Current Electricity

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)

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

Series :

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

Parallel :

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

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

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

Neglecting small terms

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

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

From (i) & (ii), we get

$W^{\prime}=4 W$

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

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

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

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

Current in resistance

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

Let x = current/division

After applying shunt

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

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

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

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

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

Where:

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

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

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

Substituting these values into the formula:

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

Perform the calculations:

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

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

Thus, the value of the shunt resistor should be:

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

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

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

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

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

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

This integral can be solved in two parts:

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

$\int 20 \, dt = 20t$

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

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

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

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

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

$Q = 400 + 600 = 1000$ C

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

From KVL,

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

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

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

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

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

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

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

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

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

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

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

Option D: $S_1$

For null point,

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

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

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