The cohesive energy is calculated using:
\[E_{\text{coh}} = -\frac{Me^2}{4\pi\epsilon_0 r} + Be^{-r/\rho}\]
where $M = 1.748$, $r = 2.018 \text{ \AA}$, $\rho = 0.32 \text{ \AA}$, and $B$ is a constant obtained experimentally.
Substituting the values and solving numerically:
\[E_{\text{coh}} \approx -7.94 \text{ eV}\]
Match List-I with List-II\[\begin{array}{|c|c|} \hline \textbf{Provision} & \textbf{Case Law} \\ \hline \text{(A) Strict Liability} & \text{(1) Ryland v. Fletcher} \\ \hline \text{(B) Absolute Liability} & \text{(II) M.C. Mehta v. Union of India} \\ \hline \text{(C) Negligence} & \text{(III) Nicholas v. Marsland} \\ \hline \text{(D) Act of God} & \text{(IV) MCD v. Subhagwanti} \\ \hline \end{array}\]
Match Fibre with Application.\[\begin{array}{|l|l|} \hline \textbf{LIST I} & \textbf{LIST II} \\ \textbf{Fibre} & \textbf{Application} \\ \hline \hline \text{A. Silk fibre} & \text{I. Fire retardant} \\ \hline \text{B. Wool fibre} & \text{II. Directional lustre} \\ \hline \text{C. Nomex fibre} & \text{III. Bulletproof} \\ \hline \text{D. Kevlar fibre} & \text{IV. Thermal insulation} \\ \hline \end{array}\]