JEE Mains · Physics · STD 12 - 3. current electricity
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
- A \(\frac{33}{5}\, \Omega \)
- B \( 6\,\Omega \)
- C \( 7\,\Omega \)
- D \(\frac{20}{3}\, \Omega \)
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…
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