NEET Solved Paper 2014 Question 28
Question: A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance, R, connected across the given cell, has values of (i) infinity, (ii) $ 9.5\Omega $ the ‘balancing lengths’, on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is [AIPMT 2014]
Options:
A) 0.25 $ \Omega $
B) 0.95 $ \Omega $
C) 0.5 $ \Omega $
D) 0.75 $ \Omega $
Show Answer
Answer:
Correct Answer: C
Solution:
- Given, e = 2V and $ l $ = 4m Potential drop per unit length $ \phi =\frac{e}{l}=\frac{2}{4}=0.5V/m $ For the first case, Þ $ e’=\phi l _1,….(i) $ ( $ e’ $ $\rightarrow$ emf of the given cell) For the second case, $ V=\phi l _2,….(ii) $ From Eqs. (i) and (ii), $ e’/V=l _1/l _2 $ $ e’=l(r+R) $ and $ V=lR $ for the second case So, $ r=R( \frac{l _1}{l _2}-1 )=9.5( \frac{3}{2.85}-1 )=9.5(1.05-1) $ $ =9.5\times 0.05=0.475\simeq 0.5\Omega $
