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

Option 1)
60 $\Omega$
Option 2)
40 $\Omega$
Option 3)
20 $\Omega$
Option 4)
30 $\Omega$

Answers (1)

best_answer

Meter bridge -

$\frac{P}{Q}=\frac{R}{S}\Rightarrow S= \frac{(100-l)}{l}R$

- wherein

$AB=l$

$BC=(100-l)$

Initially

$\frac{R_{1}}{R_{2}}=\frac{40}{60}=\frac{2}{3}-----------(1)$

When connected $10\Omega$ in series with $R_{1}$

$\Rightarrow \frac{R_{1}+10}{R_{2}}=\frac{40+10}{60-10}=\frac{50}{50}=1----(2)$

From (1) & (2)

$R_{1}+10=R_{2}$

$&3R_{1}=2R_{2}$

$2R_{1}+20=2R_{2}$

$\Rightarrow R_{1}-20=0$

$R_{1}=20\Omega$

so $R_{2}=30\Omega$

Now connect $R\Omega$ parallel to $(R_{1}+10)\Omega$

$\Rightarrow R\Omega$ parallel to $(30)\Omega$

so $\frac{30\times R}{\frac{30+R}{R_{2}}}=\frac{2}{3}$

$\Rightarrow \left ( \frac{30\times R}{30+R} \right )3=2R_{2}=2\times 30$

$\Rightarrow R=60\Omega$

Option 1)
60 $\Omega$
Option 2)

40 $\Omega$

Option 3)

20 $\Omega$

Option 4)

30 $\Omega$

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