MHT CET · Physics · Current Electricity
The deflection in moving coil galvanometer of resistance \(45 \Omega\) falls from 30 divisions to 3 divisions. The length of the shunt wire required to convert galvanometer to ammeter is [specific resistance of material of shunt wire \(=5 \times 10^{-7} \Omega \mathrm{m}\) and area of cross- section of wire \(\left.=4 \times 10^{-7} \mathrm{~m}^{2}\right]\)
- A \(4 \mathrm{~m}\)
- B \(6 \mathrm{~m}\)
- C \(8 \mathrm{~m}\)
- D \(2 \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) \(4 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
As per the Question,
\(I_{g}=\frac{3}{30} I=\frac{1}{10} I_{,}\)
\(\Rightarrow\left(I-I_{G}\right) S=I_{G}G\)
\(\Rightarrow(I-\frac{1}{10} I) S=\frac{1}{10}I G\)
\(\Rightarrow (\frac{9}{10}) I S\) \(=(\frac{1}{10})I G\)
\(\Rightarrow 9 S=G\)
\(\therefore G=45 \quad \therefore S=5\)
\(\therefore S=\frac{S L}{A}\)
\(\Rightarrow \frac{S A}{S}=L=4 \mathrm{~m}\)
\(I_{g}=\frac{3}{30} I=\frac{1}{10} I_{,}\)
\(\Rightarrow\left(I-I_{G}\right) S=I_{G}G\)
\(\Rightarrow(I-\frac{1}{10} I) S=\frac{1}{10}I G\)
\(\Rightarrow (\frac{9}{10}) I S\) \(=(\frac{1}{10})I G\)
\(\Rightarrow 9 S=G\)
\(\therefore G=45 \quad \therefore S=5\)
\(\therefore S=\frac{S L}{A}\)
\(\Rightarrow \frac{S A}{S}=L=4 \mathrm{~m}\)
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