NEET Solved Paper 2014 Question 27
Question: The resistances in the two arms of the meter bridge are 5 $ \Omega $ and R $ \Omega $ , respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6 $ l _1 $ . The resistance R, is [AIPMT 2014]
Options:
A) $ 10\Omega $
B) $ 15\Omega $
C) $ 20\Omega $
D) $ 25\Omega $
Show Answer
Answer:
Correct Answer: B
Solution:
For first case, $ \frac{5}{l _1}=\frac{R}{(100-l _1)}…(i) $ Now, by shunting resistance R by an equal resistance R, new resistance in that arm become $ \frac{R}{2} $ So
$ \frac{5}{1.6l _1}\frac{R/2}{(100-l _1)}….(ii) $ From Eqs. (i) and (ii)
$ \frac{1.6}{1}=\frac{(100-1.6l _1)}{100-l _1}\times 2 $
Þ $ 160-1.6l _1=200-3.2l _1 $ $ 1.6l _1=40 $ $ l _1=\frac{400}{1.6}=25m $ From Eq. (i),
$ \frac{5}{25}=\frac{R}{75}\Rightarrow R=15\Omega $
BETA
