Current Electricity

If $R_{\mathrm{eq}}$ be the equivalent resistance of the circuit, then

$ \begin{aligned} & \begin{aligned} \frac{1}{R_{\mathrm{eq}}} & =\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\ldots \text { nine times } \\ & =\frac{9}{1}=1 \end{aligned} \\ & \Rightarrow R_{\mathrm{eq}}=1 \Omega \end{aligned} $

$\therefore$ Current drawn from the 9 volt battery, i.e. main current

$ I=\frac{V}{R_{\mathrm{eq}}}=\frac{9}{1}=9 \mathrm{~A} $

From the circuit diagram, it is clear that current through each resistor will be 1 A and only 5 resistors on the right side of ammeter contributes for passing current through the ammeter.

Thus, reading of ammeter will be 5A.

$ \text{Cross-sectional area, } A = 4 \times 10^{-7} \text{ m}^2 $

$ \text{Electron density, } n = 8 \times 10^{28} \text{ m}^{-3} $

$ \text{Current, } I = 6.4 \text{ A} $

Electron charge,

$ e = 1.6 \times 10^{-19} \text{ C} $

Formula for drift velocity:

$ I = n e A v_d $

$ \Rightarrow v_d = \frac{I}{n e A} $

Substituting values:

$ v_d = \frac{6.4}{(8 \times 10^{28})(1.6 \times 10^{-19})(4 \times 10^{-7})} $

Compute step-by-step:

1. $ n e A = 8 \times 10^{28} \times 1.6 \times 10^{-19} \times 4 \times 10^{-7} $

$ = 8 \times 1.6 \times 4 \times 10^{28 - 19 - 7} = 51.2 \times 10^{2} = 5.12 \times 10^{3} $

2. Therefore,

$ v_d = \frac{6.4}{5.12 \times 10^{3}} = 1.25 \times 10^{-3} \text{ m/s} $

So, in units of $10^{-3} \text{ m/s}$:

$ v_d = 1.25 $

✅ Final Answer:

$ \boxed{v_d = 1.25 \times 10^{-3} \text{ m/s}} $

Correct Option: D) 1.25

$ \begin{aligned} & \text { } I_1=\frac{V}{R}=\frac{12}{40}=\frac{3}{10} \mathrm{~A} \\ & I_2=\frac{V}{R_{\mathrm{eq}}}=\frac{12}{\left(\frac{40 \times 40}{40+40}\right)}=\frac{12}{20}=\frac{6}{10} \\ & \therefore \quad \frac{I_1}{I_2}=\frac{\frac{3}{10}}{\frac{6}{10}}=\frac{1}{2} \\ & \therefore I_1: I_2=1: 2 \end{aligned} $

Initially,

$ \begin{aligned} & \frac{R}{S}=\frac{1}{100-1} \\ & \Rightarrow \frac{R}{S}=\frac{20}{100-20}=\frac{20}{80}=\frac{1}{4} \\ & \therefore S=4 R \end{aligned} $

After shunting

$ \begin{array}{ll} & \frac{1}{S^{\prime}}=\frac{1}{S}+\frac{1}{60} \Rightarrow S^{\prime}=\frac{60 S}{S+60} \\ \therefore & \frac{R}{S^{\prime}}=\frac{l^{\prime}}{100-l^{\prime}}=\frac{25}{100-25}=\frac{1}{3} \\ \Rightarrow & \quad S^{\prime}=3 R \\ \Rightarrow & \frac{60 S}{4 R+60}=3 R \Rightarrow \frac{60(4 R)}{4 R+60}=3 R \\ \Rightarrow & 4 R+60=80 \\ & \quad R=5 \Omega \\ \therefore \quad & \quad S=4 R=4 \times 5=20 \Omega \end{array} $

$ \text { By applying the KVL in loop } $

$ \begin{array}{ll} 12-I(0.6)-I(\mathrm{l})-6-4 I-0.4 I=0 \\ \,\,\,\,\,\,\,\,\,\,\,\, 6=6 I \\ \Rightarrow \quad I=1 \mathrm{~A} \end{array} $

So, reading of ammeter is 1 A and reading of voltmeter

$ \begin{aligned} & =6+I \times 1 \\ & =6+1 \times 1=7 \mathrm{~V} \end{aligned} $

Let emf of both batteries is $E$. When both are connected in series current through the circuit

$ I=\frac{2 E}{2+R} $

and power dissipated in $R$

$ P_1=I^2 R=\frac{4 E^2 R}{(2+R)^2} $

When they are connected in parallel their equivalent emf will $E$ and equivalent resistance $\frac{1}{2}$

$ \begin{aligned}So,\,\, & I=\frac{E}{\left(\frac{1}{2}+R\right)} \\ & P_2=\frac{E^2}{\left(\frac{1}{2}+R\right)^2} \cdot R \end{aligned} $

$ \begin{aligned}Now,\,\, & P_1=2.25 P_2 \\ & \frac{4 E^2 R}{(2+R)^2}=\left(\frac{E^2 R}{\left(\frac{1}{2}+R\right)^2}\right)\left(\frac{9}{4}\right) \\ & \frac{4}{(2+R)^2}=\left(\frac{1}{\left(\frac{1}{2}+R\right)^2}\right)\left(\frac{9}{4}\right) \\ & 16\left(\frac{1}{2}+R\right)^2=9(2+R)^2 \\ & =4\left(\frac{1}{2}+R\right)=3(2+R) \\ & \begin{array}{l} 2+4 R=6+3 R \\ R=4 \Omega \end{array} \end{aligned} $

We know the electron mobility ($\mu$) and want to find the drift velocity ($v_d$).

The formula is: $ \mu = \frac{v_d}{E} $

Here, $E$ stands for the electric field. For a wire, $E = \frac{V}{l}$, where $V$ is the voltage and $l$ is the length of the wire.

So, the drift velocity can be found as:

$ v_d = E \mu = \frac{V}{l} \cdot \mu $

Plug in the values: $V = 30$ V, $l = 20 \text{ cm} = 20 \times 10^{-2}$ m, and $\mu = 2 \times 10^{-6} \text{ m}^2 \text{V}^{-1} \text{s}^{-1}$.

$ v_d = \frac{30 \times 2 \times 10^{-6}}{20 \times 10^{-2}} $

Do the calculation step by step: $30 \times 2 = 60$, $20 \times 10^{-2} = 0.2$.

$ v_d = \frac{60 \times 10^{-6}}{0.2} = 300 \times 10^{-6} = 3 \times 10^{-4}~\mathrm{ms}^{-1} $

Here, $I_g=0.5 \times 10^{-3} \mathrm{~A}$

$ V_{A B}=10 \mathrm{~V}, \mathrm{G}=15 \Omega $

As, $\quad V_{A B}=I_g\left(G+R_s\right)$

$ \begin{aligned} \Rightarrow \quad 10 & =0.5 \times 10^{-3}\left(1.5+R_s\right) \\ \Rightarrow \quad R_s & =\frac{10}{0.5 \times 10^{-3}}-15 \\ & =20000-15=19985 \Omega \end{aligned} $

$ \begin{aligned} \text { } V_{A C} & =I_{A C} \times 20 \\ & =(5-3) \times 20=40 \mathrm{Volt} \\ V_{D E} & =I_{A D} \times 25=(5-2) \times 25=75 \mathrm{~V} \\ \therefore \frac{V_{A C}}{V_{A D}} & =\frac{40}{75}=\frac{8}{15} \\ \therefore V_{A C} & : V_{A D}=8: 15 \end{aligned} $

Terminal voltage of the battery during charging,

$ V=E+I r \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Here, $E=10$ volt

Net emf in the circuit

$ \begin{aligned} E^{\prime} & =V_{\text {supply }}-E=160-10 \\ & =150 \mathrm{~V} \\ \therefore \quad I & =\frac{E^{\prime}}{R+r}=\frac{150}{24+1}=\frac{150}{25}=6 \mathrm{~A} \end{aligned} $

$\therefore$ From eqn (i)

$ \begin{aligned} V & =E+I r \\ & =10+6 \times 1 \\ & =16 \mathrm{~V} \end{aligned} $

$ \begin{aligned} &\begin{aligned} R_1 & =\frac{3 R}{4} \text { and } R_2=\frac{R}{4} \\ \therefore \quad \frac{1}{R_{e q}} & =\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{\frac{3 R}{4}}+\frac{1}{\frac{R}{4}} \\ \frac{1}{R_{e q}} & =\frac{4}{3 R}+\frac{4}{R}=\frac{16}{3 R} \Rightarrow R_{e q}=\frac{3 R}{16} \end{aligned}\\ &\text { Heat generated in the wire per second }\\ &\begin{aligned} P & =\frac{H}{t}=\frac{E^2}{R_{e q} \cdot t}=\frac{E^2}{\left(\frac{3 R}{16}\right) \times 1} \quad[t=1 s] \\ & =\frac{16 E^2}{3 R} \end{aligned} \end{aligned} $

Let $R_1$ be the initial resistance of the wire in the left gap.

$R_2$ be the resistance in the right gap $l_1$ be the initial balancing length ( 40 cm )

Let $l_2$ be the remaining length $(100-40=60 \mathrm{~cm})$

The initial ratio is $\frac{R_1}{R_2}=\frac{40}{60}=\frac{2}{3}$

When the wire is stretched to double its length, its resistance becomes $4 R_1$

Let $l_1$ '' be the new balancing length $l_2$ 'be the new remaining length $\left(100-l_1\right)$

The new ratio is $\frac{4 R_1}{R_2}=\frac{l_1{ }^{\prime}}{100-l_1{ }^{\prime}}$

$ \begin{aligned} & 4 \cdot\left(\frac{2}{3}\right)=\frac{l_1^{\prime}}{100-l_1^{\prime}} \\ & 8\left(100-l_1^{\prime}\right)=3 l_1^{\prime} \\ & 800-8 l_1^{\prime}=3 l_1^{\prime} \\ & 11 l_1^{\prime}=800 \\ & l_1^{\prime}=\frac{800}{11} \end{aligned} $

The new balancing point is approximately $\left(\frac{800}{11}\right) \mathrm{cm}$ from the left end.

Given:

• Length $ L = 30 \, \mathrm{m} $

• Area of cross-section $ A = 6 \times 10^{-7} \, \mathrm{m^2} $

• Resistivity $ \rho = 1.7 \times 10^{-8} \, \Omega\mathrm{m} $

Formula:

$ R = \rho \frac{L}{A} $

Substitution:

$ R = (1.7 \times 10^{-8}) \times \frac{30}{6 \times 10^{-7}} $

Simplify:

$ \frac{30}{6 \times 10^{-7}} = 5 \times 10^7 $

So,

$ R = 1.7 \times 10^{-8} \times 5 \times 10^7 $

$ R = (1.7 \times 5) \times 10^{-1} $

$ R = 8.5 \times 10^{-1} = 0.85 \, \Omega $

✅ Final Answer:

Option C: $ 0.85 \, \Omega $

Given Data:

The current $I = 80$ A, and the length of the conductor $L = 10$ m.

Expression for Current:

Current can be written as $I = n e A V_d$, where $n$ is the number of electrons per unit volume, $e$ is the charge of an electron, $A$ is the cross-sectional area, and $V_d$ is the drift velocity of electrons.

Total Number of Electrons:

Let $N$ be the total number of free electrons in the length $L$ of the conductor. $N = n A L$.

Total Momentum:

The total momentum of all free electrons is $P_{\text{total}} = N(m_e V_d)$, where $m_e$ is the mass of an electron.

Substitute $N$:

So, $P_{\text{total}} = (n A L)(m_e V_d)$.

Rewrite with Current Expression:

Since $I = n e A V_d$, we can replace $n A V_d$ with $\frac{I}{e}$.

Therefore, $P_{\text{total}} = (I/e) L m_e$.

Simplified Formula:

So, $P_{\text{total}} = I \dfrac{L m_e}{e}$.

Plug in the Values:

$P_{\text{total}} = 80 \times \dfrac{10 \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19}}$

First, multiply $80 \times 10 = 800$.

Now $800 \times 9.1 = 7280$.

So, numerator is $7280 \times 10^{-31}$ and denominator is $1.6 \times 10^{-19}$.

Divide $7280$ by $1.6$ to get $4550$.

Power of ten: $10^{-31} / 10^{-19} = 10^{-12}$

Final Answer:

$P_{\text{total}} = 4550 \times 10^{-12}\, \mathrm{N}\!-\!\mathrm{s} = 4.55 \times 10^{-9}\, \mathrm{N}\!-\!\mathrm{s} = 455 \times 10^{-11}\, \mathrm{N}\!-\!\mathrm{s}$

We are given:

$ Q = 3t^2 + t $

and we know that current $ I = \frac{dQ}{dt} $.

Differentiate $ Q $ with respect to $ t $:

$ I = \frac{dQ}{dt} = \frac{d}{dt}(3t^2 + t) = 6t + 1 $

Now, at $ t = 3\ \text{s} $:

$ I = 6(3) + 1 = 18 + 1 = 19\ \text{A} $

✅ Answer: Option C (19 A)

Given data:

$ n = 9.1 \times 10^{28}\ \mathrm{m}^{-3} $

$ \sigma = 6.4 \times 10^{7}\ \mathrm{S/m} $

$ E = 10\ \mathrm{N/C} $

$ e = 1.6 \times 10^{-19}\ \mathrm{C} $

$ m = 9.1 \times 10^{-31}\ \mathrm{kg} $

Step 1. Relation between conductivity and relaxation time (Drude model)

$ \sigma = \frac{n e^2 \tau}{m} $

$ \Rightarrow \tau = \frac{m\sigma}{n e^2} $

Step 2. Substitute values

$ \tau = \frac{(9.1\times10^{-31})(6.4\times10^{7})}{(9.1\times10^{28})(1.6\times10^{-19})^2} $

Compute step by step:

Denominator:

$ n e^2 = 9.1\times10^{28} \times (1.6\times10^{-19})^2 $

$ = 9.1\times10^{28} \times 2.56\times10^{-38} $

$ = 2.33\times10^{-9} $

Numerator:

$ m\sigma = 9.1\times10^{-31} \times 6.4\times10^{7} = 5.82\times10^{-23} $

Now divide:

$ \tau = \frac{5.82\times10^{-23}}{2.33\times10^{-9}} = 2.5\times10^{-14}\ \text{s} $

✅ Answer:

$ \tau = 2.5 \times 10^{-14}\ \mathrm{s} $

Step 3 (check options)

Option D: $2.5\times10^{-14}\ \mathrm{s}$

✔️ Correct option: D

$ \begin{aligned} &\text { Resistance of each side of triangle, }\\ &R_1=R_2=R_3=\frac{18}{3}=6 \Omega \end{aligned} $

$ \begin{aligned} \therefore R_{A B} & =\left(R_1+R_2\right) \|\left(R_3\right) \\ & =(6+6)\|6=12\| 6 \\ & =\frac{12 \times 6}{12+6}=\frac{72}{18}=4 \Omega \end{aligned} $

$ \text { } \begin{aligned} P & =\frac{V^2}{R}=\frac{120^2}{100} \\ & =\frac{120 \times 120}{100}=144 \mathrm{~W} \end{aligned} $

$R_1=100 \Omega l_1=l$

$ \begin{aligned} R_2 & =?, l_2=l_1+20 \% \text { of } l_1 \\ & =l+0.2 l=1.2 l \\ \therefore \quad R^{\prime} & =n^2 R=(1.2)^2 R \\ & =1.44 R=1.44 \times 100=144 \Omega \end{aligned} $

The wire of resistance $144 \Omega$ is bent into a rectangle with sides in the ratio $3: 2$. Let the side be $3 x$ and $2 x$.

Total length, $l=2(3 x+2 x)=10 x$

$ \begin{aligned} \therefore \quad & R_{\text {length }}=\frac{3 x}{10 x} \times 144=43.2 \Omega \\ & R_{\text {width }}=\frac{2 x}{10 x} \times 144=28.8 \Omega \end{aligned} $

The effective resistance across the diagonal is the equivalent resistance of two parallel branches. On branch consists of the other two sides of the same resistance. Each branch has a resistance of

$ \begin{aligned} & R_{\text {length }}+R_{\text {width }}=43.2+28.8=72 \Omega \\ & \therefore \quad \frac{1}{R_{\mathrm{eq}}}=\frac{1}{72}+\frac{1}{72}=\frac{1}{36} \\ & \Rightarrow \quad R_{\mathrm{eq}}=36 \Omega \end{aligned} $

Initially $E_1+E_2=K 160\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

On reversing polarity of $E_1$

and $E_1+E_2=K 160 \times 0.25\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Dividing, we get

$ \begin{aligned} & \frac{1.2+E_2}{-1.2+E_2}=4 \\ & \Rightarrow \quad 1.2+E_2=-4.8+4 E_2 \\ & 3 E_2=6 \\ & E_2=\frac{6}{3}=2 \mathrm{~V} \end{aligned} $

The percentage error in measuring the electromotive force (emf) of a cell with an internal resistance of $2 \, \Omega$ when using a voltmeter with a resistance of $998 \, \Omega$ is calculated as follows:

$ \text{Percentage error} = \left(1 - \frac{R_V}{R_V + r}\right) \times 100 $

Where:

$ R_V $ is the resistance of the voltmeter ($998 \, \Omega$)

$ r $ is the internal resistance of the cell ($2 \, \Omega$)

Substitute the given values:

$ \text{Percentage error} = \left(1 - \frac{998}{998 + 2}\right) \times 100\% $

$ = \left(1 - 0.998\right) \times 100\% $

$ = 0.002 \times 100\% = 0.2\% $

Therefore, the error in the emf measured is $0.2\%$ of the actual emf of the cell.

$ \text { Case-1 } $

$ \frac{R}{R^{\prime}}=\frac{l_1}{100-l_1} \Rightarrow \frac{9}{R^{\prime}}=\frac{l_1}{100-l_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

$ \text { Case-2 } $

$ \begin{array}{ll} \therefore & \frac{R^{\prime}}{9}=\frac{l_1+100}{90-l_1} \\ \Rightarrow & \frac{9}{R^{\prime}}=\frac{90-l_1}{l_1+10} \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,...(ii)\end{array} $

From Eqs. (i) and (ii), we get

$ \begin{aligned} & & \frac{l_1}{100-l_1} & =\frac{90-l_1}{l_1+10} \\ & \Rightarrow & l_1^2+10 l_1 & =9000-90 l_1-100 l_1+l_1^2 \\ & \Rightarrow & 200 l_1 & =9000 \\ \Rightarrow & & l_1 & =45 \mathrm{~cm} \end{aligned} $

$\therefore$ From Eq. (i), $\frac{9}{R^{\prime}}=\frac{45}{100-45}$

$ \Rightarrow \quad R^{\prime}=11 \Omega $

Let $S_I$ be the current sensitivity (divisions per ampere) and $S_V$ the voltage sensitivity (divisions per volt). Since for a galvanometer of resistance $R$,

$ S_V \;=\;\frac{S_I}{R}\,, $

we get:

• For $G_1$:

$S_I = 10^8\;\mathrm{div/A}$, $R=100\;\Omega$

$ S_V^{(1)} = \frac{10^8}{100} = 10^6\;\mathrm{div/V}. $

• For $G_2$:

$S_I = 0.5\times10^5 = 5\times10^4\;\mathrm{div/A}$, $R=50\;\Omega$

$ S_V^{(2)} = \frac{5\times10^4}{50} = 10^3\;\mathrm{div/V}. $

Since $10^6 > 10^3$, the voltage sensitivity is greater for $G_1$.

Answer: more in $G_1$.

$ \begin{aligned} &\begin{aligned} E_{\text {eff }} & =E_{\text {source }}-E_{\text {battery }} \\ & =120-8=112 \mathrm{~V} \\ R_{\text {eq }} & =R+r=15.5+0.5=16 \Omega \\ I & =\frac{E_{\text {eff }}}{R_{\text {eq }}}=\frac{112}{16}=7 \mathrm{~A} \end{aligned}\\ &\text { Terminal voltage of battery }=V_A-V_B\\ &\begin{aligned} & V_A-E_{\text {battery }}-I r=V_B \\ & V_A-V_B=E_{\text {battery }}+I r \\ &=8+7 \times 0.5=8+3.5 \\ & V_A-V_B=11.5 \mathrm{~V} \end{aligned} \end{aligned} $

To find the final resistance of the wire, we start with the basic formula for resistance:

$ R = \rho \frac{L}{A} $

Here, $ R $ is the resistance, $ \rho $ is the resistivity, $ L $ is the length, and $ A $ is the cross-sectional area.

Given that the resistance of the wire is $ 8 \, \Omega $ and it experiences a longitudinal strain of $ 400\% $, the length of the wire increases as follows:

$ L' = L + \frac{L \times 400}{100} = L + 4L = 5L $

This implies the new length $ L' $ is $ 5 $ times the original length $ L $.

Since the volume of the wire remains constant, we have:

$ V = V' \Rightarrow AL = A'L' $

$ AL = A' \times 5L $

$ A' = \frac{A}{5} $

Now, to find the new resistance $ R' $, we use the resistance formula with the new dimensions:

$ R' = \rho \frac{L'}{A'} = \rho \frac{5L}{\frac{A}{5}} = 25 \rho \frac{L}{A} $

From the original equation, we know:

$ R = \rho \frac{L}{A} $

Thus, substituting, we get:

$ R' = 25 \times R = 25 \times 8 = 200 \, \Omega $

Therefore, the final resistance of the wire is $ 200 \, \Omega $.

Given, $R=30 \Omega$

$ X_L=25 \Omega \Rightarrow X_C=25 \Omega $

Source voltage, $V_s=240 \mathrm{~V}$

Since, $X_L=X_C=25 \Omega$

So, $V_L$ and $V_C$ are equal in magnitude but opposite in phase, i.e

Voltmeter reading $=0 \mathrm{~V}$

Where, $V_L$ and $V_C$ are voltage across inductor and capacitor respectively.

For ammeter reading,

$ \begin{aligned} & I=\frac{V}{Z}=\frac{V}{\sqrt{R^2+\left(X_L-X_C\right)^2}} \\ & I=\frac{240}{\sqrt{(30)^2+(25-25)^2}} \\ & I=\frac{240}{30}=8 \mathrm{~A} \end{aligned} $

Given the following measurements:

Voltage, $ V $: 30 V

Error in voltage, $ \Delta V $: 0.3 V

Current, $ I $: 5 A

Error in current, $ \Delta I $: ±0.1 A

The resistance, $ R $, is calculated using Ohm's Law:

$ R = \frac{V}{I} $

To find the percentage error in the resistance, we use the formula for the propagation of uncertainty for division:

$ \frac{\Delta R}{R} \times 100 = \frac{\Delta V}{V} \times 100 + \frac{\Delta I}{I} \times 100 $

Plugging in the given values:

$ \frac{\Delta V}{V} \times 100 = \frac{0.3}{30} \times 100 $

$ \frac{\Delta I}{I} \times 100 = \frac{0.1}{5} \times 100 $

Calculating these:

$ \frac{0.3}{30} \times 100 = 1\% $

$ \frac{0.1}{5} \times 100 = 2\% $

Adding these percentage errors together gives:

$ \frac{\Delta R}{R} \times 100 = 1\% + 2\% = 3\% $

Therefore, the percentage error in the resistance is ±3%.

As, the batteries are connected inverse polarities,

The net potential applied to the circuit

$ =12 \mathrm{~V}-4 \mathrm{~V}=8 \mathrm{~V} $

The net resistance in the circuit,

$ R=r_1+r_2+r_3=1+1+8=10 \Omega $

Thus, According to Ohm's law,

$ V=\frac{I}{R} \Rightarrow I=\frac{V}{R}=\frac{8}{10}=0.8 \mathrm{~A} $

and potential difference across $P$ and $Q$

$ V=I \times R=0.8 \times 8=6.4 \mathrm{~V} $

Series Arrangement

For cells in series, the total electromotive force (EMF) is doubled, and the total internal resistance is also doubled. The current $I_1$ in this setup is given by:

$ I_1 = \frac{2E}{2 + 2r} $

where $E$ is the electromotive force of each cell and $r$ is the internal resistance of each cell.

Parallel Arrangement

For cells in parallel, the EMF remains the same, but the internal resistance is halved. The current $I_2$ in this setup is given by:

$ I_2 = \frac{E}{2 + \frac{r}{2}} = \frac{2E}{4 + r} $

Equating the Currents

Since the problem states $I_1 = I_2$, we set the equations equal:

$ \frac{2E}{4 + r} - \frac{2E}{2 + 2r} = 0 $

Solving this equation:

$ \frac{2E(2 + 2r) - 2E(4 + r)}{(4 + r)(2 + 2r)} = 0 $

Simplifying further:

$ 2E(2 + 2r) - 2E(4 + r) = 0 $

$ 4E + 4Er - 8E - 2Er = 0 $

$ 2Er - 4E = 0 $

Factoring out common terms:

$ Er - 2E = 0 $

Divide by $E$ (assuming $E \neq 0$):

$ r - 2 = 0 $

Hence, solving this gives:

$ r = 2 $

Thus, the internal resistance of each cell is $2 \, \Omega$.

Given that the charge $ q $ passing through a $10 \, \Omega$ resistor as a function of time $ t $ is expressed by the equation $ q = 3t^2 - 2t + 6 $, we can determine the potential difference across the resistor.

To find the current $ I $ as a function of time, we differentiate the charge function with respect to time:

$ I = \frac{dq}{dt} = \frac{d}{dt}(3t^2 - 2t + 6) $

Solving for the derivative:

$ I = 6t - 2 $

Next, we calculate the current at $ t = 5 \, \text{s} $:

$ I = 6(5) - 2 = 30 - 2 = 28 \, \text{A} $

The voltage $ V $ across the resistor, using Ohm's Law, is given by:

$ V = IR $

With the resistance $ R = 10 \, \Omega $, the potential difference is:

$ V = 28 \times 10 = 280 \, \text{V} $

Thus, the potential difference across the resistor at $ t = 5 \, \text{s} $ is $ 280 \, \text{V} $.

Given the following values:

$E_1 = 1.2 \, \text{V}$ with an internal resistance $r_1 = 2 \, \Omega$

$E_2 = 1.5 \, \text{V}$ with an internal resistance $r_2 = 1 \, \Omega$

To find the equivalent emf of the cells connected in parallel with like poles together, we first calculate the equivalent resistance as follows:

$ r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 \times 1}{2 + 1} = \frac{2}{3} \, \Omega $

Next, we calculate the equivalent emf using the formula:

$ \text{Emf (equivalent)} = \left[\frac{E_1}{r_1} + \frac{E_2}{r_2}\right] \times r_{\text{eq}} $

Substituting the given values, we have:

$ \begin{aligned} \text{Emf (equivalent)} & = \left[\frac{1.2}{2} + \frac{1.5}{1}\right] \times \frac{2}{3} \\ & = \left[0.6 + 1.5\right] \times \frac{2}{3} \\ & = 2.1 \times \frac{2}{3} \\ & = 1.4 \, \text{V} \end{aligned} $

Thus, the equivalent emf of the combination of the two cells is $1.4 \, \text{V}$.

Given, Voltage across battery $=V$

Each resistor have resistance $=2 R$

$ i_{F C}=\text { ? } $

According to figure, it is a balanced Wheatstone bridge.

$ \text { } \begin{aligned}So, R_1 & =R_{F C}+R_{C E}=2 R+2 R=4 R \\ R_2 & =R_{F D}+R_{D E}=2 R+2 R=4 R \\ R_{\text {eq }} & =\frac{R_1 \times R_2}{R_1+R_2}=\frac{4 R \times 4 R}{4 R+4 R}=\frac{16 R R}{8 R}=2 R \end{aligned} $

We know that, voltage will be same in parallel combination. So, $R_1$ will have voltage $V$.

Current will be same in series combination. So $R_{F C}$ and $R_{C E}$ will have current same as $R_1$

So, $i_{F C}=\frac{V}{R_1} \Rightarrow i_{F C}=\frac{V}{4 R}$

Given the specifications for the lamp:

Voltage, $ V = 240 \, \text{V} $

Power, $ P = 60 \, \text{W} $

The problem states that when the lamp is in use, the resistance of the filament (hot resistance, $ R_{\text{Hot}} $) is 20 times that of when it is not in use (cold resistance, $ R_{\text{Cold}} $).

Calculating Hot Resistance

Using the power formula from Joule's law of heating:

$ P = \frac{V^2}{R} $

Solve for $ R_{\text{Hot}} $:

$ R_{\text{Hot}} = \frac{V^2}{P} = \frac{240 \times 240}{60} = 960 \, \Omega $

Relating Hot and Cold Resistance

According to the problem, the hot resistance is 20 times the cold resistance:

$ R_{\text{Hot}} = 20 \times R_{\text{Cold}} $

Solving for $ R_{\text{Cold}} $:

$ R_{\text{Cold}} = \frac{R_{\text{Hot}}}{20} = \frac{960}{20} = 48 \, \Omega $

Thus, the resistance of the lamp when not in use is $ 48 \, \Omega $.

Since, wire is bent into form of a square so each side will have resistance,

$ r=\frac{R}{4} $

Here, $A D, D C$ and $C B$ are in series and $A B$ will be parallel,

$ \therefore R_{\text {eq }}=\frac{3 R}{4} \| \frac{R}{4}=\frac{\frac{3 R}{4} \times \frac{R}{4}}{\frac{3 R}{4}+\frac{R}{4}}=\frac{\frac{3 R^2}{16}}{R}=\frac{3 R}{16} $

Current in the circuit, $I=\frac{V}{R_{\text {eq }}}=\frac{12}{\frac{3 R}{16}}=\frac{4}{4}$

So, potential difference between end $A$ and $B$

$ =\frac{V}{R / 4}=\frac{48}{R} $

$\therefore$ Current in the branch

$ A D C=\frac{64}{R}-\frac{48}{R}=\frac{16}{R} $

$\therefore$ Potential dropp across any diagnal $A C$ or $B D$ )

$ =\frac{16}{R} \times \frac{R}{2}=8 \mathrm{~V} $

Applying KVL in the loop considering current in anti-clockwise direction.

$ \begin{aligned} -36+2 i+12+3 i & =0 \\ 5 i & =24 \\ i & =\frac{24}{5} \mathrm{~A} \end{aligned} $

$ \begin{aligned} & \begin{aligned} \text { Now; } V_B-V_A & =12+3 \times i \\ 24-V_A & =12+3 \times \frac{24}{5} \\ \Rightarrow \quad V_A & =24-12-\frac{72}{5}=12-\frac{72}{5} \\ & =-\frac{12}{5}=-2.4 \mathrm{~V} \end{aligned} \end{aligned} $

Given the block dimensions of $1 \, \mathrm{cm}$, $2 \, \mathrm{cm}$, and $3 \, \mathrm{cm}$, we need to determine the ratio of the maximum to minimum resistance between any pair of opposite faces of the block.

The resistance $R$ of a conductor is given by the formula:

$ R = \rho \frac{l}{A} $

where:

$\rho$ is the resistivity of the material,

$A$ is the area of cross-section,

$l$ is the length.

Calculating Maximum Resistance:

The resistance is maximum when:

$l = 3 \, \mathrm{cm} = 3 \times 10^{-2} \, \mathrm{m}$

$A = 1 \times 10^{-2} \times 2 \times 10^{-2} = 2 \times 10^{-4} \, \mathrm{m}^2$

Thus, the maximum resistance $R_{\max}$ is:

$ R_{\max} = \rho \times \frac{3 \times 10^{-2}}{2 \times 10^{-4}} = \frac{3}{2} \times 10^2 \rho $

Calculating Minimum Resistance:

The resistance is minimum when:

$l = 1 \, \mathrm{cm} = 10^{-2} \, \mathrm{m}$

$A = 2 \times 10^{-2} \times 3 \times 10^{-2} = 6 \times 10^{-4} \, \mathrm{m}^2$

Thus, the minimum resistance $R_{\min}$ is:

$ R_{\min} = \rho \times \frac{10^{-2}}{6 \times 10^{-4}} = \frac{10^2 \rho}{6} $

Calculating the Ratio:

The ratio of the maximum resistance to the minimum resistance is:

$ \frac{R_{\max}}{R_{\min}} = \frac{\frac{3 \times 10^2 \rho}{2}}{\frac{10^2 \rho}{6}} = \frac{3}{2} \times 6 = \frac{9}{1} $

Therefore, the ratio is $9:1$.

Using Kirchhoff's current law at point $P$,

$ I_1+I_2=6 $

Using Krichhoff's voltage law to the closed circuit $P Q R P$,

$ \begin{array}{rr} \Rightarrow & -2 I_1-2 I_1+2 I_2=0 \\ \Rightarrow & -4 I_1+2 I_2=0 \\ \Rightarrow & 2 I_1-I_2=0 \end{array} $

On adding Eqs. (i) and (ii), we get

$ \begin{array}{cc} \Rightarrow & I_1+I_2+2 I_1-I_2=6+0 \\ \Rightarrow & 3 I_1=6 \\ \Rightarrow & I_1=2 \mathrm{~A} \end{array} $

Substitute the value of $I_1$ in Eq. (i)

$ \Rightarrow \quad I_2=6-I_1=6-2=4 \mathrm{~A} $

Given, Galvanometer resistance, $G=99 \Omega$

Let $I$ be the current passes through main circuit.

$ \therefore \quad I_g=\left(\frac{10}{100}\right) I=0.1 I $

Using the formula for shunt resistance,

$ S=\frac{I_g G}{\left(I-I_g\right)} $

$ \begin{aligned} & \Rightarrow \quad S=\frac{0.1 I \times 99}{(I-0.1 I)}=\frac{0.1 \times 99}{0.9} \\ & \therefore \quad S=11 \Omega \end{aligned} $

In a steady current flowing through a conductor, the quantity that remains constant is the electric current. Here's why:

For any section of the conductor, the electric current $I$ is defined as the net charge per unit time passing through that section. In a steady state (no accumulation of charge), charge conservation requires that $I$ is constant along the length of the conductor.

The current density $J$ is given by

$J = \frac{I}{A},$

where $A$ is the local cross-sectional area. If the conductor has a non-uniform cross-section, the area $A$ changes, causing $J$ to vary.

Similarly, the drift velocity $v_d$ of the charge carriers is related to the current by

$I = n e v_d A,$

where $n$ is the carrier density and $e$ is the elementary charge. Again, if $A$ varies, $v_d$ must adjust accordingly to keep $I$ constant.

Thus, it is only the electric current $I$ that remains unchanged along a conductor with non-uniform cross-section.

The correct answer is:

Option D

electric current

The circuit diagram given below represents balanced Wheatstone bridge.

$\therefore \quad R_{A C}=\frac{2 R \times 2 R}{2 R+2 R}=\frac{4 R^2}{4 R}=R$

Since, potential barrier in Si diode is less $(0.7 \mathrm{~V})$ than green, and both are forward biased. Therefore, for conduction, minimum potential difference across the junction diode $=0.7 \mathrm{~V}$.

$ \therefore \quad V_0=12-0.7=11.3 \mathrm{~V} $

To find the resistivity of the potentiometer wire, we are provided with the following information:

Area of cross-section (A): $ 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 $

Current (i): $ 1 \, \text{A} $

Potential Gradient $\left(\frac{V}{l}\right)$: $ 7.5 \, \text{V/m} $

We need to calculate the resistivity ($\rho$) of the wire.

Calculation:

From Ohm's law, we know the following relationship:

$ V = i \cdot R $

The resistance $R$ for a wire can be expressed in terms of its resistivity:

$ R = \rho \frac{l}{A} $

Thus, substituting $R$ in the equation for $V$:

$ V = i \cdot \frac{\rho l}{A} $

Rearranging this equation to solve for resistivity ($\rho$):

$ \rho = \frac{V}{l} \cdot \frac{A}{i} $

Substitute the known values:

$ \rho = \left(7.5 \, \text{V/m}\right) \cdot \left(\frac{4 \times 10^{-4} \, \text{m}^2}{1 \, \text{A}}\right) $

Calculate $\rho$:

$ \rho = \frac{7.5 \times 4 \times 10^{-4}}{1} $

$ \rho = 30 \times 10^{-4} $

$ \rho = 3 \times 10^{-3} \, \Omega\text{-m} $

Therefore, the resistivity of the potentiometer wire is $3 \times 10^{-3} \, \Omega\text{-m}$.

Drift velocity is the average velocity acquired by charged particles, such as electrons, within a material when influenced by an electric field.

The magnitude of drift velocity per unit electric field is defined as mobility. This relationship is given by:

$ m = \frac{|v_d|}{E} \Rightarrow |v_d| = m \cdot E $

From this, we understand that:

$ v_d \propto E $

Thus, the drift velocity is directly proportional to the electric field $ E $. Therefore, the correct relationship is represented by $ v \propto E $.

The given circuit diagram is shown as :

Applying KVL in loop $A B D A$, we get

$\begin{aligned} 2 I_1+4-I_2 & =0 \\ I_2 & =2 I_1+4\quad \text{... (i)} \end{aligned}$

Applying KVL in loop $BCDB$, we get

$\begin{array}{rlrl} & 1\left(I_1+I_3\right)-4\left(I_2-I_3\right)-4 =0 \\ \Rightarrow & I_1+I_3-4 I_2+4 I_3-4 =0 \\ \Rightarrow & I_1+5 I_3-4\left(2 I_1+4\right)-4 =0 \\ \Rightarrow & -7 I_1+5 I_3 =20 \\ \Rightarrow & I_3= \frac{20+7 I_1}{5}\quad \text{... (ii)} \end{array}$

By applying KVL in loop $ADCA$, we get

$ \begin{aligned} & & I_2+4\left(I_2-I_3\right)-10 & =0 \\ \Rightarrow & & 5 I_2-4 I_3 & =10 \\ \Rightarrow & & 5\left(2 I_1+4\right)-4\left(\frac{20+7 I_1}{5}\right) & =10 \quad \text{[From Eqs. (i) and (ii)]} \end{aligned}$

$\Rightarrow \quad I_1=1.364 \mathrm{~A}$

From Eq. (i), $I_2=2 \times 1.364+4=6.278 \mathrm{~A}$

From Eq (ii), $I_3=\frac{20+7 \times 1.364}{5}=5.91 \mathrm{~A}$

Resistance of wire at $0^{\circ} \mathrm{C}, R_0=20 \Omega$

Temperature coefficient,

$\begin{aligned} & \alpha=5 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1} \\ & R_t=2 R_0=2 \times 20=40 \Omega \end{aligned}$

we know that,

$\begin{aligned} R_t & =R_0(1+\alpha t) \\ 40 & =20\left(1+5 \times 10^{-3} t\right) \Rightarrow 2=1+5 \times 10^{-3} t \end{aligned}$

$\Rightarrow \quad t=\frac{1}{5 \times 10^{-3}}=\frac{1000}{5}=200^{\circ} \mathrm{C}$

Given, length of wire, $l=2 \mathrm{~m}$

Time taken to drift electron, $t=40 \times 10^3 \mathrm{~s}$

Current, $I=1.6 \mathrm{~A}, A=4 \mathrm{~mm}^2=4 \times 10^{-6} \mathrm{~m}^2$

$\therefore$ Drift velocity, $v_d=\frac{l}{t}$

$=\frac{2}{40 \times 10^3}=5 \times 10^{-5} \mathrm{~m} / \mathrm{s}$

We know that,

$\begin{aligned} I & =n e A v_d \\ \Rightarrow \quad n & =\frac{I}{e A v_d} \\ & =\frac{1.6}{1.6 \times 10^{-19}} \times 4 \times 10^{-6} \times 5 \times 10^{-5} \\ & =0.5 \times 10^{29} \\ & =5 \times 10^{28} \text { electrons } / \mathrm{m}^3 \end{aligned}$

The circuit diagram is shown as,

Applying KVL in loop 3, we get

$I_3 R+R\left(I_3-I_2\right)=0$

$\begin{aligned} 2 I_3 R-R I_2 & =0 \\ I_3 & =\frac{I_2}{2} \quad \text{... (i)} \end{aligned}$

Applying KVL in loop 2, we get

$\begin{aligned} & \text { Applying } K V L \text { in } \\ & R I_2-3 \varepsilon+R\left(I_2-I_1\right)+2 \varepsilon+R I_2-R\left(I_2-I_3\right)=0 \\ & \Rightarrow 2 R I_2-R I_1+R I_3-\varepsilon=0 \\ & \Rightarrow 2 R I_2-R I_1+R \frac{I_2}{2}-\varepsilon=0 \\ & \Rightarrow \quad 5 R I_2-2 R I_1-2 \varepsilon=0 \quad \text{... (ii)} \end{aligned}$

By applying KVL in loop 1, we get

$\begin{gathered} R I_1+\varepsilon-2 \varepsilon+R\left(I_1-I_2\right)=0 \\ 2 R I_1-R I_2-\varepsilon=0 \\ I=\frac{R I_2+\varepsilon}{2 R} \quad \text{... (iii)} \end{gathered}$

From Eqs. (ii) and(iii), we get

$\begin{aligned} 5 R I_2-2 R \frac{\left(R I_2+\varepsilon\right)}{2 R}-2 \varepsilon & =0 \\ 5 R I_2-R I_2-\varepsilon-2 \varepsilon & =0 \\ \Rightarrow \quad I_2 & =\frac{3 \varepsilon}{4 R} \end{aligned}$

Given, current density,

$J=\beta\left(r+r_0\right)^2=\beta\left(r^2+2 \pi \pi_0+r_0^2\right)$

We know that, current density, $J=\frac{\operatorname{Current}(I)}{\operatorname{Area}(A)}$

$\Rightarrow$ Current, $I=J \cdot A$

or $d I=J \cdot d A$

$\begin{aligned} & \Rightarrow \quad \int d I=\int J \cdot d A \\ & \Rightarrow \quad I=\int \beta\left(r^2+2 \pi_0+r_0^2\right) \cdot d A \end{aligned}$

Where, $d A$ is area element in cartesian coordinate.

In polar coordinates $d A=r d r d \theta$

$\begin{aligned} & \therefore \quad I=2 \beta \int_0^{\pi / 6} d \theta \int_0^R\left(r^2+2 \pi_0+r_0^2\right) r d r \\ & (\because \text { for } \theta, 0 \text { to } \pi \backslash 6 \text { and } 5 \pi \backslash 6 \text { to } \pi \text { is symmetric) } \\ & =2 \beta[\theta]_0^{\pi / 6}\left[\int_0^R r^3 d r+2 r_0 \int_0^R r^2 d r+r_0^2 \int_0^R r d r\right] \\ & 2 \beta\left(\frac{\pi}{6}\right)\left[\left(\frac{r^4}{4}\right)_0^R+2 r_0\left(\frac{r_3}{3}\right)_0^R+r_0^2\left(\frac{r^2}{2}\right)_0^R\right] \\ & =\beta \frac{\pi}{3}\left[\frac{R^4}{4}+2 r_0 \frac{R^3}{3}+\frac{r_0^2 R^2}{2}\right] \\ & =\pi \beta\left[\frac{R^4}{12}+\frac{2 r_0 R^3}{9}+\frac{r_0^2 R^2}{6}\right] \end{aligned}$

Let $r$ be the internal resistance of the cell.

for case -1

$\begin{aligned} & I=1 \mathrm{~A}, R=25 \Omega \text {, then } \\ & I=\frac{E}{r+R} \\ & \Rightarrow \quad E=I(r+R)=1(r+25) \\ & \Rightarrow \quad E=r+25 \quad \text{... (i)}\\ \end{aligned}$

For the second condition,

$\begin{aligned} & I^{\prime}=0.5 \mathrm{~A}, R^{\prime}=10 \Omega \\ & I^{\prime}=\frac{E}{R^{\prime}+r} \\ & \Rightarrow \quad E=I^{\prime}\left(R^{\prime}+r\right) \\ & =0.5(10+r) \\ & \Rightarrow \quad E=5+0.5 r \quad \text{... (ii)}\\ \end{aligned}$

Now, from Eqs. (i) and (ii), we get

$\begin{aligned} \Rightarrow & & r+25 & =5+0.5 r \\ \Rightarrow & & r-0.5 r & =5-25 \Rightarrow 0.5 r=25 \\ \Rightarrow & & r & =\frac{25}{0.5}=5 \Omega \end{aligned}$

Since, conductors are material which can conduct electricity at room temperature due to presence of valence electrons in their outermost shell and on heating valence electrons gain energy, get excited and leave the conduction band. Now, due to absence / lack of free electrons in conduction band, on heating conductors do not conduct electricity.