A) \[10\Omega \]
B) \[20\Omega \]
C) \[40\Omega \]
D) \[100\Omega \]
Correct Answer: C
Solution :
[c] Length of each wire = \[\ell \]; Area of thick wire \[\left( {{A}_{1}} \right)=3A;\] Area of thin wire \[({{A}_{2}})=A;\]and resistance of thick wire \[({{R}_{1}})=10\Omega .\] Resistance \[\left( R \right)=\rho \frac{\ell }{A}\propto \frac{1}{A}\](if l is constant) \[\therefore \frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{A}{3A}=\frac{1}{3}\] or, \[{{R}_{2}}=3{{R}_{1}}=3\times 10=30\Omega \] The equivalent resistance of these two resistors in series \[={{R}_{1}}+{{R}_{2}}=30+10=40\Omega .\]
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