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3 grooves AB,AC and AD are made in a circular disk lying in a vertical plane 3 particles B,C and D constrained to move along path AB,AC and AD respectively are released from point A.Particle B reaches point B in time tB,particle C reaches point C in time tC and particle D reaches at point D in time tD.Then which of the following statements is incorrect tB=tC=tD tB>tC=tD tC>tB>tD tD=tB>tC

3 grooves AB,AC and AD are made in a circular disk lying in a vertical plane 3 particles B,C and D constrained to move along path AB,AC and AD respectively are released from point A.Particle B reaches point B in time tB,particle C reaches point C in time tC and particle D reaches at point D in time tD.Then which of the following statements is incorrect
tB=tC=tD
tB>tC=tD
tC>tB>tD
tD=tB>tC

Grade:12

3 Answers

Saurabh Kumar
askIITians Faculty 2400 Points
9 years ago
The time taken would be same...
Calculate the relative velocity of one w.r.t. another in the same line of action.
Relative velocity of one particle would be v- v cos 60..
and length of one side would be L..

In this calculate for individual one..
You will get the nsame time for all the three particles...

Good Luck.............................
Sarthak Mehta
37 Points
7 years ago
openstudy.com/updates/525dad28e4b0ede9e44a98fe Go to this link and scroll down and see the answer. It is very clearly.
Tushar Bharti
26 Points
6 years ago
We know in this case s=1/2a(t^(2)) or t=(2s/a)^(1/2) now here a=g (for AC) and a=gcos@, s=scos@(for AD)also a=gcos@/2 and s=scos@/2 (for AB)......For AC: t=(2s/g)^(1/2)For AD: t=(2scos@/gcos@)^(1/2)For AB: t=(2scos(@/2)/gcos(@/2))^(1/2)In this way t1=t2=t3Because all the cos terms will cancel each other and we will only be left witht=(2s/g)^(1/2)

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