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) Resistance of galvanometer is 200 Ω (b) Full scale deflection current is 2 mA. (c) Current sensitivity of galvanometer is 20 μ A/division. (d) Resistance required on R.B. for a deflection of 10 divisions is 9800 Ω.

Answer: (c). G is in series with R and the 2 V cell, so I = (2)/(R + G) and deflection ∝ I. 40/20 = (4900 + G)/(2400 + G) ⇒ 4800 + 2G = 4900 + G ⇒ G = 100 Ω, so (a) is false. At 40 div: I = 2/2500 = 0.8 mA, so K = (0.8 mA)/(40) = 20 /div. Full scale = 50 × 20 = 1 mA (b false); for 10 div, I = 0.2 mA, R = 10000 - 100 = 9900 Ω (d false). So the answer is (c).

A galvanometer has a 50 division scale. Battery has no internal… | JEE Main 2016 (10 Apr, Shift 1)

Q177JEE Main 2016 · 10 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments

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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Answer: (c)

G is in series with R and the 2 V cell, so $I = \frac{2}{R + G}$ and deflection $\propto I$ .
$\frac{40}{20} = \frac{4900 + G}{2400 + G} \Rightarrow 4800 + 2G = 4900 + G \Rightarrow G = 100\ \Omega$ , so (a) is false.
At 40 div: $I = \frac{2}{2500} = 0.8\text{ mA}$ , so $K = \frac{0.8\text{ mA}}{40} = 20\ \mu\text{A/div}$ .
Full scale $= 50 \times 20\ \mu\text{A} = 1\text{ mA}$ (b false); for 10 div, $I = 0.2\text{ mA}$ , $R = 10000 - 100 = 9900\ \Omega$ (d false).
So the answer is (c).

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