Question

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 of the two. The null point shifts to $40$cm. The value of the smaller resistance in ohms is.

• A. $3$
• B. $6$
• C. $9$
• D. $12$

Answer 1

Answer: (C)

$9$

Given : $l=20$ cm $l^{\prime }=40$ cm

Using $\frac{R_1}{R_2}=\frac{l}{100-l}$

$\therefore$ $\frac{R_1}{R_2}=\frac{20}{100-20}=\frac{1}{4}$ $⟹R_2=4R_1$

Using $\frac{R_1^{\prime }}{R_2^{\prime }}=\frac{l^{\prime }}{100-l^{\prime }}$ where $R_1^{\prime }=R_1+15$ and $R_2^{\prime }=R_2=4R_1$

$\therefore$ $\frac{R_1+15}{4R_1}=\frac{40}{100-40}=\frac{2}{3}$ $⟹R_1=9\Omega$