The Moving Coil Galvanometer (Sensitivity) and Ammeter and Voltmeter Conversion Questions in English

(A) Let $G$ be the resistance of the galvanometer and $I_g$ be the full-scale deflection current.
The initial voltage range is $V = I_g G$.
To increase the range by a factor of $n$,the new voltage range becomes $V' = nV = n I_g G$.
Let $R$ be the series resistance connected to the galvanometer.
The total resistance of the voltmeter is $(G + R)$.
According to Ohm's law,$V' = I_g(G + R)$.
Substituting $V' = n I_g G$,we get $n I_g G = I_g(G + R)$.
Dividing both sides by $I_g$,we get $nG = G + R$.
Therefore,$R = nG - G = G(n - 1)$.

(N/A) Voltage sensitivity of a galvanometer is defined as the deflection produced per unit voltage applied across it.
It is denoted by $V_s$.
If a voltage $V$ is applied across a galvanometer and it produces a deflection $\theta$,then voltage sensitivity is given by:
$V_s = \frac{\theta}{V}$
Since $V = I R_g$,where $I$ is the current and $R_g$ is the resistance of the galvanometer,we can write:
$V_s = \frac{\theta}{I R_g}$
Since current sensitivity $I_s = \frac{\theta}{I}$,the equation becomes:
$V_s = \frac{I_s}{R_g}$
Where $I_s = \frac{NAB}{k}$ ($N$ is the number of turns,$A$ is the area,$B$ is the magnetic field,and $k$ is the restoring torque per unit twist).

(B) The current sensitivity of a galvanometer is given by $I_s = \frac{NBA}{k}$,where $N$ is the number of turns,$B$ is the magnetic field,$A$ is the area,and $k$ is the restoring torque per unit twist.
Voltage sensitivity is given by $V_s = \frac{I_s}{R} = \frac{NBA}{kR}$,where $R$ is the resistance of the coil.
If we increase current sensitivity by increasing the number of turns $N$,the resistance $R$ of the coil also increases proportionally because $R = \rho \frac{l}{A_{wire}}$,where $l$ is the length of the wire $(l = N \times \text{circumference})$.
Since $R$ increases as $N$ increases,the voltage sensitivity $V_s$ does not necessarily increase; it may remain constant or even decrease.
Therefore,the statement is false.

(N/A) The torque on a current-carrying coil in a magnetic field is given by $\tau = N I A B \sin \theta$. The restoring torque is $\tau_r = k \phi$. In equilibrium,$N I A B = k \phi$,which implies $\phi = (\frac{NAB}{k}) I$. The current sensitivity is $S_i = \frac{\phi}{I} = \frac{NAB}{k}$. If the number of turns $N$ is doubled to $2N$,the length of the wire $l$ also doubles,as $l = N \times (2\pi r)$. Since resistance $R = \rho \frac{l}{A_{wire}}$,doubling the length $l$ results in the resistance $R$ becoming $2R$. Therefore,to double the number of turns,the resistance of the coil becomes $2$ times the original resistance.

(N/A) To convert a galvanometer into a voltmeter,connect a high-value resistance in series with the galvanometer.
The governing equation is $I_g(G + R_{total}) = V$,where $I_g$ is the full-scale deflection current,$G$ is the galvanometer resistance,and $R_{total}$ is the total series resistance.
Given: $I_g = 1 \ mA = 10^{-3} \ A$,$G = 10 \ \Omega$.
$1$. For the $2 \ V$ range:
$I_g(G + R_1) = 2$
$10^{-3}(10 + R_1) = 2$
$10 + R_1 = 2000$
$R_1 = 1990 \ \Omega$
$2$. For the $20 \ V$ range:
$I_g(G + R_1 + R_2) = 20$
$10^{-3}(10 + 1990 + R_2) = 20$
$2000 + R_2 = 20000$
$R_2 = 18000 \ \Omega = 18 \ k\Omega$
$3$. For the $200 \ V$ range:
$I_g(G + R_1 + R_2 + R_3) = 200$
$10^{-3}(10 + 1990 + 18000 + R_3) = 200$
$20000 + R_3 = 200000$
$R_3 = 180000 \ \Omega = 180 \ k\Omega$
Thus,$R_1 = 1990 \ \Omega$,$R_2 = 18 \ k\Omega$,and $R_3 = 180 \ k\Omega$.

(D) To convert a galvanometer into an ammeter,a low-value resistance (shunt) is connected in parallel. Let $G = 10 \text{ } \Omega$ be the galvanometer resistance and $I_g = 1 \text{ mA} = 0.001 \text{ A}$ be the full-scale deflection current.
For range $I_1 = 10 \text{ mA} = 0.01 \text{ A}$ (between terminals $A$ and $B$): The shunt resistance is $S_1 + S_2 + S_3$. The current through the shunt is $I_1 - I_g = 0.01 - 0.001 = 0.009 \text{ A}$.
$I_g G = (I_1 - I_g)(S_1 + S_2 + S_3) \implies 0.001 \times 10 = 0.009(S_1 + S_2 + S_3) \implies S_1 + S_2 + S_3 = \frac{10}{9} \approx 1.11 \text{ } \Omega$.
For range $I_2 = 100 \text{ mA} = 0.1 \text{ A}$ (between terminals $A$ and $C$): The shunt resistance is $S_2 + S_3$. The current through the shunt is $I_2 - I_g = 0.1 - 0.001 = 0.099 \text{ A}$.
$I_g(G + S_1) = (I_2 - I_g)(S_2 + S_3) \implies 0.001(10 + S_1) = 0.099(S_2 + S_3)$.
For range $I_3 = 1 \text{ A}$ (between terminals $A$ and $D$): The shunt resistance is $S_3$. The current through the shunt is $I_3 - I_g = 1 - 0.001 = 0.999 \text{ A}$.
$I_g(G + S_1 + S_2) = (I_3 - I_g)S_3 \implies 0.001(10 + S_1 + S_2) = 0.999 S_3$.
Solving these equations simultaneously:
$S_3 = \frac{10}{999} \approx 0.0101 \text{ } \Omega$,
$S_2 = \frac{10}{99} - S_3 \approx 0.101 - 0.0101 = 0.0909 \text{ } \Omega$,
$S_1 = \frac{10}{9} - (S_2 + S_3) = \frac{10}{9} - \frac{10}{99} = \frac{100}{99} \approx 1.0101 \text{ } \Omega$.

(A) The figure of merit $(k)$ of a galvanometer is defined as the current required to produce a unit deflection in the galvanometer.
It is given by the formula: $k = \frac{I}{\theta}$
Given:
Current $I = 6 \, mA = 6 \times 10^{-3} \, A$
Deflection $\theta = 2^{\circ}$
Substituting the values:
$k = \frac{6 \times 10^{-3} \, A}{2^{\circ}} = 3 \times 10^{-3} \, A/\text{div}$
Therefore,the figure of merit is $3 \times 10^{-3} \, A/\text{div}$.

(B) Let $i_{g}$ be the full-scale deflection current of the galvanometer.
For the first case,the range is $0-1\, V$ with resistance $R_{1}$ in series:
$1 = i_{g}(G + R_{1}) \quad \dots(1)$
For the second case,the range is $0-2\, V$ with an additional resistance $R_{2}$ in series:
$2 = i_{g}(G + R_{1} + R_{2}) \quad \dots(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{2}{1} = \frac{i_{g}(G + R_{1} + R_{2})}{i_{g}(G + R_{1})}$
$2 = \frac{G + R_{1} + R_{2}}{G + R_{1}}$
$2(G + R_{1}) = G + R_{1} + R_{2}$
$2G + 2R_{1} = G + R_{1} + R_{2}$
$R_{2} = G + R_{1}$
Thus,the additional resistance required is $G + R_{1}$.

(C) Let $I$ be the total current,$I_g$ be the current through the voltmeter,and $R_v$ be the resistance of the voltmeter.
From the circuit diagram,the total current $I = 2 \, A$ splits into the current through the resistor $R$ and the current through the voltmeter $I_g$.
The voltage across the parallel combination is $V = 120 \, V$.
The current through the resistor $R = 75 \, \Omega$ is given by $I_R = \frac{V}{R} = \frac{120}{75} = 1.6 \, A$.
Since $I = I_R + I_g$,we have $I_g = I - I_R = 2 - 1.6 = 0.4 \, A$.
The resistance of the voltmeter is $R_v = \frac{V}{I_g} = \frac{120}{0.4} = 300 \, \Omega$.

(D) Given:
Total divisions for full scale deflection $(N)$ = $50 \, div$.
Voltage for full scale deflection $(V)$ = $50 \, mV = 50 \times 10^{-3} \, V$.
Current sensitivity $(S_i)$ = $2 \, div/mA = 2 \, div / (10^{-3} \, A) = 2000 \, div/A$.
Step $1$: Calculate the full scale current $(I_{fs})$.
The current sensitivity is defined as $S_i = N / I_{fs}$.
Therefore, $I_{fs} = N / S_i = 50 \, div / (2 \, div/mA) = 25 \, mA = 25 \times 10^{-3} \, A$.
Step $2$: Calculate the resistance of the galvanometer $(G)$.
Using Ohm's law, $V = I_{fs} \times G$.
$G = V / I_{fs} = (50 \times 10^{-3} \, V) / (25 \times 10^{-3} \, A) = 2 \, \Omega$.
Thus, the resistance of the galvanometer is $2 \, \Omega$.

(C) Let the total current be $i$.
Given that the current through the galvanometer is $I_g = 0.02i$.
Therefore,the current through the shunt resistor $S = 5\, \Omega$ is $I_s = i - 0.02i = 0.98i$.
Since the galvanometer and the shunt resistor are in parallel,the potential difference across them is equal:
$I_g R_g = I_s S$
$0.02i \times R_g = 0.98i \times 5$
$R_g = \frac{0.98 \times 5}{0.02}$
$R_g = 49 \times 5 = 245\, \Omega$.

(D) The figure of merit $K$ of a galvanometer is defined as the current required to produce a unit deflection,given by $K = \frac{I_g}{n}$,where $I_g$ is the current through the galvanometer and $n$ is the number of deflections.
Therefore,the current through the galvanometer is $I_g = Kn$.
In a converted galvanometer (ammeter),the shunt resistance $S$ is connected in parallel with the galvanometer resistance $G$.
Using the principle of current division in a parallel circuit,the total current $I$ is given by $I = I_g \left( \frac{G + S}{S} \right)$.
Substituting $I_g = Kn$ into the equation,we get $I = \frac{nK(G + S)}{S}$.

(B) In the half-deflection or fractional-deflection method,the galvanometer is connected in parallel with a shunt resistance $S$.
Since the galvanometer and shunt are in parallel,the potential difference across them is the same:
$(I - I_g) S = I_g G$
Where $I$ is the total current,$I_g$ is the current through the galvanometer,and $G$ is the galvanometer resistance.
Rearranging the terms,we get the ratio of currents as:
$\frac{I_g}{I} = \frac{S}{S + G}$
Given that the deflection is $\frac{1}{3}$ of the initial value,the current through the galvanometer becomes $I_g = \frac{1}{3} I$,which implies $\frac{I_g}{I} = \frac{1}{3}$.
Substituting this into the equation:
$\frac{1}{3} = \frac{S}{S + G}$
$S + G = 3S$
$G = 2S$
Thus,the resistance of the galvanometer is twice the value of the shunt resistance used.

(B) Let $G = 72 \; \Omega$ be the resistance of the galvanometer and $S = 8 \; \Omega$ be the shunt resistance.
The current $I$ divides into $I_g$ (current through the galvanometer) and $I_s$ (current through the shunt).
According to the principle of parallel circuits,the voltage across the galvanometer and the shunt is the same: $I_g G = I_s S$.
Since $I = I_g + I_s$,we have $I_s = I - I_g$.
Substituting this into the equation: $I_g G = (I - I_g) S$.
Rearranging to find the ratio $\frac{I_g}{I}$: $I_g G = I S - I_g S \Rightarrow I_g (G + S) = I S$.
Therefore,$\frac{I_g}{I} = \frac{S}{G + S}$.
Substituting the values: $\frac{I_g}{I} = \frac{8}{72 + 8} = \frac{8}{80} = \frac{1}{10}$.
The percentage of the total current passing through the galvanometer is $\frac{I_g}{I} \times 100 = \frac{1}{10} \times 100 = 10 \%$.

(D) The current sensitivity $(I_s)$ of a galvanometer is defined as the deflection produced per unit current,given by the formula:
$I_s = \frac{\theta}{i} = \frac{NAB}{K}$
where:
$N$ = number of turns in the coil
$A$ = area of the coil
$B$ = magnetic field strength
$K$ = torsional constant of the spring
From the formula,$I_s \propto N$,$I_s \propto A$,$I_s \propto B$,and $I_s \propto \frac{1}{K}$.
To increase the current sensitivity,we must:
$1$. Increase the number of turns $(N)$.
$2$. Increase the area of the coil $(A)$.
$3$. Increase the magnetic field $(B)$.
$4$. Decrease the torsional constant of the spring $(K)$.
Comparing these requirements with the given options,increasing the magnetic field $(B)$ and decreasing the torsional constant $(D)$ will increase the current sensitivity. Therefore,the correct option is $(B)$ and $(D)$ only.

(B) In circuit $(a)$,the voltmeter is in series with the $1 \,\Omega$ resistor,and the ammeter is in parallel with the $1 \,\Omega$ resistor. Let $R_V$ be the resistance of the voltmeter and $R_A$ be the resistance of the ammeter. The voltage across the $1 \,\Omega$ resistor is $V = I_1 \times 1$. The current through the ammeter is $I_2 = V / R_A = I_1 / R_A$. The total current $I = I_1 + I_2 = I_1(1 + 1/R_A)$. The voltmeter reading is $V_m = I \times R_V$. The ratio of voltage to current is $V_m / I = R_V = 1000 \,\Omega$ (assuming the voltmeter measures the total voltage drop across the series combination). However,based on the circuit,the ratio $V/I$ given is $1000 \,\Omega$. From the circuit $(a)$,the effective resistance is $R_{eq} = R_V + (1 \parallel R_A) \approx R_V = 1000 \,\Omega$.
In circuit $(b)$,the ammeter is in series with the $1 \,\Omega$ resistor,and the voltmeter is in parallel with the $1 \,\Omega$ resistor. The effective resistance is $R_{eq} = R_A + (1 \parallel R_V) \approx R_A + 1 = 0.999 \,\Omega$. Since $R_V$ is large,$1 \parallel R_V \approx 1 \,\Omega$. Thus $R_A + 1 = 0.999$,which implies $R_A \approx 0$ (or $10^{-3} \,\Omega$).
Thus,$R_V = 10^3 \,\Omega$ and $R_A = 10^{-3} \,\Omega$.

(A) Let the initial current sensitivity be $S_{i_1} = \frac{NAB}{k}$ and initial resistance be $R$.
The initial voltage sensitivity is $S_{v_1} = \frac{S_{i_1}}{R} = \frac{NAB}{kR}$.
According to the problem,the new current sensitivity $S_{i_2} = S_{i_1} + 0.20 S_{i_1} = 1.2 S_{i_1}$.
The new resistance is $R' = 2R$.
The new voltage sensitivity is $S_{v_2} = \frac{S_{i_2}}{R'} = \frac{1.2 S_{i_1}}{2R} = 0.6 \left( \frac{S_{i_1}}{R} \right) = 0.6 S_{v_1}$.
Since $0.6 = \frac{6}{10} = \frac{3}{5}$,the voltage sensitivity becomes $\frac{3}{5}$ times the initial value.

(A) In a moving coil galvanometer,the phosphor bronze strip is used as a suspension wire.
The restoring torque per unit twist is known as the torsional constant $(k)$.
For a sensitive galvanometer,we require a small restoring torque for a given deflection,which means the torsional constant $(k)$ should be as small as possible.
Phosphor bronze has a very low torsional constant,which allows the coil to rotate even for very small currents,thereby increasing the sensitivity of the instrument.
Additionally,phosphor bronze is non-magnetic and does not oxidize easily,making it ideal for this purpose.
Therefore,the correct option is $A$.

(C) For a voltmeter,the current $I_g$ passing through it is given by $V = I_g G$,where $G$ is the resistance of the voltmeter.
Initially,for a range of $5 \text{ V}$:
$5 = I_g \times 20000 \quad \dots(i)$
To increase the range to $20 \text{ V}$,we connect an additional resistance $R$ in series with the voltmeter. The new total resistance becomes $(G + R)$.
For the new range of $20 \text{ V}$:
$20 = I_g \times (G + R) \quad \dots(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{20}{5} = \frac{I_g(G + R)}{I_g G}$
$4 = \frac{G + R}{G}$
$4G = G + R$
$R = 3G$
Given $G = 20000\,\Omega$,we have:
$R = 3 \times 20000 = 60000\,\Omega$
Thus,an extra resistance of $60000\,\Omega$ must be connected in series.

(B) The torque acting on the coil of a moving coil galvanometer is given by $\tau = NiAB$,where $N$ is the number of turns,$i$ is the current,$A$ is the area,and $B$ is the magnetic field.
The restoring torque provided by the suspension wire is $\tau = K\theta$,where $K$ is the torsional constant and $\theta$ is the deflection.
Equating the two torques: $NiAB = K\theta$.
Rearranging for the area $A$: $A = \frac{K\theta}{NiB}$.
Given values:
$K = 4.0 \times 10^{-5}\,Nm\,rad^{-1}$
$\theta = 0.05\,rad$
$N = 200$
$i = 10\,mA = 10 \times 10^{-3}\,A = 0.01\,A$
$B = 0.01\,T$
Substituting the values:
$A = \frac{4.0 \times 10^{-5} \times 0.05}{200 \times 0.01 \times 0.01}$
$A = \frac{2.0 \times 10^{-6}}{0.02} = 1.0 \times 10^{-4}\,m^2$.
Since $1\,m^2 = 10^4\,cm^2$,we have:
$A = 1.0 \times 10^{-4} \times 10^4\,cm^2 = 1\,cm^2$.

(D) Current sensitivity $(I_s)$ of a moving coil galvanometer is given by $I_s = \frac{NBA}{k}$,where $N$ is the number of turns,$B$ is the magnetic field,$A$ is the area,and $k$ is the torsional constant.
Since $I_s \propto N$,if $I_s$ increases by $50 \%$,the new number of turns $N' = 1.5N$.
Voltage sensitivity $(V_s)$ is given by $V_s = \frac{I_s}{R}$,where $R$ is the resistance of the coil.
Resistance $R$ is proportional to the length of the wire,and length is proportional to the number of turns $(R \propto N)$.
Therefore,$V_s = \frac{I_s}{R} \propto \frac{N}{N} = \text{constant}$.
Since $V_s$ is independent of the number of turns $N$,the voltage sensitivity remains unchanged.
Thus,the percentage change in voltage sensitivity is $0 \%$.

(A) voltmeter is connected in parallel to the component across which the potential difference is to be measured.
To minimize the loading effect and ensure the measured voltage is as close to the actual voltage as possible,the resistance of the voltmeter should be as high as possible.
Assertion $A$ is incorrect because a voltmeter with $4000\,\Omega$ resistance is preferred over one with $1000\,\Omega$ resistance,as higher resistance draws less current from the circuit,thereby reducing the error in measurement.
Reason $R$ is correct because,by Ohm's law $(I = V/R)$,for a given voltage $V$,a higher resistance $R$ results in a smaller current $I$ being drawn by the voltmeter.
Therefore,$A$ is incorrect but $R$ is correct.

(C) To verify Ohm's law $(V=IR)$,we need to measure the voltage across the test resistor $R_T$ and the current flowing through it.
$1$. $A$ voltmeter is connected in parallel to the resistor. To convert a galvanometer into a voltmeter,a very high resistance must be connected in series with it. Here,$R_1=10\,M\,\Omega$ is a very high resistance,so $G_1$ in series with $R_1$ acts as a voltmeter.
$2$. An ammeter is connected in series with the resistor. To convert a galvanometer into an ammeter,a very low resistance must be connected in parallel with it. Here,$R_2=0.001\,\Omega$ is a very low resistance,so $G_2$ in parallel with $R_2$ acts as an ammeter.
$3$. In circuit $C$,$G_1$ is in series with $R_1$ (forming a voltmeter) and this combination is in parallel with $R_T$. $G_2$ is in parallel with $R_2$ (forming an ammeter) and this combination is in series with $R_T$. This is the correct configuration for verifying Ohm's law.

(B) When the coil of the galvanometer moves in the magnetic field,the magnetic flux linked with the non-magnetic metallic core changes.
This change in magnetic flux induces eddy currents in the metallic core.
According to Lenz's law,these eddy currents oppose the motion of the coil.
As a result,the coil is brought to rest quickly,providing electromagnetic damping.

(D) For a moving coil galvanometer,the torque is given by $\tau = NIAB \sin \theta = k \phi$,where $N$ is the number of turns,$I$ is the current,$A$ is the area,$B$ is the magnetic field,and $k$ is the torsional constant.
The current sensitivity is defined as $S_i = \frac{\phi}{I} = \frac{NBA}{k}$.
If $N$ is doubled,$S_i$ becomes $2 \times \frac{NBA}{k}$,so Statement $I$ is true.
The voltage sensitivity is defined as $S_v = \frac{\phi}{V} = \frac{\phi}{IR} = \frac{S_i}{R} = \frac{NBA}{kR}$.
When the number of turns $N$ is doubled,the length of the wire in the coil also doubles,which means the resistance $R$ of the coil also doubles $(R \propto N)$.
Therefore,$S_v = \frac{(2N)BA}{k(2R)} = \frac{NBA}{kR}$.
Since $S_v$ remains unchanged,Statement $II$ is false.

(A) The current sensitivity of a moving coil galvanometer is given by $I_s = \frac{NBA}{C}$,where $N$ is the number of turns,$B$ is the magnetic field,$A$ is the area,and $C$ is the torsional constant.
The voltage sensitivity is given by $V_s = \frac{I_s}{G}$,where $G$ is the resistance of the galvanometer coil.
Given that the resistance $G$ is kept constant,the relationship between voltage sensitivity and current sensitivity is $V_s \propto I_s$.
Since the current sensitivity $I_s$ increases by $25 \%$,the voltage sensitivity $V_s$ will also increase by $25 \%$ because the ratio $\frac{1}{G}$ is constant.
Therefore,the percentage change in voltage sensitivity is $25 \%$.

(C) Let $I_g$ be the full-scale deflection current of the galvanometer and $R_G$ be its resistance.
Case $1$: Shunting with $5\,\Omega$ resistance.
The total current is $I = 250\,mA = 0.25\,A$. The shunt resistance $S = 5\,\Omega$.
Using the current division rule: $I_g = I \times \frac{S}{S + R_G}$
$I_g = 0.25 \times \frac{5}{5 + R_G} \dots(i)$
Case $2$: Connecting $1050\,\Omega$ in series.
The total voltage is $V = 25\,V$. The series resistance $R = 1050\,\Omega$.
Using Ohm's law: $I_g = \frac{V}{R + R_G}$
$I_g = \frac{25}{1050 + R_G} \dots(ii)$
Equating $(i)$ and $(ii)$:
$0.25 \times \frac{5}{5 + R_G} = \frac{25}{1050 + R_G}$
$\frac{1.25}{5 + R_G} = \frac{25}{1050 + R_G}$
$1.25(1050 + R_G) = 25(5 + R_G)$
$1312.5 + 1.25 R_G = 125 + 25 R_G$
$1312.5 - 125 = 25 R_G - 1.25 R_G$
$1187.5 = 23.75 R_G$
$R_G = \frac{1187.5}{23.75} = 50\,\Omega$

(C) For the voltmeter:
The resistance $R$ to be connected in series is given by $R = \frac{V}{I_g} - G$.
Given $V = 50\,V$,$I_g = 1\,mA = 10^{-3}\,A$,and $G = 54\,\Omega$.
$R = \frac{50}{10^{-3}} - 54 = 50000 - 54 = 49946\,\Omega \approx 50\,k\Omega$.
Thus,statement $(A)$ is correct.
For the ammeter:
The shunt resistance $r$ to be connected in parallel is given by $r = \frac{I_g G}{I - I_g}$.
Given $I = 10\,mA = 10^{-2}\,A$,$I_g = 1\,mA = 10^{-3}\,A$,and $G = 54\,\Omega$.
$r = \frac{10^{-3} \times 54}{10 \times 10^{-3} - 1 \times 10^{-3}} = \frac{54 \times 10^{-3}}{9 \times 10^{-3}} = 6\,\Omega$.
Thus,statement $(C)$ is correct.
Therefore,the correct option is $(A)$ and $(C)$.

(C) In a moving coil galvanometer,the deflection $\theta$ is directly proportional to the current $i$ flowing through the coil,given by $i = k\theta$,where $k$ is the galvanometer constant.
Given:
$i_1 = 200 \ \mu A$
$\theta_1 = 60^{\circ} = 60 \times \frac{\pi}{180} \text{ radians} = \frac{\pi}{3} \text{ radians}$
$\theta_2 = \frac{\pi}{10} \text{ radians}$
Using the proportionality $i \propto \theta$,we have:
$\frac{i_2}{i_1} = \frac{\theta_2}{\theta_1}$
Substituting the values:
$\frac{i_2}{200 \ \mu A} = \frac{\pi / 10}{\pi / 3}$
$\frac{i_2}{200 \ \mu A} = \frac{3}{10}$
Solving for $i_2$:
$i_2 = 200 \ \mu A \times \frac{3}{10} = 60 \ \mu A$

(C) Given:
Galvanometer resistance,$G = 10 \ \Omega$
Full scale deflection current,$I_g = 3 \ mA = 3 \times 10^{-3} \ A$
Range of ammeter,$I = 8 \ A$
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.
The potential difference across the galvanometer and the shunt must be equal:
$I_g G = (I - I_g) S$
Rearranging for $S$:
$S = \frac{I_g G}{I - I_g}$
Substituting the values:
$S = \frac{(3 \times 10^{-3} \ A) \times 10 \ \Omega}{8 \ A - 3 \times 10^{-3} \ A}$
$S = \frac{0.03}{8 - 0.003} \ \Omega$
$S = \frac{0.03}{7.997} \ \Omega \approx 3.75 \times 10^{-3} \ \Omega$
Thus,the required shunt resistance is $3.75 \times 10^{-3} \ \Omega$.

(B) Let $k$ be the current per division of the galvanometer.
Initially,the total current $I = 25k$ flows through the galvanometer.
When a shunt resistance $S = 24\ \Omega$ is connected in parallel,the deflection becomes $5$ divisions,meaning the current through the galvanometer is $I_g = 5k$.
The remaining current flows through the shunt: $I_s = I - I_g = 25k - 5k = 20k$.
Since the galvanometer and shunt are in parallel,the potential difference across them is the same:
$I_g \times G = I_s \times S$
$(5k) \times G = (20k) \times 24$
$5G = 480$
$G = 96\ \Omega$
Thus,the resistance of the galvanometer coil is $96\ \Omega$.

(C) To convert a galvanometer into a voltmeter,a high resistance $R$ is connected in series with the galvanometer.
Given:
Resistance of galvanometer,$R_g = 50 \ \Omega$
Maximum current for full-scale deflection,$I_g = 5 \ mA = 5 \times 10^{-3} \ A$
Required voltage range,$V = 100 \ V$
The formula for the series resistance $R$ is given by:
$V = I_g(R + R_g)$
$R + R_g = \frac{V}{I_g}$
$R = \frac{V}{I_g} - R_g$
Substituting the values:
$R = \frac{100}{5 \times 10^{-3}} - 50$
$R = 20000 - 50$
$R = 19950 \ \Omega$

(A) Let the main current be $I$. The current through the galvanometer is $I_g = 5 \% \text{ of } I = \frac{5}{100} I = \frac{I}{20}$.
The current through the shunt resistance $S$ is $I_s = I - I_g = I - \frac{I}{20} = \frac{19I}{20}$.
Since the galvanometer and shunt are in parallel,the potential difference across them is equal: $I_g G = I_s S$.
Substituting the values: $(\frac{I}{20}) G = (\frac{19I}{20}) S$.
Solving for $S$: $S = \frac{G}{19}$.
The total resistance of the ammeter $R_A$ is the equivalent resistance of $G$ and $S$ in parallel: $R_A = \frac{G \cdot S}{G + S}$.
Substituting $S = \frac{G}{19}$: $R_A = \frac{G \cdot (G/19)}{G + (G/19)} = \frac{G^2/19}{20G/19} = \frac{G}{20}$.

(C) The torque acting on the coil in a magnetic field is given by $\tau = BINA \sin \phi$,where $\phi$ is the angle between the normal to the coil and the magnetic field. In a radial magnetic field,$\phi = 90^{\circ}$,so $\sin 90^{\circ} = 1$.
The restoring torque provided by the suspension wire is $\tau = C \theta$,where $C$ is the torsional constant and $\theta$ is the deflection.
Equating the two torques: $C \theta = BINA$.
Given values: $N = 100$,$A = 2.0 \,cm^2 = 2.0 \times 10^{-4} \,m^2$,$B = 0.01 \,T$,$I = 10 \,mA = 10 \times 10^{-3} \,A$,and $\theta = 0.05 \,rad$.
Substituting these values into the formula for $C$:
$C = \frac{BINA}{\theta} = \frac{0.01 \times 10 \times 10^{-3} \times 100 \times 2.0 \times 10^{-4}}{0.05}$
$C = \frac{0.01 \times 0.01 \times 100 \times 2.0 \times 10^{-4}}{0.05} = \frac{2.0 \times 10^{-6}}{0.05} = 40 \times 10^{-6} = 4 \times 10^{-5} \,N-m/rad$.
Comparing this with $x \times 10^{-5} \,N-m/rad$,we get $x = 4$.

(A) The current $I$ in the galvanometer is given by $I = K \theta$,where $K$ is the figure of merit.
From the circuit,the current is $I = \frac{E}{G+R}$,where $E = 2 \text{ V}$ is the emf of the source,$G$ is the galvanometer resistance,and $R$ is the external resistance.
Equating the two expressions: $K \theta = \frac{E}{G+R} \Rightarrow \frac{1}{\theta} = \frac{G+R}{E} = \frac{1}{E} R + \frac{G}{E}$.
Comparing this with the equation of a straight line $y = mx + c$,we have the slope $m = \frac{1}{E}$.
From the graph,the slope $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 1}{6 - 2} = \frac{1}{4} \text{ } \Omega^{-1}$.
Since $m = \frac{K}{E}$,we have $\frac{K}{2} = \frac{1}{4} \Rightarrow K = 0.5 \text{ A/division}$.
Expressing $K$ in terms of $10^{-1}$,we get $K = 5 \times 10^{-1} \text{ A/division}$.
Thus,the correct option is $A$.

(C) First,calculate the full-scale deflection current $i_g$ of the galvanometer.
Given that the galvanometer (resistance $R_g = 100 \ \Omega$) is in series with $400 \ \Omega$ to measure $10 \ V$:
$i_g = \frac{V}{R_g + R_{series}} = \frac{10}{100 + 400} = \frac{10}{500} = 20 \times 10^{-3} \ A$.
To convert the galvanometer into an ammeter to read up to $I = 10 \ A$,a shunt resistance $S$ is connected in parallel.
The condition for the shunt is $i_g R_g = (I - i_g) S$.
Substituting the values:
$20 \times 10^{-3} \times 100 = (10 - 20 \times 10^{-3}) S$.
Since $20 \times 10^{-3} = 0.02 \ A$,we have:
$2 = (10 - 0.02) S = 9.98 S$.
$S = \frac{2}{9.98} \approx 0.2004 \ \Omega$.
Given the form $x \times 10^{-2} \ \Omega$,we approximate $S \approx 20 \times 10^{-2} \ \Omega$.
Thus,$x = 20$.

(B) The ammeter consists of a $240 \Omega$ coil in parallel with a $10 \Omega$ shunt. The equivalent resistance of the ammeter $(R_A)$ is given by:
$R_A = \frac{240 \times 10}{240 + 10} = \frac{2400}{250} = 9.6 \Omega$
The total resistance of the circuit $(R_{eq})$ is the sum of the external resistor and the ammeter resistance:
$R_{eq} = 140.4 \Omega + 9.6 \Omega = 150 \Omega$
The current $(I)$ flowing through the circuit is given by Ohm's law:
$I = \frac{V}{R_{eq}} = \frac{24 \text{ V}}{150 \Omega} = 0.16 \text{ A}$
Converting the current to milliamperes $(mA)$:
$I = 0.16 \times 1000 \text{ mA} = 160 \text{ mA}$
Thus,the reading of the ammeter is $160 \text{ mA}$.

(B) Given:
Resistance of galvanometer $G = 200 \Omega$
Full scale deflection current $I_g = 20 \mu A = 20 \times 10^{-6} A$
Required range of ammeter $I = 20 mA = 20 \times 10^{-3} A$
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel.
The formula for shunt resistance is $S = \frac{I_g G}{I - I_g}$.
Substituting the values:
$S = \frac{20 \times 10^{-6} \times 200}{20 \times 10^{-3} - 20 \times 10^{-6}}$
$S = \frac{4000 \times 10^{-6}}{20 \times 10^{-3} (1 - 0.001)}$
$S = \frac{4 \times 10^{-3}}{20 \times 10^{-3} \times 0.999}$
$S = \frac{4}{20 \times 0.999} = \frac{0.2}{0.999} \approx 0.2002 \Omega$
Thus,the required shunt resistance is approximately $0.20 \Omega$.

(C) In a moving coil galvanometer,a soft iron core is placed inside the coil to increase the magnetic field strength $(B)$ within the coil,which directly increases the sensitivity of the galvanometer.
$STATEMENT-1$ is True because the high magnetic permeability of the soft iron core concentrates the magnetic field lines,thereby increasing the torque on the coil.
$STATEMENT-2$ is False because soft iron is a soft magnetic material,which means it has high magnetic permeability and can be easily magnetized and demagnetized. The statement incorrectly claims that it cannot be easily magnetized or demagnetized.

(D) For a voltmeter,the range $V = I_g(R_G + R_{ext})$. To maximize $V$,we need to maximize the total resistance $R_{total}$. Connecting all components in series gives $R_{total} = 2R_G + 2R$. This is the maximum possible resistance,so option $(A)$ is correct for voltage range.
For an ammeter,the range $I = I_g(1 + R_G/R_s)$. To maximize $I$,we need to minimize the equivalent resistance of the shunt $R_s$. Connecting all components in parallel gives the minimum equivalent resistance,which maximizes the current range. Thus,option $(C)$ is correct for current range.
Therefore,the correct options are $(A)$ and $(C)$.

(C) To verify Ohm's law,we need to measure the potential difference across the resistor $R_T$ and the current flowing through it.
$1$. $A$ voltmeter is connected in parallel to the resistor. $A$ voltmeter is formed by connecting a high resistance in series with a galvanometer. Thus,$G_1$ in series with $R_1$ connected in parallel to $R_T$ acts as a voltmeter.
$2$. An ammeter is connected in series with the resistor. An ammeter is formed by connecting a small resistance (shunt) in parallel with a galvanometer. Thus,$G_2$ in parallel with $R_2$ connected in series with $R_T$ acts as an ammeter.
$3$. Comparing this with the given options,option $C$ shows $G_1$ in series with $R_1$ (voltmeter) across $R_T$ and $G_2$ in parallel with $R_2$ (ammeter) in series with $R_T$.

(C) Given:
Number of turns $n = 50$
Area $A = 2 \times 10^{-4} \ m^2$
Magnetic field $B = 0.02 \ T$
Torsional constant $C = 10^{-4} \ N \ m \ rad^{-1}$
Full-scale deflection angle $\theta = 0.2 \ rad$
Galvanometer resistance $G = 50 \ \Omega$
The torque balance equation for a moving coil galvanometer is given by:
$C \theta = n i_g A B$
Where $i_g$ is the full-scale deflection current.
$i_g = \frac{C \theta}{n A B} = \frac{10^{-4} \times 0.2}{50 \times 2 \times 10^{-4} \times 0.02} = \frac{0.2 \times 10^{-4}}{200 \times 10^{-4} \times 0.02} = \frac{0.2}{4} = 0.05 \ A$
Wait,recalculating: $i_g = \frac{0.2 \times 10^{-4}}{50 \times 2 \times 10^{-4} \times 0.02} = \frac{0.2}{50 \times 2 \times 0.02} = \frac{0.2}{2} = 0.1 \ A$
To convert the galvanometer into an ammeter of range $I = 1.0 \ A$,a shunt resistance $S$ is connected in parallel.
The current through the shunt is $I_s = I - i_g = 1.0 - 0.1 = 0.9 \ A$.
Since the galvanometer and shunt are in parallel,the voltage across them is equal:
$i_g G = I_s S$
$0.1 \times 50 = 0.9 \times S$
$S = \frac{5}{0.9} = \frac{50}{9} \approx 5.56 \ \Omega$

(C) $1$. Voltmeter conversion: $V = I_g(R_g + R_v) \implies 0.1 = 2 \times 10^{-6} (10 + R_v) \implies 50000 = 10 + R_v \implies R_v = 49990 \Omega \approx 50 \text{ k} \Omega$. Statement $(2)$ is correct.
$2$. Ammeter conversion: $I_g R_g = (I - I_g) R_s \implies 2 \times 10^{-6} \times 10 = (10^{-3} - 2 \times 10^{-6}) R_s \implies 2 \times 10^{-5} = 0.998 \times 10^{-3} R_s \implies R_s \approx 0.02004 \Omega$. The total resistance of the ammeter is $R_A = \frac{R_g R_s}{R_g + R_s} \approx R_s \approx 0.02 \Omega$. Statement $(3)$ is correct.
$3$. Measured resistance: The voltmeter is in parallel with $R = 1000 \Omega$. The equivalent resistance is $R_{eq} = \frac{R \times R_v}{R + R_v} = \frac{1000 \times 50000}{1000 + 50000} = \frac{50000}{51} \approx 980.39 \Omega$. Statement $(1)$ is incorrect.
$4$. Internal resistance: Adding internal resistance $r = 5 \Omega$ in series with the circuit reduces the total current $x$ flowing through the ammeter,but the measured value of $R$ is determined by the ratio of voltage across the voltmeter to the current through the resistor $R$. Since the voltmeter is in parallel with $R$,the measured resistance remains $R_{eq} \approx 980.39 \Omega$,which is independent of the internal resistance of the cell. Statement $(4)$ is incorrect.
Thus,statements $(2)$ and $(3)$ are correct.

(D) Let $G$ be the resistance of the galvanometer and $I_g = 0.006 \ A$ be the current for full scale deflection.
For a voltmeter of range $V = 30 \ V$ with a series resistance $R = 4990 \ \Omega$:
$V = I_g(G + R)$
$30 = 0.006(G + 4990)$
$G + 4990 = \frac{30}{0.006} = 5000$
$G = 5000 - 4990 = 10 \ \Omega$
For an ammeter of range $I = 1.5 \ A$ with a shunt resistance $S = \frac{2n}{249} \ \Omega$:
$I_g G = (I - I_g) S$
$0.006 \times 10 = (1.5 - 0.006) \times S$
$0.06 = 1.494 \times S$
$S = \frac{0.06}{1.494} = \frac{60}{1494} = \frac{10}{249} \ \Omega$
Equating the shunt resistance:
$\frac{2n}{249} = \frac{10}{249}$
$2n = 10 \Rightarrow n = 5$

(A) $A :$ The torque $\tau = C\theta$,where $C$ is the torsional constant. Thus,$[C] = [\tau]/[\theta] = [ML^2 T^{-2}] / [1] = [ML^2 T^{-2}]$. Statement $A$ is correct.
$B :$ Current sensitivity $(I_s) = \frac{NBA}{k}$ and Voltage sensitivity $(V_s) = \frac{I_s}{R} = \frac{NBA}{kR}$. Increasing $N$ increases $I_s$,but $R$ also increases with $N$,so $V_s$ may not increase. Statement $B$ is correct.
$C :$ Since $V_s = \frac{NBA}{kR}$ and $R \propto N$,if $N \to 2N$,then $R \to 2R$. Thus,$V_s' = \frac{(2N)BA}{k(2R)} = V_s$. The voltage sensitivity remains unchanged. Statement $C$ is incorrect.
$D :$ To convert $MCG$ to an ammeter,a small shunt resistance is connected in parallel. Statement $D$ is incorrect.
$E :$ Current sensitivity $I_s = \frac{NBA}{k}$,so $I_s \propto N$. It depends directly on the number of turns. Statement $E$ is incorrect.
Therefore,only $A$ and $B$ are correct.

(C) The condition for the shunt resistance $r_s$ to convert a galvanometer into an ammeter is given by the parallel circuit relation: $I_g R_g = (I - I_g) r_s$.
Given:
Galvanometer resistance $R_g = 30 \ \Omega$
Full-scale deflection current $I_g = 20 \ \text{mA} = 20 \times 10^{-3} \ \text{A} = 0.02 \ \text{A}$
Maximum current to be measured $I = 3 \ \text{A}$
Substituting the values into the formula:
$0.02 \times 30 = (3 - 0.02) \times r_s$
$0.6 = 2.98 \times r_s$
$r_s = \frac{0.6}{2.98} \ \Omega$
We are given that $r_s = \frac{30}{X} \ \Omega$. Therefore:
$\frac{30}{X} = \frac{0.6}{2.98}$
$X = \frac{30 \times 2.98}{0.6}$
$X = 50 \times 2.98 = 149$.
Thus,the value of $X$ is $149$.

(A) The voltage sensitivity $(S_V)$ of a moving coil galvanometer is given by the formula:
$S_V = \frac{\theta}{V} = \frac{NAB}{kR}$
where $N$ is the number of turns,$A$ is the area,$B$ is the magnetic field,$k$ is the torsional constant,and $R$ is the resistance.
Given that $k$ is the same for both coils,the ratio of voltage sensitivity is:
$\frac{S_{V1}}{S_{V2}} = \frac{N_1 A_1 B_1}{N_2 A_2 B_2} \times \frac{R_2}{R_1}$
Substituting the given values:
$\frac{S_{V1}}{S_{V2}} = \frac{15 \times 3.6 \times 10^{-3} \times 0.25}{21 \times 1.8 \times 10^{-3} \times 0.50} \times \frac{7}{5}$
$\frac{S_{V1}}{S_{V2}} = \frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.50} \times \frac{7}{5}$
$\frac{S_{V1}}{S_{V2}} = \frac{13.5}{18.9} \times 1.4 = \frac{13.5}{18.9} \times \frac{7}{5} = \frac{13.5 \times 7}{18.9 \times 5} = \frac{94.5}{94.5} = 1$
Therefore,the ratio is $1: 1$.

(C) To verify Ohm's law,we need to measure the potential difference across the test resistor $R_T$ and the current flowing through it.
$1$. $A$ voltmeter is connected in parallel to the resistor $R_T$. $A$ voltmeter is formed by connecting a high resistance $R_1$ in series with a galvanometer $G_1$.
$2$. An ammeter is connected in series with the resistor $R_T$. An ammeter is formed by connecting a small resistance $R_2$ (shunt) in parallel with a galvanometer $G_2$.
$3$. Looking at the options,the circuit in option $C$ shows $G_1$ in series with $R_1$ (forming a voltmeter) connected in parallel to $R_T$,and $G_2$ in parallel with $R_2$ (forming an ammeter) connected in series with $R_T$. Thus,option $C$ is correct.

(D) Let the main current be $I$ and the current through the galvanometer be $I_g$.
Given that $I_g = 2 \% \text{ of } I = 0.02 I$.
The resistance of the galvanometer is $G$.
Let the shunt resistance be $S$.
Since the galvanometer and shunt are in parallel,the potential difference across them is equal:
$I_g G = (I - I_g) S$.
Substituting the values:
$0.02 I \cdot G = (I - 0.02 I) S$.
$0.02 I \cdot G = 0.98 I \cdot S$.
$S = \frac{0.02}{0.98} G = \frac{2}{98} G = \frac{G}{49}$.
Thus,the shunt resistance is $G / 49$.

(D) In the half-deflection method,the resistance of the galvanometer $G$ is given by the formula:
$G = \frac{RS}{R - S}$
For the first observation:
$G_1 = \frac{3300 \times 80}{3300 - 80} = \frac{264000}{3220} \approx 81.98 \ \Omega \approx 82 \ \Omega$
For the second observation:
$G_2 = \frac{5000 \times 80}{5000 - 80} = \frac{400000}{4920} \approx 81.30 \ \Omega \approx 81 \ \Omega$
The average value of the galvanometer resistance is:
$G = \frac{G_1 + G_2}{2} = \frac{82 + 81}{2} = 81.5 \ \Omega$
Comparing this with the given options,the value is closest to $80 \ \Omega$.