Reading of voltmeter = 4 V, Resistance of voltmeter  \(=10^{4} \Omega\)
Current drawn by the voltmeter, \(I=\frac{4}{10^{4}}=4 \times 10^{-4} \mathrm{~A}\)
Again, \(I=\frac{E}{X+10^{4}}\)
\( \Rightarrow \left(X+10^{4}\right) I =E \)
\(\therefore X I =E-I \times 10^{4} \)
\(\Rightarrow X =\frac{E}{I}-10^{4}=\frac{120}{4 \times 10^{-4}}-10^{4} \)
\(\therefore X =29 \times 10^{4} \Omega=290 \mathrm{k} \Omega \)

(a) (i) Presence of mind.
(ii) High degree of general awareness.
(iii) Helping and caring nature. 2
(b) For maximum current, external resistance,
R = 0
\(I=\frac{E}{R+r}=\frac{12}{0.4}=\frac{12}{0.4}=30A\)

(1) Honesty, sharing of knowledge, willingness to help.
(2) 1 kWh = 3.6 x 10⁶ J,R = V²/P = 48.45Ω

(i) Post office box and meter bridge are two electrical appliances based on the principal of Wheatstone bridge.
(ii) The post office box is used practically in post and telegraph department to locate the snapping of telephone line.
The broken telephone line will touch the ground. Using post office box, resistance S of broken line is determined.
As resistance per unit length of line is known, the length of broken line can be calculated. Therefore, snapping of line is located.

(i) Acknowledging the help from others with gratitude. Awareness of the technology, helping tendency, practical knowledge of the subject.
(ii) Vasu checked the wiring because he has practical knowledge and he can rectify the problem.
(iii) An electric fuse is a wire used as a safety device, which melts when current exceeds the limit. Fuse wire has low melting point, high resistivity 

Here, I₁ = 500\(\mu A\) = 500 x 10^-6 A,
 \({ \theta }_{ 1 }={ 60 }^{ o },{ I }_{ 2 }=?,{ \theta }_{ 2 }=\frac { \pi }{ 5 } rad.=\frac { { 180 }^{ o } }{ 5 } ={ 36 }^{ o }\) 
\({ I }_{ 1 }=\frac { k }{ nBA } { \theta }_{ 1 }\) and \({ I }_{ 2 }=\frac { k }{ nBA } { \theta }_{ 2 }\) 
\(\therefore \frac { { I }_{ 2 } }{ { I }_{ 1 } } =\frac { { \theta }_{ 2 } }{ { \theta }_{ 1 } } or\) 
\({ I }_{ 2 }=\frac { { \theta }_{ 2 } }{ { \theta }_{ 1 } } { I }_{ 1 }=\frac { 36 }{ 60 } \times 500\times { 10 }^{ -6 }=300\times { 10 }^{ -6 }\quad A\) 
Current sensitivity   \(=\frac { { \theta }_{ 2 } }{ { I }_{ 2 } } =\frac { 36^{ o } }{ 300\times { 10 }^{ -6 } } \) 
= 0.12 x 10^-6 degree/A = 0.12 degree/\(\mu A\)

Given, \({ I' }_{ s }={ I }_{ s }+\frac { 20 }{ 100 } { I }_{ s }=\frac { 120 }{ 100 } { I }_{ s }\) 
\(R'=2R\) 
Then, initial voltage sensitivity, \({ V }_{ s }=\frac { { I }_{ s } }{ R } \) 
New voltage sensitivity, 
\({ V' }_{ s }=\frac { { I' }_{ s } }{ R' } =\left( \frac { 120 }{ 100 } { I }_{ s } \right) \times \frac { 1 }{ 2R } =\frac { 3 }{ 5 } { V }_{ s }\) 
% decrease in voltage sensitivity
\(=\frac { { V }_{ s }-{ V' }_{ s } }{ { V }_{ s } } \times 100=\frac { { V }_{ s }-\frac { 3 }{ 5 } { V }_{ s } }{ { V }_{ s } } \times 100\) 
 = 40%  

Here, A = 5.0 x 10^-4m², n = 60 ;
B = 90 G = 90 x 10^-4 T, K = ?
I = 2.0 mA = 2.0 x 10^-3 A, \(\theta \) =18^o,
As, \(I=\frac { k }{ nBA } \theta \) 
\(\therefore k=\frac { nBAI }{ \theta } \) 
\(=\frac { { 60\times 90\times 10 }^{ -4 }\times 5.0{ \times 10 }^{ -4 }\times 2.0\times { 10 }^{ -3 } }{ 18 } \) 
= 3.0 x 10^-9 Nm per degree

From Kirchhoff's junction rule, we have
\( { I }_{ 1 }={ I }+{ I }_{ 2 }\)   ...........(i)
Applying Kirchhoff's loop rule to outer loop containing 10V cell, we get
\(10=IR+{ 10I }_{ 1 }\) ............(ii)
Applying Kirchhoff's loop rule to outer loop containing 2V cell, we get
\(2={ 5I }_{ 2 }-RI\)
\(2=5\left( { I }_{ 1 }-I \right) -RI\)
\(4={ 10I }_{ 1 }-10I-2RI\)
Subtracting eq (ii) from eq (i), we get
\(6=3RI+10I\)
\(2=I\left( R+\frac { 10 }{ 3 } \right) \)
From Ohm's law, we have
\(V=I\left( R+{ R }_{ off } \right) \)
Comparing eq (iii) and (iv), we get
\({ R }_{ ef }=\frac { 10 }{ 3 } \Omega \)
If \({ E }_{ eff }\) and \({ R }_{ eff }\) are the effective voltage and effective internal resistance of the combination, then the equivalent circuit is shown.

Given, potential difference = 230 V
Initial current at 27°C = I_27°C = 3.2 A
Final current at t°C = I_t°C = 2.8 A
Room temperature = 27°C
Temperature coefficient of resistance, \(\alpha=1.70 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\)
Resistance at 27°C, R_27°C\(=\frac{V}{I_{27^{\circ} \mathrm{C}}}=\frac{230}{3.2}=\frac{2300}{32} \Omega\)
Resistance at t°C, R_t°C = \(\frac{V}{I_{t^{\circ} \mathrm{C}}}=\frac{230}{2.8}=\frac{2300}{28} \Omega\)
Temperature coefficient of resistance
\(\begin{aligned} \alpha & =\frac{R_t-R_{27}}{R_{27}(t-27)} \end{aligned}\)
\(\begin{aligned} \Rightarrow 1.70 \times 10^{-4} & =\frac{\frac{2300}{28}-\frac{2300}{32}}{\frac{2300}{32}(t-27)} \\ \end{aligned}\)
\(\begin{aligned} \text { or } \quad t-27 & =\frac{82.143-71.875}{71.875 \times 1.70 \times 10^{-4}}=840.347 \end{aligned}\)
or t = 840.3 + 27 = 867.3 °C
Thus, the steady temperature of heating element is 867.3 °C

When key is open, then
\( R_{e q}=(12+6) \|(6+12)=9 \Omega \)
\(\therefore I_{1}=\frac{V}{9} \)
When key is closed, then
\( R_{e q \boldsymbol{x}} =(12 \| 6)+(6 \| 12)=8 \Omega \)
\( I_{2} =\frac{V}{8} \)
\(\therefore \ \frac{I_{2}}{I_{1}} =\frac{V}{8} \times \frac{9}{V} \)
\(I_{2}=\frac{9}{8} I_{1}, \text { i.e. current becomes } \frac{9}{8} \text { times. }\)

Here, \(\frac{l_{1}}{l_{2}}=\frac{1}{2} \text { and } \frac{R_{1}}{R_{2}}=\frac{2}{1}\)
(a) \( R =\rho \frac{l}{A} \)
\(\frac{R_{1}}{R_{2}} =\frac{l_{1}}{l_{2}} \cdot \frac{A_{2}}{A_{1}}=\frac{l_{1}}{l_{2}} \cdot\left(\frac{d_{2}}{d_{1}}\right)^{2} \)
here, d is the diameter.
\( \Rightarrow \ \frac{d_{2}}{d_{1}}=\sqrt{\frac{R_{1}}{R_{2}} \times \frac{l_{2}}{l_{1}}}=\sqrt{2 \times 2}=\frac{2}{1} \)
\(d_{1}: d_{2} =1: 2 \)
(b) \(v_{d}=\frac{e V}{m l} \tau=\frac{e I R}{m l} \tau\)
\(\frac{v_{d_{1}}}{v_{d_{2}}}=\frac{R_{1}}{R_{2}} \cdot \frac{l_{2}}{l_{1}}=\frac{2}{1} \times \frac{2}{1}=4: 1\)

Difference between emf (\(\varepsilon \)) and terminal voltage (V)

(Anyone) or any other relevant difference
Negative of slope gives internal resistance

As, the value of R is increased, the current flowing in the circuit will decrease. And the potential gradient, i.e. potential drop per unit length also decreases, so that the balance length will increase. Thus, J will shift towards B.

By Kirchhoff's first law at D,
 \({ I }_{ DC }=1A\)          \([\because { I }_{ Dc }+1=2]\)
    
Along ACDBA,                                                             
\({ V }_{ A }+1+1\times 2-2={ V }_{ B }\)  (V_A = 0)
But,  \({ V }_{ B }=1+2-2=1V\)
\({ V }_{ B }=1V\)                           

According to the question,

Applying Kirchhoff's junction rule at F,
 I₃ = \({ I }_{ 1 }\) +\({ I }_{ 2 }\)                ......(i)
Applying Kirchhoff's second rule in loop ABCF,
-30I₁+ 20-20I₃ = 0
\(\Rightarrow \) 3I₁+2I₃ = 2           ....(ii)
In loop ABDE,
-30I₁+20I₂ - 80 =0
3I₁ + 2I₂ = 8          ......(iii)  (1)
From Eqs. (i) and (ii), we have
3I1+ 2I1 + 2I2 = 2
\(\Rightarrow \) 5I1 + 2I₂ = 2         .....(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
8I₁ = -6
\(\Rightarrow \)I1 = -\(\frac { 3 }{ 4 } \) A              (1)    

Let V be the emf of the d.c. source in volt. \(R \ \Omega\) be the resistance of each resistor.
In series, total resistance = R + R + R = 3 R
Power dissipated = \(\frac{V^2}{3R}=X \ \ or \ \ \frac{V^2}{R}=3X\)
In parallel, total resistance, R' is given by
\(\frac{1}{R'}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{3}{R} \ or \ \ R'=R/3\)
Power dissipated = \(\frac{V^2}{R'}=\frac{V^2}{R/3}=\frac{3V^2}{R}\)
= 3 x 3 x = 9 

As the temperature of certain metals and alloys decreases, their resistance also decreases. When the temperature reaches a certain critical value called critical temperature, the resistance of material completely disappears, i.e., it becomes zero. Then the material behaves as a superconductor.
Thus superconductors are those material conductors whose resistances disappear at critical temperature. The critical temperature is different materials.
Superconductors are used
(i) in power transmission
(ii) to produce very high speed computers.

A thermistor differs from an ordinary resistance in the following ways.
(i) The resistivity and hence the resistance of the thermistor changes very rapidly with change of temperature.
(ii) The temperature coefficient of resistivity of the thermistor is very high.
(iii) The temperature coefficient of resistivity of the thermistor can be both, positive and negative.

The units of conductance and conductivity are simen (denoted by S) and simen metre ^-1 (denoted by S m^-1) respectively. We know that conductance = 1/resistance.
We know, electrical conductivity,  \(\sigma = \frac{1}{\rho}=\frac{l}{RA}\)
If l = 1 m, R = 1 \(\Omega\), A = 1 m² then \(\sigma\) = 1 S m^-1
Thus 1 S m^-1 is the electrical conductivity of a conductor of length 1 m, area of cross-section 1 m² and resistance 1 \(\Omega\) or conductance 1S.
Dimensional formula of conductance,
\(G=\frac{1}{R}=\frac{1}{V}=\frac{[A]}{[ML^2T^{-3}A^{-1}]}=[M^{-1}L^{-2}T^{3}A^2]\)
Dimensional formula of conductivity,
\(\sigma=[M^{-1}L^{-3}T^3A^2]\)

For manganin, the temperature coefficient of resistance is very low. Due to it, the resistance of manganin wire remains almost unchanged with a change in temperature. The resistivity of manganin is high. Therefore, for making a standard resistance of the given value, the smaller length of wire is needed. It is due to these facts, the wire of manganin is used for making standard resistance coil. The coil is double folded on itself to avoid the inductive effect and it wound over the non-conducting frame in order to avoid the conductive effect and the leakage of current.

The cause of death is not heating due to the current passing through a person. Though he may receive burns if the currents are too large. The cause of death is the interference of the external currents with our highly sensitive nervous system which is basically electrical in nature, which in turn affects the heart beating. Beyond a certain point, this interference becomes fatal.

About \(10 kΩ.\) This resistance is mainly due to skin through which current enters and leaves our body.

With an increase in temperature (i.e., on a warm day) the internal resistance of a car battery decreases. Due to it, the battery can supply large current which helps in starting the car engine easily.

Here \(R=\frac{\rho l}{A}=4\)
\(R^{\prime}=\frac{\rho(l / 2)}{2 A}=\frac{\rho l}{4 A}=\frac{R}{4}=\frac{4}{4}=1 \Omega\)

A constant current can be kept inside a conductor by maintaining a constant potential difference across the two ends of conductor.

Here, Ohm's law is valid because ratio of voltage and for different readings is same.
Also,the resistivity of alloy manganin is nearly independent of temperature.