MCQ
A turn table which is rotating uniformly has a particle placed on it. As seen from the ground, the particle goes in a circle with speed 20 cm/s and acceleration $20 cm / s ^2$ The particle is now shifted to a new position where radius is half of the original value. The new values of speed and acceleration will be
  • $10 cm / s , 10 cm / s ^2$
  • B
    $10 cm / s , 80 cm / s ^2$
  • C
    $40 cm / s , 10 cm / s ^2$
  • D
    $40 cm / s , 40 cm / s ^2$

Answer

Correct option: A.
$10 cm / s , 10 cm / s ^2$
(A)
Velocity, $v =\omega r$
$\therefore \quad v ^{\prime}=\omega r ^{\prime}=\frac{\omega r }{2}=\frac{ v }{2}=10 cm / s$
$\therefore \quad a =\omega^2 r$
$\therefore \quad a ^{\prime}=\omega^2 r ^{\prime}=\omega^2 \times \frac{ r }{2}=\frac{ a }{2}=10 cm / s ^2$

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