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 Main Online Paper (Held On 09 April 2013)

A) \[50.0\Omega \]                                              

B) \[10\Omega \]                                 

C) \[36.7\Omega \]                              

D) \[26.7\Omega \]                

Correct Answer: D

Solution :

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