Question 191: Current Electricity

Question

Two resistors $2 \Omega$ and $3 \Omega$ are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire $X Y$. When an unknown resistor is connected in parallel with $3 \Omega$ resistor, the null point is shifted by 22.5 cm toward $Y$. The resistance of unknown resistor is $\_\_\_\_$ $\Omega$.

Accepted Answer

A meter bridge works on the principle of a Wheatstone bridge. When the bridge is balanced (null point), no current flows through the galvanometer.
The relationship between the resistors in the gaps ( $R_1, R_2$ ) and the balancing lengths ( $1,100-1$ ) is:
$$ \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{l}}{100-\mathrm{l}} $$
Here, $\mathrm{R}_1=2 \Omega, \mathrm{R}_2=3 \Omega$
Length of the wire from the left end $(\mathrm{X})$ to the null point is $\mathrm{l}_1$ then remaining length of the wire (assuming standard 100 cm wire) is $\left(100-l_1ight)$.
$$ \frac{2}{3}=\frac{\mathrm{l}_1}{100-\mathrm{l}_1} $$
$$\Rightarrow $$ $2\left(100-l_1ight)=3 l_1$
$$\Rightarrow $$ $200-2 \mathrm{l}_1=3 \mathrm{l}_1$
$$\Rightarrow $$ $$ 200=5 l_1 \Rightarrow l_1=40 \mathrm{~cm} $$
So, the initial null point is at 40 cm from X.
When an unknown resistor X is connected in parallel with the $3 \Omega$ resistor in the right gap.
The new right gap resistance is $\mathrm{R}_2^{\prime}$.
For two resistors in parallel, the equivalent resistance is:
$$ \mathrm{R}_2^{\prime}=\frac{3 	imes \mathrm{X}}{3+\mathrm{X}} $$
The null point shifts by 22.5 cm towards Y (the right end).
$$ \mathrm{l}_2=\mathrm{l}_1+22.5 \mathrm{~cm} $$
$$\Rightarrow $$ $$ \mathrm{l}_2=40+22.5=62.5 \mathrm{~cm} $$
Using balanced condition,
$$ \frac{\mathrm{R}_1}{\mathrm{R}_2^{\prime}}=\frac{\mathrm{l}_2}{100-\mathrm{l}_2} $$
$$\Rightarrow $$ $\frac{2}{\mathrm{R}_2^{\prime}}=\frac{62.5}{100-62.5}$
$$\Rightarrow $$ $\frac{2}{\mathrm{R}_2^{\prime}}=\frac{62.5}{37.5}=\frac{5}{3}$
$$ \mathrm{R}_2^{\prime}=\frac{6}{5}=1.2 \Omega $$
We know that $R_2^{\prime}$ is the parallel combination of $3 \Omega$ and $X$.
$$ 1.2=\frac{3 X}{3+X} \Rightarrow 1.2(3+X)=3 X $$
$\Rightarrow 3.6+1.2 \mathrm{X}=3 \mathrm{X}$
$\Rightarrow 3.6=3 X-1.2 X$
$\Rightarrow 3.6=1.8 \mathrm{X}$
$$ \Rightarrow \mathrm{X}=2 \Omega $$
Therefore, the value of X is $2 \Omega$. Hence, the correct option is (D).

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