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Question

Physics Question on Current electricity

In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance of 40cm40\, cm from AA. If a 10Ω10 \Omega resistor is connected in series with R1R_1, the null point shifts by 10cm10\, cm. The resistance that should be connected in parallel with (R1+10)Ω(R_1 + 10)\Omega such that the null point shifts back to its initial position is :

A

40Ω40\, \Omega

B

60Ω60\, \Omega

C

20Ω20\, \Omega

D

30Ω30\, \Omega

Answer

60Ω60\, \Omega

Explanation

Solution

R1R2=23\frac{R_{1}}{R_{2}} = \frac{2}{3} ......(i)
R1+10R2=1R1+10=R2\frac{R_{1} +10}{R_{2}} = 1 \Rightarrow R_{1} + 10 = R_{2} ....(ii)
2R23+10=R2\frac{2R_{2}}{3} + 10 = R_{2}
10=R33R2=30Ω10= \frac{R_{3}}{3} \Rightarrow R_{2}= 30 \Omega
&R1=20Ω\& R_{1} = 20 \Omega
30×R30+R30=23\frac{\frac{30\times R}{30+R}}{30} = \frac{2}{3}
R=60ΩR = 60 \Omega