A ball is rolling without slipping in a spherical shallow bowl (radius $R$ ) as shown in the figure and is executing simple harmonic motion. If the radius of the ball is doubled, then the time period of oscillation
KVPY 2013, Medium
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(d)
Time period of oscillation of ball (radius $r$ ) in a bowl (radius $R$ ) rolling without slipping is
$T=2 \pi \sqrt{\frac{R-r}{g}}$
So, if $r$ increases, then time period is decreases.
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