We know that current density at any point inside the conductor is
$J=\sigma E$
=> $E=\frac{J}{\sigma}=\rho J$ ---(1)
Since current is flowing along the axis of the tube so field is also directed along the axis of the tube. Consider the tube to be made up of large number of coaxial annular discs as shown below in the figure
Consider a disc of thickness dx at a distance x from one of the ends of the cylindrical conductor, Thus we have
$E=-\frac{dV}{dx}$
And
$J=\frac{I}{A}=\frac{I}{\pi(b^2-a^2)}$
Where A is the area of cross-section of the cylinder. From equation 1
$\frac{-dV}{dx}=\frac{\rho I}{\pi(b^2-a^2)}$
Or
$\int_{v_1}^{v_2}dV=\int_{0}^{L}\frac{\rho Idx}{\pi(b^2-a^2)}$
Or
$V=V_1-V_2=\frac{\rho LI}{\pi(b^2-a^2)}$
Now Resistance
$R=\frac{V}{I}$
So,
$R=\frac{\rho L}{\pi(b^2-a^2)}$
At $t= T_0$ °C, equivalent resistance of resistors connected in series is
$R_{eq}= R_1 + R_2$
At t= T °C
$R'_{eq}= R'_1 + R'_2$
$=R_1[1+\alpha_1(T-T_0)]+R2[1+ \alpha _2(T-T_0)]$
$=R_1+R_2+(R_1\alpha_1+R_2\alpha_2)(T-T_0)$ ---(1)
If $\alpha _{eq}$ is equivalent temperature coefficient in series combination then
$R'_{eq}=R_{eq}[1+\alpha_{eq}(T-T_0)]$
$=(R_1+R_2)[1+\alpha_{eq}(T-T_0)]$ ---(2)
From equation 1 and 2,we get
$R_1+R_2+(R_1\alpha_1+R_2\alpha_2)(T-T)= (R_1+R_2)[1+\alpha_{eq}(T-T_0)]$
Solving the equation we get
$\alpha_{eq}=\frac{R_1\alpha_1+R_2\alpha_2}{R_1+R_2}$
Again find the $\alpha _{eq}$ for the parallel combination of resistors, Answer shall be
$\alpha_{eq}=\frac{R_1\alpha_2+R_2\alpha_1}{R_1+R_2}$
After stretching the wire ,its volume would remain same .If $L_1$ and $A_1$ be the length and area of the wire before stretching and $L_2$ and $A_2$ be the length and area after stretching then
$L_1A_1=L_2A_2=V(Volume)$
Now we know that
$R_1=\frac{\rho L_1}{A_1}=\frac{\rho L_1^2}{V}$
Similarly
$R_2=\frac{\rho L_2}{A_2}=\frac{\rho L_2^2}{V}$
Hence
$R \alpha L^2$
=> $\left(\frac{R_1}{R_2}\right)=\left(\frac{L_1}{L_2}\right)^2$
But As per given data
$L_2= L_1 + \frac{.2 \times L_1}{100} =1.002L_1$
So,
$\left(\frac{R_1}{R_2}\right)=\left(\frac{1}{1.002}\right)^2$
Or
$R_2=(1.002)^2R_1$
Or $R_2=1.004R_1$
Change in resistance =$R_2-R_1 =.004R_1$
% Change in resistance=$\frac{.004R_1 \times 100}{R_1} =.4$ %
Let ρ be the resistivity of the material of the wires then resistance of the three wires can be written as
$R_1=\frac{\rho L_1}{A_1}$ ,$R_2=\frac{\rho L_2}{A_2}$ ,$R_3=\frac{\rho L_3}{A_3}$
Also from the given data
L_{1}=2L ,L_{2}=3L,L_{3}=4L
And
D_{1}=3D,D_{2}=4D,D_{3}=5D
Now Areas of the resistors can be written as
$A_1=\pi r_1^2=\pi(\frac{3D}{2})^2$
$A_2=\pi r_2^2=\pi\left(\frac{4D}{2}\right)^2$
$A_3=\pi r_3^2=\pi\left(\frac{5D}{2}\right)^2$
Thus resistance of the wires A,B ,C can be written as
$R_1=\frac{8\rho L}{9\pi D^2}$
$R_2=\frac{3\rho L}{4\pi D^2}$
$R_3=\frac{16\rho L}{25\pi D^2}$
Let I_{1} , I_{2} and I_{3} be the current in the wires A,B ,C respectively. As wires are connected in parallel,
We have
$I_1+I_2+I_3=5$
As these three wires are connected between the same two points A and B ,we have
V=IR_{eq}
Now R_{eq} is given by
$\frac{1}{R_{eq}}=\left(\frac{9\pi D^2}{8\rho L}+\frac{4\pi D^2}{3\rho L}+\frac{25\pi D^2}{16\rho L}\right)$
$=\frac{\pi D^2}{\rho L}\left(\frac{9}{8}+\frac{4}{3}+\frac{25}{16}\right)$
$ =\frac{193\pi D^2}{48\rho L}$
Or
$R_{eq}=\frac{48\rho L}{193\pi D^2}$
Now
V=IR_{eq}
Or
$V=5 \times \frac{48\upsilon L}{193\pi D^2}=\frac{240\rho L}{193\pi D^2}$
Since Potential in Parallel combination is same across all the resistors. Hence current flowing through the first wire
$I_1=\frac{V_1}{R_1}=\left(\frac{240}{193}\right)\left(\frac{\pi D^2}{\rho L}\right)\left(\frac{\rho L}{\pi D^2}\right)\left(\frac{9}{8}\right)=1.4A$
Similarly I_{2}=1.65 A and I_{3}=1.94 A
Let us first label the resistors in the figure
Resistors c and d are connected in series combination
Thus $R_{cd}=R+R=2R$
Now $R_{cd}$ is connected in parallel with resistors e
So equivalent resistance
$\frac{1}{R_{cde}}=\frac{1}{2R}+\frac{1}{2R}$
Or
$R_{cde}=R$
Now the above figure is reduced to below figure
$R_{cde}$ and g are in series
So, $R_{cdeg}=2R$
Now $R_{cdeg}$ and F in parallel
$\frac{1}{R_{AB}}=\frac{1}{2R}+\frac{1}{2R}$
$R_ab=R$ ω
Let us assume an elemental cylindrical shell of thickness dx at distance x from axis.
Now we know
$J=\frac{I}{A}$
For this elemental cylindrical shell
$J=\frac{dI}{2\pi x}$
Or $dI= K(1-\frac{x}{A})(2\pi x )$
Integrating
$I=\int_{0}^{A}{K\left(1-\frac{x}{A}\right)(}2\pi x) $
$=\frac{K\pi A^2}{3}$
(a) Once the capacitors are fully charged they do not draw any current from the battery. Thus in a steady state
$I=\frac{\xi}{r+R_1+R_2}=\frac{10}{.5+2+4}=1.53$A
The potential difference across the capacitor is
$V=I\left(R_1+R_2\right)=1.53\left(2+4\right)=9.18$V
Since the capacitors are connected in series so, the equivalent capacitance is
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$
Using the values of $C_1$ and $C_2$ in above equation we find
$C_{eq}=0.0143 $ μ F
Charge on a capacitor with capacitance equal to equivalent capacitance is $Q=C_{eq}V$
Thus
$Q=C_{eq}V=9.18 \times 0.0143=0.131 $ μ C
And
$Q=Q_1=Q_2=0.131$ μ C
(b) When the switch A is closed the steady state current remains the same that is I=1.53A but capacitors are no longer connected in series combination so,
$V_1=IR_1=1.53 \times 2=3.06$V
$V_2=IR_2=1.53 \times 4=6.12$V
And charge on $C_1$ equals
$Q_1=C_1 V_1=0.02 \times 3.06=.0612$ μ C
$Q_2=C_2V_2=0.05 \times 6.12=.306$ μ C
Let X is the equivalent resistance of the entire network. Since the network is infinite, so adding more set of resistance each of resistance R=2 ω across the terminals should not affect total resistance i.e., it should still remain equal to X. Figure given below shows the network after adding one set of three resistances.
If R' is equivalent resistance of this new network then,
R'=R+(resistance equivalent to parallel combination of X and R) + R
$=2R+\frac{XR}{X+R}$
Now since addition of one set of three resistances should not alter the total resistance of the infinite network. So we have
$R'=X$
Or,
$2R+\frac{XR}{X+R}=X$
Since, R=2 ω
$4+\frac{2X}{X+2}=X$
$4X+8+2X=X^2+2X$
$X^2-4X-8=0$
Using quadratic formula
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$X=\frac{-(-4)\pm\sqrt{16-4(-8)}}{2}$
$X=\frac{4\pm4\ \sqrt3}{2}$
Ignoring X with negative value we find
X=5.46 ω
Now E=12V and r=.5 ω
Current drawn by the network is
$I=\frac{E}{X+r}$
$I=\frac {12}{5.46+.5}$
I=2.01A