COMEDK · Physics · 19. Current Electricity
Two resistances are connected in two gaps of a meterbridge. The balance point is \(20 \mathrm{~cm}\) from the zero en(d) A resistance of \(15 \Omega\) is connected in series with the smaller of the two. The null point shifts to \(40 \mathrm{~cm}\). The value of the smaller resistance \((\) in \(\Omega)\) is
- A 3
- B 6
- C 9
- D 12
Answer & Solution
Correct Answer
(C) 9
Step-by-step Solution
Detailed explanation
In first case, at balance point, \(\frac{P}{Q}=\frac{l}{100-l}=\frac{20}{100-20}=\frac{1}{4} \Rightarrow Q=4 P...(i)\) Let \(P\) be the smaller resistance, then in second case at balance point,…
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