(i) I = 12/11 A
(ii) I₁ = 2/11 A
(iii) V₂ = 246/11 V
(iv) V₁ = 168/11 V

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, \(I=\frac{E}{X+10^{4}}\)
\( \Rightarrow \left(X+10^{4}\right) I =E \)
\(\therefore X I =E-I \times 10^{4} \)
\(\Rightarrow X =\frac{E}{I}-10^{4}=\frac{120}{4 \times 10^{-4}}-10^{4} \)
\(\therefore X =29 \times 10^{4} \Omega=290 \mathrm{k} \Omega \)

Power consumption in a day, i.e. in 5 h = 10 units
or power consumption per hour = 2 units
or power consumption = 2 units = 2kW= 2000 J/s
Also, we know that, power consumption in resistor,
P = V \(\times\)I \(\Rightarrow\) 2000 W = 220 V \(\times\)I
or   I \(\approx\)9A
Now, the resistance of wire is given by\(R=\rho \frac{l}{A}\)
where, A is cross-sectional area of conductor.
Power consumption in first current carrying wire is given by P = I² R
\(=\rho \frac{l}{A} I^2=1.7 \times 10^{-8} \times \frac{10}{\pi \times 10^{-6}} \times 81 \mathrm{~J} / \mathrm{s} \approx 4 \mathrm{~J} / \mathrm{s}\)
The fractional loss due to the joule heating in first wire
\(=\frac{4}{2000} \times 100=0.2 \%\)
Power loss in aluminium wire \(=4 \frac{\rho_{\mathrm{Al}}}{\rho_{\mathrm{Cu}}}\)
=1.6 \(\times\)4 = 6.4 J/s
The fractional loss due to the joule heating in second wire
\(=\frac{6.4}{2000} \times 100=0.32 \%\)

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.

This impression is misleading. In fact, there is no special attractive force that keeps a person stuck with a high power line while touching that wire, whereas a current of few milliampere is enough to disorganise our nervous system. As a result of it, the affected person loses temporarily his ability to exercise his nervous control to get himself free from the high power line.

When a person accidently touches the line wire of a 240 V supply, a current I = 240/10000 = 0.024 A = 24 mA, flows through the body of the person. This current interferes with the nerve process related to our heart beating, which is basically electrical in nature, this, in turn, accounts for a strong shock.

We know that if two wires are in series, there will be the same current in each wire. The current I through the wire is related with drift velocity v_d by the relation
I = nAev_d  or  Av_d = a  constant
or  \(v_d \propto \frac{1}{A}\)   or   \(v_d \propto \frac{1}{\pi r^2}\)
Since in the second wire, the radius of wire becomes 2 times hence the drift velocity will become one-fourth. In the case of iron wire, the number density of free electrons will be less in comparison to copper wire and hence the drift velocity will be more in comparison to that in copper wire.

The best conductor is silver because electrical conductivity is inversely proportional to the resistivity and resistivity of silver is least.

(i) Constantan [Cu (60%) + Ni (40%)]
(ii) Manganin [Cu (84%) + Mn (12%) + Ni (4%)]

With the increase in temperature, the free electrons collide more frequently with the ions/atoms of conductor, resulting decrease in relaxation time.

Yes. A steady current in a cylindrical conductor if electric force is acting on free electrons which make the electrons to move in a particular direction. It is possible due to electric field within the conductor.

Electric current is a scalar quantity because it does not follow the laws of vectors addition, i.e., the angle between the wires carrying the current does not affect the total current in the circuit.