Current Electricity Ques 2
- During an experiment with a meter bridge, the galvanometer shows a null point when the jockey is pressed at $40.0 \mathrm{~cm}$ using a standard resistance of $90 \Omega$, as show in the scale used in the meter bridge is $1 \mathrm{~mm}$. The unknown resistance is
(2014 Adv.)
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Answer:
Correct Answer: 2.( c )
Solution:
- For balanced meter bridge
$ \begin{array}{rlrl} & & \frac{X}{R} & =\frac{l}{(100-l)} \\ \therefore & & \frac{X}{90} & =\frac{40}{100-40} \\ & & X & \\ & & \text { (where, } R=90 \Omega \text { ) } \end{array} $
$ \begin{aligned} X & =R \frac{l}{(100-l)} \\ \frac{\Delta X}{X} & =\frac{\Delta l}{l}+\frac{\Delta l}{100-l}=\frac{0.1}{40}+\frac{0.1}{60} \\ \Delta X & =0.25 \\ X & =(60 \pm 0.25) \Omega \end{aligned} $
So,
BETA
