A bullet moving with a speed of $100 \mathrm{~ms}^{-1}$ can just penetrate two planks of equal thickness. Then, the number of such planks penetrated by the same bullet when the speed is doubled will be

Given, velocity of the bullet $v=100 \mathrm{~ms}^{-1}$ Let, thickness of one plank, $t=s$ According to third equation of motion, we get $v^{2}-u^{2}=2 a s$ $(0)-(100)^{2}=2 \times a \times(2 s)$ ( $\because$ two planks are penetrated) $\Rightarrow \quad-1000=4 a s \quad \ldots$. (i) When velocity of bullet is doubled, $v=200 \mathrm{~ms}^{-1}$ Applying third equation of motion again, we get $\Rightarrow \quad 0-(200)^{2}=2 \times a \times(n s)$ (where, $n$ is number of planks) $-40000=2 a n s$ From Eqs. (i) and (ii), we get $\frac{-40000}{-10000}=\frac{2 a n s}{4 a s}$ $\Rightarrow \quad n=8$ Hence, 8 planks can be penetrated by the same bullet if speed is doubled.