Question

In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance of $40$cm from A. If a $10\Omega$ resistor is connected in series with $R_1$, the null point shifts by $10$cm. The resistance that should be connected in parallel with $(R_1+10)\Omega$ such that the null point shifts back to its initial position is?

• A. $40\Omega$
• B. $60\Omega$
• C. $20\Omega$
• D. $30\Omega$

Answer 1

Answer: (B)

$60\Omega$

The null point is at 40 cm mark. So, it divides the resistance of the wire into two segments. One of length 40cm and other of length 60 cm as the total; length of the meter bridge wire is 100cm.

So, the ratio of the resistance will be same as the ratio of the lengths

$\frac{R_1}{R_2}=\frac{40}{60}=\frac{2}{3}$ .(i)

$\frac{R_1+10}{R_2}=1\Rightarrow R_1+10=R_2$ ..(ii)

$\frac{2R_2}{3}+10=R_2$
$10=\frac{R_2}{3}\Rightarrow R_2=30\Omega$

& $R_1=20\Omega$

$\frac{\frac{30\times R}{30+R}}{30}=\frac{2}{3}$
$R=60\Omega$.