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JEE MAIN 2022
26-07-2022 S1
Question
Resistance are connected in a meter bridge circuit as shown in the figure. The balancing length $\mathrm{l}_1$ is 40cm. Now an unknown resistance x is connected in series with P and new balancing length is found to be 80cm measured from the same end. Then the value of x will be _____$\Omega$
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Solution
Initially, $\frac{P}{Q}=\frac{40 \mathrm{~cm}}{60 \mathrm{~cm}}=\frac{2}{3}$ Finally, $\frac{P+x}{Q}=\frac{80 \mathrm{~cm}}{20 \mathrm{~cm}}=\frac{4}{1}$ Divide (2) by (1) $$ \begin{aligned} & \frac{P+x}{Q}=4 \times \frac{3}{2}=6 \quad \Rightarrow 1+\frac{x}{P}=6 \Rightarrow \frac{x}{P}=5 \\ & \therefore x=5 P=5 \times 4=20 \Omega \end{aligned} $$
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