A ball is released from rest from top of a tower. The retardation due to air resistance is bv, where b is 10 per second and velocity v is ms^-1. The velocity of ball at $t=\frac{1}{10}\mathrm{s}\text{ is }\frac{n}{27}{\mathrm{ms}}^{-1}$ The value of n (Given, e=2.7) is _________.

$\text{ The acceleration of ball at instant }t\text{ is }a=g-bv$

$\because a=g-bv$

$\begin{array}{l}\text{ Or }\frac{dv}{dt}=g-bv\text{ or }{\int }_0^v \frac{dv}{g-bv}={\int }_0^t dt\\ \text{ Or }-\frac{1}{2}[\ln (g-bv){]}_0^v=[t{]}_0^t\end{array}$

$\begin{array}{l}\text{ Or }[\ln (g-bv)-\ln g]=-bt\\ \text{ Or }\ln \frac{g-bv}{g}=-bt\end{array}$

$\begin{array}{l}\text{ Or }1-\frac{bv}{g}=e^{-bt}\text{ or }\frac{bv}{g}=1-e^{-bt}\\ \text{ Or }v=\frac{g}{b}\left(1-e^{-bt}\right)=\frac{10}{10}\left(1-\frac{1}{e}\right)\end{array}$

$\begin{array}{l}=\left(\frac{e-1}{e}\right)=\frac{2.7-1}{2.7}=\frac{7}{27}=\frac{n}{27}\\ \therefore n=7\end{array}$