The heat supplied under these conditions is the change in internal energy
$$Q = \Delta U$$
The heat supplied $$Q = {i^2}RT$$
=1 × 1 × 100 × 5 × 60 = 30,000 $$J$$ = $$30\,kJ$$

The algebraic sum of the changes in potential in complete transversal of a mesh (closed loop) is zero. i.e. $$\Sigma V = 0$$
So, $${\varepsilon _1} - \left( {{i_1} + {i_2}} \right)R - {i_1}{r_1} = 0$$

According to first law of Faraday, the mass of a substance deposited on an electrode is directly proportional to the charge flowing through the electrolyte i.e. $$m \propto q$$
If a current $$i$$ passes for a time $$t,$$ then we know,
$$q = it$$
Hence, $$m \propto it\,\,{\text{or}}\,\,m \propto i$$
Thus, mass deposited on an electrode is directly proportional to current passed through it.

Effective resistance of a circuit,

$${R_{{\text{eff}}}} = 40.8 + \frac{{480 \times 20}}{{480 + 20}} = 40.8 + 19.2 = 60\,\Omega $$
So, current flowing across ammeter,
$$I = \frac{V}{R} = \frac{{30}}{{60}} = \frac{1}{2} = 0.5\,A$$
Hence, reading of ammeter $$= 0.5\,A$$

The circuit is symmetrical about the axis $$POQ.$$
The circuit above the axis $$POQ$$  represents balanced wheatstone bride. Hence the central resistance $$2R$$  is ineffective. Similarly in the lower part (below the axis $$POQ$$ ) the central resistance $$2R$$  is ineffective.

Therefore the equivalent circuit is drawn.
$$\eqalign{ & \therefore \frac{1}{{{R_{PQ}}}} = \frac{1}{{4R}} + \frac{1}{{4R}} + \frac{1}{{2r}} = \frac{{r + r + 2r}}{{4Rr}} \cr & {R_{PQ}} = \frac{{2Rr}}{{R + r}} \cr} $$

The relation between length and area is
$$R = \frac{{\rho l}}{A}\,......\left( {\text{i}} \right)$$
where, $$\rho $$ being specific resistance is the proportionality constant and depends on nature of material.
(i) Length $$ = \frac{L}{2},$$   area $$ = 2\,A$$
Putting in Eq. (i), we have
$$R = \frac{{\rho \left( {\frac{L}{2}} \right)}}{{2A}} = \frac{{\rho L}}{{4A}}$$
(ii) Length $$= L,$$  area $$ = A$$
Putting in Eq. (i), we have
$$R = \frac{{\rho l}}{A}$$
(iii) Length $$ = 2L,$$  area $$ = \frac{A}{2}$$
Putting in Eq. (i), we have
$$R = \rho \frac{{2L}}{{\frac{A}{2}}} = \frac{{4\rho L}}{A}$$
As it is understood from above, resistance is minimum only in option (B).

When current is passed through a spring then current flows parallel in the adjacent turns.
NOTE : When two wires are placed parallel to each other and current flows in the same direction, the wires attract each other.
Similarly here the various turns attract each other and the spring will compress.

Total power consumed by electrical appliances in the building, $${P_{total}} = 2500W$$
Watt = Volt × ampere
$$\eqalign{ & \Rightarrow 2500 = V \times I \cr & \Rightarrow 2500 = 2201 \cr & \Rightarrow I = \frac{{2500}}{{220}} = 11.36 \approx 12A \cr} $$
(Minimum capacity of main fuse)

As the cross-sectional area of the conductor is non-uniform so current density will be different.
As $$I = JA\,......\left( {\text{i}} \right)$$
It is clear from Eq. (i), when area increases the current density decreases so the number of flow of electrons will be same and thus the current will be constant.

There will be no current flowing in branch $$BE$$  in steady condition.
Let $$I$$ be the current flowing in the loop $$ABCDEFA.$$
Applying Kirchoff's law in the loop moving in anticlockwise direction starting from $$C.$$
$$\eqalign{ & + 2V - I\left( {2R} \right) - I\left( R \right) - V = 0 \cr & \therefore V = 3IR \cr & \Rightarrow I = \frac{V}{{3R}}\,\,......\left( 1 \right) \cr} $$
Applying Kirchoffs law in the circuit $$ABEFA$$  we get on moving in anticlockwise direction starting from $$B$$

$$\eqalign{ & + V + {V_{cap}} - IR - V = 0\left( {{\text{where}}\,{V_{cap}}\,{\text{is}}\,{\text{the}}\,p.d.\,{\text{across}}\,{\text{capacitor}}} \right) \cr & \therefore {V_{cap}} = IR = \left( {\frac{V}{{3R}}} \right) \times R = \frac{V}{3} \cr} $$