Q.6. A potentiometer wire of length 4 m and resistance 5 Ω is connected in series with a resistance of 992 Ω and a cell of e.m.f. 4 V with internal resistance 3 Ω. The length of 0.75 m on potentiometer wire balances the e.m.f. of
A. 4.00 mV
B. 3.75 mV
C. 3.00 mV
D. 2.50 mV
Answer :- B
Explanation :-
\therefore \text{ Total Resistance:} \mathrm{R} = 992 + 5 + 3 = 1000 \, \Omega \text{Voltage across 4 m wire:} \frac{5}{995 + 5} \times 4 = 0.02 \, \mathrm{V} \therefore \text{ For one metre wire:} \frac{0.02}{4} = 0.005 \, \mathrm{V} \therefore \text{ For 0.75 m wire:} 0.004 \times 0.75 = 0.00375 = 3.75 \, \mathrm{mV}