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).

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

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: $19950 \Omega$.

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

% change in power $=$

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

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

From (i) and (ii)

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