MCQ
A horizontal disc rotating freely about a vertical axis through its centre makes $90$ revolutions per minute. A small piece of wax of mass $m$ falls vertically on the disc and sticks to it at a distance $r$ from the axis. If the number of revolutions per minute reduce to $60$ , then the moment of inertia of the disc is .........
  • A
    $m r^2$
  • B
    $\frac{3}{2} m r^2$
  • $2{m r^2}^2$
  • D
    $3 mr ^2$

Answer

Correct option: C.
$2{m r^2}^2$
c
(c)

$w_1=\frac{2 \pi 90}{60}=3 \pi rps$

$w_2=\frac{2 \pi 60}{60}=2 \pi rps$

Using angular momentum conservation

$I(1.5)=\left(I+m r^2\right)(1)$

$\frac{1}{2}=m r^2$

$I=2 m r^2$

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