= 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,
$$
F_{\max}\simeq10(8\,\mathrm{nN})=80\,\mathrm{nN}.
$$
The <graphene> areal <mass> is $\sigma_m=0.75\,\mathrm{mg\,m^{-2}}=7.5\times10^{-7}\,\mathrm{kg\,m^{-2}}$, giving linear <mass> <density>
$$
\lambda=\sigma_mC=1.845\times10^{-15}\,\mathrm{kg\,m^{-1}}=1.845\times10^{-9}\,\mathrm{tex},
$$
since the <tex unit> has $1\,\mathrm{tex}=10^{-6}\,\mathrm{kg\,m^{-1}}$. Thus the <ideal carbon nanotube tenacity> is
$$
\boxed{\frac{F_{\max}}{\lambda}=43.4\,\mathrm{N/tex}.}
$$
This 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 $2\,\mathrm{N/tex}$ can be much lower than this ideal single-tube estimate.
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