**Problem (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,

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

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

**Problem (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,

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

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

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

**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, $l_1+l_2=100 \;\mathrm{cm}$. 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 (IIT JEE 2004): **
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 $C_1$

**Problem (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?

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

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 (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.

- 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. 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 $X=8\;\Omega$.