A) \[\frac{30E}{100}\]
B) \[\frac{30E}{100.5}\]
C) \[\frac{30E}{(100-0.5)}\]
D) \[\frac{30(E-0.5i)}{100}\], where i is the current in the potentiometer
Correct Answer: A
Solution :
From the principle of potentiometer \[V\propto l\] Þ \[\frac{V}{E}=\frac{l}{L}\]; where V = emf of battery, E = emf of standard cell, L = Length of potentiometer wire \[V=\frac{El}{L}=\frac{30E}{100}\].
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