Step 1: Convert the given conductivity into molar conductivity.
Molar conductivity relates to specific conductivity through the concentration expressed in mol per cubic centimetre, using $\Lambda_m = \dfrac{\kappa\times1000}{C}$. \[ \Lambda_m = \frac{1.06\times10^{-2}\times1000}{0.1} = 106\ \text{S cm}^2\text{mol}^{-1} \]
Step 2: Build the limiting molar conductivity from the individual ionic values and find the degree of dissociation.
By Kohlrausch's law of independent migration, $\Lambda_m^\circ = \lambda^\circ_{Na^+} + \lambda^\circ_{Cl^-} = 50.1 + 76.5 = 126.6\ \text{S cm}^2\text{mol}^{-1}$, and the degree of dissociation follows from $\alpha = \Lambda_m/\Lambda_m^\circ$. \[ \alpha = \frac{106}{126.6} \approx 0.837 \]
Step 3: Work out the direction of current flow in the given galvanic reaction.
In $2Ag^+ + Zn \rightarrow 2Ag + Zn^{2+}$, zinc is oxidised at the anode and loses electrons, which travel through the external wire to the silver electrode where reduction occurs. Since conventional current flows opposite to electron flow, current travels from the silver electrode to the zinc electrode.
Step 4: Distinguish primary and secondary batteries.
A primary battery, such as the ordinary dry cell, runs its chemical reaction only once and cannot be recharged once the reactants are used up. A secondary battery, such as the lead storage battery, can be recharged by passing a current through it in the reverse direction, regenerating the original reactants so it can be used again. \[ \boxed{\Lambda_m=106\ \text{S cm}^2\text{mol}^{-1},\ \alpha\approx0.837;\ \text{current: Ag}\to\text{Zn}} \]