Home
Class 11
PHYSICS
A particle of mass m is revolving in a h...

A particle of mass m is revolving in a horizontal circle on a smooth table by an in extensible massless string if the initial speed of the particle is `V_(0)` and the particle has tangential acceleration of constant magnitude a,
After what time the tension of the string will be equal to `T (gt (mv_(0)^(2))/(R))`

A

`(1)/(a)(sqrt((TR)/(m))-V_(0))`

B

`(v_(0))/(a)(sqrt((TR)/(m))-V_(0))`

C

`(v_(0))/(a)`

D

`(1)/(a)(sqrt((2TR)/(m)-V_(0)))`

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • HORIZONTAL CIRCULAR MOTION

    AAKASH SERIES|Exercise PRACTICE SHEET (EXERCISE-III (Applications of Circular Motion (LEVEL-I (Straight Objective Type Questions))))|13 Videos
  • HORIZONTAL CIRCULAR MOTION

    AAKASH SERIES|Exercise PRACTICE SHEET (EXERCISE-III (Applications of Circular Motion (LEVEL-II (Straight Objective Type Questions))))|9 Videos
  • HORIZONTAL CIRCULAR MOTION

    AAKASH SERIES|Exercise PRACTICE SHEET (EXERCISE-II (LEVEL-II (Straight Objective Type Questions)))|5 Videos
  • GRAVITATIONAL

    AAKASH SERIES|Exercise EXERCISE -3|154 Videos
  • KINEMATICS

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE (PRACTICE SHEET (ADVANCED)) (Integer Type Questions)|12 Videos

Similar Questions

Explore conceptually related problems

A particle of mass m is revolving in a horizontal circle on a smooth table by an in extensible massless string if the initial speed of the particle is V_(0) and the particle has tangential acceleration of constant magnitude a, If V_(0) =0 then the distance covered by the stone till the string ruptures

A particle of mass m is revolving in a horizontal circle on a smooth table by an in extensible massless string if the initial speed of the particle is V_(0) and the particle has tangential acceleration of constant magnitude a, If the breaking strength of the string is equal to weight of the stone. The time after which the string will break

A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to - (K//r^(2)) , where K is constant. What is the total energy of the particle ?

A particle of mass .m. is moving in a horizontal circle of radius .r. under a centripetal force equal to -(k)/(r^(2)) , where k is a constant. The potential energy of the particle