AP EAMCET · PHYSICS · Current Electricity
When a wire is connected in the left gap of a metre bridge, the balancing point is at 40 cm from the left end of the bridge wire. If the wire in the left gap is stretched so that its length is doubled and again connected in the same gap, then the balancing point from the left end of the bridge wire is
- A \(\frac{300}{11} \mathrm{~cm}\)
- B \(\frac{800}{11} \mathrm{~cm}\)
- C \(\frac{400}{11} \mathrm{~cm}\)
- D \(\frac{700}{11} \mathrm{~cm}\)
Answer & Solution
Correct Answer
(B) \(\frac{800}{11} \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
\(\frac{R_1}{R_2} = \frac{40}{100-40} = \frac{2}{3}\) \(R_1' = 4R_1\) \(\frac{R_1'}{R_2} = \frac{l}{100-l}\) \(4 \times \frac{R_1}{R_2} = \frac{l}{100-l}\) \(4 \times \frac{2}{3} = \frac{l}{100-l}\) \(8(100-l) = 3l\) \(800 - 8l = 3l\) \(11l = 800\)…
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