JEE Advanced · Physics · 19. Current Electricity
A meter bridge is set up as shown in figure, to determine an unknown resistance \(X\) using a standard \(10 \Omega\) 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
- A \(10.2 \Omega\)
- B \(10.6 \Omega\)
- C \(10.8 \Omega\)
- D \(11.1 \Omega\)
Answer & Solution
Correct Answer
(B) \(10.6 \Omega\)
Step-by-step Solution
Detailed explanation
Using the concept of balanced Wheatstone bridge, we have
\(\frac{P}{Q} =\frac{R}{S} \)
\( \therefore \frac{X}{(52+1)} =\frac{10}{(48+2)} \)
\( \therefore X =\frac{10 \times 53}{50}=10.6 \Omega\)
\(\therefore\) Correct option is (b).
Analysis of Question
Question is moderately tough, because normally end corrections are not taught in the coaching/schools in this type of problem.
\(\frac{P}{Q} =\frac{R}{S} \)
\( \therefore \frac{X}{(52+1)} =\frac{10}{(48+2)} \)
\( \therefore X =\frac{10 \times 53}{50}=10.6 \Omega\)
\(\therefore\) Correct option is (b).
Analysis of Question
Question is moderately tough, because normally end corrections are not taught in the coaching/schools in this type of problem.
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