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 Ω resistor is connected in series with R1, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R1 + 10)Ω such that the null point shifts back to its initial position is :

Question

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 Ω resistor is connected in series with R1, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R1 + 10)Ω such that the null point shifts back to its initial position is : (A) 40 Ω (B) 60 Ω (C) 20 Ω (D) 30 Ω

Tardigrade Admin · Accepted Answer

(R1/R2) = (2/3) ......(i) (R1 +10/R2) = 1 ⇒ R1 + 10 = R2 ....(ii) (2R2/3) + 10 = R2 10= (R3/3) ⇒ R2= 30 Ω R1 = 20 Ω ((30× R/30+R)/30) = (2/3) R = 60 Ω

$40 Ω$

$60 Ω$

$20 Ω$

$30 Ω$

$\frac{R _{1}}{R _{2}} = \frac{2}{3}$ ......(i)
$\frac{R _{1} + 10}{R _{2}} = 1 \Rightarrow R_{1} + 10 = R_{2}$ ....(ii)
$\frac{2 R _{2}}{3} + 10 = R_{2}$
$10 = \frac{R _{3}}{3} \Rightarrow R_{2} = 30Ω$
$& R_{1} = 20Ω$
$\frac{\frac{30 \times R}{30 + R}}{30} = \frac{2}{3}$
$R = 60Ω$

40Ω

60Ω

20Ω

30Ω

R2​R1​​=32​ ......(i) R2​R1​+10​=1⇒R1​+10=R2​ ....(ii) 32R2​​+10=R2​ 10=3R3​​⇒R2​=30Ω &R1​=20Ω 3030+R30×R​​=32​ R=60Ω

$40 Ω$

$60 Ω$

$20 Ω$

$30 Ω$

$\frac{R _{1}}{R _{2}} = \frac{2}{3}$ ......(i)
$\frac{R _{1} + 10}{R _{2}} = 1 \Rightarrow R_{1} + 10 = R_{2}$ ....(ii)
$\frac{2 R _{2}}{3} + 10 = R_{2}$
$10 = \frac{R _{3}}{3} \Rightarrow R_{2} = 30Ω$
$& R_{1} = 20Ω$
$\frac{\frac{30 \times R}{30 + R}}{30} = \frac{2}{3}$
$R = 60Ω$