Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of \( 15\,\Omega \) is connected in series with the smaller resistance, then the null point shifts to 40 cm. The value of the smaller resistance is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 9\,\Omega \)
At the balance point of a metre bridge the resistances in the two gaps are in the same ratio as the lengths of wire on the two sides, \( \displaystyle\frac{R_1}{R_2} = \frac{l}{100-l} \). Let \( R_1 \) be the smaller resistance (in the left gap) and \( R_2 \) the other.
First balance (\( l = 20\,\mathrm{cm} \))
\( \displaystyle \frac{R_1}{R_2} = \frac{20}{80} = \frac{1}{4} \Rightarrow R_2 = 4R_1 \)
Second balance (\( 15\,\Omega \) added in series with \( R_1 \), \( l = 40\,\mathrm{cm} \))
\( \displaystyle \frac{R_1+15}{R_2} = \frac{40}{60} = \frac{2}{3} \)
Putting \( R_2 = 4R_1 \):
\( \displaystyle R_1 + 15 = \frac{2}{3}\times4R_1 = \frac{8R_1}{3} \)
\( \displaystyle 15 = \frac{5R_1}{3} \Rightarrow R_1 = 9\,\Omega \)