Specific Resistance of the Material of a Wire using Meter Bridge and Post Office Box

The resistance of a wire of length l and diameter D is given by \begin{align} R=\frac{\rho l}{A}=\frac{4\rho l }{\pi D^2},\nonumber \end{align} where $\rho$ is resistivity (or specific resistance) of the material of the wire. The unit of resistance R is Ohm ($\Omega$) and that of the resistivity is Ohm-metre ($\Omega\text{-m}$). The resistivity is a material property and it is reciprocal of conductivity.

To find the resistivity of the material of the wire, we need to measure length l by using a scale, diameter D by using a screw gauge, and resistance R by using a metre bridge. The metre bridge works on the principle of balanced Wheatstone bridge.

Metre Bridge Construction

The metre bridge consists of a known resistance R, an unknown resistance X, a one metre long constantan wire AB of uniform cross-sectional area, and a galvanometer G. One terminal of the galvanometer is connected to a jockey N. The jockey is free to slide on the wire AB.

Metre Bridge Principle

A Wheatstone bridge is formed by the resistance R, resistance X, resistance of the wire segment AN and resistance of the wire segment NB. The resistances of four arms of the bridge are $R_1=X$, $R_2=R$, $R_3=\rho_c l_1/A_c$, and $R_4=\rho_c l_2/A_c$, where $l_1$ and $l_2$ are lengths AN and NB. Here, $\rho_c$ is the resistivity of the constantan wire and $A_c$ is its cross-sectional area. Note that $l_2=(100\, \mathrm{cm}-l_1)$. The bridge is balanced when galvanometer shows null deflection. In this condition, the unknown resistance is given by \begin{align} X=R \frac{\rho_c l_1/A_c}{\rho_c l_2/A_c}=R\frac{l_1}{100-l_1}.\nonumber \end{align}

Procedural Steps

1. Measure the length l of the given wire by using a scale.
2. Measure the diameter D of the given wire by using a screw gauge.
3. The resistance of the wire is measured by using a metre bridge. Connect the wire in place of the unknown resistance X.
4. Connect the circuit. Slide the jockey on the bridge wire AB to find the null position. The null position N should lie close to the mid-point of AB, say between 30 cm and 70 cm. This improves sensitivity of the metre-bridge. Adjust the known resistance R to do so. Measure the length $l_1$. Reapeat this for five different values of known resistance R.
5. Interchange the positions of X and R and repeat steps (4). This takes care of unknown resistances offered by the terminals.
6. Calculate mean value of X obtained in steps (4) and (5). Use this to calculate resistivity of the wire.
7. Do error analysis to find error in the resistivity.

Questions on Metre Bridge

Question 1: In a metre bridge experiment, it is recommended to have known and unknown resistances of the similar values. This recommendation

A. reduces error in finding position of the null point.
B. makes null point to lie in the middle of the wire.
C. improves sensitivity of the galvanometer.
D. reduces current through the galvanometer.

Question 2: In a metre bridge experiment, two students have identical setup and follow similar procedure. But one of the student uses very low source voltage V. This student gets more error because

A. it reduces sensitivity of the galvanometer.
B. it increases error in position of the null point.
C. it increases temperature of the wire.
D. it reduces current in the wire.

Question 3: In a modified metre bridge, the length of the bridge wire is reduced to 50 cm from 100 cm. This will

A. decrease the error in finding the null point.
B. increase the error in finding the null point.

Question 4: In a metre bridge experiment, the known and the unknown resistances are interchanged to remove

A. index error.
B. end correction error.
C. error due to temperature effect.
D. random error.

Question 5: The bridge wire in metre bridge construction is made of materials like constantan or manganin because

A. they have small resistivity and small temperature coefficient of resistivity.
B. they have large resistivity and large temperature coefficient of resistivity.
C. they have small resistivity and small temperature coefficient of resistivity.
D. they have large resistivity and small temperature coefficient of resistivity.

Solved Problems on Metre Bridge

Problem from IIT JEE 2014

During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of $90\; \Omega$, as shown in the figure. The least count of the scale used in the metre bridge is 1 mm. The unknown resistance is,

1. $60\pm0.15\;\Omega$
2. $135\pm0.56\;\Omega$
3. $60\pm0.25\;\Omega$
4. $135\pm0.23\;\Omega$

Solution: Let $\lambda$ be resistance per unit length (in ohm/cm) of the potentiometer wire. Total length of the wire is 100 cm and null point is obtained at $x=40\;\mathrm{cm}$. The resistances of four branches of Wheatstone bridge are, $R_1=R$, $R_2=90\;\Omega$, $R_3=\lambda x$, and $R_4=\lambda(100-x)$. The Wheatstone bridge is balanced if, \begin{align} &\frac{R_1}{R_2}=\frac{R_3}{R_4}, & &\text{or} & & \frac{R}{90}=\frac{x}{100-x}. \end{align} Solve to get $R=60\;\Omega$.

The least count of scale gives error in measurement of $x$, i.e., $\Delta x=0.1\;\mathrm{cm}$. To find error in $R$, differentiate above equation and simplify to get, \begin{alignat}{2} &\frac{\Delta R}{R}=\frac{\Delta x}{x}+\frac{\Delta x}{100-x}. \end{alignat} Substitute values and then solve to get $\Delta R=0.25\;\Omega$.

Problem from IIT JEE 2011

A metre bridge is set-up as shown, to determine an unknown resistance $X$ using a standard $10\;\Omega$ resistor. The galvanometer shows null point when tapping-key is at 52 cm mark. The end corrections are 1 cm and 2 cm respectively for the ends A and B. The determined value of $X$ is,

1. $10.2\;\Omega$
2. $10.6\;\Omega$
3. $10.6\;\Omega$
4. $11.1\;\Omega$

Solution: Let N be the null point on the wire. Given $\mathrm{AN}=52\;\mathrm{cm}$ and $\mathrm{NB}=100-52=48\;\mathrm{cm}$.

Let $A^\prime$ and $B^\prime$ represents two end points with end corrections i.e., $\mathrm{A^\prime N}=52+1=53\;\mathrm{cm}$ and $\mathrm{NB^\prime}=48+2=50\;\mathrm{cm}$. The resistance of branch $A^\prime N$ is $P=53\lambda$ and that of branch $N^\prime B$ is $Q=50\lambda$, where $\lambda$ is resistance per unit length (in ohm/cm). The Wheatstone bridge is balanced at null condition given by, \begin{align} &\frac{X}{10}=\frac{P}{Q}, & &\implies & &X=\frac{10P}{Q}=\frac{10(53\lambda)}{50\lambda}=10.6\;\Omega.\nonumber \end{align}

Problem from IIT JEE 2008

Statement 1: In a metre bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance.

Statement 2: Resistance of a metal increase with increase in temperature.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, statement 2 is true.

Solution: Let the unknown resistance be $X$ and standard resistance be $R$.

The null point is obtained when Wheatstone bridge is balanced. Thus, $\frac{X}{R}=\frac{\rho l_1/A}{\rho l_2/A}=\frac{l_1}{l_2}$. The resistance of a metal increases with temperature. Thus $X$ increases when its temperature is increased. To obtain the null point at the same location (same $l_1$ and $l_2$), $R$ should be increased as $\Delta R=\frac{l_2}{l_1} \Delta X$.

Problem from IIT JEE 2007

A resistance of $2\;\Omega$ is connected across one gap of a meter-bridge (the length of the wire is 100 cm) and an unknown resistance, greater than $2\;\Omega$, is connected across the other gap. When the resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is,

1. $3\;\Omega$
2. $4\;\Omega$
3. $5\;\Omega$
4. $6\;\Omega$

Solution: Let $X$ be the unknown resistance. The null point is obtained when Wheatstone bridge is balanced i.e., $\frac{X}{2}=\frac{\rho l_1/A}{\rho l_2/A}=\frac{l_1}{l_2}$, where $l_1$ and $l_2$ are as shown in figure. Since $X > 2\;\Omega$, we get $l_1 > l_2$. Also, $l_1+l_2=100 \;\mathrm{cm}$.

When resistances are interchanged, the null point shift by 20 cm. As $X > 2\;\Omega$, the null point will shift towards left i.e., $l_1^\prime=l_1-20$ and $l_2^\prime=l_2+20$. The balance condition gives $\frac{2}{X}=\frac{l_1^\prime}{l_2^\prime}=\frac{l_1-20}{l_2+20}$. Solve to get $X=3\;\Omega$.

Problem from IIT JEE 2004

For the post office box arrangement to determine the value of unknown resistance, the unknown resistance should be connected between,

1. B and C
2. C and D
3. A and D
4. $B_1$ and $C_1$

Solution: The unknown resistance $X$ is connected between point A and D. Let resistance of branch BC be $P$, resistance of branch CD is $Q$, and resistance of rheostat branch AB be $R$. The Wheatstone bridge is balanced when $P/Q=R/X$ which gives $X=QR/P$.

Problem from IIT JEE 2003

In the shown arrangement of the experiment of the meter bridge if AC corresponding to null deflection of galvanometer is $x$, what should be its value if the radius of the wire AB is doubled?

1. $x$
2. $x/4$
3. $4x$
4. $2x$

Solution: If radius of the wire AB is doubled then its resistance ($R=\rho l/A$) becomes one fourth and current ($i=V/R$) becomes four times. However, the potential drop per unit length, $V/l$, remains same.

Problem from IIT JEE 2002

A thin uniform wire AB of length 1 m, an unknown resistance $X$ and a resistance of $12\;\Omega$ are connected by thick conducting strips, as shown in figure. A battery and galvanometer (with a sliding jockey connected to it) are also available. Connections are to be made to measure the unknown resistance $X$ using the principle of Wheatstone bridge. Answer the following questions.

1. Are there positive and negative terminals on the galvanometer?
2. Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at appropriate points.
3. After appropriate connections are made, it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of 60 cm from A. Obtain the value of the resistance $X$.

Solution: The galvanometer does not have positive or negative terminals. The circuit diagram to measure unknown resistance $X$ is given in the figure.

Let $R$ be total resistance of the potentiometer wire of length 100 cm and $N$ be the null point. The resistance of branch AN is $R_\text{AN}=60(R/100)=0.6R$ and that of branch BN is $R_\text{BN}=0.4R$. The balancing condition of Wheatstone bridge, $R_\text{AN}/R_\text{BN}=12/X$, gives $X=8\;\Omega$.

Related

1. Screw Gauge - Principle | Problems
2. Significant Figures - Rules | Problems
3. Measurement error analysis
4. Verification of Ohm's law using voltmeter and ammeter

References and External Links

1. Concepts of Physics Part 2 by HC Verma (Link to Amazon)
2. NCERT User Manual for Experiments (pdf)
3. Metre Bridge Experiment on Amrita-CDAC Olabs