NCERT Exemplar Q & A

Hint

(b) Key concept: Current per unit area (taken normal to the current), $I / A$, is called current density and is denoted by $\vec{J}$.
The SI unit of the current density are $\mathrm{A} / \mathrm{m}^2$. The current density is also directed along $E$ and which is also a vector quantity and the relationship is given by
$\vec{J}=\sigma \vec{E}=\frac{\vec{E}}{\rho}$
where $\sigma=$ conductivity and $\rho=$ resistivity or specific resistance of the substance.
The $\vec{J}$ changes due to the electric field produced by charges accumulated on the surface of wire.

Hint

(a) The equivalent emf of this combination is given by
$\varepsilon_{\text {eq }}=\frac{\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_1}{r_2}}{\left(\frac{1}{r_1}+\frac{1}{r_2}\right)}=\frac{\varepsilon_1\left(\frac{1}{r_1}+\frac{\varepsilon_2 / \varepsilon_1}{r_2}\right)}{\left(\frac{1}{r_1}+\frac{1}{r_2}\right)}=\frac{\varepsilon_2\left(\frac{\varepsilon_1 / \varepsilon_2}{r_1}+\frac{1}{r_2}\right)}{\left(\frac{1}{r_1}+\frac{1}{r_2}\right)}$
$\text { As } \frac{\varepsilon_2}{\varepsilon_1}>1 \Rightarrow \Rightarrow \frac{\left(\frac{1}{r_1}+\frac{\varepsilon_2 / \varepsilon_1}{r_2}\right)}{\left(\frac{1}{r_1}+\frac{1}{r_2}\right)}>1 \text { or } \varepsilon_{\mathrm{eq}}>\varepsilon_1 \text { also } \frac{\varepsilon_1}{\varepsilon_2}<1 \Rightarrow \frac{\left(\frac{\varepsilon_1 / \varepsilon_2}{r_1}+\frac{1}{r_2}\right)}{\left(\frac{1}{r_1}+\frac{1}{r_2}\right)}<1 \text { or } \varepsilon_{\mathrm{eq}}<\varepsilon_2$
$\begin{aligned} &\text { Hence }\\ &\varepsilon_1<\varepsilon_{\mathrm{eq}}<\varepsilon_2 . \end{aligned}$

Hint

(c) The percentage error in $R$ can be minimised by adjusting the balance point near the middle of the bridge, i.e., when $l_1$ is close to 50 cm . This requires a suitable choice of $S$.
Since,
$\frac{R}{S}=\frac{R l_1}{R\left(100-l_1\right)}=\frac{l_1}{100-l_4}$
Since here, $R: S:: 2.9: 97.1$ imply that the $S$ is nearly 33 times to that of $R$. In orded to make this ratio $1: 1$, it is necessary to reduce the value of $S$ nearly $\frac{1}{33}$ times i.e., nearly $3 \Omega$.

Hint

(a) The resistance of wire is given by
$R=\rho \frac{l}{A}$
For greater value of $R, l$ must be higher and $A$ should be lower and it is possible only when the battery is connected across $1 \mathrm{~cm} \times\left(\frac{1}{2}\right) \mathrm{cm}$ (area of cross-section $A$ ).

Hint

(a) The relationship between current and drift speed is given by
$I=n e A v_d$
Here, $I$ is the current and $v_d$ is the drift velocity.
So, $I \propto v_d$
Thus, only drift velocity determines the current in a conductor.

Hint

(b, d)
Kirchhoff’s junction rule is also known as Kirchhoff’s current law which states that the algebraic sum of the currents flowing towards any point in an electric network is zero. i.e., charges are conserved in an electric network.
So, Kirchhoff’s junction rule is the reflection of conservation of charge.

Hint

(a, d)
Here, the potential drop is taking place across $A B$ and $r$. Since the equivalent resistance of parallel combination of $R$ and $R^{\prime}$ is always less than $R$, therefore $I \geq \frac{V}{r+R}$ always.
Note In parallel combination of resistances, the equivalent resistance is smaller than smallest resistance present in combination.

Hint

(a, b) Resistivity is the intrinsic property of the substance.
For a metallic conductor, resistivity is given by
$\rho=\frac{m}{n e^2 \tau}$
where $n$ is the number of charge carriers per unit volume (number density) which can change with temperature $T$ and $\tau$ is relaxation time (time interval between two successive collisions) which decreases with the increase of temperature $\left(T \propto \frac{1}{\tau}\right)$.

Hint

(b, c)
Given, for first student, $R_2=10 \Omega, R_1=5 \Omega, R_3=5 \Omega$
For second student, $R_1=500 \Omega, R_3=5 \Omega$
Now, according to Wheatstone bridge rule,
$\frac{R_2}{R}=\frac{R_1}{R_3} \Rightarrow R=R_3 \times \frac{R_2}{R_1} \dots(1)$
Now putting all the values in Eq. (i), we get $R=10 \Omega$ for both students. Thus, we can analyse that the Wheatstone bridge is most sensitive and accurate if resistances are of same value.
Thus, the errors of measurement of the two students depend on the accuracy and sensitivity of the bridge, which inturn depends on the accuracy with which $R_2$ and $R_1$ can be measured.
When $R_2$ and $R_1$ are larger, the currents through the arms of bridge is very weak. This can make the determination of null point accurately more difficult.

Hint

(a, c) Key concept: Meter bridge: In case of meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point $B$, bridge is balanced. If in balanced position of bridge $A B=l, B C=(100-l)$ so that $Q / P=(100-l) / l$. Also $P / Q=R / S=>S=\frac{100-l}{l} \times R$

When there is no deflection in galvanometer there is no current across the galvanometer, then points B and D are at same potential. That point at which galvanometer shows no deflection is called null point, then potential at B and neutral point D are same. When the jockey contacts a point on the meter wire to the right of D , the potential drop across AD is more than potential drop across AB, which brings the potential of point D less than that of B, hence current flows from B to D in the galvanometer wire.

Hint

(a) If the number of collisions of the free electrons with the lattice is decreased, then the drift velocity of the electrons increases.
Current $i$ is directly proportional to the drift velocity ‘ ${V}_{{d}}$ ‘ and is given by the following relation: ${i}={neAV} {V}_{{d}}$, where ‘ $n$ ‘ is the number density of electrons and ‘ $A$ ‘ is the area of the cross-section of the conductor.
So, we can easily see that current increases with increase in drift velocity.

Hint

(d) The resistance $R$ of a conductor depends on its resistivity $\rho$, length $l$ and cross-sectional area $A$. Thus,
$R=\rho \frac{l}{A}$
From the given comparison of resistances, we cannot derive the correct relation between the resistances. We also need to know the cross-sectional areas and the lengths of both the conductors before concluding about their resistivities. Only then can the relation between the resistivities be found.
The information is not sufficient to find the relation between $\rho_A$ and $\rho_B$.

Hint

The relation between the resistivity $\rho$ and conductivity $\sigma$ of a material is given by
$\begin{aligned} & \rho=\frac{1}{\sigma} \\ & \Rightarrow \rho \times \sigma=1 \end{aligned}$
The product of conductivity and resistivity is unity. So, this product does not depend on any of the given quantities.

Hint

(c) As the temperature of a metallic resistor increases, the product of its resistivity and conductivity remains constant.

Explanation: Resistivity and conductivity are inversely proportional to each other, meaning their product is always equal to 1, regardless of temperature changes.

$\begin{aligned} & \rho=\frac{1}{\sigma} \\ & \Rightarrow \rho \times \sigma=1 \end{aligned}$

Hint

(b) In the study of electric current, the direction opposite the flow of electrons is regarded as the direction of flow of positive charge. In a battery, positive charge flows from the negative terminal to the positive terminal when it is discharging (connected to external circuit). But when the battery is charged, the positive charge flows from the positive terminal to the negative terminal.

Hint

(a) As the resistance is connected to an ideal battery, it provides a constant potential difference across the two terminals.
The internal resistance of the battery is also zero.
The power dissipated in the resistor,
$\mathrm{P}=\frac{V^2}{R}$
$V$ is constant; hence $\mathrm{P} \propto \frac{1}{R}$
Thus, if the value of the resistance is decreased, the power dissipated in the resistor will increase.

Hint

(c) Kinetic energy ( $K$ ) is proportional to the square of velocity ( $v$ ) according to $K=\frac{1}{2} m v^2$, where $m$ is mass.
Since the velocity of conduction electrons is greater than that of the metal ions, their kinetic energy is also greater.
The average kinetic energy of conduction electrons ( $K_1$ ) is greater than that of the metal ions ( $\kappa_2$ ).

Hint

The heat (thermal energy) developed in a resistor can be calculated using the formula:
$H=I^2 R T$
Where $H$ is the heat developed, $I$ is the current, $R$ is the resistance, and $T$ is the time for which the current flows.
For resistor $R_1$ (which is $R$ ):
$H_1=I^2 R T$
For resistor $R_2$ (which is $2 R$ ):
$H_2=I^2(2 R) T=2 I^2 R T$
$\begin{aligned} &\frac{H_1}{H_2}=\frac{I^2 R T}{2 I^2 R T}\\ &\text { – Simplifying this gives: }\\ &\frac{H_1}{H_2}=\frac{1}{2} \end{aligned}$

Hint

(b) The thermal energy developed in resistor $R$ is:
$H_1=\frac{V^2}{R} t$
The thermal energy developed in resistor $2 R$ is:
$\mathrm{H}_2=\frac{V^2}{2 R} t$
The ratio of thermal energy developed in $R$ and $2 R$ is:
$\frac{H_1}{H_2}=\frac{\frac{V^2}{R} t}{\frac{V^2}{2 R} t}$
$\frac{H_1}{H_2}=\frac{V^2}{R} t \cdot \frac{2 R}{V^2 t}$
$\frac{H_1}{H_2}=\frac{2}{1}$
The ratio of thermal energy developed in $R$ and $2 R$ is $2: 1$.

Hint

(a) Resistance of a wire is directly proportional to its length.
So, when we cut the wire into 5 equal parts, the resistance of each part will be $10 \Omega$.
On connecting these wires in parallel, the net resistance will be
$\begin{aligned} & \frac{1}{R}=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10} \\ & \Rightarrow R=2 \Omega \end{aligned}$

Hint

(a) Kirchhoff’s junction law states that the current entering the junction is equal to the current leaving the junction, which indirectly means the charge entering the junction is equal charge leaving the junction. From the statement, it is clear that Kirchhoff’s junction law follows from the conservation of charge. Therefore, statement A is correct. Kirchhoff’s loop law states that the net potential drop in a loop is Zero $\Sigma \mathbf{V}=$ $\mathbf{0}$, which means potential changes around a closed loop is zero. Voltage is the measure of energy per unit charge in this context KVL can be restated as the total energy gained per unit charge must be equal to the amount of energy lost per unit charge (ie, Law of conservation of Energy). Therefore statement B is correct.

Hint

(b) Let the emfs of the batteries be $e_1$ and $e_2$, and their respective resistances be $r_1$ and $r_2$.
Since the batteries are connected in series, the equivalent emf will be the sum of the emf of the two batteries ( $\mathrm{e}=\mathrm{e}_1+\mathrm{e}_2$ ).
Thus, $\mathrm{e}>\mathrm{e}_1$ and $\mathrm{e}>\mathrm{e}_2$
Thus, the equivalent emf is larger than either of the two emfs. Hence, statement A is correct.
Since the batteries are connected in series, the equivalent internal resistance (r) of the combination will be the sum of the internal resistance of the two batteries ( $r=r_1+r_2$ ).
$r>r_1[latex] and [latex]r>r_2$
Thus, the equivalent internal resistance is greater that either of the two resistances. Hence, statement B is wrong.

Hint

(c) For batteries in parallel, the equivalent emf $\left(\epsilon_{e q}\right)$ can be calculated using the formula:
$\epsilon_{e q}=\frac{\epsilon_1 / r_1+\epsilon_2 / r_2}{1 / r_1+1 / r_2}$
Thus, the equivalent emf is greater than either of the emfs. Thus, statement A is wrong.
For the parallel combination of the batteries of internal resistance $r_1$ and $r_2$, the equivalent internal resistance $r$ is given as
$r=\frac{r_1 r_2}{r_1+r_2}$
Thus, the value of the resultant resistance is even smaller than either resistance.

Hint

(d) it does not appreciably change the current to be measured.

Explanation:
An ammeter is designed to measure the current flowing in a circuit. If the ammeter itself has a significant resistance, it will add to the overall resistance of the circuit, thereby reducing the current flowing through it. This means the measured current will not be the actual current flowing in the circuit without the ammeter. To minimize this interference and ensure an accurate measurement, the ammeter’s resistance should be as small as possible.

Hint

(d) It does not appreciably change the potential difference to be measured.

Explanation:
A voltmeter is designed to measure voltage without significantly affecting the circuit it’s connected to. If a voltmeter had a low resistance, it would draw a substantial current from the circuit, altering the voltage being measured. By having a very high resistance, the voltmeter draws minimal current, ensuring that the measured voltage is as close as possible to the actual voltage in the circuit.

Hint

(b) The charge $Q$ on the plates of a capacitor at time $t$ after connecting it in a charging circuit,
$Q=\epsilon C\left(1-e^{-t / R C}\right)$
where
$\epsilon=e m f \text { of the battery connected in the charging circuit }$
$\mathrm{C}=\text { capacitance of the given capacitor }$
$R=\text { resistance of the resistor connected in series with the capacitor }$
The charge developed on the plates of the capacitor in first 10 mili seconds is given by
$Q_1=\epsilon C\left(1-e^{-10 \times 10^{-3} / R C}\right)$
The charge developed on the plates of the capacitor in first 20 mili seconds is given by
$Q^{\prime}=\epsilon C\left(1-e^{-20 \times 10^{-3} / R C}\right)$
The charge developed at the plates of the capacitor in the interval $t=10$ mili seconds to 20 mili seconds is given by
$\begin{aligned} & Q_2=Q^{\prime}-Q_1 \\ & Q_2=\left[\epsilon C\left(1-e^{-20 \times 10^{-3} / R C}\right)\right]-\left[\epsilon C\left(1-e^{-10 \times 10^{-3} / R C}\right)\right] \\ & \Rightarrow Q_2=\left[\epsilon C\left(e^{-10 \times 10^{-3} / R C}-e^{-20 \times 10^{-3} / R C}\right)\right] \\ & \Rightarrow Q_2=\left[\epsilon C\left(e^{-10 \times 10^{-3} / R C}\right)\left(1-e^{-10 \times 10^{-3} / R C}\right)\right] \end{aligned}$
Comparing $\mathrm{Q}_1$ with $\mathrm{Q}_2$
$\begin{aligned} & Q_1=\epsilon C\left(1-e^{-10 \times 10^{-3} / R C}\right), Q_2=\epsilon C e^{-10 \times 10^{-3} / R C}\left(1-e^{-10 \times 10^{-3} / R C}\right) \\ & \frac{Q_1}{Q_2}=\frac{1}{e^{-10 \times 10^{-3} / R C}} \\ & e^{-10 \times 10^{-3} / R C}<1 \\ & \therefore Q_1>Q_2 \end{aligned}$
For second part of the question
$10 \mu \mathrm{C}$ charge be deposited in a time interval $t_1$ and the next $10 \mu \mathrm{C}$ charge is deposited in the next time interval $\mathrm{t}_2$.
The time taken for $10 \mu \mathrm{C}$ charge to develop on the plates of the capacitor is $t_1$
$10 \mu C=\epsilon C\left(1-e^{-t_1 / R C}\right) \dots(1)$
The time taken for $20 \mu \mathrm{C}$ charge to develop on the plates of the capacitor is $t^{\prime}$
$20 \mu C=\epsilon C\left(1-e^{-t_2 / R C}\right) \dots(2)$
Dividing (2) by (1)
$\begin{aligned} & \frac{20 \mu C}{10 \mu C}=\frac{\epsilon C\left(1-e^{-t_2 / R C}\right)}{\epsilon C\left(1-e^{-t_1 / R C}\right)} \\ & 2=\frac{\left(1-e^{-t_2 / R C}\right)}{\left(1-e^{-t_1 / R C}\right)} \\ & 2\left(1-e^{-t_1 / R C}\right)=\left(1-e^{-t_2 / R C}\right) \\ & e^{-t_2 / R C}=2 e^{-t_1 / R C}-1 \end{aligned}$
Taking natural log both side,
$\begin{aligned} & \frac{t_2}{R C}=\ln (2)+\frac{t_1}{R C} \\ & \Rightarrow \mathrm{t}_1<\mathrm{t}_2 \end{aligned}$

Hint

(a) Let the potentials at A and B be $\mathrm{V}_{\mathrm{A}}$ and $\mathrm{V}_{\mathrm{B}}$.
As potential,
$E=-\frac{d V}{d r}$
potential increases in the direction opposite to the direction of the electric field.
Thus, $\mathrm{V}_{\mathrm{A}}<\mathrm{V}_{\mathrm{B}}$
Potential energy of the electrons at points A and B :
$\mathrm{U}_{\mathrm{A}}=-\mathrm{eV}_{\mathrm{A}}$
$U_B=-e V_B$
Thus, $\mathrm{U}_{\mathrm{A}}>\mathrm{U}_{\mathrm{B}}$
Let the kinetic energy of an electron at points $A$ and $B[latex] be [latex]K_A$ and $K_B$ respectively.
Applying the principle of conservation of mechanical energy, we get:
$\mathrm{U}_{\mathrm{A}}+\mathrm{K}_{\mathrm{A}}=\mathrm{U}_{\mathrm{B}}+\mathrm{K}_{\mathrm{B}}$
As, $\mathrm{U}_{\mathrm{A}}>\mathrm{U}_{\mathrm{B}}$,
$\mathrm{K}_{\mathrm{A}}<\mathrm{K}_{\mathrm{B}}$
Therefore, the speed of the electrons is more at B than at A.

Hint

(b, c) As the capacitor is connected to the battery at $t=0$, current flows through the wire up to the time the capacitor is charged. As the capacitor is completely charged, the potential difference across the capacitor is equal to the terminal potential of the battery. So, the current stops flowing and does not pass through the point between the plates of the capacitor, as there is no medium for the flow of charge.

Hint

(c, d) When no current passes through a conductor, free electrons do not move in a net direction. While they are constantly moving randomly due to thermal energy, there is no overall movement of electrons in a particular direction. When no current is passed through a conductor, the average of the velocities of all the free electrons at an instant is zero.

Hint

(d) Resistance, resistivity and drift velocity varies with relaxation time which is dependent on temperature Number of free electrons in a conductor remain in variant even if temperature is changes.

Hint

(a) Resistivity $(\rho)$ of a conductor is the reciprocal of the its conductivity $(\sigma)$. Thus,
$\rho=\frac{1}{\sigma}$
The ratio of the resistivity and the conductivity,
$\frac{\rho}{\sigma}=\frac{\rho}{1 / \rho}=\rho^2$
As resistivity increases with temperature, the square of resistivity will also increase. Hence, the ratio of resistivity and conductivity will increase with increase in temperature.

Hint

(a, d) Drift speed and current density are inversely proportional to the area of cross-section of a wire. Thus, they are dependent on the cross-section. The charge crossing in a given time interval is independent of the area of cross-section of the wire. Free electron density is the total number of free electrons per unit volume of the wire. The density of free electrons depends on the distribution of the free electrons throughout the volume of the wire. It does not depend on the cross-section of the wire.

Note: Drift speed is the average speed of electrons in a conductor. It is related to the current density and the cross-sectional area through the equation $\mathrm{J}=nev_d$ (where J is current density, n is free-electron density, e is electron charge, and ${v}_{d}$ is drift speed). Since current density depends on cross-section area, drift speed also depends on it.

Hint

(a, d)

The ammeter is connected in series in the circuit whose current is to be measured. If the net resistance of the ammeter is high, then the amount of current that the circuit draws will have error . So, we won’t be able to accurately measure the amount of current that the circuit draws from the voltage source. Thus, it should have a small resistance.

To measure the potential difference across any circuit element, the voltmeter is connected in parallel to that circuit element. If the resistance of the voltmeter is large, then maximum voltage drop occurs across the voltmeter and it will measure the correct value of the potential.

Hint

The charge $(Q)$ on the capacitor at any instant $t$.
$Q=C V\left(1-e^{-t / R C}\right)$
where
C = capacitance of the given capacitance
$R=$ resistance of the resistor connected in series with the capacitor
$\mathrm{RC}=\left(10 \times 10^3\right) \times\left(500 \times 10^{-6}\right)=5 \mathrm{~s}$
The charge on the capacitor in the first 5 seconds.
$Q_0=C V\left(1-e^{-5 / 5}\right)=C V \times 0.632$
The charge on the capacitor in the first 10 seconds.
$\begin{aligned} & Q_1=C V\left(1-e^{-10 / 5}\right) \\ & Q_1=C V\left(1-e^{-2}\right)=0.864 \times C V \end{aligned}$
Charge developed in the next 5 seconds.
$\begin{aligned} & Q^{\prime}=Q_1-Q_0 \\ & Q^{\prime}=C V(0.864-0.632)=0.232 \mathrm{CV} \end{aligned}$
The charge on the capacitor in the first 55 seconds.
$\begin{aligned} & Q_2=C V\left(1-e^{-55 / 5}\right) \\ & Q_2=C V\left(1-e^{-11}\right)=0.99 \times C V \end{aligned}$
Charge developed in the next 50 seconds.
$\begin{aligned} & Q^{\prime}=Q_2-Q_0 \\ & Q^{\prime}=C V(0.99-0.632)=0.358 C V \end{aligned}$
Charge developed in the first 505 seconds.
$Q_3=C V\left(1-e^{-500 / 5}\right)=C V\left(1-e^{-100}\right) \approx C V$
Charge developed in the next 500 seconds.
$Q^{\prime}=C V(1-0.632)=0.368 \mathrm{CV}$
Thus, the charge developed on the capacitor in the first 5 seconds is greater than the charge developed in the next $5,50,500, 5000$ seconds.

Hint

(b, d) Let the voltage of the battery connected to the capacitors be V. Both the capacitors will charge up to the same potential (V).
The charge on the capacitors $\mathrm{C}_1[latex] is [latex]\mathrm{C}_1 \mathrm{~V}=(1 \mu \mathrm{~F}) \times \mathrm{V}$
The charge on the capacitors $\mathrm{C}_2[latex] is [latex]\mathrm{C}_2 \mathrm{~V}=(2 \mu \mathrm{~F}) \times \mathrm{V}$
The charge on the discharging circuit at an instant $t$,
$Q=C V e^{-t / R C}$
The current through the discharging circuit,
$\frac{d Q}{d t}=-\frac{C V}{R C} e^{-t / R C}=\frac{V}{R} e^{-t / R C}$
At $\mathrm{t}=0$, the current through the discharging circuit will be $\frac{V}{R}$ for both the capacitors.
Let the time taken by the capacitor $C_1$ to lose $50 \%$ of the charge be $t_1$.
$\begin{aligned} & Q_1=\frac{C_1 V}{2} \\ & \frac{C_1 V}{2}=C_1 V e^{-t_1 / R C} \\ & \frac{1}{2}=e^{-t_1 / R C} \end{aligned}$
Taking natural log on both sides, we get:
$\begin{aligned} & -\ln 2=-\frac{t_1}{R C_1} \\ & t_1=R C_1 \ln 2 \end{aligned}$
Similarly,
Time taken for capacitor $\mathrm{C}_2$:
$t_2=R C_2 \ln 2$
As, $\mathrm{C}_1<\mathrm{C}_2, \mathrm{t}_1<\mathrm{t}_2$
Thus, we can say that $C_1$ loses $50 \%$ of its initial charge sooner than $C_2$ loses $50 \%$ of its initial charge.

Hint

(a) Current,
$\begin{aligned} & I=\frac{d q}{d t}=3 t^2+2 t+5 \\ & q=\int_{t=0}^{t=2}\left(3 t^2+2 t+5\right) d t \end{aligned}$
Charge, $q=\left[t^3+t^2+5 t\right]_0^2=[8+4+10]=22 \mathrm{C}$

Hint

$\begin{aligned} &\text { Given, }\\ &\begin{aligned} I & =4-0.08 t \\ \frac{d q}{d t} & =4-0.08 t \\ q & =\int_0^{50}(4-0.08 t) d t \\ n e & =\left[4 t-\frac{0.08 t^2}{2}\right]_0^{50}=100 \\ n & =\frac{100}{e}=\frac{100}{1.6 \times 10^{-19}}=6.25 \times 10^{20} \text { electrons } \end{aligned} \end{aligned}$

Hint

$\begin{aligned} &\text { (a) Drift velocity, } v_d=\frac{e E \tau}{m}\\ &\therefore \quad v_d \propto E \end{aligned}$

Hint

$\begin{aligned} &\text { (a) Drift velocity, }\\ &\begin{aligned} & v_d=\frac{i}{n e A}=\frac{3}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}} \\ & =0.11 \times 10^{-3} \mathrm{~ms}^{-1} \\ & \therefore \text { Time, } t=\frac{\text { length of the wire }}{v_d}=\frac{3}{0.11 \times 10^{-3}}=2.73 \times 10^4 \mathrm{~s} \end{aligned} \end{aligned}$

Hint

(a) The points connected by a conducting wire are at same potential. Then, redraw the diagram by placing the points of same potential at one place and then solve for equivalent resistance.

Hence, from the new figure $A$ and $Y$ are at same potential; $B$ and $X$ are at same potential.
$\begin{aligned} \Rightarrow \quad \frac{1}{R_{\text {eq }}} & =\frac{1}{6 R}+\frac{1}{9 R}+\frac{1}{12 R}=\frac{13}{36 R} \\ R_{\text {eq }} & =\frac{36 R}{13}=2.77 R \end{aligned}$

Hint

(d) It can be seen that, this diagram is symmetrical about $P Q$, so points on the perpendicular bisector of $P Q$, i.e. $X, Y$ and $Z$ are at same potential. So, in this type of diagrams, to calculate the equivalent resistance, we can remove the resistances at the same potential, i.e. the resistances between $X$ and $Y, Y$ and $Z$, are redundant and can be removed.

$\begin{aligned} &\text { All resistances in the circuit are in parallel, }\\ &\begin{aligned} & \frac{1}{R_{\text {eq }}}=\frac{1}{8 R}+\frac{1}{4 r}+\frac{1}{8 R}=\frac{R+r}{4 R r} \\ & R_{\text {eq }}=\frac{4 R r}{R+r} \Omega \end{aligned} \end{aligned}$

Hint

(a) Here, we have infinite pairs of $R$ and $2 R$. Suppose, the equivalent resistance is $R_0$ between $C$ and $D$, i.e. excluding one pair near $A B$ (since, pairs are infinite, equivalent resistance will remain same, if we include pair near $A B$ ).

$\begin{gathered} R_{\text {eq }}=R_0=R+\frac{2 R R_0}{2 R+R_0} \\ \left(R_0-R\right)\left(2 R+R_0\right)=2 R R_0 \\ 2 R R_0-2 R^2+R_0^2-R R_0=2 R R_0 \\ R_0^2-R R_0-2 R^2=0 \\ R_0=\frac{R \pm \sqrt{R^2+8 R^2}}{2}=\frac{R \pm 3 R}{2}=2 R \text { or }-R \end{gathered}$
Equivalent resistance between $A$ and $B=2 R(\because$ equivalent resistance cannot be negative).

Hint

$\begin{aligned} &\text { (a) Resistance, } R=\rho \frac{l}{A} \text { and resistivity, } \rho=\frac{m}{n e^2 \tau}\\ &\therefore \quad R=\frac{m l}{n e^2 \tau A} \end{aligned}$

Hint

(a) The resistance $R$ of a wire is given by $R=\rho \frac{l}{A}$, where $\rho$ is the resistivity of the material, $l$ is the length, and $A$ is the cross-sectional area.
The area of a circle is given by $A=\pi r^2=\pi\left(\frac{d}{2}\right)^2=\frac{\pi d^2}{4}$, where $r$ is the radius and $d$ is the diameter.
Since, wires are made of same material, so resistivity $\rho$ of all wires will be same.
Therefore, $R \propto \frac{l}{A}$
$\begin{aligned} \Rightarrow \quad & R \propto \frac{l}{\pi r^2} \\ & R \propto \frac{l}{\left(\frac{d}{2}\right)^2} \end{aligned}$
For length $=50 \mathrm{~cm}$ and diameter $=0.5 \mathrm{~mm}$
Resistance will be maximum.

Hint

(c) Given, circuit can be redrawn as follows

Here, $I_1=\frac{I}{2}=\frac{V}{2 R_{\text {eq }}}=\frac{2}{2 \times\left(\frac{15}{2}\right)}=\frac{2}{15} \mathrm{~A}$
So, potential across each resistance,
$V^{\prime}=I_1 R=\frac{2}{15} \times 5=\frac{2}{3} \mathrm{~V}$
$\therefore[latex] Potential difference across, [latex] \mathrm{AB}=\frac{2}{3}+\frac{2}{3}=\frac{4}{3} \mathrm{~V}$

Hint

(b)

$\text { From current division rule, } i_1=i\left(\frac{R_2}{R_1+R_2}\right)$
$=1.2\left(\frac{4}{6+4}\right)=1.2\left(\frac{4}{10}\right)=0.48 \mathrm{~A}$

Hint

(b) Two resistances are short circuited.
So, only third resistance will be considered and hence,
$i=V / R=10 / 3 \mathrm{~A}$

Hint

(b) As $B$ is connected to the earth, so potential at $B$ is $V_B=0$.
Now, current in the given circuit,
$i=\frac{V}{R_{\text {net }}}=\frac{50}{5+7+10+3}=2 \mathrm{~A}$
Potential difference between $A$ and $B$ is
$\begin{aligned} V_A-V_B & =2 \times 12 \\ V_A-0 & =24 \\ V_A & =24 \mathrm{~V} \end{aligned}$

Hint

$\begin{aligned} &\text { (d) We have, } V=E \text { – ir }\\ &\begin{aligned} & =1.5-2 \times 0.15 \\ V & =1.20 \mathrm{~V} \end{aligned} \end{aligned}$

Hint

$\begin{aligned} &\text { (d) Potential difference between } A \text { and } B \text { is given by }\\ &\begin{aligned} V_A-V_B & =\frac{E_1 r_2+E_2 r_1}{r_1+r_2} \\ 4 & =\frac{5 X+2 \times 10}{X+10} \\ X & =20 \Omega \end{aligned} \end{aligned}$

Hint

(a)

$E_1=4 \mathrm{~V} \text { and } E_2=8 \mathrm{~V}$
As, $E_2>E_1$, so current flows from $Q$ to $P$.
$\therefore \quad i=\frac{8-4}{12}=\frac{1}{3} \mathrm{~A}$
$\therefore$ Potential difference across $P Q=\frac{1}{3} \times 9=3 \mathrm{~V}$

Hint

(d) The potentials of different points are as shown in figure below

$\begin{aligned} &\text { Current through } 5 \Omega \text { resistance }\\ &\begin{aligned} & =\frac{\text { Potential difference }}{\text { Resistance }} \\ & =\frac{(4-2)}{5}=0.4 \mathrm{~A} \end{aligned} \end{aligned}$