In the experimental set up of metre bridge shown in the figure, the null point is obtaine data distance of 40 ~cm from A. If a 10 resistor is connected in series with R_ 1 , the null point shifts by 10 ~cm . The

Correct answer: C. 60 Initially at null deflection R_ 1 R_ 2 =(2)/(3) (i) Finally at null deflection, when null point is shifted R_ 1 +10 R_ 2 =1 R_ 1 +10=R_ 2 ...(ii) Solving equations (i) and (ii) we get 2 R_ 2 3 +10=R_ 2 10= R_ 2 3 R_ 2 =30 \& R_ 1 =20 Now if required resistance is R then 30 × R 30+R 30 =(2)/(3) R=60

JEE Main 201911 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 11 January
Shift: Evening Shift
Subject: Physics
Chapter: Experimental Physics

JEE Main 2019 · 11 January · Evening Shift · Physics · Experimental Physics

In the experimental set up of metre bridge shown in the figure, the...

In the experimental set up of metre bridge shown in the figure, the null point is obtaine data distance of

40 \mathrm{~cm}

from A. If a

10 \Omega

resistor is connected in series with

\mathrm{R}_{1},

the null point shifts by

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

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

Options

A $20 \Omega$
B $40 \Omega$
C $60 \Omega$
D $30 \Omega$

Correct answer

C. $60 \Omega$

Step-by-step solution

Initially at null deflection

\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{2}{3}

\ldots(\mathrm{i})

Finally at null deflection, when null point is shifted

\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

\frac{2 R_{2}}{3}+10=R_{2}

10=\frac{\mathrm{R}_{2}}{3} \Rightarrow \mathrm{R}_{2}=30 \Omega

\& R_{1}=20 \Omega

Now if required resistance is

\mathrm{R}

then

\frac{\frac{30 \times \mathrm{R}}{30+\mathrm{R}}}{30}=\frac{2}{3}

\mathrm{R}=60 \Omega

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