Potentiometer Questions in English

(A) For a potentiometer,the electromotive force $(E)$ is directly proportional to the balancing length $(l)$ as $E = k l$,where $k$ is the potential gradient of the potentiometer wire.
Given:
$E_1 = k l_1$
$E_2 = k l_2$
Dividing the two equations:
$\frac{E_1}{E_2} = \frac{l_1}{l_2}$
Substituting the given values:
$\frac{E_1}{E_2} = \frac{625 \, cm}{500 \, cm} = \frac{625}{500} = \frac{25}{20} = \frac{5}{4}$
Thus,the ratio $\frac{E_1}{E_2}$ is $5:4$.

(B) The sensitivity of a potentiometer is defined as the smallest potential difference that can be measured by it.
The potential gradient $(k)$ is defined as the potential drop per unit length of the wire,given by $k = V/L$.
Sensitivity is inversely proportional to the potential gradient $(k)$. Therefore,as $k$ decreases,the sensitivity increases.
Since $k = V/L$,decreasing $k$ is equivalent to increasing the length $(L)$ of the potentiometer wire for a fixed potential difference $(V)$.
Thus,sensitivity is directly proportional to the length $(L)$ of the wire and inversely proportional to the potential gradient $(k)$.
Statements $(A)$ and $(C)$ are correct.

(B) Let $\varepsilon$ be the $EMF$ of the cell and $r$ be its internal resistance. Let $x$ be the potential gradient of the potentiometer wire.
When the cell is shunted by a resistance $R_1 = 5\,\Omega$,the terminal potential difference is $V_1 = \frac{\varepsilon R_1}{r + R_1} = \frac{5\varepsilon}{r + 5}$.
The null point is at $l_1 = 200\,cm$,so $V_1 = x l_1 = 200x$.
Thus,$\frac{5\varepsilon}{r + 5} = 200x$ --- (Equation $1$)
When the cell is shunted by a resistance $R_2 = 15\,\Omega$,the terminal potential difference is $V_2 = \frac{\varepsilon R_2}{r + R_2} = \frac{15\varepsilon}{r + 15}$.
The null point is at $l_2 = 300\,cm$,so $V_2 = x l_2 = 300x$.
Thus,$\frac{15\varepsilon}{r + 15} = 300x$ --- (Equation $2$)
Dividing Equation $1$ by Equation $2$:
$\frac{5\varepsilon / (r + 5)}{15\varepsilon / (r + 15)} = \frac{200x}{300x}$
$\frac{5}{r + 5} \times \frac{r + 15}{15} = \frac{2}{3}$
$\frac{r + 15}{3(r + 5)} = \frac{2}{3}$
$\frac{r + 15}{r + 5} = 2$
$r + 15 = 2r + 10$
$r = 5\,\Omega$.

(B) For a potentiometer,the $emf$ of a cell is directly proportional to the balancing length $(l)$: $E \propto l$,which implies $\frac{E_1}{E_2} = \frac{l_1}{l_2}$.
Given $E_1 = 1.5 \ V$ and $l_1 = 60 \ cm$.
The new length $l_2 = l_1 + 40 \ cm = 60 + 40 = 100 \ cm$.
Substituting the values: $\frac{1.5}{E} = \frac{60}{100}$.
$\frac{1.5}{E} = \frac{6}{10} = 0.6$.
$E = \frac{1.5}{0.6} = \frac{15}{6} = 2.5 \ V$.
Given $E = \frac{x}{10} \ V$,so $\frac{x}{10} = 2.5$.
$x = 25$.

(D) The formula for the internal resistance $r$ of a cell using a potentiometer is given by $r = R \left( \frac{\ell_1}{\ell_2} - 1 \right)$,where $\ell_1$ is the balancing length for the open circuit and $\ell_2$ is the balancing length with external resistance $R$.
However,in this specific problem,we are given two different external resistances $R_1 = 10 \ \Omega$ and $R_2 = 1 \ \Omega$ with corresponding balancing lengths $\ell_1 = 500 \ cm$ and $\ell_2 = 400 \ cm$.
The terminal voltage is $V = \varepsilon - Ir = I R = \lambda \ell$,where $\lambda$ is the potential gradient.
For $R_1 = 10 \ \Omega$: $V_1 = \frac{\varepsilon}{r + 10} \times 10 = \lambda \times 500 \implies \varepsilon = \lambda \times 500 \times \frac{r + 10}{10} = 50 \lambda (r + 10)$.
For $R_2 = 1 \ \Omega$: $V_2 = \frac{\varepsilon}{r + 1} \times 1 = \lambda \times 400 \implies \varepsilon = \lambda \times 400 (r + 1) = 400 \lambda (r + 1)$.
Equating the two expressions for $\varepsilon$:
$50 \lambda (r + 10) = 400 \lambda (r + 1)$
$r + 10 = 8(r + 1)$
$r + 10 = 8r + 8$
$7r = 2$
$r = \frac{2}{7} \approx 0.285 \ \Omega \approx 0.3 \ \Omega$.
Thus,the correct option is $D$.

(C) The potentiometer wire has a total resistance of $1 \Omega$. Since the sliding contact is in the middle,the wire is divided into two parts,each having a resistance of $0.5 \Omega$.
One part of the wire $(0.5 \Omega)$ is in parallel with the external resistance $R_e = 2 \Omega$.
The equivalent resistance of this parallel combination is $R_p' = \frac{0.5 \times 2}{0.5 + 2} = \frac{1}{2.5} = 0.4 \Omega$.
This parallel combination is in series with the other part of the potentiometer wire $(0.5 \Omega)$.
Therefore,the total equivalent resistance of the circuit is $R_{\text{eq}} = 0.5 \Omega + 0.4 \Omega = 0.9 \Omega$.
The total current drawn from the battery is $i = \frac{V}{R_{\text{eq}}} = \frac{0.9 \text{ V}}{0.9 \Omega} = 1.0 \text{ A}$.

(C) Let $x$ be the potential gradient of the potentiometer wire. The emf $E$ of a cell is proportional to the balancing length $l$,i.e.,$E = xl$.
For cells $A, B, C$ in series: $E_A + E_B + E_C = x(420) \quad (1)$
For cells $A, B$ in series: $E_A + E_B = x(220) \quad (2)$
For cells $B, C$ in series: $E_B + E_C = x(320) \quad (3)$
Subtracting equation $(2)$ from $(1)$: $E_C = x(420 - 220) = x(200)$.
Substituting $E_C$ into equation $(3)$: $E_B + x(200) = x(320) \implies E_B = x(120)$.
Substituting $E_B$ into equation $(2)$: $E_A + x(120) = x(220) \implies E_A = x(100)$.
Thus,the ratio $E_A : E_B : E_C = 100 : 120 : 200 = 1 : 1.2 : 2$.

(D) The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} = 20 \ \Omega + 10 \ \Omega = 30 \ \Omega$.
The current flowing through the circuit is $I = \frac{E}{R_{total}} = \frac{3 \ V}{30 \ \Omega} = 0.1 \ A$.
The potential difference across the wire is $V_{wire} = I \times R_{wire} = 0.1 \ A \times 20 \ \Omega = 2 \ V$.
The potential gradient along the wire is defined as the potential drop per unit length: $\text{Potential gradient} = \frac{V_{wire}}{L} = \frac{2 \ V}{10 \ m} = 0.2 \ V/m$.

(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 $R$ is the shunt resistance and $l$ is the balancing length.
For the first case: $R_1 = 5 \Omega$,$l_1 = 250 \ cm$.
For the second case: $R_2 = 20 \Omega$,$l_2 = 400 \ cm$.
Since the $EMF$ $E$ is constant,the balancing length is proportional to the terminal voltage $V = E \left( \frac{R}{R+r} \right)$.
Thus,$\frac{l_1}{l_2} = \frac{R_1(R_2+r)}{R_2(R_1+r)}$.
Substituting the values: $\frac{250}{400} = \frac{5(20+r)}{20(5+r)}$.
$\frac{5}{8} = \frac{20+r}{4(5+r)}$.
$20(5+r) = 8(20+r)$.
$100 + 20r = 160 + 8r$.
$12r = 60$.
$r = 5 \Omega$.

(C) The balancing length $l$ of a potentiometer is proportional to the terminal potential difference $V$ of the cell. Thus,$V = kl$,where $k$ is the potential gradient.
For a cell with $EMF$ $E$ and internal resistance $r$ shunted by an external resistance $R$,the terminal voltage is $V = E \frac{R}{R+r}$.
Therefore,$E \frac{R}{R+r} = kl$.
For the first case: $E \frac{5}{5+r} = k(200) \quad ... (1)$
For the second case: $E \frac{15}{15+r} = k(300) \quad ... (2)$
Dividing equation $(1)$ by $(2)$:
$\frac{5}{5+r} \times \frac{15+r}{15} = \frac{200}{300}$
$\frac{15+r}{3(5+r)} = \frac{2}{3}$
$\frac{15+r}{5+r} = 2$
$15 + r = 10 + 2r$
$r = 5 \ \Omega$.
Thus,the internal resistance of the cell is $5 \ \Omega$.

(C) The potential gradient $k$ is defined as $k = V/L$,where $V$ is the potential difference across the wire and $L$ is the length of the wire.
Given that $V$ is constant,if the length $L$ is increased,the potential gradient $k$ decreases.
The null point is obtained when the potential drop across the balancing length $l$ equals the $EMF$ of the cell $E$,i.e.,$E = kl$.
Since $E$ is constant and $k$ has decreased,the balancing length $l = E/k$ must increase.
Therefore,the null point is obtained at a larger distance.

(C) Let the potential difference across the potentiometer wire be $V$. The potential gradient $k_1$ for the original wire of length $L$ is $k_1 = \frac{V}{L}$.
The balance point is at $l_1 = \frac{L}{5}$,so $E = k_1 l_1 = \frac{V}{L} \cdot \frac{L}{5} = \frac{V}{5}$.
When the length of the wire is increased by $\frac{L}{2}$,the new length is $L' = L + \frac{L}{2} = \frac{3L}{2}$.
The potential difference $V$ across the wire remains the same (assuming the driver circuit is unchanged). The new potential gradient $k_2$ is $k_2 = \frac{V}{L'} = \frac{V}{3L/2} = \frac{2V}{3L}$.
For the same cell $E$,the new balance point $x$ is given by $E = k_2 x$.
Substituting the values: $\frac{V}{5} = \left( \frac{2V}{3L} \right) x$.
Solving for $x$: $x = \frac{V}{5} \cdot \frac{3L}{2V} = \frac{3L}{10}$.

(B) In a potentiometer,the e.m.f. $E$ is proportional to the balancing length $\ell$,i.e.,$E = k\ell$,where $k$ is the potential gradient.
For the first cell,$E_1 = k\ell_1$.
When the two cells are connected in opposition,the effective e.m.f. is $(E_1 - E_2)$. The new balancing length is $\ell_2$,so $(E_1 - E_2) = k\ell_2$.
Dividing the two equations: $\frac{E_1}{E_1 - E_2} = \frac{k\ell_1}{k\ell_2} = \frac{\ell_1}{\ell_2}$.
Cross-multiplying gives $E_1\ell_2 = \ell_1(E_1 - E_2) = \ell_1E_1 - \ell_1E_2$.
Rearranging terms: $\ell_1E_2 = E_1(\ell_1 - \ell_2)$.
Therefore,the ratio $\frac{E_1}{E_2} = \frac{\ell_1}{\ell_1 - \ell_2}$.

(B) Let $k$ be the potential gradient of the potentiometer wire.
When the potentiometer is connected between $A$ and $B$,the potential difference measured is $E_1$. Thus,$E_1 = k \times 3.60$.
When the potentiometer is connected between $A$ and $C$,the cells are in series opposition because they are connected with opposite polarities. The net e.m.f. is $E_1 - E_2$. Thus,$E_1 - E_2 = k \times 0.90$.
Dividing the two equations: $E_1 / (E_1 - E_2) = 3.60 / 0.90 = 4$.
This implies $E_1 = 4(E_1 - E_2) = 4E_1 - 4E_2$.
Rearranging gives $3E_1 = 4E_2$,so $E_1 / E_2 = 4/3$.

(B) Let the potential difference across the potentiometer wire be $V$. The potential gradient $k$ is given by $k = \frac{V}{L}$.
For the first case,the balance length is $l_1 = \frac{L}{4}$. The e.m.f. $E$ is given by $E = k \cdot l_1 = \frac{V}{L} \cdot \frac{L}{4} = \frac{V}{4}$.
When the length of the wire is increased by $\frac{L}{3}$,the new length becomes $L' = L + \frac{L}{3} = \frac{4L}{3}$.
The new potential gradient $k'$ is $k' = \frac{V}{L'} = \frac{V}{4L/3} = \frac{3V}{4L}$.
Let the new balance length be $l_2$. Then $E = k' \cdot l_2$.
Substituting the values,we get $\frac{V}{4} = \frac{3V}{4L} \cdot l_2$.
Solving for $l_2$,we get $l_2 = \frac{V}{4} \cdot \frac{4L}{3V} = \frac{L}{3}$.

(D) The potential gradient $(k)$ of the potentiometer wire is defined as the potential drop per unit length.
Given,total length $L = 5 \ m$ and total potential difference $V = 4 \ V$.
$k = \frac{V}{L} = \frac{4 \ V}{5 \ m} = 0.8 \ V/m$.
The balancing length is given as $l = 200 \ cm = 2 \ m$.
The e.m.f. $(E)$ of the cell is given by $E = k \times l$.
Substituting the values,$E = 0.8 \ V/m \times 2 \ m = 1.6 \ V$.
Therefore,the correct option is $D$.

(C) Let $V_{total} = 2.5 \ V$ be the voltage applied across the potentiometer wire of length $L$.
Let $V_x = 1.08 \ V$ be the e.m.f. of the cell balanced at length $l = 2.16 \ m$.
The potential gradient $k$ of the potentiometer wire is given by $k = \frac{V_{total}}{L}$.
The voltage drop across length $l$ is $V_x = k \cdot l = \left( \frac{V_{total}}{L} \right) \cdot l$.
Substituting the given values: $1.08 = \left( \frac{2.5}{L} \right) \cdot 2.16$.
Rearranging for $L$: $L = \frac{2.5 \cdot 2.16}{1.08}$.
Since $\frac{2.16}{1.08} = 2$,we get $L = 2.5 \cdot 2 = 5 \ m$.
Therefore,the length of the potentiometer wire is $5 \ m$.

(B) In a potentiometer,the $E.M.F.$ of a cell is proportional to the balancing length,i.e.,$E \propto l$ or $E = kl$,where $k$ is the potential gradient.
For the first cell,$E_1 = kl_1$.
When the two cells are connected in opposition,the effective $E.M.F.$ is $(E_1 - E_2)$. The new balancing length is $l_2$,so $(E_1 - E_2) = kl_2$.
Dividing the two equations: $\frac{E_1}{E_1 - E_2} = \frac{l_1}{l_2}$.
Cross-multiplying gives $E_1 l_2 = l_1 E_1 - l_1 E_2$.
Rearranging terms: $l_1 E_2 = E_1 (l_1 - l_2)$.
Therefore,$\frac{E_1}{E_2} = \frac{l_1}{l_1 - l_2}$.

(D) The current $I$ in the circuit is given by $I = \frac{V_{total}}{R_{total}} = \frac{2}{R + 495}$.
Given potential gradient $\phi = 0.2 \ mV/cm = 0.02 \ V/m$.
The potential drop across the wire of length $L = 1 \ m$ is $V_{wire} = I \times R = \phi \times L$.
Substituting the values: $\frac{2R}{R + 495} = 0.02 \times 1$.
$2R = 0.02(R + 495)$.
$2R = 0.02R + 9.9$.
$1.98R = 9.9$.
$R = \frac{9.9}{1.98} = 5 \ \Omega$.

(D) In a potentiometer,the balancing length $l$ is directly proportional to the e.m.f. of the cell,i.e.,$E = k \cdot l$,where $k$ is the potential gradient.
When cells are connected in series with the same polarity,the effective e.m.f. is $E_1 + E_2 = k \cdot l_1$.
Given $l_1 = 80 \ cm$,so $E_1 + E_2 = 80k$ (Equation $1$).
When the polarity of $E_2$ is reversed,the effective e.m.f. is $E_1 - E_2 = k \cdot l_2$.
Given $l_2 = 20 \ cm$,so $E_1 - E_2 = 20k$ (Equation $2$).
Dividing Equation $1$ by Equation $2$:
$\frac{E_1 + E_2}{E_1 - E_2} = \frac{80k}{20k} = \frac{4}{1}$.
Applying componendo and dividendo:
$\frac{(E_1 + E_2) + (E_1 - E_2)}{(E_1 + E_2) - (E_1 - E_2)} = \frac{4 + 1}{4 - 1}$.
$\frac{2E_1}{2E_2} = \frac{5}{3}$.
Therefore,$\frac{E_1}{E_2} = \frac{5}{3}$ or $5 : 3$.

(C) The internal resistance $r$ of a cell is given by the formula: $r = R \left( \frac{l_0 - l}{l} \right)$,where $l_0$ is the balancing length in an open circuit and $l$ is the balancing length when shunted by resistance $R$.
Case $1$: $R_1 = 5 \ \Omega$,$l_1 = 150 \ cm$.
$r = 5 \left( \frac{l_0 - 150}{150} \right) = \frac{l_0 - 150}{30} \quad \dots (1)$
Case $2$: $R_2 = 10 \ \Omega$,$l_2 = 150 + 25 = 175 \ cm$.
$r = 10 \left( \frac{l_0 - 175}{175} \right) = \frac{2(l_0 - 175)}{35} \quad \dots (2)$
Equating $(1)$ and $(2)$:
$\frac{l_0 - 150}{30} = \frac{2(l_0 - 175)}{35}$
$\frac{l_0 - 150}{6} = \frac{2(l_0 - 175)}{7}$
$7(l_0 - 150) = 12(l_0 - 175)$
$7l_0 - 1050 = 12l_0 - 2100$
$5l_0 = 1050$
$l_0 = 210 \ cm$.

(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 cells are connected in series with the same polarity,the effective e.m.f. is $E_1 + E_2 = k l_1$.
Given $l_1 = 64 \ cm$,so $E_1 + E_2 = 64k$ --- (Equation $1$).
When the polarity of $E_2$ is reversed,the effective e.m.f. is $E_1 - E_2 = k l_2$.
Given $l_2 = 32 \ cm$,so $E_1 - E_2 = 32k$ --- (Equation $2$).
Dividing Equation $1$ by Equation $2$:
$\frac{E_1 + E_2}{E_1 - E_2} = \frac{64k}{32k} = \frac{2}{1}$.
Applying componendo and dividendo:
$\frac{(E_1 + E_2) + (E_1 - E_2)}{(E_1 + E_2) - (E_1 - E_2)} = \frac{2 + 1}{2 - 1}$.
$\frac{2E_1}{2E_2} = \frac{3}{1}$.
Therefore,$\frac{E_1}{E_2} = 3: 1$.

(A) The potential gradient $K$ of the potentiometer wire is given by:
$K = \frac{V_{wire}}{L} = \frac{E_{driving} \times R}{(R + r) L}$
Given $E_{driving} = 5 \, V$, $r = 4 \, \Omega$, $L = 5 \, m$, and $R = 16 \, \Omega$:
$K = \frac{5 \times 16}{(16 + 4) \times 5} = \frac{80}{100} = 0.8 \, V/m$
Case $1$: Cells assist each other $(E_{net} = E_1 + E_2)$
$E_1 + E_2 = K l_1$
$1.3 + 1.1 = 0.8 \times l_1$
$2.4 = 0.8 \times l_1 \implies l_1 = 3 \, m$
Case $2$: Cells oppose each other $(E_{net} = E_1 - E_2)$
$E_1 - E_2 = K l_2$
$1.3 - 1.1 = 0.8 \times l_2$
$0.2 = 0.8 \times l_2 \implies l_2 = 0.25 \, m$
Thus, the balancing lengths are $3 \, m$ and $0.25 \, m$.

(C) Given,the length of the wire at which the cell balances is $l = 300 \ cm = 3 \ m$.
Resistance per unit length is $2 \ \Omega/m$.
Total resistance of the wire segment $R = 3 \ m \times 2 \ \Omega/m = 6 \ \Omega$.
Since the potentiometer is balanced,the potential difference across the wire segment is equal to the e.m.f. of the cell,which is $1.5 \ V$.
Using Ohm's law,$V = IR$,we have:
$I = V / R = 1.5 \ V / 6 \ \Omega = 0.25 \ A$.
Converting to milliamperes,$I = 0.25 \times 1000 \ mA = 250 \ mA$.

(B) The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} + r = 5 \, \Omega + 992 \, \Omega + 3 \, \Omega = 1000 \, \Omega$.
The current flowing through the potentiometer wire is $I = \frac{E}{R_{total}} = \frac{4 \, V}{1000 \, \Omega} = 0.004 \, A$.
The potential drop across the entire $4 \, m$ wire is $V_{wire} = I \times R_{wire} = 0.004 \, A \times 5 \, \Omega = 0.02 \, V$.
The potential gradient (potential drop per unit length) is $k = \frac{V_{wire}}{L} = \frac{0.02 \, V}{4 \, m} = 0.005 \, V/m$.
The e.m.f. balanced by a length of $0.75 \, m$ is $E' = k \times l = 0.005 \, V/m \times 0.75 \, m = 0.00375 \, V$.
Converting to millivolts, $E' = 0.00375 \times 1000 \, mV = 3.75 \, mV$.

(D) Let $E_1$ and $E_2$ be the e.m.f.s of the two cells.
When cells are in series,the total e.m.f. is $E_1 + E_2$,and the balancing length is $l_1 = 8 \ m$. So,$E_1 + E_2 = k l_1 = 8k$,where $k$ is the potential gradient.
When cells are in opposition,the total e.m.f. is $E_1 - E_2$,and the balancing length is $l_2 = 4 \ m$. So,$E_1 - E_2 = k l_2 = 4k$.
Dividing the two equations: $\frac{E_1 + E_2}{E_1 - E_2} = \frac{8k}{4k} = 2$.
Applying componendo and dividendo: $\frac{(E_1 + E_2) + (E_1 - E_2)}{(E_1 + E_2) - (E_1 - E_2)} = \frac{2 + 1}{2 - 1}$.
This simplifies to $\frac{2E_1}{2E_2} = \frac{3}{1}$.
Therefore,$\frac{E_1}{E_2} = 3:1$.

(C) The potential difference across the external shunt resistance $R$ is given by $V = \frac{E R}{R + r}$, where $E$ is the electromotive force (emf) and $r$ is the internal resistance of the cell.
Since the potentiometer measures the potential difference, the balance length $L$ is directly proportional to the potential difference $V$, i.e., $L \propto V$.
Therefore, $\frac{L_1}{L_2} = \frac{V_1}{V_2} = \frac{R_1(R_2 + r)}{R_2(R_1 + r)}$.
Given: $L_1 = 1 \, m$, $R_1 = 3 \, \Omega$, $L_2 = 1.5 \, m$, $R_2 = 6 \, \Omega$.
Substituting the values:
$\frac{1}{1.5} = \frac{3(6 + r)}{6(3 + r)}$
$\frac{2}{3} = \frac{6 + r}{2(3 + r)}$
$2(2)(3 + r) = 3(6 + r)$
$4(3 + r) = 18 + 3r$
$12 + 4r = 18 + 3r$
$r = 18 - 12 = 6 \, \Omega$.
Thus, the internal resistance of the cell is $6 \, \Omega$.

(D) The potential gradient $k$ is defined as the potential difference per unit length of the wire.
$k = \frac{V}{L}$
According to Ohm's law,the potential difference $V$ across a wire of resistance $R$ is $V = I R$.
Substituting the expression for resistance $R = \frac{\rho L}{A}$,we get:
$V = I \left( \frac{\rho L}{A} \right)$
Now,substituting this into the potential gradient formula:
$k = \frac{I (\rho L / A)}{L}$
$k = \frac{I \rho}{A}$
Thus,the potential gradient is $\frac{I \rho}{A}$.

(B) Let $E_0$ be the potential difference across the potentiometer wire.
In the first case,the potential gradient is $k_1 = \frac{E_0}{L}$.
The balance length is $l_1 = \frac{L}{3}$.
So,the e.m.f. of the cell is $E = k_1 l_1 = \left(\frac{E_0}{L}\right) \left(\frac{L}{3}\right) = \frac{E_0}{3} \quad \dots (1)$
In the second case,the new length of the wire is $L' = L + \frac{L}{2} = \frac{3L}{2}$.
The new potential gradient is $k_2 = \frac{E_0}{L'} = \frac{E_0}{3L/2} = \frac{2E_0}{3L}$.
Let $x$ be the new balance length for the same cell $E$.
Then,$E = k_2 x = \left(\frac{2E_0}{3L}\right) x \quad \dots (2)$
Equating $(1)$ and $(2)$,we get:
$\frac{E_0}{3} = \left(\frac{2E_0}{3L}\right) x$
$\Rightarrow x = \frac{L}{2}$.

(D) The potential gradient $k$ of the potentiometer wire is defined as the potential drop per unit length.
$k = \frac{V}{L} = \frac{3 \,V}{4 \,m} = 0.75 \,V/m$.
Given that the balancing length $l = 100 \,cm = 1 \,m$.
The e.m.f. of the cell $E$ is given by $E = k \times l$.
Substituting the values,$E = 0.75 \,V/m \times 1 \,m = 0.75 \,V$.

(D) The potential difference across $A$ and $B$ is $V_{AB} = E_1$. The balancing length is $l_{AB} = 300 \ cm$. Thus,$E_1 = k \cdot l_{AB} = k \cdot 300$,where $k$ is the potential gradient of the potentiometer wire.
The potential difference across $A$ and $C$ is $V_{AC} = E_1 - E_2$ (since the cells are connected in opposition). The balancing length is $l_{AC} = 100 \ cm$. Thus,$E_1 - E_2 = k \cdot l_{AC} = k \cdot 100$.
Taking the ratio of the two equations:
$\frac{E_1}{E_1 - E_2} = \frac{k \cdot 300}{k \cdot 100} = 3$
$E_1 = 3(E_1 - E_2)$
$E_1 = 3E_1 - 3E_2$
$3E_2 = 2E_1$
$\frac{E_1}{E_2} = \frac{3}{2}$

(B) In a potentiometer,the null point length $l$ is directly proportional to the potential difference $V$ across the points being measured,i.e.,$V = kl$,where $k$ is the potential gradient.
For the potential difference between $A$ and $B$,the $EMF$ is $E_1$. Thus,$E_1 = k(0.9)$.
For the potential difference between $A$ and $C$,the net $EMF$ is $E_1 - E_2$. Thus,$E_1 - E_2 = k(0.3)$.
Dividing the two equations:
$\frac{E_1}{E_1 - E_2} = \frac{0.9}{0.3} = 3$
$E_1 = 3(E_1 - E_2)$
$E_1 = 3E_1 - 3E_2$
$3E_2 = 2E_1$
$\frac{E_2}{E_1} = \frac{2}{3}$

(B) The potential gradient $x$ of a potentiometer wire is given by $x = \frac{V}{L} = \frac{E R}{(R + R') L}$,where $E$ is the e.m.f. of the driver cell,$R$ is the resistance of the potentiometer wire,$R'$ is the resistance in the main circuit,and $L$ is the total length of the wire.
For a given cell of e.m.f. $\epsilon$,the balancing length $l$ is given by $\epsilon = x \cdot l$,which implies $l = \frac{\epsilon}{x}$.
To shift the null point from the $7^{\text{th}}$ wire to the $9^{\text{th}}$ wire,the balancing length $l$ must increase.
Since $l = \frac{\epsilon}{x}$,increasing $l$ requires a decrease in the potential gradient $x$.
Looking at the expression $x = \frac{E R}{(R + R') L}$,to decrease $x$,we must increase the denominator term $(R + R')$.
Therefore,we must increase the resistance $R'$ in the main circuit.

(B) 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 total length of the wire.
When the length of the potentiometer wire is increased while keeping the driving source constant, the total resistance of the wire increases.
Since the driving voltage $V$ remains constant, the current $I = V/R_{total}$ decreases.
Consequently, the potential gradient $k = I \cdot \rho$ (where $\rho$ is the resistance per unit length) decreases.
For a cell of electromotive force $E$, the balancing length $l$ is given by $E = k \cdot l$, or $l = E/k$.
Since $k$ decreases, the balancing length $l$ must increase to balance the same electromotive force $E$.

(D) The balancing length $\ell_1$ is proportional to the $EMF$ $E$ of the cell,so $E = k\ell_1$.
When the cell is shunted with an external resistance $R = 2 \ \Omega$,the terminal voltage $V$ is given by $V = k\ell_2$,where $\ell_2$ is the new balancing length.
Given $\ell_1 = 240 \ cm$ and $\ell_2 = \frac{\ell_1}{2} = 120 \ cm$.
The terminal voltage is $V = E \left( \frac{R}{R+r} \right)$,where $r$ is the internal resistance.
Substituting the relations,we get $k\ell_2 = k\ell_1 \left( \frac{R}{R+r} \right)$.
This simplifies to $\frac{\ell_2}{\ell_1} = \frac{R}{R+r}$.
Substituting the values: $\frac{120}{240} = \frac{2}{2+r}$.
$\frac{1}{2} = \frac{2}{2+r} \implies 2+r = 4 \implies r = 2 \ \Omega$.

(C) In a potentiometer experiment,the balancing length $\ell$ is proportional to the e.m.f. of the cell,i.e.,$E \propto \ell$ or $E = k\ell$,where $k$ is the potential gradient.
When cells are connected in series with the same polarity,the effective e.m.f. is $E_1 + E_2 = k\ell_1$,where $\ell_1 = 64 \ cm$.
When the polarity of $E_2$ is reversed,the effective e.m.f. is $E_1 - E_2 = k\ell_2$,where $\ell_2 = 32 \ cm$.
Dividing the two equations: $\frac{E_1 + E_2}{E_1 - E_2} = \frac{\ell_1}{\ell_2} = \frac{64}{32} = 2$.
Applying componendo and dividendo: $\frac{(E_1 + E_2) + (E_1 - E_2)}{(E_1 + E_2) - (E_1 - E_2)} = \frac{2 + 1}{2 - 1}$.
This simplifies to $\frac{2E_1}{2E_2} = \frac{3}{1}$,so $\frac{E_1}{E_2} = 3: 1$.

(D) The formula for the internal resistance $r$ of a cell using a potentiometer is given by $r = R \left( \frac{\ell_1}{\ell_2} - 1 \right)$,where $\ell_1$ is the balancing length for the open circuit and $\ell_2$ is the balancing length when shunted by resistance $R$.
Given:
Case $1$: $R_1 = 3 \ \Omega$,$\ell_2 = 1 \ m$
Case $2$: $R_2 = 6 \ \Omega$,$\ell_2' = 1.5 \ m$
Since the $EMF$ of the cell remains constant,the balancing length $\ell_1$ is the same in both cases.
$r = 3 \left( \frac{\ell_1}{1} - 1 \right) = 6 \left( \frac{\ell_1}{1.5} - 1 \right)$
$3 \ell_1 - 3 = 4 \ell_1 - 6$
$\ell_1 = 3 \ m$
Substituting $\ell_1$ back into the first equation:
$r = 3 \left( \frac{3}{1} - 1 \right) = 3 \times 2 = 6 \ \Omega$.

(D) The sensitivity of a potentiometer is defined as the smallest potential difference that can be measured by it.
Sensitivity is inversely proportional to the potential gradient $(k = V/L)$.
$A$ smaller potential gradient means a higher sensitivity.
Therefore, to decrease the sensitivity, we must increase the potential gradient along the wire.
Since the potential gradient $k = I \cdot R/L$, increasing the current $(I)$ through the wire increases the potential gradient, thereby decreasing the sensitivity.

(D) Let the potential difference across the wire be $V$. The potential gradient $k$ is given by $k = \frac{V}{L_{total}}$.
For the first case,the balancing length $l_1 = \frac{L}{5}$. The e.m.f. $E$ is given by $E = k_1 l_1 = \frac{V}{L} \cdot \frac{L}{5} = \frac{V}{5}$.
When the length of the wire is increased by $\frac{L}{2}$,the new total length becomes $L' = L + \frac{L}{2} = \frac{3L}{2}$.
The new potential gradient is $k_2 = \frac{V}{L'} = \frac{V}{3L/2} = \frac{2V}{3L}$.
Let the new balancing length be $l_2$. Then $E = k_2 l_2$.
Equating the two expressions for $E$: $\frac{V}{5} = \frac{2V}{3L} \cdot l_2$.
Solving for $l_2$: $l_2 = \frac{V}{5} \cdot \frac{3L}{2V} = \frac{3L}{10}$.

(D) In a potentiometer,the balancing length $\ell$ is directly proportional to the terminal potential difference $V$ across the cell.
For an open circuit,the balancing length $\ell_{1}$ corresponds to the e.m.f. $E$ of the cell: $E \propto \ell_{1}$.
When the cell is shunted with an external resistance $R$,the terminal potential difference $V$ is given by $V = E - Ir$,where $I = \frac{E}{R+r}$.
Thus,$V = E - \left(\frac{E}{R+r}\right)r = E\left(1 - \frac{r}{R+r}\right) = E\left(\frac{R}{R+r}\right)$.
The new balancing length $\ell_{2}$ corresponds to the terminal potential difference $V$: $V \propto \ell_{2}$.
Given $\ell_{2} = \frac{\ell_{1}}{2}$,we have $\frac{V}{E} = \frac{\ell_{2}}{\ell_{1}} = \frac{1}{2}$.
Substituting the expression for $V/E$: $\frac{R}{R+r} = \frac{1}{2}$.
Solving for $r$: $2R = R + r$,which gives $r = R$.

(D) In a potentiometer experiment to find the internal resistance $(r)$ of a cell,the cell $E_1$ is connected in series with a resistance box $R$.
When the circuit is open,the balancing length $l_1$ is proportional to the $EMF$ $(E_1)$: $E_1 = k l_1$.
When the circuit is closed (resistance $R$ is connected),the balancing length $l_2$ is proportional to the terminal potential difference $(V)$: $V = k l_2$.
The internal resistance is given by the formula: $r = R \left( \frac{E_1}{V} - 1 \right)$.
Here,$E_1$ is the $EMF$ of the cell,$V$ is the terminal potential difference,and $R$ is the external resistance.

(D) Let $k$ be the potential gradient of the potentiometer wire.
When the cells are connected in series with the same polarity,the total e.m.f. is $(E_{1} + E_{2})$. The balancing condition is $(E_{1} + E_{2}) = k \ell_{1}$,where $\ell_{1} = 64 \ cm$.
When the polarity of $E_{2}$ is reversed,the total e.m.f. is $(E_{1} - E_{2})$. The balancing condition is $(E_{1} - E_{2}) = k \ell_{2}$,where $\ell_{2} = 32 \ cm$.
Dividing the two equations: $\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{64}{32} = 2$.
$E_{1} + E_{2} = 2(E_{1} - E_{2})$.
$E_{1} + E_{2} = 2E_{1} - 2E_{2}$.
$3E_{2} = E_{1}$.
Therefore,$\frac{E_{1}}{E_{2}} = 3: 1$.

(B) Total resistance $R_{total} = 495 \, \Omega + 5 \, \Omega = 500 \, \Omega$.
Current $I$ flowing through the circuit is given by $I = \frac{V}{R_{total}} = \frac{4 \, V}{500 \, \Omega} = 8 \times 10^{-3} \, A$.
The potential difference $V_{wire}$ across the potentiometer wire is $V_{wire} = I \times R_{wire} = 8 \times 10^{-3} \, A \times 5 \, \Omega = 40 \times 10^{-3} \, V$.
The potential gradient $k$ is defined as the potential drop per unit length: $k = \frac{V_{wire}}{L} = \frac{40 \times 10^{-3} \, V}{4 \, m} = 10 \times 10^{-3} \, V/m = 0.01 \, V/m$.

(D) 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 total length of the wire.
When the length of the wire is doubled $(L' = 2L)$,the potential gradient becomes $k' = V/(2L) = k/2$.
The balancing condition for an unknown e.m.f. $E$ is $E = k \cdot l$,where $l$ is the balancing length.
Since $E$ remains constant,$k \cdot l = k' \cdot l'$.
Substituting $k' = k/2$,we get $k \cdot l = (k/2) \cdot l'$,which simplifies to $l' = 2l$.
Given the initial balancing length $l = 1.5 \,m$,the new balancing length is $l' = 2 \times 1.5 \,m = 3 \,m$.

(C) The potential difference between $A$ and $B$ is $V_{AB} = E_{1}$. The null point length is $L_{AB} = 0.9 \ m$.
Since $V \propto L$,we have $E_{1} = k \times 0.9$,where $k$ is the potential gradient.
The potential difference between $A$ and $C$ is $V_{AC} = E_{1} - E_{2}$. The null point length is $L_{AC} = 0.3 \ m$.
Thus,$E_{1} - E_{2} = k \times 0.3$.
Dividing the two equations: $\frac{E_{1}}{E_{1} - E_{2}} = \frac{0.9}{0.3} = 3$.
$E_{1} = 3(E_{1} - E_{2})$
$E_{1} = 3E_{1} - 3E_{2}$
$3E_{2} = 2E_{1}$
$\frac{E_{2}}{E_{1}} = \frac{2}{3}$.

(C) The potential gradient $k$ of the potentiometer wire is given by the ratio of the total potential difference $V$ to the total length $L$ of the wire.
$k = \frac{V}{L} = \frac{3 \,V}{4 \,m} = 0.75 \,V/m$.
We are given that the cell balances at a length $l = 100 \,cm = 1 \,m$.
The e.m.f. $E$ of the cell is equal to the potential drop across the balancing length $l$, which is given by $E = k \times l$.
Substituting the values, we get $E = 0.75 \,V/m \times 1 \,m = 0.75 \,V$.

(D) direct reading instrument provides the value of the measured quantity directly on a scale or display without requiring any manual calculation or balancing process.
$A$,$B$,and $C$ are direct reading instruments because they show the result immediately.
$A$ potentiometer is a null-type instrument. It requires a balancing process (finding the null point) to measure potential difference or $EMF$,and the final value is calculated based on the balancing length.
Therefore,the potentiometer is not a direct reading instrument.
Correct option is $D$.

(C) The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} + r_{internal} = 3 \ \Omega + 8 \ \Omega + 1 \ \Omega = 12 \ \Omega$.
The current flowing through the potentiometer wire is $I = \frac{V}{R_{total}} = \frac{4 \ V}{12 \ \Omega} = \frac{1}{3} \ A$.
The potential difference across the potentiometer wire is $V_{wire} = I \times R_{wire} = \frac{1}{3} \ A \times 3 \ \Omega = 1 \ V$.
The potential gradient $K$ of the wire is $K = \frac{V_{wire}}{L} = \frac{1 \ V}{100 \ cm} = 0.01 \ V/cm$.
The e.m.f. $E$ of the cell balanced at length $\ell = 50 \ cm$ is $E = K \times \ell = 0.01 \ V/cm \times 50 \ cm = 0.50 \ V$.

(D) In a potentiometer circuit,when two cells are connected to assist each other,the balancing length $\ell_{1}$ is proportional to $(E_{1} + E_{2})$. Thus,$E_{1} + E_{2} = k \ell_{1}$,where $k$ is the potential gradient.
When they are connected to oppose each other,the balancing length $\ell_{2}$ is proportional to $(E_{1} - E_{2})$. Thus,$E_{1} - E_{2} = k \ell_{2}$.
Dividing the two equations,we get: $\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{\ell_{1}}{\ell_{2}}$.
Using the given values $\ell_{1} = 490 \ cm$ and $\ell_{2} = 90 \ cm$:
$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{490}{90} = \frac{49}{9}$.
Applying componendo and dividendo:
$\frac{E_{1}}{E_{2}} = \frac{49 + 9}{49 - 9} = \frac{58}{40} = 1.45$.

(D) The potential drop across the potentiometer wire $AB$ is given by $V_{AB} = E \cdot \frac{R_{AB}}{R + R_{AB} + r}$,where $E$ is the $EMF$ of the driver cell,$R$ is the external resistance,$R_{AB}$ is the resistance of the wire,and $r$ is the internal resistance of the driver cell.
$(i)$ When student $X$ increases $R$,the total resistance of the primary circuit increases,so the current $I = \frac{E}{R + R_{AB} + r}$ decreases. Consequently,the potential gradient $k = \frac{V_{AB}}{L}$ decreases. Since the balancing condition is $E_1 = k \cdot l$,where $l$ is the balancing length,if $k$ decreases,$l$ must increase to maintain the same $E_1$. Thus,the null point shifts towards $B$.
(ii) When student $Y$ decreases $S$,the terminal potential difference across the cell $E_1$ is $V = E_1 - I_1 r_1$,where $I_1 = \frac{E_1}{S + r_1}$. Decreasing $S$ increases the current $I_1$ drawn from the cell $E_1$,which increases the voltage drop $I_1 r_1$ across the internal resistance $r_1$. Therefore,the terminal voltage $V$ across the cell $E_1$ decreases. Since $V = k \cdot l$,a decrease in $V$ requires a smaller balancing length $l$. Thus,the null point shifts towards $A$.