Current Electricity

Given circuit is,

$ \Rightarrow \quad I_1=I+I_2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} &\text { In above circuit, }\\ &\begin{array}{ll} & I \times 12=I_2 \times 6 \\ \Rightarrow & I_2=2 I \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} \end{aligned} $

$ \begin{aligned}So,\,\, & I_1 \times 4=3 \\ or\,\,\,& I_1=\frac{3}{4} \mathrm{~A} \end{aligned} $

From Eqs. (i), (ii) and (iii),

$ \begin{gathered} I_1=I+I_2=I+2 I=3 I \\ \text { or } \quad I=\frac{I_1}{3}=\frac{\frac{3}{4}}{3}=\frac{1}{4} \mathrm{~A}=0.25 \mathrm{~A} \end{gathered} $

$ \begin{aligned} & E=\rho \frac{I}{A} \\ & \rho=\frac{E A}{I}=\frac{0.15 \times 6 \times 10^{-7}}{0.3} \\ & \rho=3 \times 10^{-7} \Omega \mathrm{~m} \end{aligned} $

1. Same material: This implies that the resistivity ($\rho$) and the number density of free electrons ($n$) are the same for both wires. Also, the charge of an electron ($e$) is a constant.

2. Lengths ratio: $L_1 : L_2 = 2 : 3$, which means $L_1/L_2 = 2/3$.

3. Radii ratio: $r_1 : r_2 = 1 : 2$.

4. Connected in parallel to a battery: This means the potential difference (voltage $V$) across both wires is the same, i.e., $V_1 = V_2 = V$.

We need to find the ratio of the drift velocities ($v_{d1} : v_{d2}$).

The relationship between current ($I$), number density of free electrons ($n$), cross-sectional area ($A$), charge of an electron ($e$), and drift velocity ($v_d$) is given by:

$I = n A e v_d$

From this, the drift velocity can be expressed as:

$v_d = \frac{I}{n A e}$ (Equation 1)

According to Ohm's Law, the current in a wire is related to the potential difference ($V$) and resistance ($R$):

$I = \frac{V}{R}$ (Equation 2)

The resistance of a wire is given by:

$R = \frac{\rho L}{A}$ (Equation 3)

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

Now, substitute Equation 3 into Equation 2:

$I = \frac{V}{\rho L / A} = \frac{V A}{\rho L}$ (Equation 4)

Finally, substitute Equation 4 into Equation 1 for $I$:

$v_d = \frac{V A / (\rho L)}{n A e}$

Notice that the cross-sectional area ($A$) appears in both the numerator and the denominator, so it cancels out:

$v_d = \frac{V}{n e \rho L}$

This derived formula shows that the drift velocity ($v_d$) depends on the applied voltage ($V$), the number density of free electrons ($n$), the electron charge ($e$), the resistivity ($\rho$), and the length of the wire ($L$).

Since the wires are made of the same material, $n$ and $\rho$ are constant. The charge of an electron $e$ is also a constant. When connected in parallel, the voltage $V$ across both wires is the same.

Therefore, for these two wires, $V$, $n$, $e$, and $\rho$ are all constant. This means the drift velocity is inversely proportional to the length of the wire:

$v_d \propto \frac{1}{L}$

So, the ratio of the drift velocities will be:

$\frac{v_{d1}}{v_{d2}} = \frac{1/L_1}{1/L_2} = \frac{L_2}{L_1}$

We are given the ratio of lengths $L_1 : L_2 = 2 : 3$.

This means $\frac{L_1}{L_2} = \frac{2}{3}$.

Therefore, $\frac{L_2}{L_1} = \frac{3}{2}$.

Substituting this into the ratio of drift velocities:

$\frac{v_{d1}}{v_{d2}} = \frac{3}{2}$

The ratio of the drift velocities is $3:2$.

Note that the information about the radii ratio was not needed for this calculation, as the cross-sectional area cancels out in the final expression for drift velocity when the voltage across the wire is constant.

The final answer is $\boxed{\text{3:2}}$.

We use the formula:

$ \frac{r}{R} = \frac{l_1 - l_2}{l_2} $

where $ r $ is internal resistance, $ R $ is the external resistance, $ l_1 $ is the first balancing length, and $ l_2 $ is the new balancing length when $ R $ is connected.

Since the balancing length decreases by 10%, $ l_2 = 0.9l_1 $.

Substitute $ l_2 = 0.9l_1 $ into the formula:

$ \frac{r}{R} = \frac{l_1 - 0.9l_1}{0.9l_1} $

Simplify the numerator:

$ l_1 - 0.9l_1 = 0.1l_1 $.

So,

$ \frac{r}{R} = \frac{0.1l_1}{0.9l_1} = \frac{0.1}{0.9} = \frac{1}{9} $

Therefore, $ r = \frac{R}{9} $

First Condition

When the current flows normally, the formula for the potential difference across the cell is: $ V = E - i r $

We are given that $ V = 20\,V $ and $ i = 2\,A $. Plug these values into the formula:

$ 20 = E - 2r \hspace{0.5cm} ...(i) $

Second Condition

When the current direction is reversed, the formula for potential difference becomes: $ V = E + i r $.

Here, $ V = 30\,V $ and $ i = 2\,A $, so:

$ 30 = E + 2r \hspace{0.5cm} ...(ii) $

Find the Internal Resistance

Now, subtract equation (i) from equation (ii):

$ (30) - (20) = (E + 2r) - (E - 2r) $

$ 10 = 4r $

So, $ r = \frac{10}{4} = 2.5\ \Omega $

Step 1: Total length of the wire

The total length of the wire is the length of the semicircle plus the diameter. That is:
$ l = \pi R + 2R $

Step 2: Find resistance per unit length

Since the total resistance is $36\ \Omega$ for the entire wire, resistance per unit length is:
$ \sigma = \frac{36}{\pi R + 2R} $

Step 3: Resistance along the diameter (straight part)

The length of the diameter is $2R$. So, resistance along the diameter is:
$ R_1 = \sigma \times 2R = \frac{36}{\pi R + 2R} \times 2R = \frac{72}{\pi + 2} $

Step 4: Resistance along the semicircle

The length of the semicircle is $\pi R$. So, resistance along the semicircle is:
$ R_2 = \sigma \times \pi R = \frac{36}{\pi R + 2R} \times \pi R = \frac{36\pi}{\pi+2} $

Step 5: Connect the two paths in parallel

The diameter and semicircular parts connect at both ends, so they are in parallel. Use the formula for parallel resistances:
$ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} $

Step 6: Substitute the values for $R_1$ and $R_2$

$ R_{eq} = \frac{\frac{72}{\pi+2} \times \frac{36\pi}{\pi+2}}{\frac{72}{\pi+2}+\frac{36\pi}{\pi+2}} = \frac{72 \times 36 \pi}{(\pi+2)(36\pi+72)} $

Step 7: Simplify the expression further

$ = \frac{72 \times 36 \times 22 \times 7}{36 \times (36 \times 22 + 72 \times 7)} = \frac{72 \times 22 \times 7}{36(22+14)} = \frac{77}{9}\ \Omega $

$ \begin{aligned} &\text { Voltage sensitivity }\\ &\begin{aligned} S_V & =\frac{N A B}{k R} \\ \frac{S_{V_A}}{S_{V_B}} & =\frac{N_A}{N_B} \cdot \frac{A_A}{A_B} \cdot \frac{R_B}{R_A} \\ \Rightarrow \quad \frac{4}{3} & =\frac{200}{N_B} \times \frac{1}{2} \times \frac{4}{3} \Rightarrow N_B=100 \end{aligned} \end{aligned} $

Using, $\Sigma I=0$, at $B$,

$ I_{B C}=4+1.8=5.8 \mathrm{~A} $

Also,

At $C$,

$ I_{C D}=5.8-1.3-1=3.5 \mathrm{~A} $

So, potential difference between $C$ and $D$ is

$ \begin{aligned} V_{C D} & =I_{C D} \times R_{C D} \\ & =3.5 \times 8=28.0 \mathrm{~V} \end{aligned} $

Total resistance, $R=242+8$

$ =250 \Omega $

∴ Current in the primary circuit,

$ I=\frac{E}{R}=\frac{2.5}{250}=0.01 \mathrm{~A} $

Potential drop across the potentiometer wire,

$ \begin{aligned} V^{\prime} & =I \times R_{\text {wire }}=0.01 \times 8 \\ & =0.08 \mathrm{~V} \end{aligned} $

∴ Potential gradient, $K=\frac{V}{L}$

$ =\frac{0.08}{2.5}=0.032 \mathrm{~V} / \mathrm{m} $

∴ Potential difference between two points separated by a distance of 20 cm ,

$ \begin{aligned} V^{\prime \prime} & =K \times l=0.032 \times 0.2 \\ & =0.0064 \mathrm{~V} \\ & =6.4 \times 10^{-3} \mathrm{~V}=6.4 \mathrm{mV} \end{aligned} $

The length of the potentiometer wire is $L=10 \mathrm{~m}$.

The resistance of the potentiometer wire is

$ R_\omega=5 \Omega $

The emf of the cell is $E=2.2 \mathrm{~V}$

The potential difference across a length of $l=660 \mathrm{~cm}$ of the wire is $V_l=1.1 \mathrm{~V}$.

$ l=660 \mathrm{~cm}=6.6 \mathrm{~m} $

The potential gradient, $K=\frac{V_l}{l}$

$ K=\frac{1.1 \mathrm{~V}}{6.6 \mathrm{~m}}=\frac{1}{6} \mathrm{~V} / \mathrm{m} $

The total potential drop across the wire $V_\omega$ is

$ \begin{aligned} & V_\omega=K \times L \\ & V_\omega=\frac{1}{6} \mathrm{~V} / \mathrm{m} \times 10 \mathrm{~m}=\frac{10}{6} \mathrm{~V}=\frac{5}{3} \mathrm{~V} \end{aligned} $

Current, $I=\frac{V_\omega}{R \omega}=\frac{\frac{5}{3} \mathrm{~V}}{5 \Omega}=\frac{1}{3} \mathrm{~A}$

The current in the main circuit is also given by

$ \begin{aligned} & I=\frac{E}{R_\omega+r} \\ & r=\frac{E}{I}-R_\omega=\frac{2.2 \mathrm{~V}}{\frac{1}{3} \mathrm{~A}}-5 \Omega \\ & r=(2.2 \times 3) \Omega-5 \Omega \\ & r=6.6 \Omega-5 \Omega=1.6 \Omega \end{aligned} $

First setup:

Let the resistance in the left gap be $R'$, and in the right gap there are two resistors with resistance $R$, connected in series. So, the right gap has a total resistance of $R + R = 2R$.

The balancing point is at 50 cm. In a metre bridge, the ratio of resistances equals the ratio of the lengths on both sides: $ \frac{R'}{2R} = \frac{50}{50} $ So, $ \frac{R'}{2R} = 1 $ which means $ R' = 2R $

Second setup:

Now, one resistor from the right gap is taken out and connected in parallel with the resistor in the left gap.

Now, the left gap contains $R'$ and $R$ in parallel. The resistance of two resistors $A$ and $B$ in parallel is: $ R'' = \frac{A \times B}{A + B} $ The parallel resistance is: $ R'' = \frac{R' \times R}{R' + R} $

From earlier, we know $R' = 2R$. Substitute this value: $ R'' = \frac{2R \times R}{2R + R} = \frac{2R^2}{3R} = \frac{2R}{3} $

Now, the right gap just has $R$.

Let the new balancing point be $l$. The bridge principle gives us: $ \frac{R''}{R} = \frac{l}{100 - l} $ Substitute $R'' = \frac{2R}{3}$: $ \frac{\frac{2R}{3}}{R} = \frac{l}{100 - l} $ $ \frac{2}{3} = \frac{l}{100 - l} $

Solve for $l$: $ 3l = 2(100 - l) $ $ 3l = 200 - 2l $ $ 3l + 2l = 200 $ $ 5l = 200 $ $ l = 40~\text{cm} $

Given

• Drift speed, $ v_d = 0.3\ \mathrm{m\,s^{-1}} $

• Electric field, $ E = 2\ \mathrm{V\,m^{-1}} $

We are to find electron mobility $ \mu $, where

$ \mu = \frac{v_d}{E} $

Calculation

$ \mu = \frac{0.3}{2} = 0.15\ \mathrm{m^2\,V^{-1}\,s^{-1}} $

$ \mu = 0.15 = 15 \times 10^{-2}\ \mathrm{m^2\,V^{-1}\,s^{-1}} $

✅ Answer: Option B

$ \boxed{15 \times 10^{-2}\ \mathrm{m^2\,V^{-1}\,s^{-1}}} $

Resistance of motor

$ R=\frac{V_1^2}{P_1}=\frac{220^2}{242}=200 \Omega $

Current drawn by the motor

$ I=\frac{V_2}{R}=\frac{200}{200}=1 \mathrm{~A} $

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} $

Given, Length of wire $L=5 \mathrm{~m}$

Resistance of wire $=5 \Omega=R$

Balancing point $=3 \mathrm{~m}$

Emf of cell $=10 V=E_1$

Current in circuit with $E_1=10 \mathrm{~V}$ is circuit with $E_1=10 \mathrm{~V}$ is given by

$ \begin{aligned} I & =\frac{E_1}{R+r}[r=4+1 \Omega=5 \Omega] \\ I & =\frac{10}{5+5}=1 \mathrm{~A} \\ V_{A B} & =I \times R=1 \times 5=5 \mathrm{~V} \end{aligned} $

Potential gradient is given by

$ \begin{aligned} & K=\frac{V_{A B}}{l(\text { for whole wire })} \\ & K=5 / 5=1 \end{aligned} $

Now, at balancing point.

$ \begin{aligned} & E=k l \Rightarrow E_1=1 \times 3 \\ & E_1=3 \mathrm{~V} \end{aligned} $

Resolving the circuit

$10 \Omega$ and $10 \Omega$ are in parallei connect and 6 and $6 \Omega$ are in parallel connection.

$ \begin{aligned} \because \text { New resistance } & =\frac{10 \times 10}{10+10}=5 \Omega \\ & =\frac{6 \times 6}{6+6}=3 \Omega \end{aligned} $

Redrawing the circuit diagram.

The ratio of resistance on both side of junction is $1: 1$, this condition is equal to wheat stone bridge condition. No, current will passes through $8 \Omega$.

New circuit will be equivalent resistance $=R_{\mathrm{eq}}=(5+3)=8 \Omega$ each side They are in parallel connection

$ R_{\mathrm{eq}}=\frac{8 \times 8}{8+8}=4 \Omega $

Given,

Wheat stone bridge is balanced.

Power dissipated through $R_3=P_3$

Power dissipated through $R_1=P_1$

As, wheat stone bridge is balanced no current will flow from galvanometer arm. The condition of balancing of bridge is

$ \frac{R_1}{R_3}=\frac{L}{K} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Power discipated by resistance circuit is given by,

$ \begin{aligned} & P=V I \\or\,\, & P=I^2 R \end{aligned} $

Power dissipation through $R_1$ and $R_3$ are

$ P_3=I^2 R_3, P_1=I^2 R_1 $

Ratio of power discipation.

$ \frac{P_3}{P_1}=\frac{I^2 R_3}{I^2 R_1} \Rightarrow \frac{P_3}{P_1}=\frac{R_3}{R_1}=\frac{K}{L} $

Given,

Resistance of wire $=2 R$

Let, Radius of wire $=r$

Length of wire $=L$

Asea of wire $=A$

When the length of wire is stretched to double volume will remains same.

$ \begin{aligned} V_{\text {intial }} & =V_{\text {final }} \\ L \times A & =L^{\prime} A^{\prime} \\ A^{\prime} & =\frac{L \times A}{2 L}=\frac{A}{2} \end{aligned} $

Initial resistance

$ R_1=2 R=\frac{\rho L}{A^{\prime}} \,\,\,\,\,\,\,[Given]$

New resistance

$ \begin{aligned} & R_2=\frac{\rho L^{\prime}}{A^{\prime}} \Rightarrow \frac{\rho 2 L}{\frac{A}{2}} \Rightarrow R_2=\frac{4 \rho L}{A}=4 \times 2 R \\ & R_2=8 R \end{aligned} $

So, the increase in resistance is $R^{\prime}-R$

$ 8 R-2 R=6 R $

${ }_{}$Given, $m_1: m_2: m_3=1: 2: 3$

$ \begin{aligned} & \therefore m_1=x, m_2=2 x \text { and } m_3=3 x \\ & \text { and } l_1: l_2: l_3=3: 2: 1 \end{aligned} $

$l_1=3 K, l_2=2 K$ and $l_3=K$

Resistance of the wire is given by

$ \begin{aligned} R & =\rho \frac{l}{A}=\rho \cdot \frac{l^2}{l \cdot A}=\rho \cdot \frac{l^2}{V} \\ & =\frac{\rho l^2}{\left(\frac{m}{d}\right)}=\frac{\rho l^2 d}{m} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad R \propto \frac{l^2}{m} \Rightarrow R_1 \propto \frac{l_1^2}{m_1} \\ & \Rightarrow \quad R_1 \propto \frac{(3 K)^2}{x} \\ & \Rightarrow \quad R_1 \propto \frac{9 K^2}{x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { Similarly, } R_2 \alpha \frac{l_2^2}{m_2} \Rightarrow R_2 \propto \frac{(2 K)^2}{2 x} \\ & \Rightarrow \quad R_2 \propto \frac{(2 K)^2}{2 x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \text { and } \quad R_3 \propto \frac{l_3^2}{m_3} \\ & \Rightarrow \quad R_3 \propto \frac{K^2}{3 x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned}\\ &\text { From Eqs. (i), (ii) and (iii), we get }\\ &R_1: R_2: R_3=9: 2: \frac{1}{3}=27: 6: 1 \end{aligned} $

$ \begin{aligned} &\text { Wheatstone's bridge is balanced if }\\ &\frac{P}{Q}=\frac{R}{S_1 \| S_2} \end{aligned} $

$ \Rightarrow \frac{P}{Q}=\frac{P}{\left(\frac{S_1 S_2}{S_1+S_2}\right)} \Rightarrow \frac{P}{Q}=\frac{R\left(S_1+S_2\right)}{S_1 S_2} $

The resistance of a wire, $R$, depends on the material, the length ($l$), and the cross-sectional area ($A$). The formula is:

$ R = \rho \cdot \frac{l}{A} $ where $\rho$ is the resistivity (which depends on the material).

The area $A$ of a wire with radius $r$ is $A = \pi r^2$. So the resistance can be rewritten as: $ R = \rho \cdot \frac{l}{\pi r^2} $ This means that resistance goes up if the length ($l$) increases, and goes down if the radius ($r$) increases.

If the length is doubled ($l_2 = 2l_1$), and the radius is also doubled ($r_2 = 2r_1$), we can find the new resistance $R_2$ compared to the old resistance $R_1 = R$.

When both length and radius change, the formula for comparing the new and old resistance becomes: $ \frac{R_2}{R_1} = \left( \frac{r_1}{r_2} \right)^2 \cdot \frac{l_2}{l_1} $

Plug in the values ($r_2 = 2r_1$, $l_2 = 2l_1$): $ \frac{R_2}{R} = \left( \frac{r_1}{2r_1} \right)^2 \cdot \frac{2l_1}{l_1} = \left( \frac{1}{2} \right)^2 \times 2 = \frac{1}{4} \times 2 = \frac{1}{2} $

This means the new resistance is half the original: $ R_2 = \frac{R}{2} $

The specific resistance (or resistivity, $\rho$) does not change when you change the size or length of the material. It only depends on what the wire is made of and its temperature.

$ \begin{aligned} &\text { We know that, in meter bridge }\\ &\begin{aligned} & & \frac{R_{\text {left }}}{R_{\text {right }}} & =\frac{l}{100-l} \\ & \Rightarrow & \frac{2}{3} & =\frac{l}{100-l} \\ & \Rightarrow & 200-2 l & =3 l \Rightarrow 200=5 l \\ & \Rightarrow & 40 & =l \Rightarrow l=40 \mathrm{~cm} \end{aligned} \end{aligned} $

According to given condition,

$ \begin{array}{ll} & I_g=5 \% \text { of } I=0.05 I \\ \therefore & I-I_g=I-0.05 I=0.95 I \end{array} $

Applying KVL in the loop, we get

$ \begin{aligned} & \left(I-I_g\right) S-I_g G=0 \\ \Rightarrow \quad & \left(I-I_g\right) S=I_g G \\ \Rightarrow \quad & (0.95 I) S=(0.05 I) G \Rightarrow S=\frac{G}{19} \end{aligned} $

Given:

Internal resistance of the cell, $ r = 1 \, \Omega $

Initial balancing length, $ l_1 = 44 \, \text{cm} $

New balancing length, $ l_2 = 40 \, \text{cm} $

Formula:

The internal resistance of the cell can be expressed as:

$ r = R \left(\frac{l_1}{l_2} - 1\right) $

Here, $ R $ is the external resistance connected parallel to the cell. We know $ r = 1 \, \Omega $.

Calculation:

$ 1 = R \left(\frac{44}{40} - 1\right) $

Simplify the expression:

$ 1 = R \left(\frac{44}{40} - \frac{40}{40}\right) $

$ 1 = R \left(\frac{4}{40}\right) $

$ 1 = \frac{R}{10} $

Solve for $ R $:

$ R = 10 \, \Omega $

Thus, the resistance that needs to be connected in parallel to the cell to achieve the desired balancing point is $ 10 \, \Omega $.

Given the problem, two wires made of the same material have lengths and radii in specified ratios, and the same potential difference is applied to both. Our goal is to find the ratio of electric currents through the wires.

Given values:

Ratio of lengths of the wires: $\frac{L_1}{L_2} = \frac{2}{3}$

Ratio of radii of the wires: $\frac{r_1}{r_2} = \frac{8}{9}$

Since both wires are made of the same material and experience the same potential difference, we use Ohm's Law:

$ V = I R $

This leads to the relationship for the wires:

For the first wire: $V = I_1 R_1$

For the second wire: $V = I_2 R_2$

Equating the two since $V$ is the same:

$ I_1 R_1 = I_2 R_2 $

This implies:

$ \frac{I_1}{I_2} = \frac{R_2}{R_1} $

The resistance $R$ is given by:

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

where:

$\rho$ is the resistivity,

$L$ is the length,

$A = \pi r^2$ is the cross-sectional area.

Thus, for the resistances:

$ R_1 = \rho \frac{L_1}{\pi (r_1)^2}, \quad R_2 = \rho \frac{L_2}{\pi (r_2)^2} $

Substituting these into the current ratio equation:

$ \frac{I_1}{I_2} = \frac{\rho \frac{L_2}{\pi (r_2)^2}}{\rho \frac{L_1}{\pi (r_1)^2}} = \frac{L_2}{L_1} \times \left(\frac{r_1}{r_2}\right)^2 $

Substitute the given ratios:

$ \frac{I_1}{I_2} = \left(\frac{3}{2}\right) \times \left(\frac{8}{9}\right)^2 = \frac{3}{2} \times \frac{64}{81} = \frac{192}{162} = \frac{32}{27} $

Thus, the ratio of current $I_1 : I_2$ is $32 : 27$.

When the potentiometer is connected between points $ A $ and $ B $, the balance point is achieved at 64 cm. When connected between $ A $ and $ C $, the balance point occurs at 8 cm. Now, we need to determine the balance point when the potentiometer is connected between $ B $ and $ C $.

Calculation

Initial Conditions:

When connected between $ A $ and $ B $:

$ V_1 = k \times l_1 = 64k $

When connected between $ A $ and $ C $:

$ V_1 - V_2 = k \times l_2 = 8k $

Solve for $ V_2 $:

Rearrange the equation:

$ 64k - V_2 = 8k $

Solve for $ V_2 $:

$ V_2 = 64k - 8k = 56k $

Determine the balance point when connected between $ B $ and $ C $:

We know $ V_2 = k \times l_2 $, where $ l_2 $ is the distance we need to find.

Therefore:

$ k \times l_2 = 56k $

Solve for $ l_2 $:

$ l_2 = 56 \text{ cm} $

Thus, the balance point when the potentiometer is connected between $ B $ and $ C $ will be 56 cm.

The direction of current is from $A$ to $B$ which means point $A$ is at higher potential.

For section $A B$,

$ \begin{aligned} & V=I R=1 \times 1.5 \\ & V=+1.5 \mathrm{~V} \text { at } A \end{aligned} $

Now for section $B C$

$ \begin{gathered} V=I R \quad \text { (at } C \text { let potential beX) } \\ (2 V+X)=I R \Rightarrow I=\frac{2 V+X}{2.5 \Omega} \end{gathered} $

We know from diagram $I=1 A$

For $I$ to be 1 the $V / R$ ratio should be $L$.

$ \begin{array}{ll} \because & 2.5=2+X \Rightarrow X=2-2.5 \\ \Rightarrow & X=-0.5 \mathrm{~V} \end{array} $

Electron mobility.

$ \begin{aligned} \mu_e & =\frac{\text { Drift velocity }}{\text { Electric field }} \\ \mu_e & =\frac{v_d}{\mathbf{E}} \\ {\left[\mu_e\right] } & =\frac{\left[v_d\right]}{[E]}=\frac{\left[\mathrm{LT}^{-1}\right]}{\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{AT}]}} \\ \Rightarrow \quad\left[\mu_e\right] & =\left[\mathrm{M}^{-1} \mathrm{~T}^2 \mathrm{~A}\right] \end{aligned} $

So, SI units of mobility is $\mathrm{kg}^{-1} \mathbf{s}^2 \mathbf{A}$

Given, density of water,

$ \rho_w=1000 \mathrm{~kg} / \mathrm{m}^3 $

Specific heat capacity of water,

$ s_w=400 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} $

Volume of water, $V=31=3 \times 10^{-3} \mathrm{~m}^3$

∴ Mass of water, $m=V \times \rho_w$

$ =3 \times 10^{-3} \times 1000=3 \mathrm{~kg} $

Change of temperature of water,

$ \begin{aligned} \Delta T & =[(273+80)-(273+0)] \mathrm{K} \\ & =80 \mathrm{~K} \end{aligned} $

Heat required, $Q=m s_w \Delta T$

$ \begin{aligned} & =3 \times 4200 \times 80=1008000 \\ & =1.008 \times 10^6 \mathrm{~J} \end{aligned} $

Heat produced by heater, $H=\frac{V^2}{R} t$

$ \begin{array}{ll} \Rightarrow & 1.008 \times 10^6=\frac{(200)^2}{50} \times t \\ \Rightarrow & t=1260 \mathrm{~s}=\frac{1260}{60} \mathrm{~min}=21 \mathrm{~min} \end{array} $

Current density in circular wire,

$ J=1 \times 10^5 r \mathrm{~A} / \mathrm{m}^3 $

∴ Current, $I=J A$

$ \begin{aligned} & =1 \times 10^5 r \times \pi r^2=10^5 \pi r^3 \\ & =10^5 \times 314 \times\left(2 \times 10^{-4}\right)^3 \\ & \quad\left[\because r=2 \mathrm{~mm}=2 \times 10^{-4} \mathrm{~m}\right] \\ & =2512 \times 10^{-7} \mathrm{~A} \end{aligned} $

The rate of thermal energy produced,

$ \begin{aligned} P & =V I \\ & =70 \times 2512 \times 10^{-7}=1.76 \times 10^{-4} \mathrm{~W} \end{aligned} $

Energy consumed, $E=P \times t$

$ \begin{aligned} & =1.76 \times 10^{-4} \times 1000 \quad[\because t=1000 \mathrm{~s}] \\ & =0.176 \mathrm{~J} \end{aligned} $

Specific resistance depends on the nature of material and also depends on temperature of material. For conductor, it increases with increase of temperature. Hence, statement $I$ is incorrect. When wire of resistance $R$ is increased to $n$ times of its original length, then its new resistance $R^{\prime}$ is given as

$ R^{\prime}=n^2 R $

Here, $n=4, R=6 \Omega$

$ \therefore R^{\prime}=4^2 \times 6=96 \Omega $

Hence, statement II is incorrect and statement III is correct.

Average collision time,

$ \begin{aligned} \tau & =91 \times 10^{-15} \mathrm{~s} \\ e & =1.6 \times 10^{-19} \mathrm{C} \\ m_e & =9.1 \times 10^{-31} \mathrm{~kg} \end{aligned} $

∴ Mobility of electron in a wire,

$ \begin{aligned} \mu & =\frac{e \tau}{m_e} \\ & =\frac{1.6 \times 10^{-19} \times 9.1 \times 10^{-15}}{9.1 \times 10^{-31}} \\ & =1.6 \times 10^{-3} \mathrm{~m}^2 / \mathrm{V}-\mathrm{s} \end{aligned} $

The temperature coefficient of resistance for most of metals in pure form is positive because on increasing temperature of metals, its resistance increases. Hence, statement I is true.

In the statement II,

Given, diameter of wire $=2 \mathrm{~mm}$

∴ radius, $r=1 \mathrm{~mm}=10^{-3} \mathrm{~m}$

Charge, $q=360 \pi \mathrm{C}$

Time, $t=2 \mathrm{~h}=2 \times 60 \times 60 \mathrm{~s}=7.2 \times 10^3 \mathrm{~s}$

Charge density, $n=5 \times 10^{22}$

electrons $/ \mathrm{cm}^3=5 \times 10^{28}$ electrons $/ \mathrm{m}^3$

Current flowing through the wire.

$ I=\frac{q}{t}=\frac{360 \pi}{7.2 \times 10^3}=\frac{\pi}{20} \mathrm{~A} $

We know that, $I=n e A v_d$

$ \begin{aligned} & \Rightarrow \quad v_d=\frac{I}{n e A} \\ & =\frac{\pi / 20}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times \pi \times\left(10^{-3}\right)^2} \\ & =6.25 \times 10^{-6} \mathrm{~m} / \mathrm{s} \end{aligned} $

Hence, statement II is also true.

Semiconductors like pure germanium are operated in non-linear region, hence these do not obey Ohm's law. Hence, statement III is also true.

Radius of cylindrical resistor,

$ r=7 \mathrm{~mm}=7 \times 10^{-3} \mathrm{~m} $

Length, $l=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}$

Resistivity, $\rho=10^{-6} \Omega-\mathrm{m}$

$ \begin{aligned} \therefore \quad R & =\rho \cdot \frac{l}{A} \\ & =10^{-6} \times \frac{4 \times 10^{-2}}{\pi r^2} \\ & =\frac{4 \times 10^{-8}}{\frac{22}{7} \times\left(7 \times 10^{-3}\right)^2} \\ & =\frac{4 \times 10^{-8}}{22 \times 7 \times 10^{-6}} \\ & =\frac{2}{77} \times 10^{-2} \Omega \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text { Given, } \frac{E}{t}=1.54 \mathrm{~W} \\ & \Rightarrow \quad \frac{I^2 R T}{t}=1.54 \\ & \Rightarrow \quad I=\sqrt{\frac{1.54}{R}}=\sqrt{\frac{1.54}{\frac{2}{77} \times 10^{-2}}}=77 \mathrm{~A} \end{aligned}\\ &\text { ∴ Current density, }\\ &\begin{aligned} I=\frac{I}{A} & =\frac{I}{\pi r^2}=\frac{77}{\frac{22}{7} \times\left(7 \times 10^{-3}\right)^2} \\ & =5 \times 10^5 \mathrm{~A} / \mathrm{m}^2 \end{aligned} \end{aligned} $

We know that, $\tau=\frac{m}{n e^2 \rho}$

∴ Mean free path, $\lambda=\tau \cdot v_d=\frac{m v_d}{n e^2 \rho}$

$ \begin{aligned} & \text { Here, } n=9 \times 10^{28} / \mathrm{m}^3, \rho=1 \times 10^{-8} \Omega-\mathrm{m} \\ & \qquad \begin{aligned} v_d & =1.6 \times 10^6 \mathrm{~m} / \mathrm{s} \\ & m=9.1 \times 10^{-31} \mathrm{~kg}, e=1.6 \times 10^{-19} \mathrm{C} \end{aligned} \\ & \therefore \lambda=\frac{9.1 \times 10^{-31} \times 1.6 \times 10^6}{9 \times 10^{28} \times\left(1.6 \times 10^{-19}\right)^2 \times 1 \times 10^{-8}} \\ & \\ & =62.5 \times 10^{-9} \mathrm{~m} \\ & = 62.5 \mathrm{~nm} \end{aligned} $

Resistivity of metal,

$ \rho=1 \times 10^{-8} \Omega-\mathrm{m} $

Number of electrons per unit volume,

$ \begin{aligned} & n=9 \times 10^{28} \text { electrons } / \mathrm{m}^3 \\ & m=9 \times 10^{-31} \mathrm{~kg} \end{aligned} $

We know that, $\rho=\frac{m}{n e^2 \tau}$

$ \begin{aligned} & \Rightarrow \quad \tau=\frac{m}{n e^2 \rho} \\ & =\frac{9 \times 10^{-31}}{9 \times 10^{28} \times\left(1.6 \times 10^{-19}\right)^2 \times 1 \times 10^{-8}} \\ & =3.9 \times 10^{-14} \mathrm{~s} \simeq 4 \times 10^{-14} \mathrm{~s} \end{aligned} $

A wire is stretched such that its resistance increase by $6 \%$.

If we assume initial resistance to be 100 $\Omega$. Then the new resistance will be given by

$ R_{\text {new }}=100 \Omega+\frac{6}{100} \times 100 \Omega=106 \Omega $

We know that, the resistance is given by

$ R=\rho \frac{l}{A} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Where, area $(A)$ can be expressed in terms of volume $(V)$ is

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

Equation (i) then gives us, $R=\rho \frac{l^2}{V}$

Since, the resistivity and volume are going to remain the same, only change in length will be observed.

So, $\quad l^2=\frac{R V}{\rho}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

(Original length)

Let new length be $l_{\text {new }}$, so

$ l_{\text {new }}^2=\frac{R_{\text {new }} V}{\rho} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Dividing Eq. (iii) by Eq. (ii), we get

$ \begin{aligned} & \frac{l_{\text {new }}^2}{l^2}=\frac{R_{\text {new }}}{R}=\frac{106}{100} \\ \Rightarrow \quad & l_{\text {new }}=1.029 l \end{aligned} $

\% change in length

$ \begin{aligned} & =\left(\frac{l_{\mathrm{new}}-l}{l}\right) \times 100 \% \\ & =\left(\frac{1.029 l-l}{l}\right) \times 100 \\ & =0.029 \times 100 \\ & =29 \% \approx 3 \% \end{aligned} $

The given Circuit diagram is shown below.

Applying KVL in loop $A B C D A$.

$ 2+i_1 R-2=0 \Rightarrow i_1=0 \mathrm{~A} $

Applying KVL in loop $B E F C B$,

$ 2+i_2 R-2-i_1 R=0 \Rightarrow i_2=i_1=0 \mathrm{~A} $

Applying KVL in loop $E G H F E$, we get

$ \begin{aligned} & 2+i_3 R-2-i_2 R=0 \Rightarrow i_3=i_2=0 \mathrm{~A} \\ & \therefore \quad i_1=i_2=i_3=0 \mathrm{~A} \end{aligned} $

The resistivity of material is too much high as $10^8 \Omega$ - m which is equivalent to the value of resistivity of insulator.

$ \begin{aligned} &\text { We know that, resistance of wire, }\\ &\begin{aligned} & R=\rho \frac{L}{A}=\rho \frac{L}{\pi r^2} \\ \Rightarrow \quad & R \propto \frac{L}{r^2} \Rightarrow \frac{R_2}{R_1}=\frac{L_2}{L_1} \times\left(\frac{r_1}{r_2}\right)^2 \end{aligned}\\ &\begin{aligned} \text { Given, } L_1 & =L, L_2=2 L \\ r_1 & =r, r_2=3 r \\ \frac{R_2}{R_1} & =\frac{2 L}{L} \times\left(\frac{r}{3 r}\right)^2=2 \times \frac{1}{9}=\frac{2}{9} \\ \Rightarrow \quad R_2 & =\frac{2}{9} R_1=\frac{2}{9} R \end{aligned} \end{aligned} $

Given, $l_1=60 \mathrm{~cm}, l_2=40 \mathrm{~cm}$,

$ R=4 \Omega $

∴ Internal resistance of the cell,

$ \begin{aligned} r & =R\left(\frac{l_1}{l_2}-1\right)=4\left(\frac{60}{40}-1\right) \\ & =4 \times \frac{1}{2}=2 \Omega \end{aligned} $