(1) Raghav: affection towards birds, taking appropriate action. Authorities: duty conscious. 
(2) The bird may get electrocuted; avoid sparking as shown in the diagrams below:

First law or Current law or Junction law: It states, "In any electrical network, the algebraic sum of currents meeting at a point (or junction) is zero."
Second law or Voltage law or Loop law: It states, 
"In a closed circuit, the algebraic sum of the products of the current and the resistance in each of the conductors in any closed path (or mesh) in a network plus sum of emf in that path is equal to zero."
Now from the given figure,
I₁+ I₂ = I₃           .....(i)
I₂ = I₃ - I₁        ........(ii)
-20I₁ - 40I₃ = 40 
-40 (I₁ + 2I₃) = 80
I₁+ 2I₃ = 2            ...(iii)
40I₃ + 20I₂ = 80 + 40
20 (2I₃ + I₂) = 120
2I₃ + I₂ = 6
2I₃+ (I₃- I₁) = 6
2I₃+ I₃ -I₁ = 6
3I₃ - I₁ = 6      ......(iv)
Now solving eqns. (ii) and (iv)
I₁+ 2I₃ = 2
-I₁ + 3I₃ = 6
\(5I_3=8\Rightarrow I_3=\frac{8}{5}=1.6 \ amp.\)
2I₃+ I₁= 2
I₁ = 2-2I₃
I₁= 2-2 x \(\frac{8}{5}\)
Now, \(I_1=6-2I_3\)
\(\Rightarrow\ \ I_2=6-2\times \frac{8}{5}\)
\(\Rightarrow\ I_2=6-\frac{16}{5}\)
\(\Rightarrow\ \ I_2=\frac{30-16}{5}=\frac{14}{5}\)
\(\Rightarrow \ I_2=2.8\ amp\)

(i) Since resistivity depends on the nature of material of wire and is independent of dimensions of wire, so the resistivity remains unchanged.
(ii) In both the cases, the volume of the wire is same.
So
V = Al = A'l'
or A' = \(\frac{lA}{l'}=\frac{lA}{3l}=\frac{A}{3}\)
Now R = \(\frac{\rho l}{A}\)
and R' = \(\frac{\rho l'}{A'}=\frac{\rho 3l}{A/3}=\frac{9 \rho l}{A}= 9 \ R\)
R' = 9 x 10 = \(90 \ \Omega\)

Here,
 R = \(5.0 \ \Omega\);
r₁ = 5 x 10^-2 m;
D = 2.0 x 10^-3 m;
\(\rho= 2.0 \times 10^{-7} \ \Omega m\).
R = \(\frac{\rho l}{\pi D^2/4}\)
or l = \(\frac{R \pi D^2}{4 \rho}\)
Let n be the number of turns in the coil. Then total length of the wire used, l = \(2 \pi r_1 n\)
or n = \(\frac{l}{2 \pi r_1}=\frac{R \pi D^2}{4 \rho \times 2 \pi r_1}=\frac{RD^2}{8 \rho r_1}\)
 = \(\frac{5.0 \times (2.0 \times 10^{-3})^2}{8 \times (2.0 \times 10^{-7})\times (5 \times 10^{-2})^2}\)
 = 250

Here,
l = 2.2 m;
A = 2.0 sq. mm = 2 x 10^-6 sq.m,
I = 6 A;
n = 8.5 x 10²⁸ m^-3
Drift velocity,
v_d = \(\frac{I}{nAe}\)
= \(\frac{6}{(8.5 \times 10^{28})\times(2\times 10^{-6})\times (1.6\times10^{-19})}\)
= 2.2 x 10^-4 ms^-1
Time of drifting the electron,
t = \(\frac{l}{v_d}=\frac{2.2}{2.2 \times 10^{-4}}\)
= 10⁴ s

Here,
n = 8.4 x 10²⁸ m^-3,
A = 10^-6 m²,
I = 5.4 A
v_d = \(\frac{I}{nAe}\)
= \(\frac{5.4}{(8.4 \times10^{28})\times(10^{-6})\times(1.6 \times10^{-19})}\)
= 0.4 x 10^-3 m/s
= 0.4 mm/s