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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.

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.

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}

- Measure the length
*l*of the given wire by using a scale. - Measure the diameter
*D*of the given wire by using a screw gauge. - The resistance of the wire is measured by using a metre bridge. Connect the wire in place of the unknown resistance
*X*. - 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*. - Interchange the positions of
*X*and*R*and repeat steps (4). This takes care of unknown resistances offered by the terminals. - Calculate mean value of
*X*obtained in steps (4) and (5). Use this to calculate resistivity of the wire. - Do error analysis to find error in the resistivity.

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

**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

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

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

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

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,

- $60\pm0.15\;\Omega$
- $135\pm0.56\;\Omega$
- $60\pm0.25\;\Omega$
- $135\pm0.23\;\Omega$

**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 $\Delta R=0.25\;\Omega$.

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,

- $10.2\;\Omega$
- $10.6\;\Omega$
- $10.6\;\Omega$
- $11.1\;\Omega$

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

*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.

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

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

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,

- $3\;\Omega$
- $4\;\Omega$
- $5\;\Omega$
- $6\;\Omega$

**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, $l_1+l_2=100 \;\mathrm{cm}$.

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

- B and C
- C and D
- A and D
- $B_1$ 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 $X=QR/P$.

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?

- $x$
- $x/4$
- $4x$
- $2x$

**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.

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.

- Are there positive and negative terminals on the galvanometer?
- Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at appropriate points.
- 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.

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