Potentiometer Questions in English

(B) Let $V$ be the potential difference across the potentiometer wire.
The potential gradient initially is $k_1 = \frac{V}{\ell}$.
The balancing length for emf $E$ is $\ell_1 = \frac{\ell}{3}$,so $E = k_1 \ell_1 = \frac{V}{\ell} \cdot \frac{\ell}{3} = \frac{V}{3}$.
When the length of the wire is increased by $\ell/2$,the new length is $\ell' = \ell + \frac{\ell}{2} = \frac{3\ell}{2}$.
The new potential gradient is $k_2 = \frac{V}{\ell'} = \frac{V}{3\ell/2} = \frac{2V}{3\ell}$.
For the same cell $E$,the new balancing length $\ell_2$ satisfies $E = k_2 \ell_2$.
Substituting the values: $\frac{V}{3} = \left( \frac{2V}{3\ell} \right) \ell_2$.
Solving for $\ell_2$: $\ell_2 = \frac{V}{3} \cdot \frac{3\ell}{2V} = \frac{\ell}{2}$.

(A) In the initial arrangement,the cells $E_1$ and $E_2$ are connected in series such that their effective $EMF$ is $E_{eff} = E_1 + E_2 = 5 \, V + 7 \, V = 12 \, V$.
Given that the balancing length $l_1 = 6 \, m$,we have $E_{eff} = k \cdot l_1$,where $k$ is the potential gradient.
So,$12 = k \cdot 6 \implies k = 2 \, V/m$.
When the terminals of $E_2$ are reversed,the effective $EMF$ becomes $E'_{eff} = |E_1 - E_2| = |5 \, V - 7 \, V| = 2 \, V$.
Let the new balancing length be $l_2$. Then $E'_{eff} = k \cdot l_2$.
$2 = 2 \cdot l_2 \implies l_2 = 1 \, m$.

(D) Let $x$ be the potential gradient of the potentiometer wire $PQ$.
When the key $K_3$ is plugged in between $2$ and $1$,the potential difference across $R_1$ is balanced:
$V_1 = I R_1 = x l_1$
When the key $K_3$ is plugged in between $3$ and $1$,the potential difference across the series combination of $R_1$ and $R_2$ is balanced:
$V_2 = I (R_1 + R_2) = x l_2$
Dividing the two equations:
$\frac{I R_1}{I (R_1 + R_2)} = \frac{x l_1}{x l_2}$
$\frac{R_1}{R_1 + R_2} = \frac{l_1}{l_2}$
Inverting both sides:
$\frac{R_1 + R_2}{R_1} = \frac{l_2}{l_1}$
$1 + \frac{R_2}{R_1} = \frac{l_2}{l_1}$
$\frac{R_2}{R_1} = \frac{l_2}{l_1} - 1 = \frac{l_2 - l_1}{l_1}$
Therefore,the ratio $\frac{R_1}{R_2} = \frac{l_1}{l_2 - l_1}$.

(D) In a potentiometer experiment,the potential difference $V$ is directly proportional to the balancing length $l$,i.e.,$V = kl$,where $k$ is the potential gradient.
Given,for the standard cell,$E = 1.1 \text{ V}$ balances at $l_1 = 440 \text{ cm}$.
Thus,$1.1 = k \times 440 \implies k = \frac{1.1}{440} \text{ V/cm}$.
The actual potential difference $V_{actual}$ across the resistance,which balances at $l_2 = 220 \text{ cm}$,is:
$V_{actual} = k \times l_2 = \left( \frac{1.1}{440} \right) \times 220 = \frac{1.1}{2} = 0.55 \text{ V}$.
The voltmeter reading is given as $V_{reading} = 0.5 \text{ V}$.
The error in the reading is defined as $\text{Error} = V_{reading} - V_{actual}$.
$\text{Error} = 0.5 - 0.55 = -0.05 \text{ V}$.

(C) potentiometer is preferred over a voltmeter for measuring the $e.m.f.$ of a cell for the following reasons:
$(i)$ When the potentiometer is in the balanced state,no current is drawn from the cell. Thus,the potential difference measured is equal to the actual $e.m.f.$ of the cell.
$(ii)$ The potentiometer wire can be made very long,which increases the potential gradient $(V/L)$,allowing for much greater precision in measurement compared to a standard voltmeter.
Therefore,statements $(i)$ and $(ii)$ are correct.

(C) When the cell is in an open circuit,the balancing length $\ell$ corresponds to the electromotive force $(E)$ of the cell.
$E = K\ell$,where $K$ is the potential gradient.
When the cell is connected to an external resistance $R$,the terminal voltage $V$ is measured.
$V = K\ell'$,where $\ell'$ is the new balancing length.
The internal resistance $r$ is given by the formula: $r = \left(\frac{E - V}{V}\right)R$.
Given that $R = r$,we substitute this into the equation:
$r = \left(\frac{E - V}{V}\right)r$
$1 = \frac{E - V}{V}$
$V = E - V$
$2V = E$
Substituting $E = K\ell$ and $V = K\ell'$:
$2(K\ell') = K\ell$
$\ell' = \ell/2$.

(C) The current $I$ flowing through the potentiometer wire $AB$ is given by $I = \frac{\varepsilon}{R_{AB} + r} = \frac{\varepsilon}{12r + r} = \frac{\varepsilon}{13r}$.
The potential drop across the length $AJ$ (where $AJ = x$) is $V_{AJ} = I \times R_{AJ}$.
Since the resistance of the wire is proportional to its length,$R_{AJ} = \frac{x}{L} \times 12r$.
Thus,$V_{AJ} = \left(\frac{\varepsilon}{13r}\right) \times \left(\frac{x}{L} \times 12r\right) = \frac{12\varepsilon x}{13L}$.
For the galvanometer to show no deflection,the potential drop across $AJ$ must equal the $emf$ of cell $C$,which is $\varepsilon/2$.
So,$\frac{12\varepsilon x}{13L} = \frac{\varepsilon}{2}$.
Solving for $x$:
$\frac{12x}{13L} = \frac{1}{2}$
$24x = 13L$
$x = \frac{13}{24}L$.

(C) The current $i$ in the primary circuit is given by $i = \frac{4}{5 + R}$.
The potential difference across the entire potentiometer wire $AB$ is $V_{AB} = i \times 5 = \frac{20}{5 + R}$.
The potential gradient along the wire is $k = \frac{V_{AB}}{L} = \frac{20}{5 + R} \times \frac{1}{1} = \frac{20}{5 + R} \, V/m$.
The potential difference across a length of $10\, cm$ $(0.1\, m)$ is $V_{AP} = k \times 0.1 = \frac{20}{5 + R} \times 0.1 = \frac{2}{5 + R}$.
Given that $V_{AP} = 5\, mV = 5 \times 10^{-3} \, V$,we have:
$\frac{2}{5 + R} = 5 \times 10^{-3}$
$5 + R = \frac{2}{5 \times 10^{-3}} = \frac{2000}{5} = 400$
$R = 400 - 5 = 395\, \Omega$.

(C) The total $EMF$ of the two cells in series is $E_{eq} = 1.5\, V + 1.5\, V = 3.0\, V$.
The total internal resistance of the two cells is $r_{int} = 0.5\, \Omega + 0.5\, \Omega = 1.0\, \Omega$.
The total resistance of the circuit is $R_{total} = R_{ext} + r_{int} + R_{wire} = 1.0\, \Omega + 1.0\, \Omega + (400\, cm \times 0.01\, \Omega/cm) = 2.0\, \Omega + 4.0\, \Omega = 6.0\, \Omega$.
The current in the circuit is $i = \frac{E_{eq}}{R_{total}} = \frac{3.0\, V}{6.0\, \Omega} = 0.5\, A$.
The voltmeter measures the potential difference across the $50\, cm$ length of the wire.
The resistance of this $50\, cm$ segment is $R_{50} = 50\, cm \times 0.01\, \Omega/cm = 0.5\, \Omega$.
Therefore, the voltmeter reading is $V = i \times R_{50} = 0.5\, A \times 0.5\, \Omega = 0.25\, V$.

(A) The current $I$ in the potentiometer wire $AC$ is given by:
$I = \frac{E}{R_{total}} = \frac{2}{10 + R}$
The potential difference $V_{AC}$ across the entire potentiometer wire is:
$V_{AC} = I \times R_{wire} = \left( \frac{2}{10 + R} \right) \times 10$
The potential gradient $k$ along the wire of length $L = 100 \, cm$ is:
$k = \frac{V_{AC}}{L} = \left( \frac{2}{10 + R} \right) \times \frac{10}{100} = \frac{2}{10(10 + R)}$
The source of $emf$ $E' = 10 \, mV = 10 \times 10^{-3} \, V$ is balanced against a length $l = 40 \, cm$. The balancing condition is:
$E' = k \times l$
$10 \times 10^{-3} = \left( \frac{2}{10(10 + R)} \right) \times 40$
$10^{-2} = \frac{8}{10 + R}$
$10 + R = \frac{8}{10^{-2}} = 800$
$R = 800 - 10 = 790 \, \Omega$.

(B) Let the potential at $B$ be $V_B = 0\,V$. Since the $6\,V$ battery is connected across $AB$,the potential at $A$ is $V_A = 6\,V$.
The potential at any point $x$ (in $cm$) from $A$ on the wire $AB$ is given by $V(x) = V_A - (V_A - V_B) \cdot \frac{x}{L} = 6 - 6 \cdot \frac{x}{100} = 6(1 - 0.01x)$.
The potential at point $C$ is determined by the $4\,V$ battery. Since no current flows through the branch containing the $4\,V$ battery and the $1\,\Omega$ resistor (as it is an open circuit at $C$),the potential at $C$ is equal to the potential of the positive terminal of the $4\,V$ battery relative to $A$.
Specifically,$V_C = V_A - 4\,V = 6\,V - 4\,V = 2\,V$.
We want to find the point $D$ at distance $x$ from $A$ such that $V(D) = V_C = 2\,V$.
Setting $6(1 - 0.01x) = 2$,we get $1 - 0.01x = \frac{2}{6} = \frac{1}{3}$.
$0.01x = 1 - \frac{1}{3} = \frac{2}{3}$.
$x = \frac{2}{3} \cdot 100 = 66.67\,cm$.
Rounding to the nearest integer,we get $x \approx 67\,cm$.

(B) In a potentiometer,the $emf$ $(E)$ of a cell is directly proportional to the balancing length $(l)$ of the wire,i.e.,$E \propto l$.
Given:
$E_1 = 1.25\,V$
$l_1 = 35.0\,cm$
$l_2 = 63.0\,cm$
Using the relation $\frac{E_1}{E_2} = \frac{l_1}{l_2}$,we get:
$E_2 = E_1 \times \frac{l_2}{l_1}$
$E_2 = 1.25 \times \frac{63.0}{35.0}$
$E_2 = 1.25 \times 1.8$
$E_2 = 2.25\,V$
Therefore,the $emf$ of the second cell is $2.25\,V$.

(A) The primary circuit consists of a battery of $E = 5 \, V$ and internal resistance $r = 0.5 \, \Omega$ connected in series with the potentiometer wire $AB$ of resistance $R = 4.5 \, \Omega$.
The total resistance of the primary circuit is $R_{total} = R + r = 4.5 \, \Omega + 0.5 \, \Omega = 5.0 \, \Omega$.
The current flowing through the potentiometer wire $AB$ is $I = \frac{E}{R_{total}} = \frac{5 \, V}{5 \, \Omega} = 1 \, A$.
The potential difference across the wire $AB$ is $V_{AB} = I \times R = 1 \, A \times 4.5 \, \Omega = 4.5 \, V$.
The potential gradient along the wire is $k = \frac{V_{AB}}{L_{AB}} = \frac{4.5 \, V}{3 \, m} = 1.5 \, V/m$.
For no deflection in the galvanometer,the potential difference across the length $AC$ (let it be $\ell$) must be equal to the $EMF$ of the secondary cell $E_1 = 3 \, V$.
Therefore,$V_{AC} = k \times \ell = 3 \, V$.
$1.5 \, V/m \times \ell = 3 \, V$.
$\ell = \frac{3}{1.5} = 2 \, m$.

(B) The current $I$ flowing through the wire of resistance $R = 5 \, \Omega$ is given by the formula $I = \frac{E}{r + R}$, where $E = 3 \, V$ is the $emf$ of the battery and $r = 1 \, \Omega$ is its internal resistance.
Substituting the values: $I = \frac{3}{1 + 5} = \frac{3}{6} = 0.5 \, A$.
The potential difference $V_{wire}$ across the entire length of the wire is $V_{wire} = I \times R = 0.5 \times 5 = 2.5 \, V$.
The length of the wire is $L = 1 \, m$.
The potential gradient $k$ is defined as the potential drop per unit length: $k = \frac{V_{wire}}{L} = \frac{2.5 \, V}{1 \, m} = 2.5 \, V/m$.

(B) Let the additional resistance be $R$.
The total resistance of the circuit is $R_{total} = R + R_{wire} = R + 3\, \Omega$.
The current flowing through the circuit is $I = \frac{E}{R_{total}} = \frac{2}{R + 3}$.
The potential gradient $\phi$ is given as $1\, mV/cm = 10^{-3}\, V/cm = 0.1\, V/m$. Since the wire length is $100\, cm = 1\, m$,the total potential drop across the wire is $V_{wire} = \phi \times L = 10^{-3}\, V/cm \times 100\, cm = 0.1\, V$.
Also,$V_{wire} = I \times R_{wire} = \left( \frac{2}{R + 3} \right) \times 3$.
Equating the two expressions for $V_{wire}$:
$0.1 = \frac{6}{R + 3}$.
$R + 3 = \frac{6}{0.1} = 60$.
$R = 60 - 3 = 57\, \Omega$.

(B) The current $I$ flowing through the potentiometer wire $AB$ is given by $I = \frac{\varepsilon}{R_{wire} + r} = \frac{\varepsilon}{9r + r} = \frac{\varepsilon}{10r}$.
The potential difference across the wire $AB$ is $V_{AB} = I \times R_{wire} = \frac{\varepsilon}{10r} \times 9r = \frac{9\varepsilon}{10}$.
The potential gradient $x$ along the wire $AB$ is $x = \frac{V_{AB}}{L} = \frac{9\varepsilon}{10L}$.
For the galvanometer $G$ to show no deflection,the potential difference across the length $AJ$ must be equal to the $emf$ of cell $C$. Let the length $AJ = \ell$.
Thus,$x \times \ell = \frac{\varepsilon}{2}$.
Substituting the value of $x$,we get $\frac{9\varepsilon}{10L} \times \ell = \frac{\varepsilon}{2}$.
Solving for $\ell$,we get $\ell = \frac{\varepsilon}{2} \times \frac{10L}{9\varepsilon} = \frac{5L}{9}$.

(A) The potential gradient $x$ is given as $0.0025\,V/cm$.
To convert this into $SI$ units $(V/m)$,we multiply by $100$:
$x = 0.0025 \times 100\,V/m = 0.25\,V/m$.
The potential difference $V$ across a length $\ell$ of the wire is given by $V = x\ell$.
Given the standard cell voltage $V = 1.025\,V$,we need to find the null point distance $\ell$:
$\ell = \frac{V}{x} = \frac{1.025}{0.25}$.
$\ell = 4.1\,m$.

(D) Let the potential gradient of the potentiometer wire be $x$.
When key $K_1$ is closed,the potential difference across the resistor $R$ is balanced. The current through this resistor is $I = \frac{E}{2R}$,so the potential difference is $V_1 = I \cdot R = \frac{E}{2R} \cdot R = \frac{E}{2}$.
Thus,$\frac{E}{2} = x \cdot 100 \implies E = 200x$.
When key $K_2$ is closed,the potential difference across the series combination of two resistors $R$ and $R$ is balanced. The current through this combination is $I' = \frac{E}{R+R} = \frac{E}{2R}$.
The potential difference across the combination is $V_2 = I' \cdot (R+R) = \frac{E}{2R} \cdot 2R = E$.
Thus,$E = x \cdot \ell'$,where $\ell'$ is the new balancing length.
Substituting $E = 200x$,we get $200x = x \cdot \ell'$,which gives $\ell' = 200 \, cm$.

(C) 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 in the open circuit and $l_2$ is the balancing length when the resistance $R$ is connected.
Given:
$l_1 = 3.4 \, m$
$l_2 = 1.7 \, m$
$R = 10 \, \Omega$
Substituting the values into the formula:
$r = 10 \left( \frac{3.4}{1.7} - 1 \right)$
$r = 10 (2 - 1)$
$r = 10 \times 1 = 10 \, \Omega$
Therefore,the internal resistance of the cell is $10 \, \Omega$.

(A) Let $V_{AB}$ be the potential difference between $A$ and $B$ and $V_{BC}$ be the potential difference between $B$ and $C$. Let $k$ be the potential gradient of the potentiometer wire.
When connected between $A$ and $B$,the balance length $l_1 = 203.6 \, cm$. Thus,$V_{AB} = k \cdot l_1 = k \cdot 203.6 \, cm$.
When the end is shifted to $C$,the total potential difference is $V_{AC} = V_{AB} + V_{BC}$. The balance length is $l_2 = 24.6 \, cm$. However,since $V_{AC} > V_{AB}$,the balance length should increase. The problem implies the polarity is reversed or the connection is $V_{AB} - V_{BC}$. Given the balance point is $24.6 \, cm$,we have $V_{AB} - V_{BC} = k \cdot 24.6 \, cm$.
Subtracting the two equations:
$V_{AB} = k \cdot 203.6 \, cm$
$V_{AB} - V_{BC} = k \cdot 24.6 \, cm$
$V_{BC} = k \cdot (203.6 - 24.6) = k \cdot 179.0 \, cm$.
Thus,the balance point for $B$ and $C$ is $179.0 \, cm$.

(B) The potential gradient $x$ is given as $0.1 \, mV/cm = 0.1 \times 10^{-3} \, V / (10^{-2} \, m) = 0.01 \, V/m$.
Total potential difference across the $10 \, m$ wire is $V_{wire} = x \times L = 0.01 \, V/m \times 10 \, m = 0.1 \, V$.
The current $I$ flowing through the potentiometer wire is $I = V_{wire} / R_{wire} = 0.1 \, V / 40 \, \Omega = 0.0025 \, A = 1/400 \, A$.
The circuit consists of a $2 \, V$ cell,a resistance box $R$,and the wire resistance $40 \, \Omega$ in series.
Using Ohm's law for the whole circuit: $I = E / (R + R_{wire}) \implies 1/400 = 2 / (R + 40)$.
$R + 40 = 800 \implies R = 760 \, \Omega$.

(A) In a potentiometer experiment,the balancing length $l_1$ corresponds to the $EMF$ of the cell,$E = k l_1$,where $k$ is the potential gradient.
When an external resistance $R$ is connected in parallel,the terminal voltage $V$ is given by $V = E - Ir = E - \frac{E}{R+r} r = E \left( \frac{R}{R+r} \right) = k l_2$.
Thus,$\frac{E}{V} = \frac{l_1}{l_2} = \frac{R+r}{R} = 1 + \frac{r}{R}$.
Given $l_1 = 560 \; cm$ and the change in length is $60 \; cm$,the new balancing length $l_2 = 560 - 60 = 500 \; cm$.
Substituting the values: $\frac{560}{500} = 1 + \frac{r}{10}$.
$1.12 = 1 + \frac{r}{10} \implies 0.12 = \frac{r}{10} \implies r = 1.2 \; \Omega$.
Since $r = \frac{N}{10} \; \Omega$,we have $\frac{N}{10} = 1.2$,which gives $N = 12$.

(D) Let $L = 1200 \; cm = 12 \; m$ be the total length of the potentiometer wire.
Let $I = 60 \; mA = 0.06 \; A$ be the current flowing through the wire.
Let $R$ be the total resistance of the wire.
The potential drop across the entire wire is $V_{wire} = I \times R = 0.06 \times R$.
The potential gradient $\lambda$ is given by $\lambda = \frac{V_{wire}}{L} = \frac{0.06 \times R}{12} = 0.005 \times R \; V/cm$.
For a cell of $emf \; E = 5 \; V$,the null point is at $\ell = 1000 \; cm$.
The balance condition is $E = \lambda \times \ell$.
Substituting the values: $5 = (0.005 \times R) \times 1000$.
$5 = 5 \times R$.
Therefore,$R = 1 \; \Omega$ is incorrect based on the calculation; let's re-evaluate: $5 = \frac{0.06 \times R}{1200} \times 1000$.
$5 = \frac{0.06 \times R}{1.2} = 0.05 \times R$.
$R = \frac{5}{0.05} = 100 \; \Omega$.

(N/A) When the sliding contact $B$ is in the middle of the potentiometer,the resistance $R_{0}$ is divided into two equal parts,each of $R_{0}/2$. The resistance $R$ is connected in parallel with the lower half of the potentiometer (between points $A$ and $B$).
The equivalent resistance $R_{1}$ between points $A$ and $B$ is given by:
$\frac{1}{R_{1}} = \frac{1}{R} + \frac{1}{R_{0}/2} = \frac{1}{R} + \frac{2}{R_{0}} = \frac{R_{0} + 2R}{R \cdot R_{0}}$
$R_{1} = \frac{R \cdot R_{0}}{R_{0} + 2R}$
The total resistance of the circuit between points $A$ and $C$ is the sum of the equivalent resistance $R_{1}$ and the remaining part of the potentiometer resistance $(R_{0}/2)$:
$R_{total} = R_{1} + \frac{R_{0}}{2}$
The total current $I$ flowing from the source $V$ is:
$I = \frac{V}{R_{total}} = \frac{V}{R_{1} + R_{0}/2} = \frac{2V}{2R_{1} + R_{0}}$
The voltage $V_{1}$ across the resistance $R$ is the same as the voltage across the parallel combination $R_{1}$:
$V_{1} = I \cdot R_{1} = \left( \frac{2V}{2R_{1} + R_{0}} \right) \cdot R_{1}$
Substituting $R_{1} = \frac{R \cdot R_{0}}{R_{0} + 2R}$ into the expression for $V_{1}$:
$V_{1} = \frac{2V \cdot \left( \frac{R \cdot R_{0}}{R_{0} + 2R} \right)}{2 \left( \frac{R \cdot R_{0}}{R_{0} + 2R} \right) + R_{0}} = \frac{2V \cdot R \cdot R_{0}}{2R \cdot R_{0} + R_{0}(R_{0} + 2R)} = \frac{2V \cdot R \cdot R_{0}}{2R \cdot R_{0} + R_{0}^{2} + 2R \cdot R_{0}} = \frac{2V \cdot R \cdot R_{0}}{R_{0}^{2} + 4R \cdot R_{0}}$
Dividing numerator and denominator by $R_{0}$:
$V_{1} = \frac{2VR}{R_{0} + 4R}$

(B) $Emf$ of the first cell,$E_{1} = 1.25\; V$.
Balance point of the potentiometer,$l_{1} = 35.0\; cm$.
The cell is replaced by another cell of $emf$ $E_{2}$.
New balance point of the potentiometer,$l_{2} = 63.0\; cm$.
The balance condition for a potentiometer is given by the relation:
$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$
Rearranging for $E_{2}$:
$E_{2} = E_{1} \times \frac{l_{2}}{l_{1}}$
Substituting the given values:
$E_{2} = 1.25 \times \frac{63.0}{35.0} = 1.25 \times 1.8 = 2.25\; V$.
Therefore,the $emf$ of the second cell is $2.25\; V$.

(A-D) The constant $emf$ of the given standard cell is $E_1 = 1.02 \; V$.
The balance point on the wire is $l_1 = 67.3 \; cm$.
When a cell of unknown $emf$ $\varepsilon$ replaces the standard cell,the new balance point on the wire is $l = 82.3 \; cm$.
The relation connecting $emf$ and balance point is $\frac{E_1}{l_1} = \frac{\varepsilon}{l}$.
Therefore,$\varepsilon = \frac{l}{l_1} \times E_1 = \frac{82.3}{67.3} \times 1.02 \approx 1.247 \; V$.
The value of the unknown $emf$ is $1.247 \; V$.
$(b)$ The purpose of using the high resistance of $600 \; k \Omega$ is to protect the galvanometer from high currents when the movable contact is far from the balance point.
$(c)$ No,the balance point is not affected by the presence of this high resistance because at the balance point,the current through the galvanometer is zero,so there is no potential drop across the high resistance.
$(d)$ The method would not work if the driver cell had an $emf$ of $1.0 \; V$ because the $emf$ of the cell to be measured ($1.02 \; V$ or $1.247 \; V$) would be greater than the potential drop across the potentiometer wire,making it impossible to find a balance point.
$(e)$ The circuit would not work well for determining an extremely small $emf$ because the balance point would be very close to end $A$,leading to a large percentage error. To modify the circuit,a series resistor should be connected with the potentiometer wire $AB$ to reduce the potential drop across $AB$ so that it is only slightly greater than the $emf$ being measured.

(A) Let the internal resistance of the cell be $r$.
The balance point of the cell in an open circuit is given by $l_1 = 76.3 \; cm$.
When an external resistance $R = 9.5 \; \Omega$ is connected in parallel to the cell,the new balance point is $l_2 = 64.8 \; cm$.
The formula for the internal resistance $r$ of a cell using a potentiometer is given by:
$r = \left( \frac{l_1 - l_2}{l_2} \right) R$
Substituting the given values:
$r = \left( \frac{76.3 - 64.8}{64.8} \right) \times 9.5$
$r = \left( \frac{11.5}{64.8} \right) \times 9.5$
$r \approx 0.17747 \times 9.5 \approx 1.686 \; \Omega$.
Rounding to two decimal places,the internal resistance of the cell is $1.68 \; \Omega$.

(N/A) potentiometer is a device used to measure potential difference or to compare electromotive forces $(EMF)$ of cells. It is also used to measure internal resistance of a cell.
Principle:
As shown in the figure,a battery with $EMF$ $\varepsilon$ and internal resistance $r$ is connected in series with a resistance box $R$ and a uniform wire $AB$ of length $L$ and resistance per unit length $\rho$.
The total resistance of the circuit is $R_{total} = R + L\rho + r$.
The current flowing through the wire $AB$ is $I = \frac{\varepsilon}{R + L\rho + r}$.
If the length of a segment $AC$ of the wire is $l$,then the resistance of segment $AC$ is $R_{AC} = \rho l$.
The potential difference $V$ across the segment $AC$ is given by:
$V = I \cdot R_{AC} = I \rho l$
Substituting the value of $I$:
$V = \left( \frac{\varepsilon \rho}{R + L\rho + r} \right) l$
Since $\varepsilon, \rho, R, L,$ and $r$ are constants for a given setup,we can write $V = \phi l$,where $\phi = \frac{\varepsilon \rho}{R + L\rho + r}$ is known as the potential gradient.
Principle Statement: The potential difference across any segment of a uniform wire carrying a constant current is directly proportional to the length of that segment.
The unit of potential gradient is $V \cdot m^{-1}$ and its dimensional formula is $[M^1 L^1 T^{-3} A^{-1}]$.

(A) As shown in the figure,a battery having emf $\varepsilon$ and internal resistance $r$,a variable resistance $R$,and a switch $K_{1}$ are connected between the two ends $A$ and $B$ of the potentiometer wire.
To compare the emf of two cells $\varepsilon_{1}$ and $\varepsilon_{2}$,the positive terminals of both cells are connected to point $A$. The negative terminals of the cells are connected to points $1$ and $2$ of a two-way switch. Terminal $3$ of the switch is connected to a galvanometer $(G)$ and a jockey in series. The jockey can be moved along the potentiometer wire.
First,points $1$ and $3$ of the switch are connected,placing cell $\varepsilon_{1}$ in the circuit. By sliding the jockey along the wire,a balance point $N_{1}$ is obtained such that the galvanometer shows zero deflection. Let $AN_{1} = l_{1}$.
By applying Kirchhoff's second law for the loop $AN_{1}G31A$:
$\phi l_{1} - \varepsilon_{1} = 0$
$\therefore \varepsilon_{1} = \phi l_{1} \quad .....(1)$
where $\phi$ is the potential gradient of the wire.
Next,points $2$ and $3$ of the switch are connected,placing cell $\varepsilon_{2}$ in the circuit. By sliding the jockey,a balance point $N_{2}$ is obtained such that the galvanometer shows zero deflection. Let $AN_{2} = l_{2}$.
By applying Kirchhoff's second law for the loop $AN_{2}G32A$:
$\phi l_{2} - \varepsilon_{2} = 0$
$\therefore \varepsilon_{2} = \phi l_{2} \quad .....(2)$
Dividing equation $(1)$ by $(2)$:
$\frac{\varepsilon_{1}}{\varepsilon_{2}} = \frac{\phi l_{1}}{\phi l_{2}} = \frac{l_{1}}{l_{2}}$
Thus,the ratio of the emfs of the two cells is equal to the ratio of their respective balancing lengths.

(N/A) To measure the internal resistance $(r)$ of a cell $(\varepsilon)$,a potentiometer circuit is set up as shown in the figure.
The primary circuit consists of a battery $(B)$,a variable resistor $(R)$,and a key $(K_1)$ connected in series with the potentiometer wire $AC$.
The cell $(\varepsilon)$ whose internal resistance $(r)$ is to be measured is connected in parallel with a resistance box $(R_{ext})$ and a key $(K_2)$. The positive terminal of the cell is connected to point $A$,and the negative terminal is connected to a galvanometer $(G)$,which is then connected to a jockey.
$1$. With key $K_2$ open,the cell $(\varepsilon)$ is in an open circuit. The null point $N_1$ is found on the wire $AC$ such that the galvanometer shows zero deflection. Let the balancing length be $AN_1 = l_1$. Since the cell is in an open circuit,the potential difference across it is equal to its $EMF$ $(\varepsilon)$.
$\varepsilon = \phi l_1$ ... $(1)$,where $\phi$ is the potential gradient along the wire.
$2$. Now,close the key $K_2$ so that a current flows through the resistance box $(R_{ext})$ and the cell. The null point $N_2$ is found on the wire $AC$. Let the balancing length be $AN_2 = l_2$. In this condition,the potential difference across the cell is equal to its terminal voltage $(V)$.
$V = \phi l_2$ ... $(2)$.
Dividing equation $(1)$ by $(2)$,we get:
$\frac{\varepsilon}{V} = \frac{l_1}{l_2}$
We know that for a cell,$\varepsilon = I(R_{ext} + r)$ and $V = IR_{ext}$,so $\frac{\varepsilon}{V} = \frac{R_{ext} + r}{R_{ext}} = 1 + \frac{r}{R_{ext}}$.
Equating the two expressions for $\frac{\varepsilon}{V}$:
$1 + \frac{r}{R_{ext}} = \frac{l_1}{l_2}$
$\frac{r}{R_{ext}} = \frac{l_1}{l_2} - 1 = \frac{l_1 - l_2}{l_2}$
$r = R_{ext} \left( \frac{l_1 - l_2}{l_2} \right)$

(N/A) The potential gradient is defined as the fall of potential per unit length along the current-carrying wire.
Mathematically,it is given by $k = \frac{V}{L}$,where $V$ is the potential difference across the wire and $L$ is the length of the wire.
The $SI$ unit of potential gradient is $\text{volt per meter}$ $(V/m)$.

(N/A) potentiometer is a versatile instrument used for measuring potential differences and internal resistance. Its main advantages are:
$1$. It works on the principle of null deflection,meaning no current is drawn from the source of $EMF$ being measured. Therefore,it measures the true $EMF$ of the cell.
$2$. It does not affect the circuit conditions because it does not draw any current from the circuit under test.
$3$. It is highly sensitive and can be used to measure very small potential differences.
$4$. It can be used to compare the EMFs of two cells and to determine the internal resistance of a cell.

(C) In a potentiometer, instead of using a single very long wire, multiple wires of $1 \,m$ length are connected in series using thick metallic strips.
These thick metallic strips have a very low cross-sectional resistance compared to the potentiometer wire.
By using thick strips, the resistance of the joints becomes negligible, ensuring that the potential drop occurs only across the potentiometer wire itself.
This design allows for a compact arrangement while maintaining the accuracy of the potential gradient along the wire.

(B) In a potentiometer circuit,the potential difference across the wire $AB$ is given by $V_{AB} = I \cdot R_{AB}$,where $I$ is the current in the primary circuit and $R_{AB}$ is the resistance of the wire $AB$.
The current $I$ in the primary circuit is given by $I = \frac{E}{R + R_{AB}}$,where $E$ is the $EMF$ of the primary cell and $R$ is the external resistance.
When the value of $R$ is increased,the total resistance of the primary circuit increases,which causes the current $I$ to decrease.
Since $V_{AB} = I \cdot R_{AB}$,a decrease in $I$ leads to a decrease in the potential difference $V_{AB}$ across the wire $AB$.
The potential gradient $k$ is defined as $k = \frac{V_{AB}}{L}$,where $L$ is the length of the wire $AB$. Thus,the potential gradient $k$ decreases.
The balance point $J$ is obtained when the $EMF$ of the secondary cell $\varepsilon$ equals the potential drop across the length $AJ$,i.e.,$\varepsilon = k \cdot AJ$.
Therefore,$AJ = \frac{\varepsilon}{k}$. Since $k$ decreases,the length $AJ$ must increase to maintain the balance condition.
As the length $AJ$ increases,the balance point $J$ shifts towards $B$.

(N/A) $(i)$ When the jockey is moved from $A$ to $B$,if the galvanometer deflection decreases,it implies that the potential difference across the galvanometer circuit is decreasing. This happens because the potential drop across the length of the potentiometer wire $A$-jockey opposes the $EMF$ $E_1$. For the deflection to decrease as the jockey moves towards $B$ (where potential increases),the terminal of $E_1$ connected to $X$ must be the positive terminal,and $E_1 > E$ must hold true so that the net current is in a direction that opposes the initial deflection.
$(ii)$ When the jockey is moved from $A$ to $B$,if the deflection increases,it implies that the potential difference across the galvanometer circuit is increasing. This occurs when the terminal of $E_1$ connected to $X$ is the negative terminal,causing the $EMF$ $E_1$ to add to the potential drop across the wire segment,thereby increasing the current through the galvanometer.

(D) Let the resistance of the potentiometer wire be $R'$ and its total length be $L = 4 \, m$ (assuming $4$ segments of $1 \, m$ each).
$1$. When $R = 50 \, \Omega$,the potential drop across the wire is $V_{wire} = \frac{10 \times R'}{50 + R'}$. Since no null point is found for $E_1 \approx 8 \, V$,the potential drop across the entire wire must be less than $E_1$.
$\frac{10 R'}{50 + R'} < 8 \Rightarrow 10 R' < 400 + 8 R' \Rightarrow 2 R' < 400 \Rightarrow R' < 200 \, \Omega$.
$2$. When $R = 10 \, \Omega$,the null point is on the $4^{th}$ segment,meaning the balance length $l$ is between $3 \, m$ and $4 \, m$. The potential drop across the wire is $V'_{wire} = \frac{10 \times R'}{10 + R'}$.
The condition for the null point to be on the $4^{th}$ segment is that the potential drop across $3 \, m$ is less than $8 \, V$ and the potential drop across $4 \, m$ is greater than $8 \, V$.
$\frac{3}{4} V'_{wire} < 8 < V'_{wire} \Rightarrow \frac{3}{4} \left( \frac{10 R'}{10 + R'} \right) < 8 < \frac{10 R'}{10 + R'}$.
From $8 < \frac{10 R'}{10 + R'}$,we get $80 + 8 R' < 10 R' \Rightarrow 2 R' > 80 \Rightarrow R' > 40 \, \Omega$.
From $\frac{7.5 R'}{10 + R'} < 8$,we get $7.5 R' < 80 + 8 R' \Rightarrow -0.5 R' < 80$ (always true for positive $R'$).
Combining with the first condition $R' < 200 \, \Omega$,the resistance $R'$ is in the range $40 \, \Omega < R' < 200 \, \Omega$.
The potential gradient $\phi = \frac{V'_{wire}}{4} = \frac{10 R'}{4(10 + R')} \, V/m$.

(A) In a potentiometer,the balancing length $l$ is always measured from the high potential end $P$.
Given the total length of the wire $PQ = 100\,cm$.
The null point is obtained at $49\,cm$ from $Q$.
Therefore,the balancing length from $P$ is $l = 100\,cm - 49\,cm = 51\,cm$.
The emf of the cell $E_{2}$ is balanced by the potential drop across the length $l$ of the potentiometer wire.
The formula for the emf is $E_{2} = \phi \times l$,where $\phi$ is the potential gradient.
Substituting the given values: $1.02\,V = \phi \times 51\,cm$.
Solving for $\phi$: $\phi = \frac{1.02}{51}\,V/cm = 0.02\,V/cm$.
Thus,the potential gradient in the potentiometer wire is $0.02\,V/cm$.

(C) The potential drop per unit length of the potentiometer wire is $k = \frac{V_{AB}}{L}$,where $L = 10\, m = 1000\, cm$.
For the first battery $E_{1}$,the balancing length $l_{1}$ is measured from point $A$. The wire consists of $10$ segments of $1\, m$ each. $J_{1}$ is at $20\, cm$ on the second segment,so $l_{1} = 100 + 20 = 120\, cm$.
For the second battery $E_{2}$,the balancing length $l_{2}$ is measured from point $A$. $J_{2}$ is at $60\, cm$ on the eighth segment,so $l_{2} = 700 + 60 = 760\, cm$.
Using the principle of the potentiometer,$E_{1} = k l_{1}$ and $E_{2} = k l_{2}$.
Therefore,$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}} = \frac{120}{760} = \frac{12}{76} = \frac{3}{19}$.
Given $\frac{E_{1}}{E_{2}} = \frac{a}{b}$,we have $a = 3$ and $b = 19$.

(A) In a potentiometer, the potential drop per unit length $(\phi)$ is constant.
The balance length $(l)$ is directly proportional to the $EMF$ $(E)$ of the cell, i.e., $E = \phi l$.
Therefore, the ratio is given by: $\frac{E_1}{E_2} = \frac{l_1}{l_2}$.
Given: $E_1 = 1.5\, V$, $l_1 = 36\, cm$, $E_2 = 2.5\, V$.
Substituting the values: $\frac{1.5}{2.5} = \frac{36}{l_2}$.
Simplifying the ratio: $\frac{3}{5} = \frac{36}{l_2}$.
Solving for $l_2$: $l_2 = \frac{36 \times 5}{3} = 12 \times 5 = 60\, cm$.

(C) The maximum voltage that can be measured by a potentiometer is equal to the potential drop across the entire length of the potentiometer wire $AB$.
First,calculate the total resistance of the wire $AB$:
Length of wire $AB = 10 \, m = 1000 \, cm$.
Resistance per unit length = $0.1 \, \Omega/cm$.
Total resistance $R_{AB} = 1000 \, cm \times 0.1 \, \Omega/cm = 100 \, \Omega$.
Now,calculate the potential drop across $AB$ using the voltage divider rule:
The circuit consists of a $6 \, V$ battery,an internal resistance of $20 \, \Omega$,and the potentiometer wire resistance $R_{AB} = 100 \, \Omega$ in series.
The current in the primary circuit is $I = \frac{V}{R_{total}} = \frac{6 \, V}{20 \, \Omega + 100 \, \Omega} = \frac{6}{120} \, A = 0.05 \, A$.
The potential drop across $AB$ is $V_{AB} = I \times R_{AB} = 0.05 \, A \times 100 \, \Omega = 5 \, V$.
Thus,the maximum emf that can be measured is $5 \, V$.

(C) The total resistance of the potentiometer wire is $10\, \Omega$. When the sliding contact is in the middle,the wire is divided into two parts of $5\, \Omega$ each.
Let the potential at the node where the $2\, \Omega$ resistor and the potentiometer wire meet be $V_0$. The potential at the start of the wire is $20\, \text{V}$ and at the end is $0\, \text{V}$.
Applying Kirchhoff's Current Law $(KCL)$ at node $V_0$:
$\frac{V_0 - 20}{5} + \frac{V_0 - 0}{5} + \frac{V_0 - 20}{2} = 0$
Multiply by $10$ to simplify:
$2(V_0 - 20) + 2(V_0) + 5(V_0 - 20) = 0$
$2V_0 - 40 + 2V_0 + 5V_0 - 100 = 0$
$9V_0 = 140$
$V_0 = \frac{140}{9}\, \text{V}$
The potential drop across the $2\, \Omega$ resistor is the difference between the potential at the start $(20\, \text{V})$ and the node $V_0$:
$\Delta V = 20 - V_0 = 20 - \frac{140}{9} = \frac{180 - 140}{9} = \frac{40}{9}\, \text{V}$

(C) In a potentiometer,the balancing length $l$ is directly proportional to the emf $\varepsilon$ of the cell,i.e.,$\varepsilon = kl$,where $k$ is the potential gradient.
When the galvanometer is connected to point $(1)$,only cell $\varepsilon_{1}$ is in the circuit:
$\varepsilon_{1} = k l_{1} = k(250) \ldots (i)$
When the galvanometer is connected to point $(2)$,both cells $\varepsilon_{1}$ and $\varepsilon_{2}$ are in the circuit in series:
$\varepsilon_{1} + \varepsilon_{2} = k l_{2} = k(400) \ldots (ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{\varepsilon_{1}}{\varepsilon_{1} + \varepsilon_{2}} = \frac{250}{400} = \frac{5}{8}$
Cross-multiplying gives:
$8 \varepsilon_{1} = 5 \varepsilon_{1} + 5 \varepsilon_{2}$
$3 \varepsilon_{1} = 5 \varepsilon_{2}$
Therefore,the ratio is:
$\frac{\varepsilon_{1}}{\varepsilon_{2}} = \frac{5}{3}$

(C) The total resistance of the primary circuit is $R_{total} = R_{wire} + R_{external} = 20 \,\Omega + 30 \,\Omega = 50 \,\Omega$.
The current in the primary circuit is $I = \frac{V}{R_{total}} = \frac{25 \,V}{50 \,\Omega} = 0.5 \,A$.
The potential drop across the entire potentiometer wire is $V_{wire} = I \times R_{wire} = 0.5 \,A \times 20 \,\Omega = 10 \,V$.
The potential gradient along the wire is $k = \frac{V_{wire}}{L} = \frac{10 \,V}{10 \,m} = 1 \,V/m$.
The balancing length is $l = 250 \,cm = 2.5 \,m$.
The emf of the cell is $E = k \times l = 1 \,V/m \times 2.5 \,m = 2.5 \,V$.
Given $E = \frac{x}{10}$,we have $2.5 = \frac{x}{10}$,which implies $x = 25$.

(A) In a potentiometer,the balancing length $\ell$ is directly proportional to the emf $\varepsilon$ of the cell,given by $\varepsilon \propto \ell$ or $\frac{\varepsilon_1}{\varepsilon_2} = \frac{\ell_1}{\ell_2}$.
Given $\varepsilon_1 : \varepsilon_2 = 3 : 2$ and $\ell_1 = 75 \, cm$.
Substituting the values: $\frac{3}{2} = \frac{75}{\ell_2}$.
Solving for $\ell_2$: $\ell_2 = \frac{75 \times 2}{3} = 50 \, cm$.
The difference in the balancing lengths is $\Delta \ell = |\ell_1 - \ell_2| = |75 - 50| = 25 \, cm$.

(B) Let $E$ be the $EMF$ of the cell and $r$ be its internal resistance. The terminal potential difference $V$ across the cell when shunted by a resistance $R$ is given by $V = E \left( \frac{R}{R+r} \right)$.
Since the balancing length $l$ is directly proportional to the terminal potential difference $V$,we have $V \propto l$.
For the first case: $V_1 \propto l_1 = 3 \; m$ with $R_1 = 8 \; \Omega$. Thus,$V_1 = k \cdot 3 = E \left( \frac{8}{8+r} \right)$.
For the second case: $V_2 \propto l_2 = 2 \; m$ with $R_2 = 4 \; \Omega$. Thus,$V_2 = k \cdot 2 = E \left( \frac{4}{4+r} \right)$.
Dividing the two equations: $\frac{3}{2} = \frac{E \left( \frac{8}{8+r} \right)}{E \left( \frac{4}{4+r} \right)} = \frac{8}{8+r} \times \frac{4+r}{4} = \frac{2(4+r)}{8+r}$.
Solving for $r$: $3(8+r) = 4(4+r) \implies 24 + 3r = 16 + 4r \implies r = 8 \; \Omega$.

(B) Let the potential gradient of the potentiometer wire be $k$.
For the first cell,the balance point is at $l_1 = 36\, cm$ for emf $E_1 = 1.20\, V$.
$E_1 = k \cdot l_1 \implies 1.20 = k \times 36 \implies k = \frac{1.20}{36} = \frac{1}{30}\, V/cm$.
For the second cell,the balance point is at $l_2$ for emf $E_2 = 1.80\, V$.
$E_2 = k \cdot l_2 \implies 1.80 = \frac{1}{30} \times l_2$.
$l_2 = 1.80 \times 30 = 54\, cm$.
The difference in balancing lengths is $\Delta l = l_2 - l_1 = 54 - 36 = 18\, cm$.

(D) Let the resistance of the potentiometer wire be $R$.
The current $i$ flowing through the potentiometer wire is given by $i = \frac{4}{R + 780}$.
The potential difference across the entire potentiometer wire $AB$ is $V_{AB} = iR = \frac{4R}{R + 780}$.
The potential gradient along the wire is $k = \frac{V_{AB}}{L} = \frac{4R}{(R + 780) \times 300}$.
The null point is obtained at a length $l = 60\,cm$ for a cell of emf $E = 20\,mV = 20 \times 10^{-3}\,V$.
At the null point,the potential difference across the length $l$ is equal to the emf of the cell:
$E = k \times l$
$20 \times 10^{-3} = \left( \frac{4R}{(R + 780) \times 300} \right) \times 60$
Simplifying the equation:
$20 \times 10^{-3} = \frac{4R \times 60}{(R + 780) \times 300}$
$0.02 = \frac{4R}{5(R + 780)}$
$0.1(R + 780) = 4R$
$0.1R + 78 = 4R$
$3.9R = 78$
$R = \frac{78}{3.9} = 20\,\Omega$.

(A) The potential difference across the potentiometer wire $AB$ is given by $V_{AB} = I \times R_{AB}$,where $I$ is the current in the primary circuit.
The current $I$ is given by $I = \frac{E}{R + R_{AB}}$,where $E = 4\,V$ and $R_{AB} = 20\,\Omega$.
So,$V_{AB} = \left( \frac{4}{R + 20} \right) \times 20 = \frac{80}{R + 20}$.
The potential gradient along the wire is $k = \frac{V_{AB}}{L}$,where $L = 300\,cm$.
The emf of the secondary cell is $E' = 20\,mV = 20 \times 10^{-3}\,V$,and the null point length is $l = 60\,cm$.
At the null point,$E' = k \times l = \left( \frac{V_{AB}}{L} \right) \times l$.
Substituting the values: $20 \times 10^{-3} = \left( \frac{80}{R + 20} \right) \times \left( \frac{60}{300} \right)$.
$20 \times 10^{-3} = \left( \frac{80}{R + 20} \right) \times \left( \frac{1}{5} \right)$.
$20 \times 10^{-3} = \frac{16}{R + 20}$.
$R + 20 = \frac{16}{20 \times 10^{-3}} = \frac{16}{0.02} = 800$.
$R = 800 - 20 = 780\,\Omega$.

(A) In a potentiometer,the balancing length $l$ is directly proportional to the $e.m.f.$ of the cell,i.e.,$E \propto l$ or $E = kl$,where $k$ is the potential gradient.
When the cells are connected in series with the same polarity,the total $e.m.f.$ is $E_1 + E_2$. Thus,$E_1 + E_2 = k(625)$.
When the terminals of $E_1$ are reversed,the effective $e.m.f.$ becomes $E_2 - E_1$ (since $E_2 > E_1$). Thus,$E_2 - E_1 = k(125)$.
Dividing the two equations: $\frac{E_1 + E_2}{E_2 - E_1} = \frac{625}{125} = 5$.
$E_1 + E_2 = 5(E_2 - E_1) = 5E_2 - 5E_1$.
$6E_1 = 4E_2$.
$\frac{E_1}{E_2} = \frac{4}{6} = \frac{2}{3}$.
Therefore,the ratio $E_1 : E_2$ is $2 : 3$.

(B) The potential gradient $k$ of a potentiometer wire is given by $k = \frac{V}{L} = \frac{I \cdot R}{L} = \frac{I \cdot \rho \cdot L}{A \cdot L} = \frac{I \cdot \rho}{A}$,where $I$ is the current,$\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.
However,in a standard potentiometer circuit,the potential drop across the wire is $V_{wire} = I \cdot R_{wire} = I \cdot \frac{\rho \cdot L}{A}$.
When the cell is balanced,the emf of the cell $E = k \cdot l$,where $l$ is the balancing length.
If the length $L$ of the potentiometer wire is kept constant and the cross-section $A$ is doubled,the resistance of the wire $R = \rho \frac{L}{A}$ becomes half.
Since the battery voltage is steady,the current $I$ through the wire increases,but the potential drop across the wire remains equal to the battery voltage.
Thus,the potential gradient $k = \frac{V_{battery}}{L}$ remains unchanged because both $V_{battery}$ and $L$ are constant.
Since $E = k \cdot l$ and both $E$ and $k$ are constant,the balancing length $l$ remains $4 \,m$.