circular motion

Answer given is correct.. I got θ ≈ 150°

Resolve it.. may be you'' get it. Somewhere silly mistake you might have done.

Btw, How much are you expecting in 12th Feb's test.. Seeing your past performance , I'm presuming around ≈ 290 for you .. :)

I had a request.. Since you have a scanner at hand, could you upload one or more papers of RTPF.. these are completely inaccessible to me.

getting θ=120(cos^-1-1/2)
Ans:(c)

even i got 120 ,
@vivek - hoiw are u getting 150 ??

I take θ from the vertical.

So we have, By energy Conservation,

=> mV² = 2mgl(1+cos θ) + mV²4

=> 3mV²4 = 2mgl(1+cos θ)

V² = 5gl

Putting, cos θ = 7/8 ≈ 28.96°

So, α = Π- θ = 180° - 28.96° = 151.04° > 3Π4

yes, i m also getting cos(theta)= -7/8. Ans is d.

@vivek - it is given that velocity imparted is such that the particle just reaches B (not that it completes the circular motion) .

so shouldnt it be v²=4gl ??

It means that the string just on the verge of becoming slack at the topmost point... It however successfully completes the HALF CIRCLE... However it it was what you said, the string would slack at during time before going to top..

[ You didn't answer my hided post. Please answer ]

hmm.. ok i got my mistake . thnx

Interesting observation.
When θ<90
velocity becomes 0 before tension does.
when θ=90
Velocity and tension become zero at the same time.
When θ>90
tension becomes 0 before velocity does.