A straw of mass 1g and length 4cm lies on a smooth horizontal surface. A bug of mass 3 g starts to move from one end of the straw. The coefficient of friction between straw and the bug is $\mathrm{\mu } =$0.05. The minimum time in which it can cross the straw length is $\sqrt{\frac{\mathrm{n}}{100}}$ sec. The value of n is ______. $\left[\mathrm{Take} \mathrm{g}=10{\mathrm{ms}}^{-2}\right]$

The acceleration of the bug is ${\mathrm{a}}_2 = \frac{\mathrm{f}}{\mathrm{m}}=\frac{\mathrm{f}}{\mathrm{m}}=\frac{\mathrm{f}}{3}$ And the acceleration of the straw is ${\mathrm{a}}_2 = \frac{\mathrm{f}}{1}$ Since, bug and straw are moving in opposite direction.$\therefore$ The relative acceleration of bug with respect to straw is
${\mathrm{a}}_{\mathrm{rel}}={\mathrm{a}}_1+{\mathrm{a}}_2=\frac{\mathrm{f}}{3}+\mathrm{f}=\frac{4\mathrm{f}}{3}\therefore {\mathrm{S}}_{\mathrm{rel}}=\frac{1}{2}{\mathrm{a}}_{\mathrm{rel}}{\mathrm{t}}^2$

$\therefore t=\sqrt{\frac{2{\mathrm{s}}_{\mathrm{rel}}}{{\mathrm{a}}_{\mathrm{rel}}}}=\sqrt{\frac{8}{\frac{4\mathrm{f}}{3}}}$

Substituting values in CGS unit, we get
$\therefore {\mathrm{t}}_{min}=\sqrt{\frac{24}{4{\mathrm{f}}_{max}}}=\sqrt{\frac{24}{4\times {\mathrm{\mu }}_{\mathrm{s}}\mathrm{mg}}}=\sqrt{\frac{24}{4\times 0.05\times 3\times 1000}}$

$\Rightarrow {\mathrm{t}}_{min}=\sqrt{\frac{4}{100}}\mathrm{s}$

$\Rightarrow \mathrm{n}=4$