Question

Resistors are connected in the two gaps of a meter bridge. The balance point is 20 cm from the zero end. When a $15\Omega$ resistor is connected in series with the smaller resistance then the null point shifts to 40 cm. The smaller resistance is :

• A. $80\Omega$
• B. $9\Omega$
• C. $10\Omega$
• D. $12\Omega$

Answer 1

The correct option is B $9\Omega$

for Case 1
$\frac{R_1}{R_2}=\frac{x}{100-x}$
$x=20cm$ Given
$\Rightarrow \frac{R_1}{R_2}=\frac{20}{80}=1/4\Rightarrow R_2=4R_1$
$\Rightarrow R_1$ is small
Case 2
$x=40cm$
$\frac{R_1+15}{R_2}=\frac{40}{60}\Rightarrow \frac{R_1+15}{{4R}_1}=2/3$
$3R_1+45=8R_1$
$\Rightarrow 5R_1=45$
$R_1=9\Omega$