JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2017
- Hard
- MCQ
In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance \(P\, = 4\,\Omega \) and the neutral point \(N\) is at \(60\,cm\) from \(A\). Now an unknown resistance \(R\) is connected in series to \(P\) and the new position of the neutral point is at \(80\,cm\) from \(A\) . The value of unknown resistance \(R\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\frac{20}{3}\, \Omega \)
Step-by-step Solution
Detailed explanation
In balance position of bridge, \(\frac{P}{Q}=\frac{l}{(100-l)}\)
Initially neutral position is \(60\, \mathrm{cm} .\) from \(\mathrm{A}\),
so
\(\frac{4}{60}=\frac{Q}{40} \Rightarrow Q=\frac{16}{6}=\frac{8}{3}\, \Omega\)
Now, when unknown resistance \(\mathrm{R}\) is connected in series to \(P\), neutral point is \(80 \,cm\) from Athen,
\(\frac{4+R}{80}=\frac{Q}{20}\)
\(\frac{4+R}{80}=\frac{8}{60}\)
\(R=\frac{64}{6}-4=\frac{64-24}{6}=\frac{40}{6} \,\Omega\)
Hence, the value of unknown resistance \(\mathrm{R}\)
is \(=\frac{20}{3} \,\Omega\)
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