Current Electricity

$\begin{aligned} & \mathrm{P}=\mathrm{i}^2 \mathrm{R} \\ & \mathrm{P}_{\text {int }}=\mathrm{I}_{\text {int }}^2 \mathrm{R} \\ & \mathrm{P}_{\text {final }}=\left(0.8 \mathrm{I}_{\text {int }}\right)^2 \mathrm{R} \end{aligned}$

% change in power $=$

$\frac{P_{\text {final }}-P_{\text {int }}}{P_{\text {int }}} \times 100=(0.64-1) \times 100=-36 \%$

$\begin{aligned} & \frac{25}{\mathrm{r} \ell_1}=\frac{\mathrm{X}}{\mathrm{r} \ell_2} \quad \text{.... (i)}\\ & \frac{25}{2 \mathrm{r} \ell_1^{\prime}}=\frac{\mathrm{X}}{2 \mathrm{r} \ell^{\prime}{ }_2} \quad \text{.... (ii)} \end{aligned}$

From (i) and (ii)

$\ell_2^{\prime}=\ell_2=40 \mathrm{~cm}$

Series :

$\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2 \\ & 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=\mathrm{R}\left(1+\alpha_1 \Delta \theta\right)+\mathrm{R}\left(1+\alpha_2 \Delta \theta\right) \\ & 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=2 \mathrm{R}+\left(\alpha_1+\alpha_2\right) \mathrm{R} \Delta \theta \\ & \alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2} \end{aligned}$

Parallel :

$\begin{aligned} & \frac{1}{R_{\text {eq }}}=\frac{1}{R_1}+\frac{1}{R_2} \\ & \frac{1}{\frac{R}{2}\left(1+\alpha_{\text {eq }} \Delta \theta\right)}=\frac{1}{R\left(1+\alpha_1 \Delta \theta\right)}+\frac{1}{\mathrm{R}\left(1+\alpha_2 \Delta \theta\right)} \end{aligned}$

$\begin{aligned} & \frac{2}{1+\alpha_{\mathrm{eq}} \Delta \theta}=\frac{1}{1+\alpha_1 \Delta \theta}+\frac{1}{1+\alpha_2 \Delta \theta} \\ & \frac{2}{1+\alpha_{\mathrm{eq}} \Delta \theta}=\frac{1+\alpha_2 \Delta \theta+1+\alpha_1 \Delta \theta}{\left(1+\alpha_1 \Delta \theta\right)\left(1+\alpha_2 \Delta \theta\right)} \\ & 2\left[\left(1+\alpha_1 \Delta \theta\right)\left(1+\alpha_2 \Delta \theta\right)\right] \\ & =\left[2+\left(\alpha_1+\alpha_2\right) \Delta \theta\right]\left[1+\alpha_{\mathrm{eq}} \Delta \theta\right] \\ & 2\left[1+\alpha_1 \Delta \theta+\alpha_2 \Delta \theta+\alpha_1 \alpha_2 \Delta \theta\right] \end{aligned}$

$\begin{aligned} & = \\ & 2+2 \alpha_{\mathrm{eq}} \Delta \theta+\left(\alpha_1+\alpha_2\right) \Delta \theta+\alpha_{\mathrm{eq}}\left(\alpha_1+\alpha_2\right) \Delta \theta^2 \end{aligned}$

Neglecting small terms

$\begin{aligned} & 2+2\left(\alpha_1+\alpha_2\right) \Delta \theta=2+2 \alpha_{\mathrm{eq}} \Delta \theta+\left(\alpha_1+\alpha_2\right) \Delta \theta \\ & \left(\alpha_1+\alpha_2\right) \Delta \theta=2 \alpha_{\mathrm{eq}} \Delta \theta \\ & \alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2} \end{aligned}$

$\begin{aligned} & \frac{\mathrm{v}^2}{\mathrm{R}}=\mathrm{W} \qquad \text{.... (i)}\\ & \frac{\mathrm{v}^2}{\frac{1}{2}\left(\frac{\mathrm{R}}{2}\right)}=\mathrm{W}^{\prime} \quad \text{.... (ii)} \end{aligned}$

From (i) & (ii), we get

$W^{\prime}=4 W$

$\begin{aligned} & R_{e q}=4000 \Omega \\ & i=\frac{4}{4000}=\frac{1}{1000} \mathrm{~A} \\ & V_0=i . R=\frac{1}{1000} \times 500=0.5 \mathrm{~V} \end{aligned}$

$\begin{aligned} & \mathrm{R}_{\mathrm{T}-27}=60 \Omega, R_T=\frac{220}{2.75}=80 \Omega \\ & \mathrm{R}=\mathrm{R}_0(1+\alpha \Delta \mathrm{T}) \\ & 80=60\left[1+2 \times 10^{-4}(\mathrm{~T}-27)\right] \\ & \mathrm{T} \approx 1694^{\circ} \mathrm{C} \end{aligned}$

$\mathrm{R}_{\mathrm{eq}}=2 \Omega+2 \Omega+1 \Omega=5 \Omega$

$\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{10}{5}=2 \mathrm{~A}$

Current in resistance

$\begin{aligned} R_3 & =2 \times\left(\frac{4}{4+4}\right) \\ & =2 \times \frac{4}{8} \\ & =1 \mathrm{~A} \end{aligned}$

Let x = current/division

After applying shunt

Now $5 \mathrm{x} \times \mathrm{G}=20 \mathrm{x} \times 24$

$\begin{aligned} & \mathrm{G}=4 \times 24 \\ & \mathrm{G}=96 \Omega \end{aligned}$

To determine the value of the shunt resistor required to convert the galvanometer into an ammeter capable of measuring a current of $8 \mathrm{~A}$, we need to use the concept of shunting where the additional resistor (shunt) is placed in parallel with the galvanometer.

The formula for calculating the value of the shunt resistor $ R_s $ is given by:

$ R_s = \frac{R_g \cdot I_g}{I - I_g} $

Where:

$ R_g $ = resistance of the galvanometer = $10 \Omega$

$ I_g $ = full-scale deflection current of the galvanometer = $3 \mathrm{~mA}$ = $0.003 \mathrm{~A}$

$ I $ = total current to be measured = $8 \mathrm{~A}$

Substituting these values into the formula:

$ R_s = \frac{10 \Omega \cdot 0.003 \mathrm{~A}}{8 \mathrm{~A} - 0.003 \mathrm{~A}} $

Perform the calculations:

$ R_s = \frac{0.03 \Omega \cdot \mathrm{~A}}{7.997 \mathrm{~A}} $

$ R_s \approx 3.75 \times 10^{-3} \Omega $

Thus, the value of the shunt resistor should be:

Option A: $3.75 \times 10^{-3} \Omega$

To calculate the amount of electric charge $Q$ that crosses through a section of the wire over a period of $20$ seconds, given the current varies with time as $I = I_0 + \beta t$, where $I_0 = 20$ A (initial current) and $\beta = 3$ A/s (rate of change of current with time), we use the concept of integration from calculus because the current is not constant but changes linearly with time.

The electric charge $Q$ is the integral of current $I$ over the time interval from $0$ to $20$ seconds. The formula for $Q$ is given by:

$Q = \int_{0}^{T} I(t) \, dt$

Where $I(t) = I_0 + \beta t$ and $T = 20$ s. Substituting the given values:

$Q = \int_{0}^{20} (20 + 3t) \, dt$

This integral can be solved in two parts:

1. The integral of the constant term $20$:

$\int 20 \, dt = 20t$

2. The integral of the linear term $3t$:

$\int 3t \, dt = \frac{3}{2}t^2$

Thus, combining these and evaluating from $0$ to $20$ seconds:

$Q = \left[20t + \frac{3}{2}t^2\right]_{0}^{20}$

$Q = \left[20(20) + \frac{3}{2}(20)^2\right] - \left[20(0) + \frac{3}{2}(0)^2\right]$

$Q = 400 + 600 = 1000$ C

Therefore, the amount of electric charge that crosses through a section of the wire in $20$ seconds is $1000$ C.

From KVL,

$\mathbf{V}_{\mathbf{1}}+\mathbf{V}_{\mathbf{2}}-\mathbf{V}_{\mathbf{3}}=\mathbf{0} \Rightarrow \mathrm{V}_1+\mathrm{V}_2=\mathrm{V}_3$

$\mathrm{i} \propto \theta$ (angle of deflection)

$\begin{aligned} & \therefore \frac{\mathrm{i}_2}{\mathrm{i}_1}=\frac{\theta_2}{\theta_1} \Rightarrow \frac{\mathrm{i}_2}{200 \mu \mathrm{A}}=\frac{\pi / 10}{\pi / 3}=\frac{3}{10} \\ & \Rightarrow \mathrm{i}_2=60 \mu \mathrm{A} \end{aligned}$

The specific resistance (or resistivity) of a material is a fundamental property that describes how much the material resists the flow of electric current. The resistivity is typically denoted by the symbol $\rho$ (rho), and it can be calculated by using the resistance $X$ of a uniform specimen of the material, along with its physical dimensions. In the case of a wire, the resistivity formula in terms of its resistance $X$, length $L$, and cross-sectional area $A=\pi r^2$ is given by:

This formula is a reinterpretation of Ohm's law, and it states that the specific resistance is proportional to the area of the cross-section of the wire and inversely proportional to its length.

Given that the specific resistance of the wire $S_1$ is determined using the formula:

Now, let's see what happens to the specific resistance if the length of the wire is doubled. If we denote the new length as $2L$, the resistance of the wire with the new length will change because resistance is directly proportional to the length of the wire. However, resistivity (specific resistance) is an intrinsic property of the material and does not depend on its length or shape, only on its temperature. Therefore, even if we change the length of the wire, the specific resistance should remain the same.

If we calculate the new resistance $X'$ with the doubled length, it would be:

Thus, we would use the new resistance $X'$ and the new length $2L$ to calculate the specific resistance again:

Since this formula is essentially the same as our original formula for $S_1$, we can conclude that:

Therefore, the value of the specific resistance $S_1$ will remain the same even if the length of the wire gets doubled. The correct answer to the question is:

Option D: $S_1$

For null point,

$\begin{aligned} & \frac{4.5}{60}=\frac{R}{40} \\ & \text { Also, } R=\frac{\rho \ell}{A}=\frac{\rho \ell}{\pi r^2} \\ & 4.5 \times 40=\rho \times \frac{0.1}{\pi \times 7 \times 10^{-8}} \times 60 \\ & \rho=66 \times 10^{-7} \Omega \times \mathrm{m} \end{aligned}$

Resistance of each part $=\frac{R}{5}$

Total resistance $=\frac{1}{5} \times \frac{\mathrm{R}}{5}=\frac{\mathrm{R}}{25}$

Given, $ R=400 \Omega $

$\begin{aligned} r & =4 \Omega \\\\ R_{\mathrm{eq}} & =R+r=404 \Omega\end{aligned}$

From definition of terminal potential difference,

$ V=E-I r $

We know that, current drawn from the cell,

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

Putting value in Eq. (i),

$ \begin{array}{l} V=E-\frac{E}{R+r} \times r \\\\ V=\frac{E R+E r-E r}{R+r} \\\\ V=\frac{E R}{R+r} \\\\ V=\frac{E \cdot 400}{404}=\frac{100 E}{101} \\\\ \frac{V}{E}=\frac{100}{101} \end{array} $

Percentage error in measurement of emf of the cell is,

% error in

$ E=\left(\frac{E-V}{E}\right) \times 100=\left(1-\frac{V}{E}\right) \times 100 $

$\begin{aligned} & =\left(1-\frac{100}{101}\right) \times 100 \quad \text { [From Eq. (ii)] } \\\\ & =\frac{1 \times 100}{101}=0.99 \%\end{aligned}$

Given, balancing length, $ l=50 \mathrm{~cm} $

At balancing situating of meter bridge,

$ \begin{array}{l} \frac{R}{S}=\frac{1}{100-1}=\frac{50}{100-50} \\\\ \frac{R}{S}=\frac{1}{1} \Rightarrow R=S \end{array} $

We know that,

$ R=\rho l / A $

When the wire in the right gap is stretched to double its length, its resistance changes.

If the length is doubled, assuming the volume of the wire remains constant, the cross-sectional area will be halved.

$ V=A \cdot l \Rightarrow A=\frac{V}{l} $

$ \begin{array}{l} \text { Given, } \\\\ \therefore \quad l^{\prime}=2 l, A^{\prime}=\frac{V}{2 l} \Rightarrow A^{\prime}=\frac{A}{2} \end{array} $

So, the new resistance $ S^{\prime} $ becomes

$ S^{\prime}=\rho \frac{l^{\prime}}{A^{\prime}}=\frac{\rho 2 l}{\frac{A}{2}}=\frac{\rho \cdot 4 l}{A}=4 R \text { [from Eq (i)] } $

Now, the new balancing length is,

$ \begin{array}{l} \frac{R}{S^{\prime}}=\frac{l}{100-l} \Rightarrow \frac{R}{4 R}=\frac{l}{100-l} \\\\ 100-l=4 l \\\\ 5 l=100 \Rightarrow l=20 \mathrm{~cm} \end{array} $

Given:

The resistance of the wire at temperature $ T_1 = 373 \, \text{K} $ is $ 2.5 \, \Omega $.

The temperature coefficient of resistance, $ \alpha = 3.6 \times 10^{-3} \, \text{K}^{-1} $.

We need to determine the resistance of the wire at a temperature of $ 273 \, \text{K} $.

To find this, we use the formula for resistance change with temperature:

$ R = R_0 (1 + \alpha \Delta T) $

Where the temperature difference $ \Delta T = T_1 - T_2 = 373 \, \text{K} - 273 \, \text{K} = 100 \, \text{K} $.

Given:

$ 2.5 = R_0 \left(1 + 3.6 \times 10^{-3} \times 100 \right) $

Calculating $ R_0 $:

$ R_0 = \frac{2.5}{1.36} $

Thus, the resistance at $ 273 \, \text{K} $ is:

$ R_0 ( \text{at } 273 \, \text{K} ) = 1.84 \, \Omega $

Current in the figure 1 using Ohm's law,

$ i=\frac{V(\text { cell })}{\text { Total resistance }} $

$ i=\frac{V}{2 R+r} $ [as ideal cell, $ r=0 $ (internal resistance)]

$ \begin{array}{l} 2=\frac{V}{2 R} \\\\ \frac{V}{R}=4 \end{array} $

Now, when resistor are connect parallely

$ \begin{array}{l} i=\frac{V(\text { cell })}{\frac{R}{2}} \\\\ i=2 \frac{V}{R} \\\\ i=2 \times 4 \Rightarrow i=8 \mathrm{~A} \end{array} $

Current through each resistor $ =4 \mathrm{~A} $

Given, $ L=1.5 \mathrm{~m} $

Area of cross-section

$ \begin{array}{l} A=3 \times 10^{-5} \mathrm{~m}^{2} \\\\ R=15 \Omega, E=21 \mathrm{~V} / \mathrm{m} \end{array} $

Curren density, $ J= $ ?

$ J=\sigma E=\frac{E}{\rho} $

Charge in term of current is given

$ \begin{array}{l} d Q=I d t \\\\ Q=\int_{t_{1}}^{t_{2}} I d t \\\\ I=12 t+9 t^{2}, t_{1}=2 \mathrm{~s}, t_{2}=10 \mathrm{~s} \\\\ Q=\int_{2}^{10} 12 t d t+\int_{2}^{10} 9 t^{2} d t \\\\ {\left[\frac{12 t^{2}}{2}\right]_{2}^{10}+\left[\frac{9 t^{3}}{3}\right]_{2}^{10}} \\\\ \\\\ =6\left[10^{2}-2^{2}\right]+3\left[10^{3}-2^{3}\right] \\\\ \\\\ =6 \times[100-4]+3[1000-8] \\\\ \\\\ =6 \times 96+3 \times 992 \\\\ \\\\ =3552 \mathrm{C} \end{array} $

Given, length of conductor,

$ l=20 \mathrm{~cm}=0.2 \mathrm{~m} $

potential difference, $\mathrm{PD}=16 \mathrm{~V}$

drift speed $\left(v_d\right)=2.4 \times 10^{-4} \mathrm{~m} / \mathrm{s}$

We know that, $E=-\frac{d V}{d r}=\frac{16}{0.2}=80 \mathrm{~V} / \mathrm{m}$

Electron mobility, $\mu=\frac{v_d}{E}=\frac{2.4 \times 10^{-4}}{80}$

$ =3 \times 10^{-6} \mathrm{~m}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1} $

For a balanced Wheatstone bridge

$ \begin{aligned} & \frac{4 \Omega}{R_{\text {bulb }}}=\frac{8 \Omega}{12 \Omega} \\\\ & R_{\text {bulb }}=6 \Omega \end{aligned} $

Current in upper branch, $i=\frac{V_a}{10} \mathrm{~A}$

Potential across bulb, $V=i(2 i+1)$

$ \begin{aligned} i \times 6 & =i(2 i+1) \\\\ \Rightarrow \quad 6 & =(2 i+1) \Rightarrow i=\frac{5}{2} \mathrm{~A} \end{aligned} $

Now total voltage across circuit,

$ \begin{aligned} V_a & =i \times(6+4)=i \times(10) \mathrm{V} \\\\ & =\frac{5}{2} \times 10 \mathrm{~V}=25 \mathrm{~V} \end{aligned} $

Initial temperature of wire,

$ T_1=303 \mathrm{~K} $

Final temperature of wire, $T_2=356 \mathrm{~K}$

Resistance increases % = 10 %

Temperature coefficient $\alpha=$ ?

Here, $\alpha=\frac{\left(\frac{\Delta R}{R_0}\right)}{\Delta T} \quad \ldots \text { (i) }$

$\Delta R=$ change in resistance

$R=$ original resistance

$\Delta T=$ change in temperature

$ \Delta T=T_2-T_1=356-303=53 \mathrm{~K} \quad \ldots \text { (ii) } $

from Eq. (i),

$ \begin{aligned} & \alpha=\frac{0.10}{53}=1.88 \times 10^{-3} \mathrm{C}^{-1} \\\\ & \alpha \equiv 2 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1} \end{aligned} $

Thus, temperature coefficient of resistance of the material of the wire is approximately $2 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}$.

Applying KVL from $A$ to $B$,

$ \begin{aligned} V_A-10 I-20 I_1 & =V_B \\\\ 10-10 I-20 I_1 & =6 \\\\ 4 & =10\left[I+2 I_1\right] \quad \ldots \text { (i) } \end{aligned} $

Applying KVL from $A$ to $C$,

$ \begin{aligned} & V_A-10 I-30 I+30 I_1=V_C \\\\ & 10=40 I+30 I_1+5 \\\\ & 5=40 I-30 I_1 \\\\ & 1=8 I-6 I_1 \\\\ & I_1=\frac{8 I-1}{6} \quad \ldots \text { (ii) } \end{aligned} $

Putting the value of $I_1$ in Eq. (i),

$ \begin{aligned} & \begin{aligned} 4 & =10\left[I+\frac{2}{6}(8 I-1)\right] \\\\ 4 & =10\left[I+\frac{8}{3} I-\frac{1}{3}\right] \Rightarrow 4=10\left[\frac{11 I}{3}-\frac{1}{3}\right] \\\\ 1.2 & =11 I-1 \\\\ 11 I & =2.2 \Rightarrow I=\frac{1}{5} \end{aligned} \end{aligned} $

Let current in $10 \Omega$ is $I$,

Current in $20 \Omega$ is $I_1$ and

Current in $30 \Omega$ is $I-I_1$

$ \begin{aligned} & I_1=\frac{8 I-1}{6}=\frac{\frac{8}{5}-1}{6}=\frac{3}{5 \times 6}=\frac{1}{10} \\\\ & I-I_1=\frac{1}{5}-\frac{1}{10}=\frac{1}{10} \\\\ & \text { Ratio } \frac{I}{I-I_1}=\frac{1}{5 \times \frac{1}{10}}=2: 1 \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} $