KCET · Physics · Current Electricity
Two tangent galvanometers \(A\) and \(B\) have coils of radii \(8 \mathrm{~cm}\) and \(16 \mathrm{~cm}\) respectively and resistance \(8 \Omega\) each. They are connected in parallel with a cell of emf \(4 \mathrm{~V}\) and negligible internal resistance. The deflections produced in the tangent galvanometers \(A\) and \(B\) are \(30°\) and \(60°\) respectively. If A has 2 turns, then B must have
- A 18 turns
- B 12 turns
- C 6 turns
- D 2 turns
Answer & Solution
Correct Answer
(B) 12 turns
Step-by-step Solution
Detailed explanation
Current in tangent galvanometer
\(I=\frac{2 r H}{\mu_{0} N} \tan \theta\)
Here, \(R_{1}\) and \(R_{2}\) are in parallel
\(\therefore \frac{1}{R_{n e t}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \Rightarrow=\frac{R_{2}+R_{1}}{R_{1} R_{2}}=\frac{8+8}{8 \times 8}\)
\(R_{n e t}=4 \Omega\)
Hence, \(I=\frac{V}{R}=\frac{4}{4}=1 \mathrm{~A}\)
From Eq. (i), we get
\(\frac{r \tan \theta}{N} =\frac{\mu_{0} I}{2 H} \)
\(\therefore \frac{r_{A} \tan \theta_{A}}{N_{A}} =\frac{r_{B} \tan \theta_{B}}{N_{B}} \)
\(\Rightarrow \frac{8 \times 1}{\sqrt{3} \times 2} =\frac{16 \times \sqrt{3}}{N_{B}} \)
\(N B =12 \text { turns }\)
\(I=\frac{2 r H}{\mu_{0} N} \tan \theta\)
Here, \(R_{1}\) and \(R_{2}\) are in parallel
\(\therefore \frac{1}{R_{n e t}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \Rightarrow=\frac{R_{2}+R_{1}}{R_{1} R_{2}}=\frac{8+8}{8 \times 8}\)
\(R_{n e t}=4 \Omega\)
Hence, \(I=\frac{V}{R}=\frac{4}{4}=1 \mathrm{~A}\)
From Eq. (i), we get
\(\frac{r \tan \theta}{N} =\frac{\mu_{0} I}{2 H} \)
\(\therefore \frac{r_{A} \tan \theta_{A}}{N_{A}} =\frac{r_{B} \tan \theta_{B}}{N_{B}} \)
\(\Rightarrow \frac{8 \times 1}{\sqrt{3} \times 2} =\frac{16 \times \sqrt{3}}{N_{B}} \)
\(N B =12 \text { turns }\)
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