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A small sphere of radius $r$ is placed as a concave surface of radius of curvature $R$ a little away from the centre. When the sphere is released, it oscillates. Assuming, the oscillation to be simple harmonic motion and $r< < R$, then the time period is
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The correct answer is:
$2 \pi \sqrt{\frac{R}{g}}$
The motion of sphere is as shown in figure
As, the sphere has simple harmonic motion on a curved surface, which is equivalent to the motion of pendulum. So, its time period is
$T=2 \pi \sqrt{\frac{l}{g}}$
Here,
$l=R$
$\therefore \quad T=2 \pi \sqrt{\frac{R}{g}}$
As, the sphere has simple harmonic motion on a curved surface, which is equivalent to the motion of pendulum. So, its time period is
$T=2 \pi \sqrt{\frac{l}{g}}$
Here,
$l=R$
$\therefore \quad T=2 \pi \sqrt{\frac{R}{g}}$
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