question_answer

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 \[{{60}^{0}}\]. The resistance between the ends of the legs is close to: [JEE ONLINE 09-04-2013]

A)             \[50.0\Omega \]

B) \[10\Omega \]

C)             \[36.7\Omega \]

D) \[26.7\Omega \]

Correct Answer: D

Solution :

[d]

\[\Rightarrow \]   A and B are in parallel

\[\Rightarrow \]   \[R'=10+10=20\,\Omega \]

\[\Rightarrow \]   \[\frac{1}{R'}=\frac{1}{R}+\frac{1}{10}=\frac{1}{20}+\frac{1}{10}\]

\[\frac{1}{R'}=\frac{1+2}{20}\]

\[R'=\frac{20}{3}\Omega \]

Now resistances KE, R’ and FL are in series

\[\Rightarrow \]\[R''=10+10+\frac{20}{3}=\frac{80}{3}=26.7\,\Omega \]