JEE PYQ: Current Electricity Question 74
Question 74 - 2019 (12 Jan 2019 Shift 2)
The galvanometer deflection, when key $K_1$ is closed but $K_2$ is open, equals $\theta_0$ (see figure). On closing $K_2$ also and adjusting $R_2$ 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]:
(1) $5,\Omega$
(2) $22,\Omega$
(3) $25,\Omega$
(4) $12,\Omega$
Show Answer
Answer: (2)
Solution
When $K_1$ closed, $K_2$ open: $i_g = \frac{E}{220 + R_g} = C\theta_0$ …(i). When both closed: $i_g’ = \frac{E}{220 + \frac{5R_g}{5+R_g}} \times \frac{5}{R_g + 5} = \frac{C\theta_0}{5}$ …(ii). Dividing: $\frac{22R_g + 1100}{1100 + 5R_g} = 5$. $5500 + 25R_g = 22R_g + 1100$… Actually: $200R_g = 4400$. $R_g = 22,\Omega$.
BETA
