Current Electricity Charging Discharging Of Capacitors Question 209
Question: A potentiometer is used for the comparison of e.m.f. of two cells $ E _{1} $ and $ E _{2} $ . For cell $ E _{1} $ the no deflection point is obtained at 20 $ cm $ and for $ E _{2} $ the no deflection point is obtained at 30 $ cm $ . The ratio of their e.m.f.’s will be
[MP PET 1984]
Options:
A) 2/3
B) 1/2
C) 1
D) 2
Show Answer
Answer:
Correct Answer: A
Solution:
Ratio will be equal to the ratio of no deflection lengths i.e. $ \frac{E _{1}}{E _{2}}=\frac{l _{1}}{l _{2}}=\frac{2}{3} $
BETA
