question_answer

The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5 V and internal resistance of \[20,\Omega \]the null point on it is found to be at 1000 cm. The resistance of whole wire is : [JEE MAIN Held On 08-01-2020 Morning]

A) \[80\,\Omega \]

B) \[100\,\Omega \]

C) \[60\,\Omega \]

D) \[120\,\Omega \]

Correct Answer: B

Solution :

[b]

\[R\to \]Resistance

Potential gradient for the potentiometer wire

‘AB’ is \[-\frac{dV}{d\ell }=\left[ \frac{60\times R}{{{\ell }_{AB}}} \right]mv/m\]

\[\therefore {{V}_{AP}}=\left( -\frac{dV}{d\ell } \right){{\ell }_{AP}}=\frac{60\times R}{1200}\times 1000mV\]

\[\therefore {{V}_{AP}}=50RmV\]

Also, \[{{V}_{AP}}=5\text{ }V\](for balance point at P)

\[\therefore R=\frac{5}{50\times {{10}^{-3}}}=100\Omega \]