Current Electricity Ques 2 — JEE PYQ Chapterwise

Answer:

Correct Answer: 2.( c )

Solution:

2. For balanced meter bridge experiment

$ \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,