In meter bridge given below, when X Ω of resistor is connected in parallel to 3W, null point shifts by 10 cm. Find x

Correct Answer: 6 Ω

Step-by-step Explanation:

From the given circuit diagram of the meter bridge:
- The resistance in the left gap is $R_1=2\Omega$.
- The resistance in the right gap is $R_2=3\Omega$.

Let the initial distance of the null point from the left end of the meter bridge wire be $l$ (in cm). The total length of the meter bridge wire is 100 cm.

Using the balancing condition of the meter bridge:

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

Substitute the values of $R_1$ and $R_2$ into the equation:

$\frac{2}{3}=\frac{l}{100-l}$

Cross-multiply to solve for $l$:

$2(100-l)=3l$
$200-2l=3l$
$5l=200$
$l=40\text{cm}$

Thus, the initial null point is at 40 cm from the left end.

Now, a resistor of resistance $X\Omega$ is connected in parallel with the $3\Omega$ resistor in the right gap. This parallel combination reduces the equivalent resistance in the right gap, which is now:

${R_2}^{\text{'}}=\frac{3X}{3+X}$

Since the resistance in the right gap decreases while the resistance in the left gap remains unchanged, the balance point must shift to the right (increase in value) to maintain equilibrium.
Given that the null point shifts by 10 cm, the new balance length $l^{\text{'}}$ becomes:

$l^{\text{'}}=l+10=40+10=50\text{cm}$

Using the balancing condition again with the new values:

$\frac{R_1}{{R_2}^{\text{'}}}=\frac{l^{\text{'}}}{100-l^{\text{'}}}$

Substitute $R_1=2\Omega$ and $l^{\text{'}}=50\text{cm}$:

$\frac{2}{{R_2}^{\text{'}}}=\frac{50}{100-50}=\frac{50}{50}=1$

This gives:

${R_2}^{\text{'}}=2\Omega$

Substitute the expression for ${R_2}^{\text{'}}$ to solve for $X$:

$\frac{3X}{3+X}=2$
$3X=2(3+X)$
$3X=6+2X$
$X=6\Omega$