Current Electricity and Circuits

Since the phase angle is $45^{\circ}$,

As per mechanical-electrical analog, displacement is analogous to charge and force analogous to voltage.

The drift velocity of electrons in a conductor is of the order of $10^{−4} m/s. It is very small compared to the thermal speed which is of the order of 105 m/s.

Given
Initial current through the wire is $I = 1A$
Initial drift velocity is, $v _d = 1.2 \times 10^{-4} ms^{-1}$
Increased current is, $I' = 5A$
The current, I through the wire is given by
$I = \mu _e.e.A.v _d$
where, $\mu _e$ is the free electron density, e is the charge on electron, A is the area of cross section of the wire and v_d is the drift velocity of electrons. 
Since,  the free electron density, the charge on electron and the area of cross section of the wire are constant, hence
$I \propto v _d$.................(1)
Now, current through the wire is increased to 5 A, if the new drift velocity of electrons is $v' _d$ then
$I' \propto v' _d$................(2)

$V _d = \mu E = \mu \dfrac{V}{\ell}$
$V _d = \dfrac{\mu. I R}{\ell} \dfrac{\mu. I _{\rho} \ell}{A \ell} = \dfrac{\mu . I _{\rho}}{A}$
$\mu = \dfrac{V _d . A}{I _{\rho}} = \dfrac{1.1 \times 10^{-3} \times \lambda \times 25 \times 10^{-6}}{5 \times 1.1 \times 10^{-8}}$
$\mu = 1.57 \,  m^2 $ volt/sec.

Considering law of conservation of momentum ,electrons possess drift velocity which is greater than velocity of ions.  Thus $K _1>K _2$. hence correct option is option C.

We know 
$I = \eta eAV$
$2 = 1 \times {10^{24}} \times 1.6 \times {10^{ - 24}} \times 1 \times v$
$\boxed{v = 1.25\,m/s}$

Drift velocity v_d = I / (neA). In series, current I is the same, so v_d is inversely proportional to A. In parallel, voltage V is the same, so I = V/R, where R = rho*L/A. Thus v_d = V / (ne*rho*L). Calculating these ratios yields 3:2 for series and 2:1 for parallel.

If L is the length of wire so velocity is given by $v=\dfrac{L}{t}$

Total number of free electrons in the wire, $Q=nqLA$

Current,

$ I=\dfrac{Q}{t} $

$ I=\dfrac{nqLA}{t} $

$ I=nqvA $

$ v=\dfrac{I}{nqA} $

Where, n is the number of electron, $n={{10}^{29}}$

q is the charge of an electron, $q=1.6\times {{10}^{-19\,}}C$

A is area, $A={{10}^{-16}}\,{{m}^{2}}$

I is current, $I=40\,A$

So, drift velocity,

$ v=\dfrac{40}{{{10}^{29}}\times 1.6\times {{10}^{-19}}\times {{10}^{-16}}} $

$ v=25\times {{10}^{6}}\,m/s $

We know, for a conductor carrying current,
Drift speed $= neI =ne \dfrac{V}{R}$
Resistivity $= \dfrac {RA}{L}$
where,
$n$ is no. of electrons, 
$e$  is charge on electrons,
$V$  is applied voltage,
$R$  is resistance and 
$A$  and $L$  are area of cross-section and length of resistor.
From above equations it is clear that drift speed, resistivityand resistance of resistor will be affected due to heating of resistor

We know that Drift velocity $v _d = \displaystyle \dfrac{eE}{m} \tau = \dfrac{e}{m} \tau \left ( \dfrac{V}{l} \right ) $ ($\because E = \dfrac{V}{l})$