MCQ
A solid cube of side $l$ is made to oscillate about a horizontal axis passing through one of its edges. Its time period will be
  • $2\pi \sqrt {\frac{{2\sqrt 2 }}{3}\frac{l}{g}} $
  • B
    $2\pi \sqrt {\frac{2}{3}\frac{l}{g}} $
  • C
    $2\pi \sqrt {\frac{{\sqrt 3 }}{2}\frac{l}{g}} $  
  • D
    $2\pi \sqrt {\frac{2}{{\sqrt 3 }}\frac{l}{g}} $

Answer

Correct option: A.
$2\pi \sqrt {\frac{{2\sqrt 2 }}{3}\frac{l}{g}} $
a
We have the moment of inertia of a cube about its centroidal axis as

$I=\frac{1}{6} m l^{2}$

its moment of inertia about one of its edge is

$\frac{1}{6} m l^{2}+\frac{1}{2} m l^{2}=\frac{2}{3} m l^{2}$

Now time period of any compound pendulum is

$T=2 \pi \sqrt{\frac{I}{m g R}}$

or $T=2 \pi \sqrt{\frac{\frac{2}{3} m l^{2}}{m g \frac{l}{\sqrt{2}}}}$

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

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