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A smooth circular tube of radius R is f...

A smooth circular tube of radius `R` is fixed in a vertical plane. A particle is projectid from its lowest point with a velocity just sufficient to carry it to th ehigest point. Show that the time taken by the particle to rach the end of the horizontal diameter is `(sqrt(R)/(g))In(1+sqrt(2))` .
Hint : `intsectheta.dtheta=In(sectheta+tantheta)`

Text Solution

Verified by Experts

The correct Answer is:
A
Minimum velocity of particle at the lowest position to comlete the circle should be `sqrt(4gR)` inside a tube.
So, `u=sqrt(4gR)`
`h=R(1-costheta)`
:. `v^(2)=4gR-2gR(1-costheta)` ltBRgt `=2gR(1+costheta)`
or `v^(2)=2gR(2cos^(2)(theta)/(2))`

or `v=2sqrt(gR)cos(theta)/(2)`
From `ds=v.dt`
We get `Rdtheta=2sqrt(gR)cos(theta)/(2).dt`
or `int_(0)^(t)dt=(1)/(2)sqrt((R)/(g)int_(0)^(pi//2)sec((theta)/(2))dtheta`
or `t=sqrt((R)/(g))[In(sec(theta)/(2)+tan(theta)/(2))]_(0)^(pi//2)`
or `t=sqrt((R)/(g))In(1+sqrt(2))` Hence proved.
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