CUET · PHYSICS · PYQ PAPER 2025
In a galvanometer, there is a deflection of 10 divisions per mA. The resistance of the galvanometer is \( 60 \Omega \). If a shunt of \( 1 \Omega \) is connected to the galvanometer and there are 100 divisions in all, on the scale of the galvanometer, the maximum current that this galvanometer can read is
- A 600 mA
- B 610 mA
- C 10 mA
- D 100 mA
Answer & Solution
Correct Answer
(B) 610 mA
Step-by-step Solution
Detailed explanation
\(I_g = \frac{100 \text{ divisions}}{10 \text{ divisions/mA}} = 10 \text{ mA}\) \(I = I_g \left(1 + \frac{G}{S}\right)\) \(I = 10 \text{ mA} \left(1 + \frac{60 \, \Omega}{1 \, \Omega}\right)\) \(I = 10 \text{ mA} (1 + 60)\) \(I = 10 \text{ mA} \times 61\) \(I = 610 \text{ mA}\)
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