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} $

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} $

From symmetry we can remove two middle resistance.

New circuit is

To convert a galvanometer into an ammeter, we need to add a shunt resistance in parallel with the galvanometer's coil. The purpose of the shunt resistance is to bypass the majority of the current while allowing only a small fraction of it to pass through the galvanometer, thereby preventing it from being damaged by high currents.

Given parameters:

• Resistance of the galvanometer's coil, $R_g = 200 \Omega$


• Full-scale deflection current of the galvanometer, $I_g = 20 \mu \mathrm{A} = 20 \times 10^{-6} \mathrm{A}$


• Desired ammeter range, $I = 20 \mathrm{mA} = 20 \times 10^{-3} \mathrm{A}$

The shunt resistance, $R_s$, can be calculated using the formula:

$ \frac{R_s}{R_g + R_s} = \frac{I_g}{I} $

Solving for $R_s$:

$ R_s = R_g \left(\frac{I_g}{I - I_g}\right) $

Substituting the given values:

$ R_s = 200 \left(\frac{20 \times 10^{-6}}{20 \times 10^{-3} - 20 \times 10^{-6}}\right) $

Simplifying the expression:

$ R_s = 200 \left(\frac{20 \times 10^{-6}}{19.98 \times 10^{-3}}\right) $

$ R_s \approx 200 \left(\frac{20 \times 10^{-6}}{20 \times 10^{-3}}\right) = 200 \left(\frac{1}{1000}\right) = 0.20 \Omega $

Therefore, the value of the resistance to be added to use the galvanometer as an ammeter of range $(0-20) \mathrm{mA}$ is $0.20 \Omega$. Thus, the correct answer is:

Option C: $0.20 \Omega$

Redrawing the circuit

$\Rightarrow R_{\mathrm{eq}}=6+5+8=19 \Omega$

When an electric kettle is used to heat water, the time taken to boil the water depends on the power of the heating element. The power supplied to the heating element is inversely proportional to the heating time. If we want to reduce the boiling time, the power needs to be increased. The power of the heating element is given by:

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

where $P$ is the power, $V$ is the voltage, and $R$ is the resistance of the heating element. The resistance $R$ of the heating element is proportional to its length $L$ while the material and cross-sectional area remain constant.

So, we can write:

$R \propto L$

To achieve boiling in 15 minutes instead of 20 minutes, the power needs to increase, which implies the resistance must decrease. Let the initial length of the heating element be $L_0$, and let the new length needed be $L$. The time taken to heat is inversely proportional to the power:

$\frac{T_1}{T_2} = \frac{P_2}{P_1}$

Given that:

$T_1 = 20 \text{ minutes}$

$T_2 = 15 \text{ minutes}$

We need to find the ratio:

$\frac{20}{15} = \frac{P_2}{P_1}$

Simplifying:

$\frac{4}{3} = \frac{P_2}{P_1}$

The power is inversely proportional to the resistance:

$\frac{P_2}{P_1} = \frac{R_1}{R_2}$

Thus:

$\frac{4}{3} = \frac{R_1}{R_2}$

Since resistance is proportional to length:

$\frac{R_1}{R_2} = \frac{L_1}{L_2}$

Hence:

$\frac{4}{3} = \frac{L_1}{L_2}$

Simplifying, we find:

$L_2 = \frac{3}{4} L_1$

This means the length of the heating element should be decreased to $\frac{3}{4}$ of its initial length.

Thus, the correct option is:

Option C: decreased, $\frac{3}{4}$

$V_T=\left(\frac{3}{1+2}\right) \times 2=2 \mathrm{~V}$

$\begin{aligned} & P=v \times i \Rightarrow i=\frac{110}{220}=\frac{1}{2} \mathrm{~A} \\ & i=n e \Rightarrow n=\frac{1}{2 \times 1.6 \times 10^{-19}}=31.25 \times 10^{17} \end{aligned}$

Balanced wheatstone bridge.

$\Rightarrow \quad 12 \times 0.5=(x+6) \times \frac{1}{2} \Rightarrow x=6 \Omega$

To determine the ratio of heat dissipated per second through resistances $ 5 \Omega $ and $ 10 \Omega $ in the given parallel circuit, let's break it down step by step.

Understanding the Problem

When resistors are connected in parallel:

1. The voltage across each resistor is the same.


2. The power dissipated in each resistor can be found using the formula:

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

Applying the Power Formula in Parallel Resistors

Given:

• $ R_1 = 5 \Omega $


• $ R_2 = 10 \Omega $

Let $ V $ be the voltage across the resistors.

Power Dissipated in Each Resistor

For the $ 5 \Omega $ resistor:

$ P_{5} = \frac{V^2}{5} $

For the $ 10 \Omega $ resistor:

$ P_{10} = \frac{V^2}{10} $

Finding the Ratio of Powers

Now, let's find the ratio of the power dissipated through the $ 5 \Omega $ resistor to the power dissipated through the $ 10 \Omega $ resistor:

$ \frac{P_{5}}{P_{10}} = \frac{\frac{V^2}{5}}{\frac{V^2}{10}} = \frac{10}{5} = 2 $

So, the ratio of power dissipated through the $ 5 \Omega $ resistor to the $ 10 \Omega $ resistor is $ 2:1 $.

To convert a galvanometer into an ammeter to measure larger currents, a low resistance known as the shunt resistance ($R_{\text{sh}}$) is connected in parallel with the galvanometer. The value of this shunt resistance can be calculated using the principles of parallel circuits and the desired maximum current the ammeter should read.

The original configuration of the galvanometer allows it to measure up to $10\,\text{V}$, and it has a resistance of $100\,\Omega$. When connected in series with a $400\,\Omega$ resistor, the total resistance in the circuit is $100\,\Omega + 400\,\Omega = 500\,\Omega$. Given this configuration measures up to $10\,\text{V}$, we can calculate the maximum current it is designed to measure using Ohm's law:

$I = \frac{V}{R} = \frac{10\,\text{V}}{500\,\Omega} = 0.02\,\text{A}$

Now, to recalibrate the device to measure up to $10\,\text{A}$, we require the calculation of the shunt resistor $R_{\text{sh}}$ that needs to be connected in parallel with the galvanometer. The total current $I$ will now be $10\,\text{A}$, and the part of this current flowing through the galvanometer ($I_g$) remains $0.02\,\text{A}$ (as before, to ensure we do not exceed the device's original maximum measuring capability), leaving the rest to flow through the shunt. Thus, $I - I_g$ flows through the shunt.

Since the voltage drop across both the shunt and the galvanometer must be the same for parallel components, we use Ohm’s Law $V = IR$ for both and set up an equation to calculate $R_{\text{sh}}$:

$I_g R_g = (I - I_g) R_{\text{sh}}$

Substituting known values ($R_g = 100\,\Omega$, $I = 10\,\text{A}$, and $I_g = 0.02\,\text{A}$):

$0.02\,\text{A} \times 100\,\Omega = (10\,\text{A} - 0.02\,\text{A}) R_{\text{sh}}$

This simplifies to:

$2\,\text{V} = 9.98\,\text{A} \times R_{\text{sh}}$

Solving for $R_{\text{sh}}$ gives:

$R_{\text{sh}} = \frac{2\,\text{V}}{9.98\,\text{A}} \approx 0.2004\,\Omega$

Expressing this in terms of $\times 10^{-2} \Omega$ gives $R_{\text{sh}} \approx 20.04 \times 10^{-2} \Omega$. Therefore, the value of $x$ is approximately $20.04$, and rounding it according to the provided options leads to the closest value:

Option B: 20

$\begin{aligned} R_{\mathrm{eq}}= & 12 \Omega \\ \text { and, } I & =\frac{E}{R_{\mathrm{eq}}} \\ & =\frac{12}{12} \\ & =1 \mathrm{~A} \end{aligned}$

To find the power dissipation of the bulb when it's connected to a $100 \mathrm{V}$ supply instead of its rated $200 \mathrm{V}$ supply, we can use the relation between power (P), voltage (V), and resistance (R), which is given by $P = \frac{V^2}{R}$. The resistance of the bulb can be considered constant in this case, allowing us to calculate the change in power dissipation due to the change in voltage.

First, let's find the resistance of the bulb based on its rated conditions:

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

Substituting the rated values, we get:

$R = \frac{(200)^2}{50} = \frac{40000}{50} = 800 \, \Omega$

Now, using this resistance, we can find the power dissipation when the bulb is connected to a $100 \mathrm{V}$ supply:

$P = \frac{V^2}{R} = \frac{(100)^2}{800} = \frac{10000}{800} = 12.5 \, W$

This means the power dissipation of the bulb when connected to a $100 \mathrm{V}$ supply is $12.5 \, W$.

Therefore, the correct answer is Option C: 12.5 W.

To measure the internal resistance of a battery using a potentiometer, we need to understand the principle behind it. The potentiometer is used to measure the voltage across the battery (emf) under different conditions. The balance length corresponds to the emf of the battery when no current is drawn, whereas under load, it corresponds to the terminal voltage.

Let's denote the emf of the battery by $E$ and the internal resistance by $r$. According to Ohm's law, when a resistance $R$ is connected across the battery, the terminal voltage $V$ is given by:

$ V = E - Ir $

where $I$ is the current through the circuit.

From the problem, the balance length $l$ is proportional to the voltage across the potentiometer wire. Thus, we can write:

$ \frac{V_1}{V_2} = \frac{l_1}{l_2} $

In the first condition, when $R = 10 \Omega$ and the balance length $l_1 = 500 \, \text{cm}$:

$ V_1 = E - I_1 r $

For the second condition, when $R = 1 \Omega$ and the balance length $l_2 = 400 \, \text{cm}$:

$ V_2 = E - I_2 r $

Given that:

$ \frac{V_1}{V_2} = \frac{500}{400} = \frac{5}{4} $

Now let's denote the internal emf of the battery as $E$. We can use Ohm’s Law in the calculation of $I_1$ and $I_2$:

$ I_1 = \frac{E}{R_1 + r} = \frac{E}{10 + r} $

$ I_2 = \frac{E}{R_2 + r} = \frac{E}{1 + r} $

Now, rewriting the values of $V_1$ and $V_2$ we have:

$ V_1 = E - I_1 r = E \left(1 - \frac{r}{10 + r}\right) = E \frac{10}{10 + r} $

$ V_2 = E - I_2 r = E \left(1 - \frac{r}{1 + r}\right) = E \frac{1}{1 + r} $

Using the ratio:

$ \frac{V_1}{V_2} = \frac{\frac{10E}{10 + r}}{\frac{E}{1 + r}} = \frac{10 \cdot (1 + r)}{1 \cdot (10 + r)} = \frac{10 + 10r}{10 + r} $

Given:

$ \frac{V_1}{V_2} = \frac{5}{4} $

So, we can set up the equation:

$ \frac{10 + 10r}{10 + r} = \frac{5}{4} $

Cross multiplying gives:

$ 4(10 + 10r) = 5(10 + r) $

Expanding both sides:

$ 40 + 40r = 50 + 5r $

Rearranging terms to solve for $r$:

$ 40r - 5r = 50 - 40 $

$ 35r = 10 $

$ r = \frac{10}{35} = \frac{2}{7} \approx 0.285 \, \Omega $

Since this value is closest to $0.3 \Omega$, the correct option is:

Option B: $0.3 \Omega$

To solve this problem, let's first understand that a meter bridge setup is based on the principle of a Wheatstone bridge, in which two unknown resistances are in such a configuration that if the bridge is balanced, the ratio of the resistances on one side is equal to the ratio of resistances on the other side. In a balanced condition, no current flows through the galvanometer that is connected diagonally across the bridge.

We can express the condition of the balance as follows:

$ \frac{R_1}{R_2} = \frac{L_1}{L_2} $

where $R_1$ is the known resistance $2 Ω$, $R_2$ is the unknown resistance, $L_1$ is the balance length $40 cm$ and $L_2$ is the length of the remaining wire on the meter bridge (100 cm - 40 cm = 60 cm).

Let's calculate the initial unknown resistance ($R_2$) using the balance condition:

$ \frac{2}{R_2} = \frac{40}{60} $

$\Rightarrow R_2 = \frac{2 \times 60}{40} = 3 \Omega $

Now, when the unknown resistance $R_2$ is shunted with a 2 Ω resistor, the new combined resistance ($R'_2$) can be calculated using the parallel resistance formula:

$ \frac{1}{R'_2} = \frac{1}{R_2} + \frac{1}{2} $

$ \Rightarrow \frac{1}{R'_2} = \frac{1}{3} + \frac{1}{2} = \frac{2 + 3}{6} = \frac{5}{6} $

Therefore, the new combined resistance ($R'_2$) is:

$ R'_2 = \frac{6}{5} \Omega $

If $L'_1$ is the new balance length and $L'_2$ is the remaining length, we now have:

$ \frac{2}{\frac{6}{5}} = \frac{L'_1}{100 - L'_1} $

$ \Rightarrow \frac{L'_1}{100 - L'_1} = \frac{5}{3} $

Now, solve for (L'_1):

$ 3L'_1 = 5(100 - L'_1) $

$ \Rightarrow 3L'_1 = 500 - 5L'_1 $

$ \Rightarrow 8L'_1 = 500 $

$ \Rightarrow L'_1 = 62.5 \mathrm{~cm} $

The balance length has changed from 40 cm to 62.5 cm, so the change by $L'_1 - L_1$ is:

$ 62.5 - 40 = 22.5 \mathrm{~cm} $

Therefore, the balance length changes by 22.5 cm, which corresponds to Option B.

When an ammeter is designed, a shunt resistor ($R_s$) is placed in parallel with the galvanometer to ensure that only a small fraction of the total current passes through the galvanometer itself. This is done because the galvanometer is usually a sensitive instrument designed to measure small currents, and passing a large current through it could damage it.

In this scenario, we are told that $5\%$ of the main current passes through the galvanometer. This means that the remaining $95\%$ of the current must pass through the shunt. Let's denote the total current as $I$, the current through the galvanometer as $I_g$, and the current through the shunt as $I_s$. Therefore, we have:

$ I_g = \frac{5}{100} I $

$ I_s = I - I_g = I - \frac{5}{100} I = \frac{95}{100} I $

Since the galvanometer and shunt are in parallel, the voltage across each must be the same:

$ V_g = V_s $

According to Ohm's law, $V = IR$, where $V$ is the voltage, $I$ is the current, and $R$ is the resistance. Hence, for the galvanometer and the shunt:

$ I_g G = I_s R_s $

By substituting $I_g$ and $I_s$ from the above proportionality, we get:

$ \left(\frac{5}{100} I\right) G = \left(\frac{95}{100} I\right) R_s $

Let's solve for $R_s$:

$ R_s = \frac{5}{95} G $

Further simplifying this:

$ R_s = \frac{G}{19} $

The total resistance of the ammeter $R_a$ can be found using the parallel resistance formula:

$ \frac{1}{R_a} = \frac{1}{G} + \frac{1}{R_s} $

Substitute $R_s$ with $\frac{G}{19}$:

$ \frac{1}{R_a} = \frac{1}{G} + \frac{19}{G} $

$ \frac{1}{R_a} = \frac{20}{G} $

Thus the resistance of the ammeter $R_a$ is:

$ R_a = \frac{G}{20} $

Hence, the correct answer is Option C: $\frac{G}{20}$.

To convert a galvanometer into a voltmeter to measure higher voltages, you need to add a series resistance to it. Let's figure out the required resistance value.

First, let's find the maximum voltage that can be directly measured by the galvanometer without any additional resistance. We know the maximum current, $I$, that the galvanometer can safely measure is $5 \text{ mA}$, and the resistance of the galvanometer, $R_g$, is $50 \Omega$.

Using Ohm's law $ V = I \times R $, the maximum voltage $V_g$ the galvanometer can measure is:

$ V_g = I \times R_g $

$ V_g = 5 \times 10^{-3} \text{ A} \times 50 \Omega $

$ V_g = 0.25 \text{ V} $

Next, to measure up to $100 \text{ V}$, we need to add a series resistor $R_s$ so that the total voltage drop when the maximum current is flowing is $100 \text{ V}$. The voltage drop across the additional resistor $R_s$ when the maximum current flows would be the total voltage minus the voltage across the galvanometer:

$ V_s = V_{\text{total}} - V_g $

$ V_s = 100 \text{ V} - 0.25 \text{ V} $

$ V_s = 99.75 \text{ V} $

Applying Ohm's law to the series resistor to find its resistance value:

$ R_s = \frac{V_s}{I} $

$ R_s = \frac{99.75 \text{ V}}{5 \times 10^{-3} \text{ A}} $

$ R_s = 19950 \Omega $

Therefore, the resistance of the series resistor required to convert the galvanometer into a voltmeter that can measure up to $100 \text{ V}$ is $19950 \Omega$.

The correct option is D: $19950 \Omega$.