Question:
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 :
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 :
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Deflection is proportional to current \(I = \frac{2}{R+G}\); take the ratio of the two readings to get \(G\).
Updated On: Nov 11, 2026
- Resistance of galvanometer is 200 \(\Omega\).
- Full scale deflection current is 2 mA.
- Current sensitivity of galvanometer is 20 \(\mu\)A/division.
- Resistance required on R.B. for a deflection of 10 divisions is 9800 \(\Omega\).
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The Correct Option is C
Solution and Explanation
Let the galvanometer resistance be \(G\) and current per division \(k\). Circuit current \(I = \frac{2}{R+G}\) and \(I = k\theta\).
Case 1: \(\frac{2}{2400+G} = 40k\)
Case 2: \(\frac{2}{4900+G} = 20k\)
Dividing: \(\frac{4900+G}{2400+G} = 2 \Rightarrow 4900 + G = 4800 + 2G \Rightarrow G = 100\ \Omega\). So (1) is false.
\(40k = \frac{2}{2500} = 0.8\) mA \(\Rightarrow k = 0.02\) mA \(= 20\ \mu\)A/division. So (3) is true.
Full scale (50 div): \(50\times20\ \mu\mathrm{A} = 1\) mA, so (2) is false.
For 10 divisions: \(I = 0.2\) mA, \(R + 100 = \frac{2}{0.2\times10^{-3}} = 10000\), \(R = 9900\ \Omega\), not 9800, so (4) is false.
Answer: (3) Current sensitivity of galvanometer is 20 \(\mu\)A/division.
Case 1: \(\frac{2}{2400+G} = 40k\)
Case 2: \(\frac{2}{4900+G} = 20k\)
Dividing: \(\frac{4900+G}{2400+G} = 2 \Rightarrow 4900 + G = 4800 + 2G \Rightarrow G = 100\ \Omega\). So (1) is false.
\(40k = \frac{2}{2500} = 0.8\) mA \(\Rightarrow k = 0.02\) mA \(= 20\ \mu\)A/division. So (3) is true.
Full scale (50 div): \(50\times20\ \mu\mathrm{A} = 1\) mA, so (2) is false.
For 10 divisions: \(I = 0.2\) mA, \(R + 100 = \frac{2}{0.2\times10^{-3}} = 10000\), \(R = 9900\ \Omega\), not 9800, so (4) is false.
Answer: (3) Current sensitivity of galvanometer is 20 \(\mu\)A/division.
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