The galvanometer deflection, when key K1 is closed but K2 is open, equals θ 0 (see figure). On closing K2 also and adjusting R2 to 5 Ω, the deflection in galvanometer becomes (θ₀)/(5). The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]: (a) 5 Ω (b) 25 Ω (c) 12 Ω (d) 22 Ω

Answer: (d). K₂ open: Ig0 = (E)/(220 + G) K₂ closed: I = (E)/(220 + (5G)/(G + 5)), Ig = I(5)/(G + 5) = (5E)/(225G + 1100) Ig = Ig05 ⇒ 25(220 + G) = 225G + 1100 ⇒ 200G = 4400 G = 22 Ω So the answer is (d) 22 Ω.

The galvanometer deflection, when key K1 is closed but K2 is open… | JEE Main 2019 (12 Jan, Shift 1)

Q170JEE Main 2019 · 12 JAN · SHIFT 1Difficult

Kirchhoff's Laws & Measuring Instruments

The galvanometer deflection, when key K1 is closed but K2 is open, equals $\theta$ 0 (see figure). On closing K2 also and adjusting R2 to 5 $\Omega$ , the deflection in galvanometer becomes $\frac{\theta_0}{5}$ . The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]:

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Answer: (d)

$K_2$ open: $I_{g0} = \frac{E}{220 + G}$
$K_2$ closed: $I = \frac{E}{220 + \frac{5G}{G + 5}}$ , $I_g = I\frac{5}{G + 5} = \frac{5E}{225G + 1100}$
$I_g = \frac{I_{g0}}{5} \Rightarrow 25(220 + G) = 225G + 1100 \Rightarrow 200G = 4400$
$G = 22\ \Omega$
So the answer is (d) $22\ \Omega$ .

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