MCQ
A ring of diameter $2m$ oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre move in a plane which is perpendicular to the plane of the ring. The equivalent length of the simple pendulum is .... $m$
  • A
    $2$
  • B
    $4$
  • $1.5$
  • D
    $3$

Answer

Correct option: C.
$1.5$
c
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{mgl}}}=\sqrt{\frac{3 \mathrm{mr}^{2}}{2 \mathrm{mgr}}}$

or $T=2 \pi \sqrt{\frac{3 r}{2 m}}$

or $2 \pi \sqrt{\frac{l}{\mathrm{g}}}=2 \pi \sqrt{\frac{3 \mathrm{r}}{2 \mathrm{g}}}$

$\therefore \quad l=\frac{3}{2} \mathrm{r}=\frac{3}{2} \times 1=1.5 \mathrm{m}$

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