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With two resistances R1 and R2(> R1)  in the two gaps of a metre bridge, the balanced point was found to be \frac{1}{3}mfrom the zero end. When a 6 \ \Omega resistance is connected in series with the smaller of the two resistances, the point is shifted to \frac{2}{3}m from the same end. The value of R_1 and R_2 respectively is

Option: 1

R_1=2 \Omega \text{ and } R_2=4 \Omega.


Option: 2

R_1=6 \Omega \text{ and } R_2=4 \Omega.


Option: 3

R_1=4 \Omega \text{ and } R_2=2 \Omega.


Option: 4

R_1=2 \Omega \text{ and } R_2=6 \Omega.


Answers (1)

best_answer

\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\ell}{1-\ell}, where \ell is in metre.
\frac{R_1}{R_2}=\frac{1 / 3}{1-1 / 3}=\frac{1}{2}
or, \mathrm{R}_2=2 \mathrm{R}_1
Again, \quad \frac{\mathrm{R}_1+6}{\mathrm{R}_2}=\frac{2 / 3}{1-2 / 3}=2
or, R_1+6=2 R_2
From (i) and (ii)

R_1=2 \Omega \text{ and } R_2=4 \Omega.

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