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When two resistances R_1 and R_2 connected in series and introduced into the left gap of a meter bridge and a resistance of 10 \Omega is introduced into the right gap, a null point is found at 60 \mathrm{~cm} from left side. When R_1 and R_2 are connected in parallel and introduced into the left gap, a resistance of  3 \Omega  is introduced into the right-gap to get null point at 40 \mathrm{~cm} from left end. The product of R_1 R_2 is ______________ \Omega^2

Option: 1

30


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \frac{R_1+R_2}{10}=\frac{60}{40} \\ & R_1+R_2=15 \, \, \, \, \, .................(1)\\ & \frac{R_1 R_2}{\left(R_1+R_2\right) \times 3}=\frac{40}{60} \\ & R_1 R_2=30 \end{aligned}

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shivangi.shekhar

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