 Solution

For a $(10,0)$ tube, $C=10a=2.46\mathrm{nm}$. A cut normal to its axis crosses ten axial load-bearing C-C bonds in the ideal unstrained honeycomb geometry. Assuming equal load sharing and simultaneous failure at $8\mathrm{nN}$ per bond,The graphene areal mass is ${\sigma }_m=0.75\mathrm{mg}{\mathrm{m}}^{-2}=7.5\times 10^{-7}\mathrm{kg}{\mathrm{m}}^{-2}$, giving linear mass densitysince the tex unit has $1\mathrm{tex}=10^{-6}\mathrm{kg}{\mathrm{m}}^{-1}$. Thus the ideal carbon nanotube tenacity isThis neglects vacancies, curvature-induced changes in bond strength, thermal activation, imperfect load sharing and structural relaxation at failure. An aligned macroscopic fibre is a network of finite tubes or bundles: sliding, weak intertube shear transfer, imperfect alignment, defects and unequal load participation limit its strength. The nominal $10\mu \mathrm{m}$ fibre diameter alone does not turn it into one defect-free covalent cylinder, explaining why a value near $2N/tex$ can be much lower than this ideal single-tube estimate.