A particle is moving on a circle of radius `R` such that at every instant the tangential and radial accelerations are equal in magnitude. If the velocity of the particle be `v_(0)` at `t=0`, the time for the completion of the half of the first revolution will beA. `R//v_(0)`B. `(R//v_(0))(1-e^(-pi))`C. `(R//v_(0))e^(-pi)`D. `(R//v_(0))(1-e^(-2pi))`

A particle is moving on a circle of radius `R` such that at every inst

1.	A particle is moving on a circle of radius `R` such that at every instant the tangential and radial accelerations are equal in magnitude. If the velocity of the particle be `v_(0)` at `t=0`, the time for the completion of the half of the first revolution will beA. `R//v_(0)`B. `(R//v_(0))(1-e^(-pi))`C. `(R//v_(0))e^(-pi)`D. `(R//v_(0))(1-e^(-2pi))`
Answer» Correct Answer - B Speed just before collision = `sqrt(2gl(1-costheta_(0)))` Speed just after collision = `e sqrt(2gl(1-cos theta_(0)))` From COE, `(1)/(2)me^(2)[2gl(1-costheta_(0))]=mgl(1-costheta)` `therefore theta = cos^(-1){1-e^(2)(1-costheta_(0))}`

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