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Q: Three particles starts from rest simaltaneously along three smooth chords of a vertical circle joining at the lowest point of the circle . If the inclination of the chord is 30 degree, 60 degree , 90 degree respectively. Find the ratio of the time taken by them to move down the chords Q: the velocity 'v' and displacement 's' a body is moving in a 1-D is given by v 2 = ks where k is a const. if displacement at t=0 is zero.The velocity after 2 sec will be

Q: Three particles starts from rest simaltaneously along three smooth chords of a vertical circle joining at the lowest point of the circle . If the inclination of the chord is 30 degree, 60 degree , 90 degree respectively. Find the ratio of the time taken by them to move down the chords

  Q:  the velocity 'v' and displacement 's' a body is moving in a 1-D is given by v2 = ks where k is a const. if displacement at t=0 is zero.The velocity after 2 sec will be

Grade:12

1 Answers

AskiitiansExpert Milanshu
9 Points
12 years ago

Ans :: let the time taken by three particles on chord 900 , 600, and 300 are t1, t2 and t3 and let the radius of circle is R

initial velocity of particles are 0

therefore, 1/2(g)(t1)2 = 2*R                           eq (1)

similarly 1/2(g)(t2)2 = 2*R*cos300                           eq (2)

and 1/2(g)(t3)2 = 2*R*cos600                           eq (3)

from eq (1), eq (2) and eq (3), we get

t1 : t2 : t3   = (2)1/2 : 1: (3)1/4


Ans:: since intial displacement and velocity are zero

therefore, s =1/2 gt2 and v2 = 2as,  where a is the acceleration

so comparing with v2=ks we get new eqs as

s =1/4kt2
therefore after 2 sec, s= 1/4k(2)2

s=k

therefore v = k1/2

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