A bullet moving with a velocity of 100 m/s can just penetrate two planks of equal thickness. The number of such planks penetrated by the same bullet, when the velocity is doubled, will be

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Solution

The correct option is C
8

Let initial velocity of bullet be and it can just penetrate number of planks. Final velocity should be zero. Let thickness of each plank be x.

$v^{2} = u^{2} - 2 a s w h e r e s = n x T h u s, 2 a n x = u^{2} o r n α u^{2} t h e r e f o r e, \frac{n_{1}}{n_{2}} = {(\frac{u_{1}}{u_{2}})}^{2} = \frac{1}{4} o r n_{2} = 4 n_{1} = 4 \times 2 = 8$

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A bullet moving with a velocity of 100 m/s can just penetrate two planks of equal thickness. The number of such planks penetrated by the same bullet, when the velocity is doubled, will be

The correct option is C 8 Let initial velocity of bullet be and it can just penetrate number of planks. Final velocity should be zero. Let thickness of each plank be x. v2=u2−2as where s=nxThus,2 anx=u2 or nα u2therefore,n1n2=(u1u2)2=14or n2=4n1=4×2=8