructed of a uniform wire with resistance 1.0 Ω… | JEE Main 2013 (9 Apr, Shift 1)

Q71JEE Main 2013 · 9 APR · SHIFT 1Difficult

Combination of Resistors

A letter 'A' is constructed of a uniform wire with resistance 1.0 $\Omega$ 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:

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Answer: (d)

With apex angle $60^\circ$ , the $10$ cm cross piece joins the midpoints of the $20$ cm legs. Each half-leg $= 10\ \Omega$ , cross piece $= 10\ \Omega$ .
Between the midpoints: cross piece $10\ \Omega$ in parallel with the upper halves $10 + 10 = 20\ \Omega$ : $\dfrac{10 \times 20}{30} = \dfrac{20}{3}\ \Omega$ .
$R = 10 + \dfrac{20}{3} + 10 = \dfrac{80}{3} \approx 26.7\ \Omega$
So the answer is (d) $26.7\ \Omega$ .

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Combination of Resistors question from JEE Main 2013 (9 Apr, Shift 1)

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