Current Electricity

Resistance is a physical property of a conductor that opposes current. It is easily measured using an Ohmmeter or by calculating the ratio of voltage to current ( $\mathrm{R}=\mathrm{V} / \mathrm{I}$ ). Hence, option (A) is not the correct option.

The voltage of a single point is representing the electric potential of that point which is defined as the work done to move a unit charge from infinity to a specific point.

Since infinity is an arbitrary reference point, the absolute value of voltage at a single point cannot be measured directly. It is a mathematical construct. So, voltage cannot be measured, i.e., not a measurebale quantity. Hence, option is (D) is the correct option.

Voltage difference (Potential difference) is the difference in electric potential between two specific points. This is measured by a voltmeter. So, it is a measurable quantity. Hence, option (B) is not the correct option.

Displacement current is a quantity appearing in Maxwell's equations, representing a changing electric field (like between the plates of a capacitor). While it isn't a flow of charges like normal current, its magnetic effects are real and can be measured using magnetic sensors. Hence, it is a measurable quantity. So option (C) is not the correct option.

Therefore, the correct option is (D).

Initially : $\mathrm{R}_1=\mathrm{R}_2=\mathrm{R}_3=\mathrm{R}_4=\mathrm{R}$.

After heating $\mathrm{R}_3$ :

• 1. $\mathrm{R}_1=\mathrm{R}$

• 2. $\mathrm{R}_2=\mathrm{R}$

• 3. $\mathrm{R}_3=\mathrm{R}+10 \%$ of $\mathrm{R}=1.1 \mathrm{R}$

• 4. $\mathrm{R}_4=\mathrm{R}$

Supply voltage $\mathrm{V}=40 \mathrm{~V}$.

The voltage across $R_1$ and $R_2$ is 40 V . So the current in this branch is $i_a=\frac{40 \mathrm{~V}}{R_1+R_2}$.

The voltage across $\mathrm{R}_1=\mathrm{V}_1=\mathrm{i}_{\mathrm{a}} \mathrm{R}_1=\left(\frac{40 \mathrm{~V}}{\mathrm{R}_1+\mathrm{R}_2}\right) \mathrm{R}_1=\left(\frac{40 \mathrm{~V}}{\mathrm{R}+\mathrm{R}}\right) \mathrm{R}=20 \mathrm{~V}$

The voltage across $\mathrm{R}_2=\mathrm{V}_2=\mathrm{i}_{\mathrm{a}} \mathrm{R}_2=\left(\frac{40 \mathrm{~V}}{\mathrm{R}_1+\mathrm{R}_2}\right) \mathrm{R}_2=\left(\frac{40 \mathrm{~V}}{\mathrm{R}+\mathrm{R}}\right) \mathrm{R}=20 \mathrm{~V}$

Assuming the potential of negative terminal of battery as 0 V then $\mathrm{v}_{\mathrm{a}}=\mathrm{V}_2=20 \mathrm{~V}$

Similarly for the branch containing $\mathrm{R}_3$ and $\mathrm{R}_4$,

The voltage across $R_3$ and $R_4$ is 40 V . So the current in this branch is $i_b=\frac{40 \mathrm{~V}}{R_3+R_4}$.

The voltage across $R_3=V_3=i_b R_3=\left(\frac{40 \mathrm{~V}}{R_3+R_4}\right) R_3=\left(\frac{40 \mathrm{~V}}{1.1 R+R}\right)(1.1 R)=20.95 \mathrm{~V}$

The voltage across $\mathrm{R}_4=\mathrm{V}_4=\mathrm{i}_{\mathrm{b}} \mathrm{R}_4=\left(\frac{40 \mathrm{~V}}{\mathrm{R}_3+\mathrm{R}_4}\right) \mathrm{R}_4=\left(\frac{40 \mathrm{~V}}{1.1 \mathrm{R}+\mathrm{R}}\right) \mathrm{R}=19.05 \mathrm{~V}$

Assuming the potential of negative terminal of battery as 0 V then $\mathrm{v}_{\mathrm{b}}=\mathrm{V}_4=19.05 \mathrm{~V}$

So, the potential difference between points a and b is,

$ \left(\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}\right)=20 \mathrm{~V}-19.05 \mathrm{~V}=0.95 \mathrm{~V} $

Therefore, the potential difference is 0.95 V . Hence, the correct option is (B).

Case 1 : when cells are in series :

When two identical cells ( $\mathrm{E}, \mathrm{r}$ ) are connected in series with an external resistor $\mathrm{R}=6 \Omega$ :

Total $\mathrm{EMF} \mathrm{E}_{\mathrm{eq}}=\mathrm{E}+\mathrm{E}=2 \mathrm{E}$

Total internal resistance $\mathrm{r}_{\text {eq }}=\mathrm{r}+\mathrm{r}=2 \mathrm{r}$

So, the current in the circuit is,

$ i_s=\frac{E_{e q}}{R+r_{e q}}=\frac{2 E}{6+2 r} $

2. Case 2 : Cells in Parallel

When two identical cells ( $\mathrm{E}, \mathrm{r}$ ) are connected in parallel with an external resistor $\mathrm{R}=6 \Omega$ :

Equivalent EMF ( $\mathrm{E}_{\mathrm{eq}}$ ) in parallel is given as $\frac{\frac{\mathrm{E}_1}{r_1}+\frac{\mathrm{E}_2}{\mathrm{r}_2}+\cdots+\frac{\mathrm{E}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}}}{\frac{1}{\mathrm{r}_1}+\frac{1}{\mathrm{r}_2}+\cdots+\frac{1}{\mathrm{r}_{\mathrm{n}}}}$

Here, $\mathrm{E}_1=\mathrm{E}_2=\mathrm{E}$ and $\mathrm{r}_1=\mathrm{r}_2=\mathrm{r}$,

$ E_{e q}=\frac{\frac{E}{r}+\frac{E}{r}}{\frac{1}{r}+\frac{1}{r}}=\frac{\frac{2 E}{r}}{\frac{2}{r}}=E $

As the internal resistances are in parallel, equivalent internal resistance $\left(\mathrm{r}_{\mathrm{eq}}\right)$ :

$ \frac{1}{\mathrm{r}_{\mathrm{eq}}}=\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}=\frac{2}{\mathrm{r}} \Rightarrow \mathrm{r}_{\mathrm{eq}}=\frac{\mathrm{r}}{2} $

Hence, the current through external resistor is

$ i_p=\frac{E_{e q}}{R+r_{e q}}=\frac{E}{6+\frac{r}{2}} $

According to the question, $\mathrm{i}_{\mathrm{s}}=\mathrm{i}_{\mathrm{p}}$ :

$ \begin{gathered} \frac{2 E}{6+2 r}=\frac{E}{6+\frac{r}{2}} \Rightarrow \frac{1}{6+2 r}=\frac{1}{12+r} \\ \Rightarrow 12+r=6+2 r \Rightarrow 12-6=2 r-r \\ \Rightarrow r=6 \Omega \end{gathered} $

So, the value of internal resistance r is $6 \Omega$.

Hence, the correct option is (B).

Let the current in the primary circuit is $i_p$. If the resistance of potentiometer wire of length $L$ is $R$, then the resistance of the balancing length $l$ will be $R_l=\left(\frac{R}{L}\right) l$

The voltage drop at point $A$ is. $V=i_p R_l=\left(\frac{{ }^i p R}{L}\right) l=k \cdot l$, where $k$ is constant.

the balancing length l is proportional to the terminal potential difference V :

$ \mathrm{V}=\mathrm{k} \cdot \mathrm{l} \ldots \text { (i) } $

When a cell of EMF E and internal resistance $r$ is shunted (connected in parallel) with an external resistance $R$,

The current in the secondary circuit is, $\mathrm{i}_{\mathrm{s}}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}}$

Then terminal voltage is, $\mathrm{V}=\mathrm{iR}=\frac{\mathrm{ER}}{\mathrm{R}+\mathrm{r}} \ldots$ (ii)

From (i) and (ii),

$\mathrm{k} \cdot \mathrm{l}=\frac{\mathrm{E} \cdot \mathrm{R}}{\mathrm{R}+\mathrm{r}}$

$\Rightarrow $ $ l=\frac{1}{k}\left(\frac{R}{R+r}\right) $

Case 1 when the external resistance is $\mathrm{R}_1=4 \Omega$ and the balancing length $\mathrm{l}_1=120 \mathrm{~cm}$

$ 120=\frac{1}{\mathrm{k}}\left[\frac{4}{4+\mathrm{r}}\right] \ldots $

Case 2 when the external resistance is $\mathrm{R}_2=12 \Omega$ and the balancing length $\mathrm{l}_2=180 \mathrm{~cm}$

$ 180=\frac{1}{\mathrm{k}}\left[\frac{12}{12+\mathrm{r}}\right] . . $

Dividing equation (iv) by equation (iii) :

$ \frac{180}{120}=\frac{\frac{12}{12+\mathrm{r}}}{\frac{4}{4+\mathrm{r}}} $

$\Rightarrow $ $\frac{3}{2}=\frac{3 \cdot(4+r)}{12+r}$

$\Rightarrow $ $12+r=2(4+r)$

$\Rightarrow $ $12+r=2(4+r)$

$\Rightarrow $ $ \mathrm{r}=4 \Omega $

Therefore, the internal resistance of the cell is $4 \Omega$. Hence, the correct option is (A).

In a DC circuit, a capacitor reaches a steady state after being connected for a long time. In this state the capacitor is fully charged and it acts as an infinite resistance (an open circuit). The means no current flows through the branch containing the $10 \mu \mathrm{~F}$ capacitor and the $8 \Omega$ resistor connected in series with it.

The two $8 \Omega$ resistors $\mathrm{R}_4$ and $\mathrm{R}_5$ are in parallel.

$ \mathrm{R}_{4,5}=\frac{8 \times 8}{8+8}=\frac{64}{16}=4 \Omega $

This $\mathrm{R}_{4,5}=4 \Omega$ equivalent is in series with the $\mathrm{R}_3=4 \Omega$ resistor.

$ \mathrm{R}_{3,4,5}=4 \Omega+4 \Omega=8 \Omega $

This $\mathrm{R}_{3,4,5}=8 \Omega$ combination is now in parallel with the $\mathrm{R}_2=8 \Omega$ vertical resistor in the middle.

$ \mathrm{R}_{2,3,4,5}=\frac{8 \times 8}{8+8}=4 \Omega $

Finally, this $\mathrm{R}_{2,3,4,5}=4 \Omega$ is in series with the $\mathrm{R}_1=1 \Omega$.

$ \mathrm{R}_{\text {total }}=1 \Omega+4 \Omega=5 \Omega $

Using Ohm's Law for the whole circuit:

$ \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {total }}}=\frac{10}{5} \mathrm{~A}=2 \mathrm{~A}=\mathrm{i}_1 $

The reading of the ammeter is equal to the value of $\mathrm{i}_3$.

Since the resistances $\mathrm{R}_2$ and $\mathrm{R}_{3,4,5}$ are equal to $8 \Omega$, the current splits equally, $\mathrm{i}_2=\mathrm{i}_3$

Using Kirchhoff's current law at junction, $\mathrm{i}_1=\mathrm{i}_2+\mathrm{i}_3=2 \mathrm{i}_2=2 \mathrm{i}_3 \Rightarrow \mathrm{i}_2=\mathrm{i}_3=\frac{\mathrm{i}_1}{2}$

So, the reading of the ammeter is $\mathrm{i}_3=\frac{\mathrm{i}_1}{2}=\frac{\mathrm{i}}{2}=1 \mathrm{~A}$

Therefore, the ammeter reading is 1A in steady state. Hence, the correct option is (B).

To convert a galvanometer into an ammeter of a higher range, a small shunt resistance $(S)$ must be connected in parallel with the galvanometer.

Galvanometer resistance $=\mathrm{G}=100 \Omega$

Full-scale deflection current $=\mathrm{I}_{\mathrm{g}}=1 \mathrm{~mA}=1 \times 10^{-3} \mathrm{~A}$

The required ammeter range $=\mathrm{I}=5 \mathrm{~mA}=5 \times 10^{-3} \mathrm{~A}$

In a parallel circuit, the potential difference across the galvanometer is equal to the potential difference across the shunt:

$ V_g=V_s $

$\Rightarrow $ $\mathrm{I}_{\mathrm{g}} \times \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \times \mathrm{S}$

$\Rightarrow \mathrm{S}=\frac{\mathrm{I}_{\mathrm{g}} \times \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}$

$ \Rightarrow S=\frac{1 \times(100)}{5-1}=25 \Omega $

To measure a current of 5 mA with a $100 \Omega$ galvanometer that deflects at 1 mA , a shunt resistance of $25 \Omega$ is required. Hence, the correct option is (A).

Because the hexagon is regular and the entry/exit points (A and D ) are symmetrical, we can identify points with equal potential relative to the axis AD :

• 1. Points B and F are at the same potential due to symmetry relative to the axis $\mathrm{AD} \Rightarrow \mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{F}}$.

• 2. Points C and E are also at the same potential $\Rightarrow \mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{E}}$.

• 3. The centre point O lies exactly halfway between the entry and exit points on the direct path AOD.

$R_{B{O_1} C}=r+r=2 r$ and this is in parallel with point $B$ and $C$,

$ R_{B C}=\frac{r \times 2 r}{r+2 r}=\frac{2 r}{3} $

Now, $R_{A B}, R_{B C}$ and $R_{C D}$ are in series,

$ R_{A B C D}=r+\frac{2 r}{3}+r=\frac{8 r}{3} $

Similarly, due to symmetry, $\mathrm{R}_{\text {AFED }}=\frac{8 \mathrm{r}}{3}$

$R_{A O}$ and $R_{O D}$ are in series, $R_{O D}=r+r=2 r$

The resulting circuit is,

Now, all are in parallel, so the equivalent resistance between A and D is,

$ \frac{1}{R}=\frac{1}{R_{A B C D}}+\frac{1}{R_{A O B}}+\frac{1}{R_{A F E D}}=\frac{1}{(8 r / 3)}+\frac{1}{(2 r)}+\frac{1}{(8 r / 3)}=\frac{3}{8 r}+\frac{1}{2 r}+\frac{3}{8 r}=\frac{3+4+3}{8 r} $

$\Rightarrow $ $ \frac{1}{R}=\frac{10}{8 r} \Rightarrow R=\frac{4 r}{5} $

Therefore, the equivalent resistance between any two corners opposite to each other is $\frac{4 \mathrm{r}}{5}$.

Hence, the correct option is (B).

The initial circuit is,

When the voltmeter is connected to measure the voltage across one resistor, it is connected in parallel with that resistor.

Resistance of Voltmeter $\mathrm{R}_{\mathrm{V}}=400 \Omega$

The voltmeter $\left(\mathrm{R}_{\mathrm{V}}\right)$ and the resistor $\left(\mathrm{R}_1\right)$ are in parallel.

$ \frac{1}{R_{1, V}}=\frac{1}{R_1}+\frac{1}{R_V} $

$\Rightarrow $ $\frac{1}{R_{1, V}}=\frac{1}{100}+\frac{1}{400}=\frac{5}{400}$

$ \mathrm{R}_{1, \mathrm{~V}}=\frac{400}{5}=80 \Omega $

Now, $R_2$ is in series with the equivalent resistance $\left(R_{1, V}\right)$. So, the total resistance of the circuit is,

$ \mathrm{R}_{\text {total }}=\mathrm{R}_{1, \mathrm{~V}}+\mathrm{R}_2 $

$\Rightarrow $ $ \mathrm{R}_{\text {total }}=100 \Omega+80 \Omega=180 \Omega $

Using Ohm's Law ( $\mathrm{V}=\mathrm{I} \cdot \mathrm{R}$ ) for the entire circuit :

$ \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\text {total }}} $

$\Rightarrow $ $ I=\frac{9}{180}=\frac{1}{20}=0.05 \mathrm{~A} $

The voltmeter measures the potential difference across the parallel section ( $\mathrm{R}_{1, \mathrm{~V}}$ ).

Using Ohm's Law for that specific section :

$ V_{\text {reading }}=I \times R_{1, V} $

$\Rightarrow $ $\mathrm{V}_{\text {reading }}=0.05 \mathrm{~A} \times 80 \Omega$

$\Rightarrow $ $ \mathrm{V}_{\text {reading }}=4 \mathrm{~V} $

Therefore, the voltmeter reading is 4 V. Hence, the correct option is (C).

A meter bridge works on the principle of a Wheatstone bridge. When the bridge is balanced (null point), no current flows through the galvanometer.

The relationship between the resistors in the gaps ( $R_1, R_2$ ) and the balancing lengths ( $1,100-1$ ) is:

$ \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{l}}{100-\mathrm{l}} $

Here, $\mathrm{R}_1=2 \Omega, \mathrm{R}_2=3 \Omega$

Length of the wire from the left end $(\mathrm{X})$ to the null point is $\mathrm{l}_1$ then remaining length of the wire (assuming standard 100 cm wire) is $\left(100-l_1\right)$.

$ \frac{2}{3}=\frac{\mathrm{l}_1}{100-\mathrm{l}_1} $

$\Rightarrow $ $2\left(100-l_1\right)=3 l_1$

$\Rightarrow $ $200-2 \mathrm{l}_1=3 \mathrm{l}_1$

$\Rightarrow $ $ 200=5 l_1 \Rightarrow l_1=40 \mathrm{~cm} $

So, the initial null point is at 40 cm from X.

When an unknown resistor X is connected in parallel with the $3 \Omega$ resistor in the right gap.

The new right gap resistance is $\mathrm{R}_2^{\prime}$.

For two resistors in parallel, the equivalent resistance is:

$ \mathrm{R}_2^{\prime}=\frac{3 \times \mathrm{X}}{3+\mathrm{X}} $

The null point shifts by 22.5 cm towards Y (the right end).

$ \mathrm{l}_2=\mathrm{l}_1+22.5 \mathrm{~cm} $

$\Rightarrow $ $ \mathrm{l}_2=40+22.5=62.5 \mathrm{~cm} $

Using balanced condition,

$ \frac{\mathrm{R}_1}{\mathrm{R}_2^{\prime}}=\frac{\mathrm{l}_2}{100-\mathrm{l}_2} $

$\Rightarrow $ $\frac{2}{\mathrm{R}_2^{\prime}}=\frac{62.5}{100-62.5}$

$\Rightarrow $ $\frac{2}{\mathrm{R}_2^{\prime}}=\frac{62.5}{37.5}=\frac{5}{3}$

$ \mathrm{R}_2^{\prime}=\frac{6}{5}=1.2 \Omega $

We know that $R_2^{\prime}$ is the parallel combination of $3 \Omega$ and $X$.

$ 1.2=\frac{3 X}{3+X} \Rightarrow 1.2(3+X)=3 X $

$\Rightarrow 3.6+1.2 \mathrm{X}=3 \mathrm{X}$

$\Rightarrow 3.6=3 X-1.2 X$

$\Rightarrow 3.6=1.8 \mathrm{X}$

$ \Rightarrow \mathrm{X}=2 \Omega $

Therefore, the value of X is $2 \Omega$. Hence, the correct option is (D).

The EMF (E) of a cell in a potentiometer is directly proportional to its balancing length (1).

Therefore, the ratio of the EMFs of two cells is given by :

$ \mathrm{R}=\frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{l}_1}{\mathrm{l}_2} $

Balancing length for first cell $\mathrm{l}_1=200 \mathrm{~cm}$

Balancing length for second cell $\mathrm{l}_2=150 \mathrm{~cm}$

Absolute error in measurement $\Delta \mathrm{l}_1=\Delta \mathrm{l}_2=1 \mathrm{~cm}$ (the least count of the scale)

For a ratio $\mathrm{R}=\frac{\mathrm{l}_1}{\mathrm{l}_2}$, the relative error $\left(\frac{\Delta \mathrm{R}}{\mathrm{R}}\right)$ is the sum of the relative errors of the individual components :

$ \frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \mathrm{l}_1}{\mathrm{l}_1}+\frac{\Delta \mathrm{l}_2}{\mathrm{l}_2} $

Percentage Error $=\left(\frac{\Delta \mathrm{l}_1}{\mathrm{l}_1}+\frac{\Delta \mathrm{l}_2}{\mathrm{l}_2}\right) \times 100=\left(\frac{1}{200}+\frac{1}{150}\right) \times 100 \approx 1.17 \%$

The calculated percentage error is approximately $1.17 \%$.

For the equivalent resistance between points A and B we can identify three distinct paths for the current to flow from node A to node B. These paths are in parallel.

The resistance per unit for the wire is $\lambda \Omega / \mathrm{m}$

The resistance is given by $\mathrm{R}=$ Resistance per unit length × Length $=\lambda \times \mathrm{L}$.

Path 1 (Minor Arc AB):

The shorter curved wire along the circle's circumference. Since $\angle \mathrm{AOB}=90^{\circ}$ (a quarter circle), the length is $\frac{1}{4}$ of the circumference.

$ \mathrm{L}_1=\frac{1}{4} \times 2 \pi \mathrm{r}=\frac{\pi \mathrm{r}}{2} $

So, the resistance of path 1 is,

$ \mathrm{R}_1=\lambda\left(\frac{\pi \mathrm{r}}{2}\right) $

Path 2 (Major Arc AOB):

The longer curved wire along the remaining part of the circle. This is $\frac{3}{4}$ of the circumference.

$ \mathrm{L}_2=\frac{3}{4} \times 2 \pi \mathrm{r}=\frac{3 \pi \mathrm{r}}{2} $

So, the resistance of path 2 is,

$ \mathrm{R}_2=\lambda\left(\frac{3 \pi \mathrm{r}}{2}\right) $

Path 3 (Wire AOB):

The straight wire sections connecting A to the centre O and then O to B . Thus, the two segments AO and OB are in series. It is given that the length of this wire is $2 r$.

$ \mathrm{L}_3=\mathrm{r}+\mathrm{r}=2 \mathrm{r} $

So, the resistance of path 3 is,

$ \mathrm{R}_{34}=\mathrm{R}_3+\mathrm{R}_4=\lambda(2 \mathrm{r}) $

Since the three paths are in parallel, the equivalent resistance $\left(R_{e q}\right)$ is,

$ \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_{34}} $

$\Rightarrow $ $\frac{1}{R_{e q}}=\frac{1}{\lambda \pi r / 2}+\frac{1}{3 \lambda \pi r / 2}+\frac{1}{2 \lambda r}$

$\Rightarrow $ $\frac{1}{R_{e q}}=\frac{2}{\lambda \pi r}+\frac{2}{3 \lambda \pi r}+\frac{1}{2 \lambda r}$

$\Rightarrow $ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\lambda \mathrm{r}}\left[\frac{2}{\pi}+\frac{2}{3 \pi}+\frac{1}{2}\right]$

$\Rightarrow $ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\lambda \mathrm{r}}\left[\frac{12+4+3 \pi}{6 \pi}\right]$

$\Rightarrow $ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\lambda \mathrm{r}}\left[\frac{16+3 \pi}{6 \pi}\right]$

$\Rightarrow $ $ \mathrm{R}_{\mathrm{eq}}=\lambda \mathrm{r}\left[\frac{6 \pi}{16+3 \pi}\right]=\frac{6 \lambda \pi \mathrm{r}}{3 \pi+16} $

Therefore, the equivalent resistance between points A and B is $\frac{6 \lambda \pi \mathrm{r}}{3 \pi+16}$.

Hence, the correct option is (B).

The resistance of the transmission line, $\mathrm{R}=2 \Omega$

The power delivered (Output Power), $\mathrm{P}_{\text {out }}=1 \mathrm{~kW}=1000 \mathrm{~W}$

The given power is delivered at voltage, $\mathrm{V}=250 \mathrm{~V}$

The power delivered to the load is the product of the voltage across the load and the current flowing through it.

$ \mathrm{P}_{\mathrm{out}}=\mathrm{V} \cdot \mathrm{I} \Rightarrow \mathrm{I}=\frac{\mathrm{P}_{\mathrm{out}}}{\mathrm{~V}} $

$ \Rightarrow I=\frac{1000}{250} A=4 A $

So, a current of 4 A flows through the transmission line.

As the current flows through the transmission wires, some electrical energy is lost as heat due to the line's resistance. This power loss ( $\mathrm{P}_{\text {loss }}$ ) is calculated using Joule's heating formula :

$ \mathrm{P}_{\text {loss }}=\mathrm{I}^2 \cdot \mathrm{R} $

$\Rightarrow $ $ \mathrm{P}_{\mathrm{loss}}=(4)^2 \times 2 \mathrm{~W}=32 \mathrm{~W} $

The total power supplied at the source (Input Power, $\mathrm{P}_{\mathrm{in}}$ ) must equal the power successfully delivered to the load plus the power lost in the transmission line.

$ \mathrm{P}_{\mathrm{in}}=\mathrm{P}_{\mathrm{out}}+\mathrm{P}_{\text {loss }} $

$\Rightarrow $ $\mathrm{P}_{\mathrm{in}}=1000+32$

$\Rightarrow $ $ \mathrm{P}_{\mathrm{in}}=1032 \mathrm{~W} $

The efficiency $(\eta)$ of the transmission line is the ratio of the useful power delivered to the total power supplied, expressed as a percentage :

$ \eta=\left(\frac{P_{\text {out }}}{P_{\text {in }}}\right) \times 100 \% $

$\Rightarrow $ $ \eta=\left(\frac{1000}{1032}\right) \times 100 \% \approx 96.9 \% $

Therefore, the percentage efficiency of the transmission line is $96.9 \%$. Hence, the correct option is (A).

A meter bridge operates on the principle of a Wheatstone bridge.

Let the total length of the wire be $\mathrm{L}=100 \mathrm{~cm}$.

If the null point is found at a distance l from the left end, the length of the right segment is $(100-\mathrm{l})$.

The balancing condition is given by:

$ \frac{\mathrm{R}_{\text {left gap }}}{\mathrm{R}_{\text {right gap }}}=\frac{\mathrm{l}}{100-\mathrm{l}} $

The resistance in left gap is $R_{\text {left gap }}=R_1$ and the resistance in the right gap is $R_{\text {right gap }}=R_2$

Case 1

We are given that initially; the bridge is balanced at a distance of $\mathrm{l}_1=40 \mathrm{~cm}$ from point P .

The remaining length is $100-40=60 \mathrm{~cm}$.

Applying the balancing condition :

$ \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{40}{60} $

$\Rightarrow $ $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{2}{3}$

$\Rightarrow $ $ \mathrm{R}_1=\frac{2}{3} \mathrm{R}_2 \ldots (i)$

Case 2

When a $16 \Omega$ resistance is connected in parallel to $R_2$.

So, the new resistance in the right gap be $\mathrm{R}_2^{\prime}$ :

$ \frac{1}{\mathrm{R}_2^{\prime}}=\frac{1}{\mathrm{R}_2}+\frac{1}{16} \Rightarrow \mathrm{R}_2^{\prime}=\frac{16 \mathrm{R}_2}{16+\mathrm{R}_2} $

With this new resistance, the null point shifts to $\mathrm{l}_2=50 \mathrm{~cm}$ from point P .

The remaining length is now $100-50=50 \mathrm{~cm}$.

Applying the balancing condition again,

$ \frac{\mathrm{R}_1}{\mathrm{R}_2^{\prime}}=\frac{50}{50}=1 $

$ \mathrm{R}_1=\mathrm{R}_2^{\prime} \ldots (ii)$

Substituting the expression for $\mathrm{R}_2^{\prime}$ into equation (ii):

$ R_1=\frac{16 R_2}{R_2+16} \ldots \text { (iii) } $

Now, substituting the expression for $R_1$ from equation into equation (iii) :

$ \frac{2}{3} R_2=\frac{16 R_2}{R_2+16} $

$\Rightarrow $ $\frac{1}{3}=\frac{8}{R_2+16} \Rightarrow R_2+16=24$

$\Rightarrow $ $ \mathrm{R}_2=8 \Omega $

And, $\mathrm{R}_1=\frac{2}{3} \times 8 \Omega=\frac{16}{3} \Omega$

Therefore, the values are $\mathrm{R}_2=8 \Omega$ and $\mathrm{R}_1=\frac{16}{3} \Omega$.

Hence, the correct option is (A).

Resistors 2R and $X$ are connected in series.

$ \mathrm{R}_1=2 \mathrm{R}+\mathrm{X} $

And the combination of $(2 R+X)$ is connected across $R$ in parallel.

$ \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}}=\frac{1}{2 \mathrm{R}+\mathrm{X}}+\frac{1}{\mathrm{R}} $

$\Rightarrow $ $\frac{1}{X}=\frac{R+(2 R+X)}{R(2 R+X)}$

$\Rightarrow $ $X(3 R+X)=R(2 R+X)$

$\Rightarrow $ $3 R X+X^2=2 R^2+R X$

$\Rightarrow $ $ X^2+2 R X-2 R^2=0 $

Using the quadratic equation formula,

$ X=\frac{-2 R \pm \sqrt{(2 R)^2-4(1)\left(-2 R^2\right)}}{2 \times 1} $

$\Rightarrow $ $X=\frac{-2 R \pm \sqrt{4 R^2+8 R^2}}{2}$

$\Rightarrow $ $X=\frac{-2 R \pm \sqrt{12 R^2}}{2}$

$\Rightarrow $ $X=\frac{-2 R \pm 2 \sqrt{3} R}{2}$

$\Rightarrow $ $X=-R \pm \sqrt{3} R$

$\Rightarrow $ $ \mathrm{X}=(-1 \pm \sqrt{3}) \mathrm{R} $

Since resistance cannot be negative, we take the positive root:

$ \mathrm{X}=(\sqrt{3}-1) \mathrm{R} $

Hence, the correct option is (C).

When the galvanometer (G) shows a zero reading, it means no current is flowing through the galvanometer branch. This occurs when the circuit is in a balanced state.

Resistance $\mathrm{R}_1=6 \Omega$.

Resistance $\mathrm{R}_2=4 \Omega$.

Let the length be l . Its resistance is $\mathrm{R}_{\mathrm{AP}}$.

Since the total length AB is 50 cm , the length of PB is ( $50-\mathrm{l}$ ). Its resistance is $\mathrm{R}_{\mathrm{PB}}$.

The resistance R of a uniform wire is given by the formula :

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

Where $\rho$ is resistivity, L is length, and A is the cross-sectional area. Since the wire is uniform, the resistance is directly proportional to the length ( $\mathrm{R} \propto \mathrm{L}$ ).

$ \mathrm{R}_{\mathrm{AP}}=\mathrm{k} \cdot \mathrm{l} $

$\Rightarrow $ $ \mathrm{R}_{\mathrm{PB}}=\mathrm{k} \cdot(50-\mathrm{l}) $

where $\mathrm{k}=\frac{\rho}{\mathrm{A}}$ is a constant.

For a balanced bridge:

$ \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{R}_{\mathrm{AP}}}{\mathrm{R}_{\mathrm{PB}}} $

Substituting the known values :

$ \frac{6}{4}=\frac{\mathrm{k} \cdot \mathrm{l}}{\mathrm{k} \cdot(50-\mathrm{l})} $

$\Rightarrow $ $\frac{3}{2}=\frac{1}{50-1}$

$\Rightarrow $ $3(50-l)=2 l$

$\Rightarrow $ $150-3 l=2 l$

$\Rightarrow $ $150=5l$

$\Rightarrow $ $ \mathrm{l}=\frac{150}{5}=30 \mathrm{~cm} $

Therefore, the length of AP is 30 cm . Hence, the correct option is (B).

The total resistance of the circuit is $(R+r)$.

Using Ohm's Law, the current I flowing through the circuit is:

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

The power consumption P in the external resistor R is given by the formula:

$ \mathrm{P}=\mathrm{I}^2 \mathrm{R} $

Substituting the expression for I:

$ P=\left(\frac{E}{R+r}\right)^2 R=\frac{E^2 R}{(R+r)^2} $

To find the value of R for which P is maximum, we differentiate P with respect to R and set the derivative to zero ( $\frac{\mathrm{dP}}{\mathrm{dR}}=0$ ).

$ \frac{d P}{d R}=E^2\left[\frac{(R+r)^2 1-R \cdot 2(R+r)}{(R+r)^4}\right]=0 $

$\Rightarrow $ $(R+r)^2-2 R(R+r)=0$

$\Rightarrow $ $(R+r)-2 R=0$

$\Rightarrow $ $r-R=0$

$\Rightarrow $ $ \mathrm{R}=\mathrm{r} $

Therefore, the power consumption in the external load R is at its maximum when the external resistance is equal to the internal resistance of the battery, $\mathrm{r}=\mathrm{R}$.

Hence, the correct option is (D).

For a voltmeter, to increase its range, we connect a high resistance in series. Let us calculate it.

The voltmeter has internal resistance $x \, \Omega$ and can measure up to $20 \, \text{V}$.

So, the full-scale current of the voltmeter is

$ I = \frac{V}{R} = \frac{20}{x} $

Now we want the new range to be $30 \, \text{V}$.

Let the additional resistance connected in series be $R$.

Then for the same full-scale current,

$ 30 = I(x + R) $

Substitute $I = \frac{20}{x}$:

$ 30 = \frac{20}{x}(x + R) $

$ 30x = 20(x + R) $

$ 30x = 20x + 20R $

$ 10x = 20R $

$ R = \frac{x}{2} $

So, we must connect a resistor of $\frac{x}{2}\,\Omega$ in series with the voltmeter.

Therefore, the correct option is:

$ \boxed{\text{Option A}} $

First, find the resistance of the bulb using its rating.

Given for the bulb:

• Voltage, $V = 200\ \text{V}$

• Power, $P = 100\ \text{W}$

We know,

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

So,

$ R_b = \frac{200^2}{100} = \frac{40000}{100} = 400\ \Omega $

So the bulb has resistance $400\ \Omega$.

Now the bulb is connected across the $400\ \Omega$ resistor, so these two are in parallel.

Hence we have:

• one resistor of $200\ \Omega$ in series with

• parallel combination of $400\ \Omega$ and $400\ \Omega$

Now calculate the equivalent resistance of the parallel part:

$ R_p = \frac{400 \times 400}{400 + 400} = \frac{160000}{800} = 200\ \Omega $

So total resistance of the circuit is:

$ R_{\text{total}} = 200 + 200 = 400\ \Omega $

Now total current from the battery:

$ I = \frac{V}{R_{\text{total}}} = \frac{100}{400} = 0.25\ \text{A} $

The voltage drop across the series $200\ \Omega$ resistor is:

$ V_1 = IR = 0.25 \times 200 = 50\ \text{V} $

Therefore, the remaining voltage across the parallel combination is:

$ V_p = 100 - 50 = 50\ \text{V} $

Since the bulb is connected in parallel with the $400\ \Omega$ resistor, the potential difference across the bulb is also:

$ \boxed{50\ \text{V}} $

So, the correct option is B.

As $r=\frac{R}{6}$

(As balanced wheat stone bridge is formed) Now, Equivalent resistance between A and B can be written as

$\begin{aligned} & \frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{2 \mathrm{r}}+\frac{1}{2 \mathrm{r}}+\frac{1}{\mathrm{r}}=\frac{2}{\mathrm{r}} \\ & \mathrm{R}_{\mathrm{AB}}=\frac{\mathrm{R}}{12} \end{aligned}$

$\begin{aligned} & \mathrm{R}_{\mathrm{s}}=\mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3+\ldots \ldots \ldots+\mathrm{R}_{\mathrm{n}} \\ & \frac{\mathrm{~V}^2}{\mathrm{P}_{\mathrm{s}}}=\frac{\mathrm{V}^2}{\mathrm{P}}+\frac{\mathrm{V}^2}{\mathrm{P}}+\ldots \ldots . .+\frac{\mathrm{V}^2}{\mathrm{P}_{\mathrm{n}}} \\ & \mathrm{P}_{\mathrm{s}}=\frac{\mathrm{P}}{\mathrm{n}} \end{aligned}$

$\frac{1}{\mathrm{R}_{\mathrm{P}}}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{1}{6}$

To find the total charge $ q $ that flows through the wire from $ t = 1 \, \text{s} $ to $ t = 2 \, \text{s} $, we can integrate the current function $ I(t) = 0.02t + 0.01 \, \text{A} $ over this interval.

The formula for the charge $ q $ flowing through the wire is given by the integral of the current with respect to time:

$ q = \int I(t) \, dt $

We integrate the function from $ t = 1 $ to $ t = 2 $:

$ q = \int_1^2 (0.02t + 0.01) \, dt $

To solve this, first find the antiderivative:

$ q = \left[ 0.02 \frac{t^2}{2} + 0.01t \right]_1^2 $

Calculate the expression at the upper and lower bounds:

$ = \left[ 0.01t^2 + 0.01t \right]_1^2 $

Plug in the values:

For $ t = 2 $:

$ = 0.01(2)^2 + 0.01(2) = 0.04 + 0.02 = 0.06 $

For $ t = 1 $:

$ = 0.01(1)^2 + 0.01(1) = 0.01 + 0.01 = 0.02 $

Subtract the result at $ t=1 $ from the result at $ t=2 $:

$ q = 0.06 - 0.02 = 0.04 \, \text{C} $

Therefore, the charge that flows through the wire from $ t = 1 \, \text{s} $ to $ t = 2 \, \text{s} $ is $ 0.04 \, \text{C} $.

The motor operates at 100 V and draws 1 A, so the total electrical power input is:

$P_{\text{in}} = IV = 100 \, \text{V} \times 1 \, \text{A} = 100 \, \text{W}.$

With an efficiency of 91.6%, only 91.6% of the input power is used for the motor's work, while the rest is lost as heat. The output power is:

$P_{\text{out}} = 0.916 \times 100 \, \text{W} = 91.6 \, \text{W}.$

The power lost is:

$P_{\text{loss}} = P_{\text{in}} - P_{\text{out}} = 100 \, \text{W} - 91.6 \, \text{W} = 8.4 \, \text{W}.$

Since $1 \, \text{W} = 1 \, \text{J/s}$ and $1 \, \text{cal} \approx 4.2 \, \text{J}$, we convert the lost power into calories per second:

$P_{\text{loss}} (\text{in cal/s}) = \frac{8.4 \, \text{J/s}}{4.2 \, \text{J/cal}} = 2 \, \text{cal/s}.$

Thus, the correct answer is:

Option B: 2

Length of the wire, $L = 25\rm\,m$

Cross‐sectional area, $A = 5\rm\,mm^2 = 5\times10^{-6}\,m^2$

Resistivity, $\rho = 2\times10^{-6}\,\Omega\cdot m$

Resistance of the entire wire:

$R_{\rm total}=\frac{\rho\,L}{A} =\frac{2\times10^{-6}\times25}{5\times10^{-6}} =\frac{50\times10^{-6}}{5\times10^{-6}} =10\ \Omega$

When bent into a circle, points diametrically opposite divide the ring into two equal halves.

Each half-circle has length $L/2 = 12.5\rm\,m$, so its resistance is half of $R_{\rm total}$:

$R_{\rm half}=\frac{R_{\rm total}}{2} =\frac{10}{2} =5\ \Omega$

These two $5\,\Omega$ halves are in parallel between the opposite points.

Equivalent resistance $R$ is given by:

$\frac{1}{R}=\frac{1}{R_{\rm half}}+\frac{1}{R_{\rm half}} =\frac{1}{5}+\frac{1}{5} =\frac{2}{5}$

$R=\frac{5}{2}=2.5\ \Omega$

Answer: 2.5 Ω (Option B)

To find the energy stored in the battery, follow these steps:

Convert mAh to Ah:

The battery capacity is given as 5800 mAh. Since

$1\,\text{Ah} = 1000\,\text{mAh},$

we have

$5800\,\text{mAh} = 5.8\,\text{Ah}.$

Calculate energy in watt-hours (Wh):

The energy in Wh is given by the product of the voltage and the capacity in Ah. Thus,

$\text{Energy (Wh)} = 4.2\,\text{V} \times 5.8\,\text{Ah} = 24.36\,\text{Wh}.$

Convert watt-hours to joules (J):

Since

$1\,\text{Wh} = 3600\,\text{J},$

the total energy in joules is:

$24.36\,\text{Wh} \times 3600\,\text{J/Wh} = 87,696\,\text{J} \approx 87.7\,\text{kJ}.$

Thus, the energy stored in the battery when fully charged is about 87.7 kJ.

The correct answer is Option A: 87.7 kJ.

Assertion (A):

A choke coil is simply a coil having large inductance but small resistance.

Choke coils are used with fluorescent mercury-tube fittings.

If household electric power is directly connected to a mercury tube, the tube will be damaged.

A choke (or ballast) does have high inductance and low resistance.

In a fluorescent lamp (or mercury‐vapor lamp) circuit, the choke limits the current through the tube.

If we connect the tube directly to the full line voltage without any current-limiting device, excessive current will flow and damage the tube.

Therefore, (A) is true.

Reason (R):

By using the choke coil, the voltage across the tube is reduced by a factor

$\displaystyle \frac{R}{\sqrt{R^2 + (\omega L)^2}}$,

where $\omega$ is the supply frequency, and $R$ and $L$ are the resistance and inductance in series.

If the choke coil were not used, the voltage across the resistor (tube) would be the same as the applied voltage.

In a simple series R–L circuit (where $R$ represents the lamp’s effective resistance once conducting, and $L$ is the choke’s inductance), the total impedance is

$ Z = \sqrt{\,R^2 + (\omega L)^2\,}. $

The current is the same through $R$ and $L$. The fraction of the total supply voltage appearing across $R$ alone is

$ \frac{V_R}{V_{\text{total}}} \;=\; \frac{I\,R}{I\,Z} \;=\; \frac{R}{\sqrt{R^2 + (\omega L)^2}}. $

This shows that using a large inductance $L$ (the choke) can substantially reduce (or limit) the voltage—and hence the current—through the lamp.

Without the choke coil, the tube (treated here as a resistor once it starts conducting) would indeed see nearly the full line voltage, causing a large current that could destroy it.

Therefore, (R) is also true.

Does (R) correctly explain (A)?

(A) states why a choke coil is needed (to prevent the lamp from being damaged by excessive current).

(R) shows how the choke coil limits the voltage/current through the lamp—by giving the fraction of the supply voltage that appears across the lamp in a series R–L circuit.

Thus, (R) does provide the correct (physics-based) explanation: the choke’s inductive reactance drops a significant portion of the supply voltage, thereby protecting the lamp.

Hence, the correct option is:

$ \boxed{\text{Option B: Both (A) and (R) are true and (R) is the correct explanation of (A).}} $

$\mathrm{V}=\mathrm{V}_0 \sin \omega \mathrm{t}$

Here, ${V_A} = {V_C}$ & ${V_B} = {V_D}$

So,

So resistances are in parallel combination.

We know, ${1 \over {{R_p}}} = {1 \over {{R_1}}} + {1 \over {{R_2}}} + {1 \over {{R_3}}}$

$ \Rightarrow {1 \over {{R_p}}} = {3 \over r} + {3 \over r} + {3 \over r} = {9 \over r}$

$ \Rightarrow {R_p} = {r \over 9}$

We know, $R = \rho {l \over A} \Rightarrow R \propto l$

Side length of triangle $ = {l \over 3}$

Side length of square $ = {l \over 4}$

So,

$ \Rightarrow {\left( {{R_{eq}}} \right)_1} = {{{{2R} \over 3} \times {R \over 3}} \over {{{2R} \over 3} + {R \over 3}}}$ and ${\left( {{R_{eq}}} \right)_2} = {{{{3R} \over 4} \times {R \over 4}} \over {{{3R} \over 4} + {R \over 4}}}$

$ = {{2{R^2}} \over 9} \times {3 \over {3R}} = {{2R} \over 9}$ and ${\left( {{R_{eq}}} \right)_2} = {{3{R^2}} \over {16}} - R = {{3R} \over {16}}$

Hence, ${{{{\left( {{R_{eq}}} \right)}_1}} \over {{{\left( {{R_{eq}}} \right)}_2}}} = {{{{2R} \over 9}} \over {{{3R} \over {16}}}} = {{32} \over {27}}$

Given:

Galvanometer resistance:

$ R_g = 30 \, \Omega $

Full-scale deflection current:

$ I_g = 20 \, mA = 0.02 \, A $

Maximum current to be measured:

$ I = 3 \, A $

When using the galvanometer to measure 3 A, the extra current passing through the shunt resistor ($R_s$) is:

$ I_s = I - I_g = 3 - 0.02 = 2.98 \, A. $

Since the galvanometer and the shunt resistor are connected in parallel, their voltage drops must be equal. Therefore:

$ I_g R_g = I_s R_s. $

Substitute the given values:

$ 0.02 \times 30 = 2.98 \times R_s \quad \Rightarrow \quad R_s = \frac{0.6}{2.98} \approx 0.2013 \, \Omega. $

The shunt resistance is represented as:

$ R_s = \frac{30}{X} \, \Omega. $

Equate the two expressions:

$ \frac{30}{X} = 0.2013. $

Solving for $X$:

$ X = \frac{30}{0.2013} \approx 149. $

Thus, the value of $X$ is approximately $149.$

Let's analyze each statement step by step.

Statement A: “The torsional constant in moving coil galvanometer has dimensions $[ML^2T^{-2}]$.”

The torsional constant (often denoted by $k$) relates the restoring torque to the angular displacement via $\tau = k \theta$. Since torque has dimensions of force × distance with dimensions $[MLT^{-2} \times L = ML^2T^{-2}]$ and the angular displacement (in radians) is dimensionless, the torsional constant indeed has the dimensions $[ML^2T^{-2}]$.

Thus, Statement A is correct.

Statement B: “Increasing the current sensitivity may not necessarily increase the voltage sensitivity.”

The current sensitivity of a galvanometer is defined as its deflection per unit current (i.e. $\theta/I$).

The voltage sensitivity, on the other hand, is the deflection per unit applied voltage when the galvanometer is used as a voltmeter. This sensitivity is proportional to the current sensitivity divided by the galvanometer’s internal resistance.

Often, if you try to increase current sensitivity (for instance, by increasing the number of turns or enhancing the magnetic field), the internal resistance may also change (typically increase when number of turns is increased). Therefore, the improvement in current sensitivity does not automatically translate to a proportional increase in voltage sensitivity.

Hence, Statement B is true.

Statement C: “If we increase number of turns ($N$) to its double ($2N$), then the voltage sensitivity doubles.”

The deflection (current sensitivity) is proportional to $N$ since $\theta \propto N I$.

However, when you double $N$, the length of the wire increases, and so does the internal resistance (approximately doubling it, assuming the wire diameter and material remain the same).

Since voltage sensitivity is roughly given by $\frac{\text{current sensitivity}}{\text{internal resistance}} \propto \frac{N}{R_g}$, if both $N$ and $R_g$ double, the ratio remains essentially unchanged.

Therefore, doubling $N$ does not double the voltage sensitivity, so Statement C is incorrect.

Statement D: “MCG can be converted into an ammeter by introducing a shunt resistance of large value in parallel with galvanometer.”

To convert a galvanometer into an ammeter, a shunt resistor is used to bypass most of the current so that only a small fraction (the full‐scale deflection current) passes through the galvanometer.

For this to work, the shunt resistance should be very low compared to the galvanometer’s resistance, not of “large value.”

Therefore, Statement D is false.

Statement E: “Current sensitivity of MCG depends inversely on number of turns of coil.”

As mentioned before, the electromagnetic torque in the coil is given by $\tau_e \propto N I$, so the deflection per unit current is $\theta/I \propto N$.

This shows that the current sensitivity increases with $N$ (i.e. it is directly proportional to $N$), not inversely.

Hence, Statement E is false.

Only Statements A and B are correct.

Looking at the given options, the correct choice is:

Option B: A, B Only.

When two non-ideal batteries are connected in parallel, their equivalent emf is

${E_{eq}} = {{{{{E_1}} \over {{r_1}}} + {{{E_2}} \over {{r_2}}}} \over {{1 \over {{r_1}}} + {1 \over {{r_2}}}}} \Rightarrow {E_{eq}} = {{{E_1}{r_2} + {E_2}{r_1}} \over {{r_1} + {r_2}}}$

If emf of both cells is same say E,

${E_{eq}} = {{E{r_2} + E{r_1}} \over {{r_1} + {r_2}}} = {{E({r_1} + {r_2})} \over {({r_1} + {r_2})}} = E$

Hence, equivalent emf is equal to emf of the single cell.

Also, when two non-ideal batteries are connected in parallel, their equivalent resistance is ${r_{eq}} = {{{r_1}{r_2}} \over {{r_1} + {r_2}}}$

Hence, the equivalent internal resistance is smaller than either of the two internal resistances. Therefore 1 is correct.

We use Nichrome, manganin and constantan to make standard (reference) wire bound resistors their resistivity is less dependent on temperature. So $\rho$ vs T curve will not show any sharp change.

Hence, option 4 is correct.

We know, $R = \int {{l \over A} \Rightarrow R \propto l} $

As sliding contact of a potentiometer is in the middle of the potentiometer wire.

$\therefore$ ${R_{eq}} = 0.5 + {{0.5 \times 2} \over {0.5 + 2}} = 0.5 + {1 \over {2.5}}$

$ = {1 \over 2} + {2 \over 5} = {{5 + 4} \over {10}} = {9 \over {10}} = 0.9\Omega $

Hence, $I = {v \over {{R_{eq}}}} = {{0.9v} \over {0.9\Omega }} \Rightarrow I = 1A$

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}}} $