Specific volume of cylindrical virus particle is $6.02\times {10}^{-2}\mathrm{cc}/\mathrm{g}$ whose radius and length $7{\text{\hspace{0.17em}\hspace{0.17em}A}}^0\mathrm{and}10{\text{\hspace{0.17em}\hspace{0.17em}A}}^0$ respectively. If $N_A=6.02\times {10}^{23},$ find molecular weight of virus

Specific volume (volume of g) of cylindrical virus particle 

$=6.02\times {10}^{-2}\mathrm{cc}/\mathrm{g}$

Radius of virus (r) = $7{\text{\hspace{0.17em}\hspace{0.17em}A}}^0=7\times {10}^{-8}cm$

$\begin{array}{l}\mathrm{Length}\mathrm{of}\mathrm{virus}=10\times {10}^{-8}\mathrm{cm}\\ \mathrm{Volume}\mathrm{of}\mathrm{virus}\\ {\mathrm{\pi r}}^2\text{\hspace{0.17em}L}=\frac{22}{7}\times {\left(7\times {10}^{-8}\right)}^2\times 10\times {10}^{-8}=154\times {10}^{-23}\mathrm{cc}\\ \mathrm{Wt}.\mathrm{of}\mathrm{one}\mathrm{virus}\mathrm{particle}=\frac{\mathrm{volume}}{\mathrm{specific}\mathrm{volume}}\\ \therefore \mathrm{Mol}.\mathrm{wt}.\mathrm{of}\mathrm{virus}=\mathrm{wt}.\mathrm{of}{\mathrm{N}}_{\mathrm{A}}\mathrm{particle}\\ =\frac{154\times {10}^{-23}}{6.02\times {10}^{-2}}\times 6.02\times {10}^{23}\\ =15400\mathrm{g}/\mathrm{mol}=15.4\mathrm{kg}/\mathrm{mol}\end{array}$