JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2018
- Hard
- MCQ
In a meter bridge, as shown in the figure, it is given that resistance \(Y=12.5\, \Omega \) and that the balance is obtained at a distance \(39.5\, cm\) from end \(A\) (by jockey \(J\)) . After interchanging the resistances \(X\) and \(Y\), a new balance point is found at a distance \(l_2\) from end \(A\). What are the values of \(X\)and \(l_2\) ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(8.16\,\Omega \) and \(60.5\,cm\)
Step-by-step Solution
Detailed explanation
For a balanced meter bridge,
\(\frac{X}{39.5}=\frac{Y}{(100-39.5)}\)
\(\Rightarrow Y=39.5=X \times(100-39.5)\)
or, \(\quad {\rm{X}} = \frac{{12.5 \times 39.5}}{{60.5}} = 8.16\,\Omega \)
\({\text { When } X \text { and } Y \text { are interchanged } l_{1} \text { and }(100} \)
\({\left.-l_{1}\right) \text { will also interchange so, } l_{2}=60.5\, \mathrm{cm}}\)
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