JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2016
- Hard
- MCQ
A galvanometer has a \(50\) \(division\) scale. Battery has no internal resistance. It is found that there is deflection of \(40\) \(divisions\) when \(R = 2400\,\Omega \). Deflection becomes \(20\) \(divisions\) when resistance taken from resistance box is \(4900\,\Omega \) . Then we can conclude
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) Current sensitivity of galvanometer is \(20\,\mu A\) \(/ division\)
Step-by-step Solution
Detailed explanation
Let full scale deflection of current \(=1\)
In case \(1,\) when \(\mathrm{R}=2400 \Omega\) and deflection of 40 divisions present.
\(\therefore \quad \frac{40}{50} \mathrm{I}=\frac{V}{G+R}\)
\(\Rightarrow \quad \frac{4}{5} \mathrm{I}=\frac{2}{G+2400}\dots (1)\)
In case \(2,\) when \(\mathrm{R}=4900 \Omega\) and deflection of 20 divisions present
\(\therefore \quad \frac{20}{50} \mathrm{I}=\frac{V}{G+R}\)
\(\Rightarrow \quad \frac{2}{5} \mathrm{I}=\frac{2}{G+4900}\dots (2)\)
From \((1)\) and \((2)\) we get,
\(\frac{4}{2}=\frac{G+4900}{G+2400}\)
\(\Rightarrow 2 G+4800=G+4900\)
\(\Rightarrow \mathrm{G}=100 \Omega\)
Putting value of G in equation (1), we get.
\(\frac{4}{5} \mathrm{I}=\frac{2}{100+2400}\)
\(\Rightarrow \mathrm{I}=1 \mathrm{mA}\)
Current sensitivity \(=\frac{\mathrm{I}}{\text { number of divisions }}\)
\(=\frac{1}{50}\)
\(=0.02 \mathrm{mA} /\) division
\(=20 \mu \mathrm{A} /\) division
Resistance required for deflection of 10 divisions
\(\frac{10}{50} \mathrm{I}=\frac{V}{G+R}\)
\(\Rightarrow \frac{1}{5} \times 1 \times 10^{-3}=\frac{2}{100+R}\)
\(\Rightarrow \mathrm{R}=9900 \Omega\)
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