Temperature, T₁ = 27.5°C
Resistance of the silver wire at T₁, R₁ = 2.1 Ω
Temperature, T₂ = 100°C
Resistance of the silver wire at T₂, R₂ = 2.7 Ω
Temperature coefficient of silver = α
It is related with temperature and resistance as 
\(\alpha=\frac{R_{2}-R_{1}}{R_{2}\left(T_{2}-T_{1}\right)}\)
\(=\frac{2.7-2.1}{2.1(100-27.5)}=0.0039^{\circ} \mathrm{C}^{-1}\)
Therefore, the temperature coefficient of silver is 0.0039°C⁻¹.

Given: \(I=(2 t+4) \mathrm{A}\)
As we know dq = Idt
\( \Rightarrow \ q=\int(2 t+4) d t \)
\(q=\left(2 \cdot \frac{t^{2}}{2}+4 t\right) \)
\(\text { At } t=2 \mathrm{~s}, q=2 \cdot \frac{4}{2}+4.2=12 \mathrm{C} \)

(i) Care for elderly people, empathy, willingness to gain knowledge.
(ii) A parallel circuit can run several devices using the full voltage.
The two advantages are as follows:
(a) If one appliance fails, then others can run normally.
(b) If one of the device shorts, the other devices will receive no voltage preventing overload damage.

Since drift velocity \({ v }_{ d }\) and current, I flowing in a conductor are related by the relation:
\({ v }_{ d }=\frac { I }{ neA } ......(1)\)
Also drift velocity in terms of average relaxation time τ is given by
\({ v }_{ d }=\frac { eE\tau }{ m } .....(2)\)
From (1) and (2), we have
\(\frac { eE\tau }{ m } =\frac { I }{ neA }\)
\( \\ or\quad \frac { E }{ I } =\frac { m }{ n{ e }^{ 2 }A\tau } \)
\(or\quad \frac { V }{ lI } =\frac { m }{ n{ e }^{ 2 }A\tau }\)
\(or\quad \frac { V }{ I } =\frac { ml }{ n{ e }^{ 2 }A\tau } .......(3)\)
The R.H.S. of Eq(3) is constant
\(\frac { V }{ I } = \ Constant\)
This is Ohm′s law
\((But \ \frac { V }{ I } =R,\ \)the resistance of the conductor)
So Equation (3) becomes
\(R=\frac { ml }{ n{ e }^{ 2 }A\tau }\)
\(But\quad R=\rho \frac { l }{ A } \)
\(\therefore \quad \rho \frac { l }{ A } =\frac { ml }{ n{ e }^{ 2 }A\tau }\)
\(\rho =\frac { m }{ n{ e }^{ 2 }A\tau } )\)

Consider that a wire of length l and area of cross-section A be subjected to an electric field of strength E.
If V = Potential difference applied across the end of the wire,
\(E=\frac { V }{ l } or \ V=El\)

Let n = number of free electrons per unit volume of the conductor
\({ v }_{ d }\) = drift velocity of electrons charge flowing through the conductor wire,
q = nAle
Time taken by the electrons to cross the conductor,
\(t=\frac { Distance\quad }{ Velocity } =\frac { 1 }{ { v }_{ d } } \) 
\( Current \ I=\frac { charge }{ time } =\frac { nAle }{ \frac { l }{ { v }_{ d } } } \)

To establish that electrons carry current in a conductor, the \(\left( \frac { e }{ m } \right) \) of charge is measured from its angular momentum.
Let r = radius of circular loop
n = no. of electrons per unit length
I = strength of current flowing through the loop
v = drift velocity of charge flowing in the conductor
No.of charges flowing per unit length of the loop
= (nv)
If e = charge on each charged particle or current carrierTotal charge flowing per sec through any cross-section
= (nv) e
This is called current I = (nv) e
When charged particle goes along a circular loop its angular momentum is given by
L = mvr
Total angular momentum due to all current carriers is
L = (mvr)n = m(nv) r
\(L=m\times \frac { I }{ e } r\)
\( L=I\left( \frac { m }{ e } \right) r\)
\(\left( \frac { e }{ m } \right) \) can be measured by measuring L, I and r.

The angular momentum is measured by relating it to the twist produced in a suspension from which loop is hung. On reversing the direction of current, a twist is produced in suspension due to sudden change in angular momentum.
The value of twist depends upon the angular momentum.
It was found that \(\left( \frac { e }{ m } \right) \) determined by this experiment corresponds to electron from other experiment.
It thus experimentally established that current in metal is carried by negatively charged ions.

It states that current flowing through a conductor is proportional to the potential difference across its two ends provided the physical conditions of the conductor remain unchanged.
If V is the potential difference between two ends of a conductor and I is the current flowing through it, then
\(V\alpha I\)
\(V=RI\)
Where R is the constant of proportionality and is called the resistance of the conductor. Its value depends upon
(i) Shape of the conductor
(ii) Length of conductor
(iii) Nature of the material

If a graph is plotted between V and I, the graph will be a straight line passing through the origin.
(b) Resistance is the property of a material by virtue of which it opposes the flow of current through it and quantitatively it is given by,
\(R=\frac { V }{ I }\)
\( =\frac { Potential \ difference }{ Current } \)
definition of ohm
Hence a conductor has a resistance of one ohm if a current of one ampere flows through it when a potential difference of one volt is maintained across its two ends.

Emf of the storage battery, E = 8.0 V
Internal resistance of the battery, r = 0.5 Ω
DC supply voltage, V = 120 V
Resistance of the resistor, R = 15.5 Ω
Effective voltage in the circuit = V¹
R is connected to the storage battery in series. Hence, it can be written as
V¹ = V - E
V¹ = 120 - 8 = 112 V
Current flowing in the circuit = I, which is given by the relation,
\(I=\frac{V^{1}}{R+r}\)
\(=\frac{112}{15.5+5}=\frac{112}{16}=7 A\)
Voltage across resistor R given by the product, IR = 7 x 15.5 = 108.5 V
DC supply voltage = Terminal voltage of battery + Voltage drop across R
Terminal voltage of battery = 120 - 108.5 = 11.5 V
A series resistor in a charging circuit limits the current drawn from the external source. The current will be extremely high in its absence. This is very dangerous.