JEE Main 2016 · 10 Apr · Shift 1 · Physics Q27

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PhysicsSingle correct+4 / -1Question 27 of 90

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 Ω. Deflection becomes 20 divisions when resistance taken from resistance box is 4900 Ω. Then we can conclude :

A 2 V battery, a key and a galvanometer G in series with a resistance R from a resistance box.

Resistance of galvanometer is 200 Ω.

Full scale deflection current is 2 mA.

Current sensitivity of galvanometer is 20 μA/division.

Resistance required on R.B. for a deflection of 10 divisions is 9800 Ω.

Answer and worked solution

Official answer: C

Current is proportional to deflection: $\frac2{2400+G}=40k$ and $\frac2{4900+G}=20k$ . Dividing, $4900+G=2(2400+G)$ , so G = 100 Ω. Then 40 divisions correspond to $\frac2{2500}=0.8$ mA, i.e. 20 μA/division (full scale 1 mA). For 10 divisions, $R+G=10^4$ Ω, so R = 9900 Ω. Only (3) is correct.

Topics:Galvanometer experiments Moving coil galvanometer

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JEE Main 2016 · 10 Apr · Shift 1, Physics question 27