Mix Examples-Current Electricity Questions in English

(A) thermocouple is a device that converts temperature difference into electric potential difference. This phenomenon is known as the Seebeck effect,where an electromotive force $(EMF)$ is generated in a circuit consisting of two dissimilar conductors when their junctions are maintained at different temperatures. Therefore,the correct option is $A$.

(C) In a thermocouple,the neutral temperature $\theta_n$ is the arithmetic mean of the cold junction temperature $\theta_c$ and the inversion temperature $\theta_i$. The relation is given by: $\theta_n = \frac{\theta_i + \theta_c}{2}$.
Given that there is no deflection in the galvanometer,the hot junction temperature is at the inversion temperature $\theta_i$.
Given: $\theta_n = 270\,^\circ C$ and $\theta_c = 20\,^\circ C$.
Substituting these values into the formula: $270 = \frac{\theta_i + 20}{2}$.
$540 = \theta_i + 20$.
$\theta_i = 540 - 20 = 520\,^\circ C$.

(A) The thermo $e.m.f.$ $(E)$ produced in a thermocouple is given by the relation $E = at + \frac{1}{2}bt^2$,where $t$ is the temperature difference between the junctions.
Here,$a$ and $b$ are constants that depend solely on the nature of the metals used to form the thermocouple.
Therefore,the thermo $e.m.f.$ depends on the nature of the metals and the temperature difference between the junctions,not on the length or cross-section of the wires.

(A) In an iron-copper thermocouple,the thermoelectric current flows from the metal at a higher thermoelectric potential to the metal at a lower thermoelectric potential at the hot junction.
For an iron-copper $(Fe-Cu)$ thermocouple,iron is at a lower thermoelectric potential compared to copper.
Therefore,the current flows from copper to iron at the hot junction.

(A) The Peltier effect describes the heat evolved or absorbed at the junction of two dissimilar metals when an electric current passes through it.
The Peltier heat $Q$ is given by $Q = \pi I t$, where $\pi$ is the Peltier coefficient, $I$ is the current, and $t$ is the time.
According to the Kelvin relations for thermoelectric effects, the Peltier coefficient $\pi$ is directly proportional to the absolute temperature $T$ of the junction, expressed as $\pi = S T$, where $S$ is the Seebeck coefficient.
Therefore, the Peltier coefficient is proportional to the absolute temperature $T$ of the junction.

(A) For a thermocouple,the neutral temperature ${T_n}$ is the arithmetic mean of the temperature of the cold junction ${T_c}$ and the temperature of inversion ${T_i}$.
Mathematically,this is expressed as: ${T_n} = \frac{{T_i} + {T_c}}{2}$.
Rearranging this equation to solve for the temperature of inversion ${T_i}$:
${2{T_n} = {T_i} + {T_c}}$
${{T_i} = 2{T_n} - {T_c}}$
Thus,the correct relation is ${T_i} = 2{T_n} - {T_c}$.

(A) The thermo $e.m.f.$ in a thermocouple is given by the relation $E = \alpha \theta + \frac{1}{2} \beta \theta^2$,where $\theta$ is the temperature difference between the junctions.
At the neutral temperature,the $e.m.f.$ is maximum.
At the temperature of inversion,the thermo $e.m.f.$ becomes zero,and beyond this temperature,the $e.m.f.$ changes its sign.

(D) When the temperature of the hot junction increases while keeping the cold junction at $0 \ ^\circ C$,the thermo $e.m.f.$ increases up to a maximum positive value at the neutral temperature.
Beyond the neutral temperature,the thermo $e.m.f.$ decreases and becomes zero at the inversion temperature.
Beyond the inversion temperature,the direction of the thermo $e.m.f.$ is reversed (becomes negative).
Therefore,for a given thermocouple,the thermo $e.m.f.$ can be positive,negative,or zero.

(A) The phenomenon described is the $Peltier$ effect,where heat is absorbed or evolved at the junction of two dissimilar metals when an electric current passes through it. In an antimony-bismuth thermocouple,when the current flows from antimony to bismuth,the junction absorbs heat (becomes cold),and when it flows from bismuth to antimony,the junction evolves heat (becomes hot). Therefore,the junction becomes hot when the current flows from bismuth to antimony.

(C) When a current flows through a circuit consisting of two different metals joined at two junctions,heat is absorbed at one junction and released at the other. This phenomenon is known as the $Peltier$ effect. Therefore,when a battery is connected to a $Cu-Fe$ thermocouple,one junction will become hotter than the other due to the heat exchange at the junctions.

(B) The heat $H$ evolved or absorbed in a conductor due to the Thomson effect is given by the formula: $H = \sigma i t \Delta \theta$,where $\sigma$ is the Thomson coefficient,$i$ is the current,$t$ is the time,and $\Delta \theta$ is the temperature difference between the ends of the conductor.
Given the conditions: $i = 1\, A$,$\Delta \theta = 1\, ^oC$,and $t = 1\, s$.
Substituting these values into the equation:
$H = \sigma \times 1 \times 1 \times 1 = \sigma$.
Thus,the amount of heat evolved or absorbed per unit current per unit time for a unit temperature difference is defined as the Thomson coefficient.

(A) The electromotive force $(E)$ in a thermocouple is given by the relation: $E = \alpha \theta + \frac{1}{2} \beta \theta^2$,where $\theta$ is the temperature difference between the hot and cold junctions.
Neutral temperature $(\theta_n)$ is the temperature of the hot junction at which the thermo-emf is maximum and the thermoelectric power becomes zero. It is a constant property of the pair of metals used and is independent of the temperature of the cold junction $(\theta_c)$.
In contrast,the temperature of inversion $(\theta_i)$ depends on the temperature of the cold junction $(\theta_c)$ according to the relation: $\theta_i - \theta_n = \theta_n - \theta_c$.

(D) The temperature of inversion $(T_i)$ is the temperature at which the thermo-electromotive force (emf) becomes zero and then reverses its direction.
At the temperature of inversion,the net emf in the circuit is zero.
Consequently,the current flow in the circuit becomes zero.
Since the current is zero,it does not flow in any direction.
Therefore,none of the given options describing a direction of current flow are correct.

(B) In a thermocouple,the direction of the thermoelectric current is determined by the Seebeck series.
For any two metals in the Seebeck series,the current flows from the metal appearing earlier in the series to the metal appearing later at the cold junction.
Since $Sb$ appears before $Bi$ in the Seebeck series,the current flows from $Sb$ to $Bi$ at the cold junction and from $Bi$ to $Sb$ at the hot junction.
Therefore,the correct option is $B$.

(A) The magnitude of the thermo $e.m.f.$ generated in a thermocouple depends on the position of the metals in the thermoelectric series.
According to the thermoelectric series,the further apart two metals are in the series,the greater the potential difference $(e.m.f.)$ generated for a given temperature difference.
Antimony and Bismuth are located at opposite ends of the thermoelectric series,which results in the largest thermo $e.m.f.$ among the given pairs.

(B) The thermo $e.m.f.$ $(E)$ in a thermocouple is given by the relation $E = \alpha \theta + \frac{1}{2} \beta \theta^2$,where $\theta$ is the temperature difference between the junctions.
For a thermocouple,the neutral temperature $(T_n)$ is the temperature of the hot junction at which the thermo $e.m.f.$ becomes zero when the cold junction is at a specific temperature $(T_c)$.
The relationship is given by $T_n = \frac{T_i + T_c}{2}$,where $T_i$ is the temperature of the hot junction at which $e.m.f.$ is zero.
Given: Cold junction temperature $T_c = 10\,^{\circ}C$ and $T_i = 530\,^{\circ}C$.
Substituting the values: $T_n = \frac{530 + 10}{2} = \frac{540}{2} = 270\,^{\circ}C$.

(C) The thermo $e.m.f.$ $(E)$ in a thermocouple is given by the relation $E = \alpha \theta + \frac{1}{2} \beta \theta^2$, where $\theta$ is the temperature difference between the hot and cold junctions.
As the temperature of the hot junction increases, the thermo $e.m.f.$ increases parabolically with temperature.
Initially, the rate of increase is higher, but as the temperature approaches the neutral temperature, the rate of increase slows down.
At the neutral temperature, the thermo $e.m.f.$ reaches its maximum value.
Beyond the neutral temperature, the $e.m.f.$ starts to decrease and eventually becomes zero at the temperature of inversion.

(C) The relationship between the neutral temperature $(t_n)$,the temperature of the cold junction $(t_c)$,and the temperature of inversion $(t_i)$ is given by the formula: $t_n = \frac{t_i + t_c}{2}$.
Given: $t_c = 15\,^{\circ}C$ and $t_n = 280\,^{\circ}C$.
Substituting the values into the formula: $280 = \frac{t_i + 15}{2}$.
Multiplying both sides by $2$: $560 = t_i + 15$.
Solving for $t_i$: $t_i = 560 - 15 = 545\,^{\circ}C$.
Therefore,the temperature of inversion is $545\,^{\circ}C$.

(A) In a thermocouple,the thermo $e.m.f.$ increases with the temperature difference between the junctions until it reaches the neutral temperature,where it is maximum.
After exceeding the neutral temperature,the thermo $e.m.f.$ starts to decrease.
It continues to decrease until it becomes $zero$ at the inversion temperature,after which it changes sign.

(A) Statement $A$ is false because,in a thermocouple,the thermo $e.m.f.$ reaches its maximum value at the neutral temperature $(T_n)$.
Statement $B$ is true because this describes the Seebeck effect,where a temperature difference between two junctions of dissimilar metals creates an electromotive force,resulting in an electric current in the circuit.
Therefore,$A$ is false and $B$ is true.

(D) The heat absorbed due to the Thomson effect is given by the formula: $H = \sigma Q \Delta T$,where $\sigma$ is the Thomson coefficient,$Q$ is the charge,and $\Delta T$ is the temperature difference.
Given:
$\sigma = 10\,\mu V/K = 10 \times 10^{-6}\,V/K$
$Q = 10\,C$
$\Delta T = 60\,^{\circ}C - 50\,^{\circ}C = 10\,K$
Substituting the values:
$H = (10 \times 10^{-6}) \times 10 \times 10$
$H = 1000 \times 10^{-6}\,J$
$H = 10^{-3}\,J = 1\,mJ$
Therefore,the correct option is $D$.

(C) The neutral temperature $(t_n)$ is a property of the thermocouple materials and is independent of the cold junction temperature $(t_c)$. Therefore,the neutral temperature remains $270^{\circ}C$.
The temperature of inversion $(t_i)$ is related to the neutral temperature and cold junction temperature by the formula: $t_i = 2t_n - t_c$.
Given $t_n = 270^{\circ}C$ and the new cold junction temperature $t_c = 40^{\circ}C$:
$t_i = 2 \times 270^{\circ}C - 40^{\circ}C$
$t_i = 540^{\circ}C - 40^{\circ}C = 500^{\circ}C$.
Thus,the neutral temperature is $270^{\circ}C$ and the temperature of inversion is $500^{\circ}C$.

(D) The neutral temperature $(T_n)$ of a thermocouple is a characteristic property of the materials used to form the thermocouple junction.
It depends solely on the nature of the metals used and is independent of the temperature of the cold junction $(T_c)$ and the temperature of the hot junction $(T_h)$.
Therefore,if the temperature of the cold junction is lowered,the neutral temperature remains the same.

(A) The thermo-electromotive force $(E)$ in a thermocouple is given by the equation $E = at + bt^2$,where $t$ is the temperature difference between the junctions.
At the temperature of inversion,the thermo-electromotive force $(E)$ reaches its maximum or minimum value,meaning the rate of change of $E$ with respect to $t$ is zero.
Therefore,we set the derivative $\frac{dE}{dt} = 0$.
$\frac{d}{dt}(at + bt^2) = 0$
$a + 2bt = 0$
Solving for $t$,we get $t = -\frac{a}{2b}$.
Thus,the temperature of inversion is $t = -\frac{a}{2b}$.

(D) thermocouple works on the Seebeck effect,where a temperature difference between two junctions of dissimilar metals produces an electromotive force $(e.m.f.)$.
Antimony $(Sb)$ and bismuth $(Bi)$ are chosen for thermocouples because they are located far apart in the thermoelectric series.
This large separation in the thermoelectric series results in the production of a higher thermal $e.m.f.$ for a given temperature difference compared to most other metal pairs.
Therefore,option $(D)$ is the correct answer.

(B) The thermoelectric $EMF$ is given by $e = \alpha t - \frac{1}{2}\beta t^2$.
For the neutral temperature $t_n$,the $EMF$ is maximum,so $\frac{de}{dt} = 0$.
$\frac{de}{dt} = \alpha - \beta t = 0 \Rightarrow t_n = \frac{\alpha}{\beta}$.
Substituting the given values: $t_n = \frac{500}{5} = 100 \, ^\circ C$.
The relationship between neutral temperature $t_n$,inversion temperature $t_i$,and cold junction temperature $t_c$ is $t_n = \frac{t_i + t_c}{2}$.
Given $t_c = 0 \, ^\circ C$,we have $100 = \frac{t_i + 0}{2}$.
Therefore,$t_i = 200 \, ^\circ C$.

(A) The current $I_1$ through the silver voltameter is given by Faraday's law of electrolysis: $I_1 = \frac{m_1}{Z_{Ag} t} = \frac{1}{11.2 \times 10^{-4} \times 30 \times 60} \approx 0.496\, A$.
The current $I_2$ through the copper voltameter is: $I_2 = \frac{m_2}{Z_{Cu} t} = \frac{1.8}{6.6 \times 10^{-4} \times 30 \times 60} \approx 1.515\, A$.
The total current $I$ supplied by the battery is $I = I_1 + I_2 = 0.496 + 1.515 = 2.011\, A$.
The power supplied by the battery is $P = I \times V = 2.011 \times 12 = 24.132\, J/s$.

(D) The thermo $e.m.f.$ $E$ in a thermocouple is given by the equation $E = a\theta + b\theta^2$.
Comparing the given equation $E = 40\theta - \frac{\theta^2}{20}$ with the standard form,we get $a = 40$ and $b = -\frac{1}{20}$.
The neutral temperature $\theta_n$ is the temperature at which the thermo $e.m.f.$ is maximum,which occurs when $\frac{dE}{d\theta} = 0$.
$\frac{dE}{d\theta} = a + 2b\theta = 0$.
Therefore,$\theta_n = -\frac{a}{2b}$.
Substituting the values: $\theta_n = -\frac{40}{2 \times (-1/20)} = -\frac{40}{-1/10} = 400\,^oC$.
Thus,the neutral temperature is $400\,^oC$.

(A) The thermoelectric power $S$ is defined as the rate of change of thermo electromotive force $E$ with respect to temperature $T$,given by $S = \frac{dE}{dT}$.
Given the expression $E = K(T - T_r) \left[ T_0 - \frac{1}{2}(T + T_r) \right]$.
Let $x = (T - T_r)$,then $T = x + T_r$. The expression becomes $E = Kx \left[ T_0 - \frac{1}{2}(x + 2T_r) \right] = K \left[ T_0 x - \frac{1}{2}x^2 - T_r x \right]$.
Differentiating with respect to $T$ (noting $\frac{dx}{dT} = 1$):
$S = \frac{dE}{dT} = K \left[ T_0 - x - T_r \right] = K \left[ T_0 - (T - T_r) - T_r \right] = K(T_0 - T)$.
Substituting $T = \frac{1}{2}T_0$ into the expression for $S$:
$S = K \left( T_0 - \frac{1}{2}T_0 \right) = \frac{1}{2} K T_0$.

(B) The e.m.f. of a thermocouple is given by the equation $E = \alpha T + \frac{1}{2}\beta T^2$.
Comparing this with the given equation $E = 16T - 0.04T^2$,we get $\alpha = 16$ and $\frac{1}{2}\beta = -0.04$,which implies $\beta = -0.08$.
The neutral temperature $T_n$ is the temperature at which the e.m.f. is maximum,given by $T_n = -\frac{\alpha}{\beta}$.
$T_n = -\frac{16}{-0.08} = 200\,^{\circ}C$.
The temperature of inversion $T_i$ is related to the neutral temperature $T_n$ and the cold junction temperature $T_c$ by the relation $T_i = 2T_n - T_c$.
Given $T_c = 0\,^{\circ}C$,we have $T_i = 2(200) - 0 = 400\,^{\circ}C$.

(C) The thermo e.m.f. $E$ is given by $E = AT - \frac{1}{2}BT^2$.
At the temperature of inversion $(T_i)$,the thermo e.m.f. becomes zero.
Setting $E = 0$:
$0 = AT_i - \frac{1}{2}BT_i^2$
$AT_i = \frac{1}{2}BT_i^2$
$T_i = \frac{2A}{B}$
Given $A = 16$ and $B = 0.08$:
$T_i = \frac{2 \times 16}{0.08} = \frac{32}{0.08} = \frac{3200}{8} = 400\,^oC$.
Therefore,the temperature of inversion is $400\,^oC$.

(B) The total thermo-emf generated between two temperatures $T_1$ and $T_3$ is the sum of the thermo-emfs generated between $T_1$ to $T_2$ and $T_2$ to $T_3$.
Mathematically,$e_{T_1}^{T_3} = e_{T_1}^{T_2} + e_{T_2}^{T_3}$.
Given:
$e_{0}^{100} = 200\,\mu V$
$e_{0}^{32} = 64\,\mu V$
$e_{32}^{70} = 76\,\mu V$
We need to find $e_{70}^{100}$.
Using the additive property:
$e_{0}^{100} = e_{0}^{32} + e_{32}^{70} + e_{70}^{100}$
$200 = 64 + 76 + e_{70}^{100}$
$200 = 140 + e_{70}^{100}$
$e_{70}^{100} = 200 - 140 = 60\,\mu V$.

(A) The heat $H$ developed at a junction due to the Peltier effect is given by the formula: $H = \pi I t$,where $\pi$ is the Peltier coefficient,$I$ is the current,and $t$ is the time.
Given: $\pi = 2 \times 10^{-9} \, V$,$I = 2.5 \, A$,and $t = 2 \, min = 120 \, s$.
Substituting the values: $H = (2 \times 10^{-9}) \times 2.5 \times 120 = 600 \times 10^{-9} \, J = 6 \times 10^{-7} \, J$.
Since $1 \, J = 10^7 \, ergs$,we convert the heat into ergs: $H = (6 \times 10^{-7} \, J) \times (10^7 \, ergs/J) = 6 \, ergs$.
Therefore,the correct option is $A$.

(D) The mass deposited in a voltameter is given by Faraday's law of electrolysis: $m = Z i t$,where $m$ is the mass,$Z$ is the electrochemical equivalent $(E.C.E.)$,$i$ is the current,and $t$ is the time.
Given time $t = 0.5 \text{ hours} = 1800\,s$.
For the silver voltameter: $i_1 = \frac{m_1}{Z_1 t} = \frac{2}{1.118 \times 10^{-3} \times 1800} \approx 0.994\,A$.
For the copper voltameter: $i_2 = \frac{m_2}{Z_2 t} = \frac{1}{3.294 \times 10^{-4} \times 1800} \approx 1.687\,A$.
Since the voltameters are in parallel,the total current $I = i_1 + i_2 = 0.994 + 1.687 = 2.681\,A$.
The power supplied by the battery is $P = V \times I = 6 \times 2.681 \approx 16.086\,W$.
Thus,the power is approximately $16\,W$.

(D) The heat produced by the coil per second is given by $H = \frac{V^2}{R}$.
Since the heat is used to melt the ice,we have $H = \frac{m}{t} \times L$,where $L$ is the latent heat of fusion of ice $(L = 80\,cal/g = 336\,J/g)$.
To convert electrical energy in Joules to calories,we use the conversion factor $J = 4.2\,J/cal$.
The rate of melting is $\frac{m}{t} = \frac{V^2}{R \times L \times 4.2}$.
Substituting the values: $\frac{m}{t} = \frac{210^2}{50 \times 80 \times 4.2} = \frac{44100}{16800} = 2.625\,g/s$.

(D) The sensitivity of the thermocouple $(S)$ is the difference between the Seebeck coefficients of the two metals:
$S = S_{Te} - S_{Bi}$
$S = 500\,\mu V/{^oC} - (-72\,\mu V/{^oC})$
$S = 500\,\mu V/{^oC} + 72\,\mu V/{^oC} = 572\,\mu V/{^oC}$
Given the temperature difference $(\Delta T)$ is $100\,{^oC}$, the thermo e.m.f. $(E)$ is calculated as:
$E = S \times \Delta T$
$E = 572\,\mu V/{^oC} \times 100\,{^oC}$
$E = 57200\,\mu V$
To convert $\mu V$ to $mV$, we divide by $1000$:
$E = 57200 \times 10^{-6}\,V = 57.2 \times 10^{-3}\,V = 57.2\,mV$.

(D) Let $E_{12}$ be the thermo-emf generated by the junction pair $(1, 2)$ and $E_{23}$ be the thermo-emf generated by the junction pair $(2, 3)$.
According to the law of intermediate temperatures in thermoelectric circuits,the total emf $E$ in the circuit is the algebraic sum of the emfs generated by individual junctions.
When junction $1$ is at $50\,^{\circ}\text{C}$ and junctions $2$ and $3$ are at $0\,^{\circ}\text{C}$,the deflection is proportional to the emf $E_{12} = 14$ divisions.
When junction $3$ is at $50\,^{\circ}\text{C}$ and junctions $1$ and $2$ are at $0\,^{\circ}\text{C}$,the deflection is proportional to the emf $E_{23} = 11$ divisions.
When junction $2$ is at $50\,^{\circ}\text{C}$ and junctions $1$ and $3$ are at $0\,^{\circ}\text{C}$,the circuit effectively acts as two thermocouples in series,where the emf generated is the sum of the emfs of the two junctions,i.e.,$E = E_{12} + E_{23}$.
Therefore,the total deflection is $14 + 11 = 25$ divisions.

(C) Let the energy dissipated in each resistor be $H$. Since the resistors ${R_2}$ and ${R_3}$ are connected in parallel,the potential difference across them is the same. Therefore,the energy dissipated in them is given by $H = \frac{V^2}{R}t$. For the energy to be the same,we must have ${R_2} = {R_3}$.
Let the total current flowing through the circuit be $i$. The current ${R_1}$ is $i$. The current splits into ${i_1}$ and ${i_2}$ at the junction. Since ${R_2} = {R_3}$,the current splits equally,so ${i_1} = {i_2} = \frac{i}{2}$.
The energy dissipated in ${R_1}$ is $H = i^2 {R_1} t$.
The energy dissipated in ${R_2}$ is $H = i_1^2 {R_2} t = (\frac{i}{2})^2 {R_2} t = \frac{i^2}{4} {R_2} t$.
Since the energy dissipated in all resistors is the same,we equate the energy in ${R_1}$ and ${R_2}$:
$i^2 {R_1} t = \frac{i^2}{4} {R_2} t$
${R_1} = \frac{{R_2}}{4}$.

(D) The thermo $e.m.f.$ $(E)$ in a thermocouple is given by the relation: $E = \alpha \,t + \frac{1}{2}\beta \,{t^2}$,where $t$ is the temperature difference between the junctions,and $\alpha$ and $\beta$ are constants characteristic of the metals used.
This equation represents a parabola.
As the temperature difference $t$ increases,the $e.m.f.$ first increases,reaches a maximum value at the neutral temperature,and then decreases as the temperature increases further.
Therefore,the graph between $E$ and $t$ is a parabolic curve that opens downwards,which is represented by option $D$.

(A) The resistance $(R)$ of a metallic conductor (like the filament of a heater) increases with an increase in temperature according to the relation $R = R_0(1 + \alpha \Delta T)$.
Initially, when the heater is switched on, the filament is at room temperature, so its resistance is low. According to Ohm's law, $I = V/R$, a low resistance results in a high initial current.
As the filament heats up, its resistance increases. Consequently, the current flowing through the heater decreases over time.
As the heater reaches a steady operating temperature, the resistance also stabilizes, and the current reaches a steady state value.
Among the given options, the graph that shows a high initial current that decreases and then levels off to a constant value is represented by the curve in option $(A)$.

(B) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = Z \cdot q$,where $Z$ is the $E.C.E.$ and $q$ is the total charge passed.
The total charge $q$ is equal to the area under the current-time $(I-t)$ graph.
The time is given in minutes,so we convert it to seconds: $t = 6 \text{ min} = 6 \times 60 \text{ s} = 360 \text{ s}$.
The area under the graph consists of a triangle (from $t = 0$ to $2 \text{ min}$) and a rectangle (from $t = 2 \text{ min}$ to $6 \text{ min}$).
Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 \times 60 \text{ s}) \times 1 \text{ A} = 60 \text{ C}$.
Area of rectangle $= \text{length} \times \text{width} = (6 - 2) \times 60 \text{ s} \times 1 \text{ A} = 4 \times 60 \text{ C} = 240 \text{ C}$.
Total charge $q = 60 \text{ C} + 240 \text{ C} = 300 \text{ C}$.
Using the formula $m = Z \cdot q$,we get $Z = \frac{m}{q} = \frac{m}{300} \text{ g/C}$.

(D) $1$. For the terminal voltage $V$: The relation is $V = E - Ir$. This is a linear equation of the form $y = mx + c$,where the slope is $-r$ and the intercept is $E$. Thus,the graph of $V$ versus $I$ is a straight line with a negative slope.
$2$. For the external power $P$: The power dissipated in the external circuit is $P = VI = (E - Ir)I = EI - I^2r$. This is a downward-opening parabola with respect to $I$,passing through the origin $(0,0)$ and $(E/r, 0)$.
$3$. For the fraction $\eta$: The thermal power generated in the cell is $P_{thermal} = I^2r$. The total electrical power generated in the cell is $P_{total} = EI$. The fraction $\eta$ is given by $\eta = \frac{P_{thermal}}{P_{total}} = \frac{I^2r}{EI} = \frac{r}{E}I$. This is a linear equation of the form $y = mx$,representing a straight line passing through the origin.
Since all three relationships are correctly described by their respective graphs,the correct option is $(d)$.

(B) To verify Ohm's law,the following conditions must be met:
$1$. An ammeter $(A)$ must be connected in series with the resistor to measure the current flowing through it.
$2$. $A$ voltmeter $(V)$ must be connected in parallel across the resistor to measure the potential difference across it.
$3$. $A$ rheostat $(Rh)$ is used to vary the current in the circuit.
Analyzing the given options:
- In option $(A)$,the voltmeter is in parallel with the ammeter,which is incorrect.
- In option $(B)$,the voltmeter is connected in parallel across the resistor,and the ammeter is in series with the resistor. This is the correct configuration for verifying Ohm's law.
- In option $(C)$,the voltmeter is in series with the resistor,which is incorrect.
- In option $(D)$,the ammeter is in series with the resistor,but the voltmeter is not connected across the resistor to measure the potential difference across it correctly.
Therefore,the circuit in option $(B)$ is the correct one.

(A) When an electric heater is switched on,the filament is initially at room temperature,where its resistance is low.
According to Ohm's law,$I = V/R$. Since $R$ is low initially,the current $I$ is high.
As the filament heats up,its temperature increases,which causes its resistance $R$ to increase (since $R = R_0(1 + \alpha \Delta T)$).
As the resistance $R$ increases,the current $I$ decreases.
Eventually,the temperature of the filament reaches a steady state where the heat generated equals the heat lost to the surroundings,and the current becomes constant.
Graph $A$ shows the current starting at a high value and decreasing to a steady state value,which correctly represents this behavior.

(A) The power dissipated in a circuit is given by $P = \frac{E^2}{R_{eq}}$,where $E$ is the $EMF$ of the source and $R_{eq}$ is the equivalent resistance of the circuit. To maximize power $P$,we need to minimize the equivalent resistance $R_{eq}$.
For circuit $A$: Two resistors $R$ are in parallel. $R_{eq,A} = \frac{R \cdot R}{R + R} = \frac{R}{2} = 0.5R$.
For circuit $B$: Two resistors $R$ are in series. $R_{eq,B} = R + R = 2R$.
For circuit $C$: Two resistors $R$ are in parallel,and this combination is in series with another resistor $R$. $R_{eq,C} = \frac{R}{2} + R = 1.5R$.
For circuit $D$: Two resistors $R$ are in series,and this combination is in parallel with another resistor $R$. $R_{eq,D} = \frac{(2R) \cdot R}{2R + R} = \frac{2R^2}{3R} = \frac{2}{3}R \approx 0.67R$.
Comparing the equivalent resistances: $R_{eq,A} (0.5R) < R_{eq,D} (0.67R) < R_{eq,C} (1.5R) < R_{eq,B} (2R)$.
Since circuit $A$ has the minimum equivalent resistance,it will dissipate the maximum power.

(C) The phenomenon where heat is evolved or absorbed at the junction of two dissimilar metals when an electric current passes through it is known as the $Peltier$ effect.
This is a reversible process,unlike the $Joule$ heating effect which is irreversible.

(C) The mass deposited $m$ is given by Faraday's law of electrolysis: $m = Z i t$.
Also,the mass can be expressed as $m = \rho \times V = \rho \times A \times x$,where $\rho$ is density,$A$ is area,and $x$ is thickness.
Equating the two expressions: $Z i t = \rho A x$.
Solving for thickness $x$: $x = \frac{Z i t}{A \rho}$.
Given: $Z = 0.00033\ g/C = 0.00033 \times 10^{-3}\ kg/C$,$i = 1.5\ A$,$t = 20 \times 60\ s = 1200\ s$,$A = 50\ cm^2 = 50 \times 10^{-4}\ m^2$,$\rho = 9000\ kg/m^3$.
Substituting the values: $x = \frac{0.00033 \times 10^{-3} \times 1.5 \times 1200}{50 \times 10^{-4} \times 9000}$.
$x = \frac{0.00033 \times 1.5 \times 1200}{50 \times 9000} \times 10^{-3} = \frac{0.594}{450000} \times 10^{-3} = 1.32 \times 10^{-6} \times 10 = 1.3 \times 10^{-5}\ m$.

(B) As we move along the conductor in the direction of the current,the potential $V$ decreases linearly due to the resistance of the conductor $(V = IR)$.
When we cross the cell from the negative terminal to the positive terminal,the potential increases by an amount equal to the terminal voltage of the cell,which is given by $V = E - Ir$,where $r$ is the internal resistance.
Since $Ir > 0$,the increase in potential is strictly less than the $e.m.f. \ E$.
Therefore,the graph must show a linear decrease,followed by a sharp increase of magnitude less than $E$,and then another linear decrease.
Comparing this with the given options,Graph $B$ correctly depicts this behavior.

(A) According to Faraday's law of electrolysis, the mass $m$ deposited is given by $m = ZIt$, where $Z$ is the electrochemical equivalent, $I$ is the actual current, and $t$ is the time in seconds.
Given: $m = 2.0124 \ g$, $Z = 1.118 \times 10^{-3} \ g \ C^{-1}$, $t = 1 \ hour = 3600 \ s$.
The actual current $I$ is calculated as:
$I = \frac{m}{Zt} = \frac{2.0124}{1.118 \times 10^{-3} \times 3600}$
$I = \frac{2.0124}{4.0248} = 0.5 \ A$.
The ammeter reading is $I_{obs} = 0.54 \ A$.
The error in the ammeter reading is defined as $I_{obs} - I_{actual}$.
Error $= 0.54 \ A - 0.5 \ A = +0.04 \ A$.

(A) Given:
Resistance of the coil,$R = 0.1\,\Omega$
Time,$t = 20\,\text{minutes} = 20 \times 60 = 1200\,\text{s}$
Mass of copper deposited,$m = 0.99\,\text{g}$
Electrochemical equivalent,$Z = 0.00033\,\text{g/C}$
Using Faraday's law of electrolysis,$m = Z \times i \times t$:
$i = \frac{m}{Z \times t} = \frac{0.99}{0.00033 \times 1200} = \frac{0.99}{0.396} = 2.5\,\text{A}$
The heat produced in the coil is given by Joule's law of heating,$H = i^2Rt$:
$H = (2.5)^2 \times 0.1 \times 1200$
$H = 6.25 \times 0.1 \times 1200$
$H = 0.625 \times 1200 = 750\,\text{J}$