Question

Two resistances are connected in the two gaps of a metre bridge. The balance point is $20cm$ from the zero end. When a resistance $15$ $\Omega$ is connected in series with the smaller of two resistances, the null point shifts to $40cm$. The smaller of the two resistances has the value.

• A. $8$$\Omega$
• B. $9$$\Omega$
• C. $10$$\Omega$
• D. $12$$\Omega$

Answer 1

The correct option is B $9$$\Omega$

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

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

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

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

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