Experimental Physics

JEE Main level, single correct, medium

53 of 70

Question 53 of 70

In the experimental set up of metre bridge shown in the figure, the null point is obtaine data distance of 40 cm40 \mathrm{~cm} from

A. If a 10Ω10 \Omega resistor is connected in series with R1,\mathrm{R}_{1}, the null point shifts by 10 cm10 \mathrm{~cm}. The resistance that should be connected in parallel with (R1+10)Ω\left(\mathrm{R}_{1}+10\right) \Omega such that the null point shifts back to its initial position is

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Answer: (C) 60Ω60 \Omega

Initially at null deflection R1R2=23\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{2}{3}…(i)\ldots(\mathrm{i})

Finally at null deflection, when null point is shifted R1+10R2=1⇒R1+10=R2\frac{\mathrm{R}_{1}+10}{\mathrm{R}_{2}}=1 \Rightarrow \mathrm{R}_{1}+10=\mathrm{R}_{2}...(ii)

Solving equations (i) and (ii) we get 2R23+10=R2\frac{2 R_{2}}{3}+10=R_{2}

10=R23⇒R2=30Ω10=\frac{\mathrm{R}_{2}}{3} \Rightarrow \mathrm{R}_{2}=30 \Omega

&R1=20Ω\& R_{1}=20 \Omega

Now if required resistance is R\mathrm{R} then 30×R30+R30=23\frac{\frac{30 \times \mathrm{R}}{30+\mathrm{R}}}{30}=\frac{2}{3}

R=60Ω\mathrm{R}=60 \Omega

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