JEE PYQ: Current Electricity Question 70
Question 70 - 2019 (11 Jan 2019 Shift 2)
In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance of 40 cm from A. If a $10,\Omega$ resistor is connected in series with $R_1$, the null point shifts by 10 cm. The resistance that should be connected in parallel with $(R_1 + 10),\Omega$ such that the null point shifts back to its initial position is:
(1) $20,\Omega$
(2) $40,\Omega$
(3) $60,\Omega$
(4) $30,\Omega$
Show Answer
Answer: (3)
Solution
Initially: $\frac{R_1}{R_2} = \frac{2}{3}$ …(i). After adding $10,\Omega$: $\frac{R_1+10}{R_2} = 1 \Rightarrow R_1 + 10 = R_2$ …(ii). From (i) and (ii): $\frac{2R_2}{3} + 10 = R_2$. $R_2 = 30,\Omega$, $R_1 = 20,\Omega$. Required parallel resistance R: $\frac{30 \times R}{30 + R} = 20$. Actually need $\frac{(R_1+10) \cdot x}{R_1+10+x} = R_1 = 20$. $\frac{30x}{30+x} = 20$. $30x = 600 + 20x$. $x = 60,\Omega$.
BETA
