Potentiometer Questions in English

(D) In a potentiometer circuit,the balance point is obtained when the potential difference across the portion of the potentiometer wire equals the $e.m.f.$ of the cell being measured.
At the balance point,no current flows through the galvanometer connected to the cell.
Since no current flows through the cell circuit at the balance point,the internal resistance of the cell and any external resistance connected in series with the cell do not affect the potential difference across the cell terminals.
Therefore,the balance point remains unchanged regardless of the resistance connected in series with the cell.

(A) The total resistance of the circuit is $R_{total} = R_{resistor} + R_{potentiometer} = 15\,\Omega + 5\,\Omega = 20\,\Omega$.
The current flowing through the circuit is $I = \frac{V}{R_{total}} = \frac{2\,V}{20\,\Omega} = 0.1\,A$.
The potential drop across the potentiometer wire is $V_{wire} = I \times R_{potentiometer} = 0.1\,A \times 5\,\Omega = 0.5\,V$.
The potential gradient $k$ is defined as the potential drop per unit length: $k = \frac{V_{wire}}{L} = \frac{0.5\,V}{100\,cm} = 0.005\,V/cm$.

(D) The correct answer is $(D)$. $A$ potentiometer is preferred because it works on the null deflection method. In the balanced condition,no current flows through the galvanometer branch,meaning the potentiometer does not draw any current from the source circuit being measured. Consequently,it measures the true potential difference without causing a voltage drop due to internal resistance,unlike a voltmeter which requires some current to operate.

(A) The potential gradient $(k)$ is defined as the potential drop per unit length of the wire.
Given: $k = 10\,V/m$.
The distance between points $B$ and $C$ is $\Delta l = 60\,cm - 30\,cm = 30\,cm$.
Converting the distance to meters: $\Delta l = 30\,cm = 0.3\,m$.
The potential difference $(V_{BC})$ is given by the formula: $V_{BC} = k \times \Delta l$.
Substituting the values: $V_{BC} = 10\,V/m \times 0.3\,m = 3\,V$.
Therefore,the potential difference between $B$ and $C$ is $3\,V$.

(A) galvanometer has a low internal resistance,but a voltmeter must have a very high resistance to ensure it draws minimal current from the circuit it is measuring.
To convert a galvanometer into a voltmeter,we need to increase its effective resistance. This is achieved by connecting a high resistance in series with the galvanometer coil.
As shown in the diagram,the combination of the galvanometer and the series resistor acts as a voltmeter.
Note: The resistance of an ideal voltmeter should be infinite.

(A) The formula for the internal resistance $r$ of a cell using a potentiometer is given by $r = R \left( \frac{l_1}{l_2} - 1 \right)$,where $R$ is the external resistance,$l_1$ is the balancing length without the external resistance,and $l_2$ is the balancing length with the external resistance.
Given:
$l_1 = 125\,cm$
$l_2 = 100\,cm$
$R = 2\,\Omega$
Substituting the values into the formula:
$r = 2 \left( \frac{125}{100} - 1 \right)$
$r = 2 \left( 1.25 - 1 \right)$
$r = 2 \times 0.25 = 0.5\,\Omega$.
Therefore,the internal resistance of the Daniel cell is $0.5\,\Omega$.

(B) The sensitivity of a potentiometer is inversely proportional to the potential gradient $(P.G.)$.
$P.G. = \frac{V}{L}$,where $V$ is the potential difference across the wire and $L$ is the length of the wire.
To increase sensitivity,we need to decrease the potential gradient.
Since $P.G. \propto \frac{1}{L}$,increasing the length of the potentiometer wire decreases the potential gradient,thereby increasing the sensitivity of the instrument.
Therefore,the correct option is $(b)$.

(B) potentiometer is considered an ideal device for measuring potential difference because,at the balance condition (null point),it does not draw any current from the source being measured.
Since no current is drawn,there is no voltage drop across the internal resistance of the source,and the potential difference measured is equal to the actual electromotive force $(EMF)$ of the source.
Therefore,it does not disturb the potential difference it measures.

(A) The potential drop across a uniform wire is directly proportional to its length,as $V = IR$ and $R = \rho \frac{l}{A}$.
Since the current $I$ and the resistance per unit length are constant,we have $V \propto l$.
Given: Total length $L = 3\,m = 300\,cm$,Total potential $V_{total} = 6\,V$,and the length segment $l = 50\,cm$.
Using the ratio: $\frac{V_{segment}}{V_{total}} = \frac{l}{L}$.
Substituting the values: $\frac{V_{segment}}{6} = \frac{50}{300}$.
$\frac{V_{segment}}{6} = \frac{1}{6}$.
Therefore,$V_{segment} = 1\,V$.

(A) In a potentiometer,the $e.m.f.$ of a cell is directly proportional to the balancing length $(l)$ at which the null point is obtained,i.e.,$E \propto l$.
Therefore,for two cells with $e.m.f.$s $E_1$ and $E_2$ and balancing lengths $l_1$ and $l_2$,the ratio is given by $\frac{E_1}{E_2} = \frac{l_1}{l_2}$.
Given $l_1 = 20 \ cm$ and $l_2 = 30 \ cm$.
Substituting the values: $\frac{E_1}{E_2} = \frac{20}{30} = \frac{2}{3}$.
Thus,the ratio of their $e.m.f.$s is $2/3$.

(A) The potential gradient $k$ of a potentiometer wire is given by $k = V/L$,where $V$ is the potential difference across the wire and $L$ is the length of the wire.
When the length $L$ of the potentiometer wire is increased,the potential gradient $k$ decreases.
The balance point length $l$ is given by the relation $E = kl$,where $E$ is the $EMF$ of the cell being measured.
Since $E$ remains constant and $k$ decreases,the balance point length $l = E/k$ must increase.

(B) In a potentiometer,the balance point is reached when the potential difference $(p.d.)$ across the portion of the potentiometer wire between the positive terminal and the jockey is exactly equal to the $e.m.f.$ of the experimental cell being measured.
At this point,the potential difference across the galvanometer becomes zero,resulting in no current flowing through the galvanometer branch,which is indicated by a null deflection.

(B) In a potentiometer experiment,the balance point is reached when the potential difference across the segment of the potentiometer wire is exactly equal to the electromotive force $(EMF)$ of the cell in the secondary circuit.
Since these two potentials are equal and connected in opposition,the net potential difference across the galvanometer branch becomes zero.
Consequently,no current flows through the galvanometer circuit,and the galvanometer shows a null deflection.

(B) Let the $e.m.f.$'s of the two cells be $E_1$ and $E_2$.
When connected in series with the same polarity,the total $e.m.f.$ is $E_1 + E_2$. The balancing length is $l_1 = 8 \; m$. Thus,$E_1 + E_2 = k l_1 = 8k$,where $k$ is the potential gradient of the potentiometer wire.
When the polarity of one cell is reversed,the total $e.m.f.$ is $E_1 - E_2$ (assuming $E_1 > E_2$). The balancing length is $l_2 = 2 \; m$. Thus,$E_1 - E_2 = k l_2 = 2k$.
Dividing the two equations: $\frac{E_1 + E_2}{E_1 - E_2} = \frac{8k}{2k} = 4$.
Applying componendo and dividendo: $\frac{(E_1 + E_2) + (E_1 - E_2)}{(E_1 + E_2) - (E_1 - E_2)} = \frac{4 + 1}{4 - 1}$.
This simplifies to $\frac{2E_1}{2E_2} = \frac{5}{3}$,so $\frac{E_1}{E_2} = \frac{5}{3}$.

(B) The total resistance of the circuit is $R_{total} = R_{wire} + R_{external} = 15\, \Omega + 5\, \Omega = 20\, \Omega$.
The current flowing through the potentiometer wire is $I = \frac{E}{R_{total}} = \frac{2\, V}{20\, \Omega} = 0.1\, A$.
The potential drop across the potentiometer wire is $V_{wire} = I \times R_{wire} = 0.1\, A \times 15\, \Omega = 1.5\, V$.
The potential gradient $x$ is defined as the potential drop per unit length of the wire: $x = \frac{V_{wire}}{L} = \frac{1.5\, V}{1000\, cm} = \frac{1.5}{1000}\, V/cm = \frac{3}{2000}\, V/cm$.

(D) In a potentiometer circuit,the potential drop across the potentiometer wire must be greater than the $e.m.f.$ of the cells being compared.
If the $e.m.f.$ of the standard cell (which determines the potential gradient) is smaller than the $e.m.f.$ of the experimental cells,the potential drop across the wire will be insufficient to balance the $e.m.f.$ of the test cells.
Consequently,no balance point (null point) will be obtained on the potentiometer wire,making the experiment unsuccessful.
Therefore,option $D$ is the correct answer.

(A) In a potentiometer,the potential drop across a length $l$ of the wire is given by $V = kl$,where $k$ is the potential gradient.
At the null point,the $e.m.f.$ of the cell is balanced by the potential drop across the balancing length $l$,so $E = kl$.
Since $E_1 > E_2$,the balancing length $l_1$ for $E_1$ will be greater than the balancing length $l_2$ for $E_2$ $(l_1 > l_2)$.
Since the null point $C$ corresponds to $l_1$,and $l_2 < l_1$,the new null point for $E_2$ will be at a shorter distance from $A$ than $C$.
Therefore,the null point will shift to the left of $C$.

(B) The formula for the internal resistance $r$ of a cell using a potentiometer is given by $r = R \left( \frac{l_1}{l_2} - 1 \right)$,where $l_1$ is the balancing length without shunt and $l_2$ is the balancing length with shunt $R$.
Given:
Case $1$: $R_1 = 5\,\Omega$,$l_1 = 2\,m$
Case $2$: $R_2 = 10\,\Omega$,$l_2 = 3\,m$
Let $L$ be the balancing length of the cell without any shunt.
For the first case: $r = 5 \left( \frac{L}{2} - 1 \right) = 2.5L - 5$ ... $(i)$
For the second case: $r = 10 \left( \frac{L}{3} - 1 \right) = \frac{10}{3}L - 10$ ... $(ii)$
Equating $(i)$ and $(ii)$:
$2.5L - 5 = \frac{10}{3}L - 10$
$5 = \frac{10}{3}L - 2.5L$
$5 = \frac{10L - 7.5L}{3} = \frac{2.5L}{3}$
$15 = 2.5L \implies L = 6\,m$
Substituting $L = 6$ into equation $(i)$:
$r = 5 \left( \frac{6}{2} - 1 \right) = 5(3 - 1) = 5 \times 2 = 10\,\Omega$.

(A) In a potentiometer, the potential drop across the length $XP$ is given by $V_{XP} = I \cdot R_{XP}$, where $I$ is the current in the potentiometer wire $XY$.
For a balance point to exist, the potential drop across $XP$ must be equal to the $e.m.f.$ of the cell $E$ $(V_{XP} = E)$.
Since the galvanometer deflection decreases as $P$ moves towards $Y$, it implies that the potential drop across $XP$ is still less than $E$ even at the end $Y$ $(V_{XY} < E)$.
To obtain a balance point between $X$ and $Y$, we need to increase the potential gradient along the wire $XY$.
This can be achieved by increasing the current $I$ in the wire $XY$.
Since $I = V / (R_{wire} + R)$, decreasing the external resistance $R$ will increase the current $I$, thereby increasing the potential drop across any segment $XP$ for a given position of $P$.
Thus, decreasing $R$ allows the balance point to be reached within the length $XY$.

(B) $1$. Calculate the current $I$ in the primary circuit:
$I = \frac{V_{total}}{R_{total}} = \frac{10\,V}{4\,\Omega + 5\,\Omega + 1\,\Omega} = \frac{10}{10} = 1\,A$.
$2$. Calculate the potential drop across the potentiometer wire:
The potential difference across the $5\,m$ wire is $V_{wire} = I \times R_{wire} = 1\,A \times 5\,\Omega = 5\,V$.
$3$. Calculate the potential gradient $x$:
$x = \frac{V_{wire}}{L} = \frac{5\,V}{5\,m} = 1\,V/m$.
$4$. The balancing length is $l = 300\,cm = 3\,m$.
The potential difference across the balancing length is $V_{bal} = x \times l = 1\,V/m \times 3\,m = 3\,V$.
$5$. Since the two identical cells of $e.m.f.$ $E$ are connected in parallel,their equivalent $e.m.f.$ is $E_{eq} = E$.
Therefore,$E = 3\,V$.

(A) The potential gradient $x$ is defined as the potential drop per unit length of the wire,given by $x = \frac{V}{L}$.
Since $V = iR$ and resistance $R = \rho \frac{L}{A}$,we can substitute these into the expression:
$x = \frac{iR}{L} = \frac{i (\rho L / A)}{L} = \frac{i \rho}{A}$.
Given:
$i = 0.2 \, A$
$\rho = 40 \times 10^{-8} \, \Omega \cdot m$
$A = 8 \times 10^{-6} \, m^2$
Substituting the values:
$x = \frac{0.2 \times 40 \times 10^{-8}}{8 \times 10^{-6}}$
$x = \frac{8 \times 10^{-8}}{8 \times 10^{-6}} = 1 \times 10^{-2} \, V/m = 10^{-2} \, V/m$.

(A) The potential difference across the entire length of the potentiometer wire is equal to the $e.m.f.$ of the connected cell,which is $2\, V$.
The potential gradient (potential difference per unit length) is given by the formula:
Potential gradient $= \frac{V}{L}$
Where $V = 2\, V$ and $L = 4\, m$.
Therefore,the potential difference per unit length $= \frac{2\, V}{4\, m} = 0.5\, V/m$.

(A) The potential gradient $x$ is defined as the potential drop per unit length of the wire.
First,calculate the total current $I$ flowing through the circuit:
$I = \frac{V}{R_{total}} = \frac{2.5\,V}{20\,\Omega + 80\,\Omega} = \frac{2.5}{100} = 0.025\,A$.
The potential drop across the potentiometer wire of resistance $R = 20\,\Omega$ is:
$V_{wire} = I \times R = 0.025\,A \times 20\,\Omega = 0.5\,V$.
The potential gradient $x$ is given by:
$x = \frac{V_{wire}}{L} = \frac{0.5\,V}{10\,m} = 0.05\,V/m$.
To convert this to $V/mm$,we know $1\,m = 1000\,mm$:
$x = \frac{0.05\,V}{1000\,mm} = 5 \times 10^{-5}\,V/mm$.

(C) Let $E$ be the $EMF$ of the cell and $r$ be its internal resistance. The balance length $l_1 = 60 \ cm$ corresponds to the $EMF$ of the cell: $E = k l_1$,where $k$ is the potential gradient.
When the cell is shunted by a resistance $R' = 6 \ \Omega$,the terminal potential difference $V$ is given by $V = E - Ir = E - \frac{E}{R' + r} r = E \left( \frac{R'}{R' + r} \right)$.
The new balance length $l_2 = 50 \ cm$ corresponds to this terminal potential difference: $V = k l_2$.
Dividing the two equations: $\frac{E}{V} = \frac{l_1}{l_2} = \frac{R' + r}{R'}$.
Substituting the values: $\frac{60}{50} = \frac{6 + r}{6}$.
$1.2 = 1 + \frac{r}{6} \implies 0.2 = \frac{r}{6}$.
$r = 0.2 \times 6 = 1.2 \ \Omega$.

(C) The potential gradient $x$ is defined as the potential drop per unit length of the wire,given by $x = \frac{V_{wire}}{L}$.
Given: Length of wire $L = 10\, m$,Resistance per unit length $\lambda = 1\,\Omega/m$,so total resistance of wire $R = 10\,\Omega$.
$e.m.f.$ of the cell $E = 2.2\, V$.
Let the resistance taken from the resistance box be $R_h$.
The current $I$ in the circuit is $I = \frac{E}{R + R_h} = \frac{2.2}{10 + R_h}$.
The potential drop across the wire is $V_{wire} = I \times R = \frac{2.2 \times 10}{10 + R_h}$.
The potential gradient $x = \frac{V_{wire}}{L} = \frac{2.2 \times 10}{(10 + R_h) \times 10} = \frac{2.2}{10 + R_h}$.
We are given $x = 2.2\, mV/m = 2.2 \times 10^{-3}\, V/m$.
Equating the two: $2.2 \times 10^{-3} = \frac{2.2}{10 + R_h}$.
$10 + R_h = \frac{2.2}{2.2 \times 10^{-3}} = 10^3 = 1000$.
$R_h = 1000 - 10 = 990\,\Omega$.

(A) The potential gradient $x$ is defined as the potential drop per unit length of the wire,given by $x = \frac{V}{L}$.
From Ohm's law,the potential drop $V$ across a wire of length $L$ is $V = IR$,where $R$ is the resistance of the wire.
The resistance $R$ is given by the formula $R = \frac{\rho L}{A}$,where $\rho$ is the resistivity and $A$ is the area of cross-section.
Substituting the expression for $R$ into the potential drop equation: $V = I \left( \frac{\rho L}{A} \right)$.
Now,substituting $V$ into the potential gradient formula: $x = \frac{I \left( \frac{\rho L}{A} \right)}{L}$.
Simplifying the expression,we get $x = \frac{I\rho}{A}$.

(B) The $e.m.f.$ of the cell is $E = k l_1$,where $k$ is the potential gradient and $l_1 = 50\, cm$.
When the cell is shunted by a resistor $R = 2\,\Omega$,the terminal potential difference $V$ is given by $V = k l_2$,where $l_2 = 40\, cm$.
We know that $V = E - Ir$ and $I = \frac{E}{R+r}$,so $V = E \left( \frac{R}{R+r} \right)$.
Substituting the expressions for $E$ and $V$,we get $k l_2 = k l_1 \left( \frac{R}{R+r} \right)$.
This simplifies to $\frac{l_2}{l_1} = \frac{R}{R+r}$,which gives $r = R \left( \frac{l_1}{l_2} - 1 \right)$.
Substituting the given values: $r = 2 \left( \frac{50}{40} - 1 \right) = 2 \left( 1.25 - 1 \right) = 2 \times 0.25 = 0.5\,\Omega$.

(A) The circuit shown in the figure consists of a voltage source connected across a fixed resistor. $A$ sliding contact (wiper) is used to tap a portion of the total potential difference across the resistor. By moving the sliding contact,the output voltage (Variable $P.D.$) can be adjusted from $0$ to the total voltage (Total $P.D.$) across the resistor. This specific arrangement is known as a Potential divider.

(B) The total resistance of the circuit is $R_{total} = R + R_{wire} + r = R + 10 + 1 = R + 11\,\Omega$.
The current $I$ flowing through the circuit is given by $I = \frac{E}{R_{total}} = \frac{2}{R + 11}$.
The potential difference $V$ across the potentiometer wire is $V = I \times R_{wire}$.
Given $V = 1\,mV = 10^{-3}\,V$ and $R_{wire} = 10\,\Omega$.
Substituting the values: $10^{-3} = \left( \frac{2}{R + 11} \right) \times 10$.
$10^{-3} = \frac{20}{R + 11}$.
$R + 11 = \frac{20}{10^{-3}} = 20 \times 10^3 = 20000$.
$R = 20000 - 11 = 19989\,\Omega$.

(A) Let $V$ be the potential difference across the potentiometer wire per unit length (potential gradient).
When the battery $E$ is balanced,its $EMF$ is given by $E = k \cdot l_1$,where $l_1 = 55 \, cm$.
When a resistance $R = 10 \, \Omega$ is connected in parallel with the battery,the terminal voltage $V'$ is balanced,so $V' = k \cdot l_2$,where $l_2 = 50 \, cm$.
The terminal voltage is given by $V' = E \left( \frac{R}{R + r} \right)$.
Dividing the two equations: $\frac{E}{V'} = \frac{l_1}{l_2} = \frac{R + r}{R}$.
Substituting the values: $\frac{55}{50} = \frac{10 + r}{10}$.
$1.1 = 1 + \frac{r}{10} \Rightarrow 0.1 = \frac{r}{10} \Rightarrow r = 1 \, \Omega$.

(D) The potential difference $V$ across the resistor is given by the product of the potential gradient $x$ and the balancing length $l$.
$V = x \times l$
Given,potential gradient $x = 2\, mV/cm = 2 \times 10^{-3}\, V/cm$.
Balancing length $l = 50\, cm$.
Therefore,$V = (2 \times 10^{-3}\, V/cm) \times (50\, cm) = 100 \times 10^{-3}\, V = 0.1\, V$.
Since $V = iR$,where $R = 10\, \Omega$ is the resistance:
$i = V / R = 0.1\, V / 10\, \Omega = 0.01\, A$.
Converting to $mA$,$i = 0.01 \times 1000\, mA = 10\, mA$.

(D) $1$. First,calculate the current $I$ flowing through the potentiometer wire $AB$. The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} = 10\,\Omega + 40\,\Omega = 50\,\Omega$.
$2$. The current $I$ in the circuit is given by $I = \frac{V}{R_{total}} = \frac{2\,V}{50\,\Omega} = 0.04\,A$.
$3$. The potential drop across the potentiometer wire $AB$ is $V_{AB} = I \times R_{wire} = 0.04\,A \times 10\,\Omega = 0.4\,V$.
$4$. The potential gradient $k$ (potential drop per unit length) is $k = \frac{V_{AB}}{L} = \frac{0.4\,V}{100\,cm} = 0.004\,V/cm$.
$5$. The unknown $e.m.f.$ $E$ is balanced by a length $l = 40\,cm$. Therefore,$E = k \times l = 0.004\,V/cm \times 40\,cm = 0.16\,V$.

(C) Let $k$ be the potential gradient of the potentiometer wire.
When the cells are in series and in conjunction,the total $e.m.f.$ is $(E_1 + E_2)$. The balancing length is $l_1 = 58 \ cm$.
So,$(E_1 + E_2) = k \cdot l_1 = 58k$ --- $(1)$
When the polarity of $E_2$ is reversed,the total $e.m.f.$ is $(E_1 - E_2)$. The balancing length is $l_2 = 29 \ cm$.
So,$(E_1 - E_2) = k \cdot l_2 = 29k$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{E_1 + E_2}{E_1 - E_2} = \frac{58k}{29k} = 2$
$E_1 + E_2 = 2(E_1 - E_2)$
$E_1 + E_2 = 2E_1 - 2E_2$
$3E_2 = E_1$
Therefore,$\frac{E_1}{E_2} = \frac{3}{1}$.

(D) An ideal voltmeter has infinite resistance.
In the given circuit,the voltmeter is connected in series with the parallel combination of the $6\,\Omega$ and $4\,\Omega$ resistors.
Because the voltmeter has infinite resistance,no current flows through the circuit $(I = 0)$.
Therefore,there is no potential drop across the resistors.
The reading of an ideal voltmeter connected across a source in an open circuit is equal to the electromotive force $(e.m.f.)$ of the cell.
Thus,the voltmeter reading is $6\,V$.

(C) The potential gradient $x$ is defined as the potential drop per unit length of the wire.
Given,$x = 1\, mV/cm = 10^{-3}\, V/cm = 0.1\, V/m$.
Total length of the wire $L = 100\, cm = 1\, m$.
The potential drop across the wire is $V_{wire} = x \times L = 0.1\, V/m \times 1\, m = 0.1\, V$.
Let the additional resistance be $R_h$. The total resistance in the circuit is $R_{total} = R + R_h = 3 + R_h$.
Using Ohm's law,the current in the circuit is $I = \frac{E}{R_{total}} = \frac{2}{3 + R_h}$.
The potential drop across the wire is also given by $V_{wire} = I \times R$.
Substituting the values: $0.1 = \left( \frac{2}{3 + R_h} \right) \times 3$.
$0.1(3 + R_h) = 6$.
$3 + R_h = 60$.
$R_h = 57\, \Omega$.

(C) The potential gradient $x$ is defined as the potential drop per unit length of the wire,given by $x = \frac{V_{wire}}{L}$.
First,calculate the total resistance of the circuit: $R_{total} = R_{wire} + R_{series} + r = 20 \, \Omega + 10 \, \Omega + 0 \, \Omega = 30 \, \Omega$.
The current $I$ flowing through the circuit is $I = \frac{E}{R_{total}} = \frac{3 \, V}{30 \, \Omega} = 0.1 \, A$.
The potential drop across the wire is $V_{wire} = I \times R_{wire} = 0.1 \, A \times 20 \, \Omega = 2 \, V$.
The potential gradient $x = \frac{V_{wire}}{L} = \frac{2 \, V}{10 \, m} = 0.2 \, V/m$.

(D) Let $E_1$ and $E_2$ be the $e.m.f.$s of the two cells,and $x$ be the potential gradient of the potentiometer wire.
When the cells support each other,the total $e.m.f.$ is $(E_1 + E_2) = x \cdot l_1$,where $l_1 = 6 \ m$.
When the cells oppose each other,the total $e.m.f.$ is $(E_1 - E_2) = x \cdot l_2$,where $l_2 = 2 \ m$.
Dividing the two equations: $\frac{E_1 + E_2}{E_1 - E_2} = \frac{6}{2} = 3$.
Applying componendo and dividendo: $\frac{(E_1 + E_2) + (E_1 - E_2)}{(E_1 + E_2) - (E_1 - E_2)} = \frac{3 + 1}{3 - 1}$.
This simplifies to: $\frac{2E_1}{2E_2} = \frac{4}{2} = 2$.
Therefore,the ratio $\frac{E_1}{E_2} = 2:1$.

(C) The wire of a potentiometer is typically made of alloys like $Manganin$ or $Constantan$.
These materials are chosen because they have a high resistivity and a very low temperature coefficient of resistance.
This ensures that the resistance of the wire remains practically constant even if the temperature changes slightly during the experiment,providing accurate measurements.

(B) The potential difference $V$ across the wire is given by the voltage divider rule: $V = E \times \frac{R}{R + r}$, where $E = 10 \, V$, $R = 3 \, \Omega$, and $r = 3 \, \Omega$.
$V = 10 \times \frac{3}{3 + 3} = 10 \times \frac{3}{6} = 5 \, V$.
The potential gradient $k$ is defined as the potential drop per unit length: $k = \frac{V}{L}$.
Given $L = 500 \, cm = 5 \, m$.
$k = \frac{5 \, V}{500 \, cm} = 0.01 \, V/cm$.
Converting to $mV/cm$: $0.01 \times 1000 = 10 \, mV/cm$.

(C) The potential gradient $(x)$ is defined as the potential drop per unit length of the wire, given by $x = \frac{V}{L}$.
Since $V = IR$ and $R = \frac{\rho L}{A}$, we can substitute these into the expression for the potential gradient:
$x = \frac{I \cdot (\rho L / A)}{L} = \frac{I \rho}{A}$.
Given values are:
Current $(I) = 0.1 \, A$,
Resistivity $(\rho) = 10^{-7} \, \Omega \cdot m$,
Area $(A) = 10^{-6} \, m^2$.
Substituting these values into the formula:
$x = \frac{0.1 \times 10^{-7}}{10^{-6}} = \frac{10^{-1} \times 10^{-7}}{10^{-6}} = \frac{10^{-8}}{10^{-6}} = 10^{-2} \, V/m$.

(A) The principle of a potentiometer states that the potential drop across a length $l$ of the wire is directly proportional to the length,provided the current $i$ in the potentiometer wire remains constant.
$V = kl$,where $k$ is the potential gradient $(k = \frac{E}{L})$.
Given: $L = 100 \, cm$,$E = emf$ of the standard cell,$l = 30 \, cm$.
The $emf$ of the battery being measured is $V = \frac{E}{L} \times l$.
Substituting the values: $V = \frac{E}{100} \times 30 = \frac{30E}{100}$.
Note: At the balance point,no current flows through the battery being measured,so its internal resistance does not affect the reading.

(B) Let the resistance of the potentiometer wire be $R_w = 10 \, \Omega$ and its length be $L = 100 \, cm = 1 \, m$.
The total resistance in the primary circuit is $R_{total} = R + R_w = R + 10$.
The current in the primary circuit is $I = \frac{V}{R + R_w} = \frac{2}{R + 10}$.
The potential drop across the potentiometer wire is $V_w = I \times R_w = \frac{2 \times 10}{R + 10} = \frac{20}{R + 10}$.
The potential gradient $k$ is given by $k = \frac{V_w}{L} = \frac{20}{R + 10} \times \frac{1}{100} = \frac{0.2}{R + 10} \, V/cm$.
The null point is obtained at $l = 40 \, cm$ for a source of $E = 10 \, mV = 10 \times 10^{-3} \, V$.
Using the principle of the potentiometer,$E = k \times l$.
$10 \times 10^{-3} = \left( \frac{0.2}{R + 10} \right) \times 40$.
$0.01 = \frac{8}{R + 10}$.
$R + 10 = \frac{8}{0.01} = 800$.
$R = 800 - 10 = 790 \, \Omega$.

(B) The formula for the internal resistance $r$ of a cell using a potentiometer is given by $r = R \left( \frac{l_1}{l_2} - 1 \right)$,where $R$ is the external shunt resistance,$l_1$ is the balancing length without shunt,and $l_2$ is the balancing length with shunt.
Given:
$l_1 = 150 \ cm$
$l_2 = 100 \ cm$
$R = 2 \ \Omega$
Substituting these values into the formula:
$r = 2 \left( \frac{150}{100} - 1 \right)$
$r = 2 \left( 1.5 - 1 \right)$
$r = 2 \times 0.5 = 1 \ \Omega$.
Therefore,the internal resistance of the cell is $1 \ \Omega$.

(D) The current $I$ flowing through the potentiometer wire $AB$ is given by $I = \frac{V_{source}}{R_{total}} = \frac{5 \, V}{R_{AB} + R_{series}} = \frac{5}{5 + 45} = \frac{5}{50} = 0.1 \, A$.
The potential drop across the wire $AB$ is $V_{AB} = I \times R_{AB} = 0.1 \times 5 = 0.5 \, V$.
The potential gradient $k$ along the wire is $k = \frac{V_{AB}}{L} = \frac{0.5 \, V}{10 \, m} = 0.05 \, V/m$.
For the balancing length $l$ corresponding to an $emf$ of $E = 0.4 \, V$,we have $E = k \times l$.
$0.4 = 0.05 \times l$.
$l = \frac{0.4}{0.05} = 8 \, m$.

(B) The balancing length $l_1$ is proportional to the electromotive force $(E)$ of the cell: $E \propto l_1 = 240 \ cm$.
When the cell is shunted with an external resistance $R = 2 \ \Omega$,the terminal potential difference $V$ is balanced at length $l_2 = 120 \ cm$,so $V \propto l_2$.
The formula for internal resistance $r$ of a cell is given by $r = R \left( \frac{E}{V} - 1 \right)$.
Since $E/V = l_1/l_2$,we substitute the values:
$r = R \left( \frac{l_1}{l_2} - 1 \right)$
$r = 2 \left( \frac{240}{120} - 1 \right)$
$r = 2 \left( 2 - 1 \right)$
$r = 2 \ \Omega$.
Therefore,the internal resistance of the cell is $2 \ \Omega$.

(D) The potential gradient $k$ of a potentiometer is defined as $k = \frac{E}{l}$,where $E$ is the $emf$ and $l$ is the balancing length.
Since the potential gradient remains constant for a given circuit,we have $\frac{E_1}{l_1} = \frac{E_2}{l_2}$.
Given $E_1 = 1.1 \, V$,$l_1 = 140 \, cm$,and $l_2 = 180 \, cm$.
Substituting the values: $\frac{1.1}{140} = \frac{E_2}{180}$.
Solving for $E_2$: $E_2 = \frac{1.1 \times 180}{140}$.
$E_2 = \frac{198}{140} \approx 1.41 \, V$.

(A) The potential gradient $x$ is given by the formula: $x = \frac{V_{wire}}{L} = \frac{I \cdot R}{L}$,where $I = \frac{E}{R + R_{ext}}$.
Given: $L = 1 \, m = 100 \, cm$,$E = 2 \, V$,$R_{ext} = 490 \, \Omega$,and $x = 0.2 \, mV/cm = 0.2 \times 10^{-3} \, V/cm = 0.02 \, V/m$.
The potential drop across the wire is $V_{wire} = x \cdot L = 0.02 \, V/m \times 1 \, m = 0.02 \, V$.
The current in the circuit is $I = \frac{V_{wire}}{R} = \frac{0.02}{R}$.
Also,$I = \frac{E}{R + R_{ext}} = \frac{2}{R + 490}$.
Equating the two expressions for $I$: $\frac{0.02}{R} = \frac{2}{R + 490}$.
$0.02(R + 490) = 2R$.
$0.02R + 9.8 = 2R$.
$1.98R = 9.8$.
$R = \frac{9.8}{1.98} \approx 4.95 \, \Omega$. Given the options,the closest value is $4.9 \, \Omega$.

(A) The potential difference across the balancing length $l$ of the potentiometer wire is equal to the $emf$ of the thermocouple $(E_{th})$.
Given: Balancing length $l = 5 \, m$, resistance per unit length $\rho = 1.2 \, \Omega/m$, and $E_{th} = 2.4 \, mV = 2.4 \times 10^{-3} \, V$.
The potential drop across the balancing length is $V = i \times R_{wire} = i \times (\rho \times l)$.
Since the galvanometer deflection is zero, $V = E_{th}$.
Therefore, $i \times (1.2 \, \Omega/m \times 5 \, m) = 2.4 \times 10^{-3} \, V$.
$i \times 6 \, \Omega = 2.4 \times 10^{-3} \, V$.
$i = \frac{2.4 \times 10^{-3}}{6} \, A = 0.4 \times 10^{-3} \, A = 4 \times 10^{-4} \, A$.

(C) The sensitivity of the galvanometer is $1\,\mu A/\text{div}$. To detect the smallest temperature difference,we need a minimum deflection of $1\,\text{div}$.
Therefore,the minimum current required is $I = 1\,\mu A$.
The resistance of the galvanometer is $R = 10\,\Omega$.
The voltage (e.m.f.) required to produce this current is $V = I \times R = 1\,\mu A \times 10\,\Omega = 10\,\mu V$.
Given that the thermocouple produces an $e.m.f.$ of $40\,\mu V/^{\circ}C$,the temperature difference $\Delta T$ corresponding to $10\,\mu V$ is calculated as:
$\Delta T = \frac{V}{\text{sensitivity of thermocouple}} = \frac{10\,\mu V}{40\,\mu V/^{\circ}C} = 0.25^{\circ}C$.
Thus,the smallest temperature difference that can be detected is $0.25^{\circ}C$.