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Q.

A particle is projected from the ground at an angle of 600  with horizontal with speed u = 20m/s. The radius of curvature of the path of the particle, when its velocity makes an angle of 300  with horizontal is (g=10m/s2)

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a

10.6 m

b

12.8 m

c

15.4 m

d

24.2 m

answer is C.

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Detailed Solution

Let v be the velocity of particle when it makes  300  with horizontal. Then vcos300=ucos600 Or v=ucos600cos300 =(20)(12)(3/2)=203m/s Now g cos300=v2R or               R=v2gcos300=(203)2(10)(32)=15.4m
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