| Figure shows a \[2.0\,V\] potentiometer used for the determination of internal resistance of a \[1.5\,V\,\]cell. The balance point of the cell in open circuit is\[75\,cm\]. When a resistor of of \[0.5\Omega \] is used in the external circuit of the cell, the balance point shifts to \[60\,cm\] length of the potentiometer wire. Length of wire AB is \[100\,cm\]. |
 |
| When \[0.5\,W\] is used in the external resistance then terminal voltage of cell is |
A) \[1.5\,V\]
B) \[1.2\,V\]
C) \[1.0\,V\]
D) \[1.5\,V\]
Correct Answer: B
Solution :
When the circuit is closed, the balance point is at 60 cm terminal voltage of the cell \[=\frac{60}{100}\times 2=1.2V\]
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