A bullet, incident normally on a wooden plank, loses one-tenth of its speed in passing through the plank. The least number of such planks required to stop the bullet is

Let v be the speed of the bullet incident on the first plank. Its speed after it passes the plank $=\frac{9\mathrm{v}}{10}$. If x is the thickness of the plank, the deceleration a due to the resistance of the plank is given by

              $2\mathrm{ax}={\mathrm{v}}^2-{\left(\frac{9\mathrm{v}}{10}\right)}^2=\frac{19{\mathrm{v}}^2}{100} \left(1\right)$

Suppose the bullet is stopped after passing through n such planks. Then the distance covered by the bullet is s = nx. Thus, we have 

              ${\mathrm{v}}^2-0=2\mathrm{as}=2\mathrm{anx}$

$\begin{array}{l}\mathrm{or} \mathrm{n}=\frac{{\mathrm{v}}^2}{2\mathrm{ax}}=\frac{{\mathrm{v}}^2\times 100}{19{\mathrm{v}}^2}\text{ [use Eq. (i) above] }\\ =\frac{100}{19}=5.26\end{array}$

Thus the minimum number of planks required is 6. Hence the correct choice is (2).