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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
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Then, the number of such planks penetrated by the same bullet when the speed is doubled will be
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The correct answer is:
8
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.
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.
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