A bullet traveling horizontally looses $1 / 20^{t h}$ of its velocity while piercing a wooden plank. Then the number of such planks required to stop the bullet is

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Solution

The correct option is C 11
Let the thickness of plank $= d$
The acceleration provided by plank $= a$
$v^{2} = v_{0}^{2} + 2 a d 0^{2} = v_{0}^{2} + 2 a (h d) 2 a (h d) = - v_{0}^{2} n = v_{0}^{2} / (- 2 a d) \to (1)$
$v = v_{0} - v_{0} / 20 = 19 v_{0} / 20 (19 v_{0} / 20)^{2} = v_{0}^{2} + 2 a d - 2 a d = v_{0}^{2} (1 - 361 / 400) - 2 a d = v_{0}^{2} (39 / 400)$
Substituting $- 2 a d$
$n = v_{0}^{2} / (v_{0}^{2} \times 39 / 400) = 400 / 39$
The minimum no. of plank needed
$= \frac{400}{39} = 11 (a p p r o x)$

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Similar questions

{\frac{1}{20}}^{t h}

When a bullet passed through a wooden plank it looses 1/20th of its velocity. The number of such bullets required to stop the bullet is?

1 / 20^{t h}

{\frac{1}{20}}^{t h}

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