Question

In the given arrangement of experiment on metre bridge, if AD corresponding to null deflection of the galvanometer is X, what would be its value if the radius of the wire AB is doubled?

• A. X
• B. \(\frac{X}{4}\)
• C. 4 X
• D. 2 X

Answer 1

The Correct Option is A

In a meter bridge experiment, the condition for null deflection in the galvanometer is given by:

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

where:

• $R_1$ and $R_2$ are the resistances in the two arms of the bridge,
• $x$ is the distance from point A to point D where the galvanometer shows null deflection, and
• $l$ is the total length of the wire AB (usually 1 meter).

Initially, the null deflection occurs at a distance $x = X$. If the radius of the wire AB is doubled, its cross-sectional area becomes 4 times the original area ($A' = 4A$), since the area is proportional to the square of the radius ($A = \pi r^2$).

The resistance of a wire is given by $R = \frac{\rho l}{A}$, where $\rho$ is the resistivity. Since the resistivity and length of the wire remain constant, the resistance is inversely proportional to the area. Therefore, when the area becomes 4 times larger, the resistance of the wire becomes $\frac{1}{4}$ times the original resistance. Let the initial resistance of wire AD be $R_{AD}=\frac{\rho X}{A}$ and the resistance of wire DB be $R_{DB}=\frac{\rho (l-X)}{A}$. Then $R_1/R_2 = \frac{R_{AD}}{R_{DB}}=\frac{\frac{\rho X}{A}}{\frac{\rho (l-X)}{A}} = \frac{X}{l-X}$.

Now let $x'$ be the distance corresponding to null deflection when the radius is doubled. Let $r'$ be the new radius $r'=2r$ Then $A'=\pi(r')^2 = \pi(2r)^2 = 4\pi r^2=4A$ The resistance of the wire segment AD changes to $\frac{\rho x'}{A'}=\frac{\rho x'}{4A}$, and the resistance of segment DB changes to $\frac{\rho (l-x')}{4A}$ We have $\frac{R_1}{R_2} = \frac{\frac{\rho x'}{4A}}{\frac{\rho (l-x')}{4A}}=\frac{x'}{l-x'}$

Since the resistances connected to the gaps are not changed, we have $\frac{R_1}{R_2} = \frac{X}{l-X} = \frac{x'}{l-x'}$

Therefore, $x'=X$.

The correct answer is (A) X.

Answer 2

Step 1: Resistance of Wire Segments in terms of Length and Radius:

The resistance of a wire is given by $R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$, where $\rho$ is resistivity, $L$ is length, and $r$ is radius.

Let the initial radius of wire AB be $r$. Let the length AD corresponding to null deflection be X. Then length DB is $(L_{AB} - X)$, where $L_{AB}$ is the total length of wire AB.

$R_{AD} = \rho \frac{X}{\pi r^2}$

$R_{DB} = \rho \frac{(L_{AB} - X)}{\pi r^2}$

Step 2: Ratio of Resistances:

$\frac{R_{AD}}{R_{DB}} = \frac{\rho \frac{X}{\pi r^2}}{\rho \frac{(L_{AB} - X)}{\pi r^2}} = \frac{X}{L_{AB} - X}$

So, the null deflection condition is $\frac{R_1}{R_2} = \frac{X}{L_{AB} - X}$.

Step 3: Effect of Doubling the Radius:

Now, the radius of the wire AB is doubled to $r' = 2r$. The resistivity $\rho$ remains the same as the material of the wire is unchanged.

New resistance of AD, $R'_{AD} = \rho \frac{X}{\pi (r')^2} = \rho \frac{X}{\pi (2r)^2} = \rho \frac{X}{4\pi r^2} = \frac{1}{4} R_{AD}$

New resistance of DB, $R'_{DB} = \rho \frac{(L_{AB} - X)}{\pi (r')^2} = \rho \frac{(L_{AB} - X)}{\pi (2r)^2} = \rho \frac{(L_{AB} - X)}{4\pi r^2} = \frac{1}{4} R_{DB}$

Step 4: New Ratio of Resistances:

$\frac{R'_{AD}}{R'_{DB}} = \frac{\frac{1}{4} R_{AD}}{\frac{1}{4} R_{DB}} = \frac{R_{AD}}{R_{DB}} = \frac{X}{L_{AB} - X}$

The ratio of the resistances of the segments AD and DB remains the same even when the radius of the wire is doubled, for the same length X.

Step 5: Conclusion:

Since the ratio $\frac{R'_{AD}}{R'_{DB}}$ is still equal to $\frac{R_1}{R_2}$ for the same length X, the condition for null deflection is still satisfied at the same position D. 

Final Answer:  Therefore, the value of AD corresponding to null deflection will remain X.