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Physics Question on distance and displacement

In alpha particle scattering experiment, if v is the initial velocity of the particle, then the distance of closest approach is d. If the velocity is doubled, then the distance of closest approach becomes

A

4 d

B

2 d

C

d2\frac{d}{2}

D

d4\frac{d}{4}

Answer

d4\frac{d}{4}

Explanation

Solution

The correct answer is Option (D) : d4\frac{d}{4}