Current Electricity Charging Discharging Of Capacitors Question 313
Question: With a potentiometer null point were obtained at 140 cm and 180 cm with cells of emf 1.1 V and one unknown X volts. Unknown emf is
[DCE 2002]
Options:
A) 1.1 V
B) 1.8 V
C) 2.4 V
D) 1.41 V
Show Answer
Answer:
Correct Answer: D
Solution:
$ E=\frac{V}{l} $ ; E is constant (volt. gradient).
Therefore $ \frac{V _{1}}{l _{1}}=\frac{V _{2}}{l _{2}} $
Therefore $ \frac{1.1}{140}=\frac{V}{180} $
Therefore $ V=\frac{180\times 1.1}{140}=1.41,V $
BETA
