Search any question & find its solution
Question:
Answered & Verified by Expert
A meter bridge is set-up as shown, to determine an unknown resistance $X$ using a standard $10 \mathrm{ohm}$ resistor. The galvanometer shows null point when tapping-key is at $52 \mathrm{~cm}$ mark. The end-corrections are $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$ respectively for the ends $A$ and $B$. The determined value of $X$ is
Options:
Solution:
2615 Upvotes
Verified Answer
The correct answer is:
$10.6 \mathrm{ohm}$
Applying the condition of balanced Wheatstone bridge, we get
$\begin{aligned} & \frac{X}{10 \Omega}=\frac{(52+1) \mathrm{cm}}{(48+2) \mathrm{cm}}=\frac{53}{50}, \\ & X=10 \Omega \times \frac{53}{50}=10.6 \Omega\end{aligned}$
$\begin{aligned} & \frac{X}{10 \Omega}=\frac{(52+1) \mathrm{cm}}{(48+2) \mathrm{cm}}=\frac{53}{50}, \\ & X=10 \Omega \times \frac{53}{50}=10.6 \Omega\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.