In the experimental setup 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 $\mathrm{R}_{1}$, the null point shifts by 10 cm . The resistance that should be connected in parallel with $\left(\mathrm{R}_{1}+10\right) \Omega$ such that the null point shifts back to its initial position is:

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Current Electricity

In the experimental setup of metre bridge shown in the figure, the null point is obtained at a distance of 40 cm from A . If a

10Ω

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) Ω

such that the null point shifts back to its initial position is:

Answer: (C)

At the balance point of a metre bridge the ratio of the two resistances equals the ratio of the two lengths. Initially the null is at

40

cm, so

\frac{R _{1}}{R _{2}} = \frac{40}{60} = \frac{2}{3}

. Adding

10 Ω

in series with

R_{1}

moves the null to

50

cm, so

\frac{R _{1} + 10}{R _{2}} = \frac{50}{50} = 1

. Solving the two relations gives

R_{1} = 20 Ω

and

R_{2} = 30 Ω

. To return the null to

40

cm, a resistance

R

placed in parallel with

(R_{1} + 10) = 30 Ω

must make that combination

20 Ω

so the ratio is again

\frac{2}{3}

. From

\frac{30 R}{30 + R} = 20

R = 60 Ω

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option