Using the formula,
$R_t = R_O (1 + \alpha t)$

Therefore,
5 = R_o ( 1+ 50$\mathrm{\alpha }$)
6 = R_o ( 1+ 100$\mathrm{\alpha }$ ) 

On solving the above two equations we get,
                    $$\begin{gathered} \frac{5}{6} =\frac{1 + 50\alpha }{1 + 100\alpha } \\ \Rightarrow \alpha =\frac{1}{200} \end{gathered}$$ 
Now, putting the value of $\mathrm{\alpha }$ in one of the equations above , we get  

               5= R_O [1+ 50$\left(\frac{1}{200}\right)$ ] = 4 $\mathrm{\Omega }$ .
Therefore, the value of resistance at 0^o C is 4Ω .  

Given, the metallic wire is stretched such that, it's length increases to 5%. 

Percentage change in length, dl/l = 5% 

Resistance of wire is given by R = $\mathrm{\rho }\frac{\mathrm{l}}{\mathrm{A}}$ = $\mathrm{\rho }\frac{{\mathrm{l}}^2}{\mathrm{V}}$ 

Therefore, percentage change in the resistance of wire is given by,

$\frac{dR}{R} = 2. \frac{dl}{l } = 2 \times 5 \% = 10 \%$  

Resistance is given by R= $\mathrm{\rho }\frac{\mathrm{l}}{\mathrm{A}}$
The length and resistance of the wires are equal.  

$$\begin{gathered} \Rightarrow {\mathrm{R}}_1 = {\mathrm{R}}_2 \\ \Rightarrow {\mathrm{\rho }}_1\frac{\mathrm{l}}{{\mathrm{A}}_1} ={\mathrm{\rho }}_2\frac{\mathrm{l}}{{\mathrm{A}}_2} \\ \Rightarrow \frac{{\mathrm{\rho }}_1}{{\mathrm{\rho }}_2} = \frac{{\mathrm{A}}_1}{{\mathrm{A}}_2} \end{gathered}$$  

Resistivity of Copper wire is less than that of the resistivity of magnanin wire. Therefore, the area of cross- section of copper wire is less than that of magnanin. 
Hence, Manganin wire is thicker. 

Wheatstone bridge principle states that if four resistances P, Q, R and S are arranged to form a bridge as shown in the fig. below with a cell E and one way key, put between the points A and C and a galvanometer is connected in between the points B and D such that there is no current through G. The balance point is obtained when, galvanometer shows no deflection.
  

In this case, $\frac{\mathrm{P}}{\mathrm{Q}} = \frac{R}{S}$ 

It is usually used to find the value of an unknown resistance.

Kirchoff's rule states that :

a) Junction rule : At any junction, the sum of currents entering the junction is equal to the sum of currents leaving the junction. 

b) Loop rule : The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero.