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Physics-

General

Easy

Question

A particle originally at a rest at the highest point of a smooth circle in a vertical plane, is gently pushed and starts sliding along the circle. It will leave the circle at a vertical distance $h$ below the highest point such that

$h = 2 R$
$h = \frac{R}{2}$
$h = R$
$h = \frac{R}{3}$

The correct answer is: $h equals fraction numerator R over denominator 3 end fraction$

From law of conservation of energy, potential energy of fall gets converted to kinetic energy.

$therefore blank P E equals K E$
$m g h equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent$
$o r blank v equals square root of 2 g h end root blank open parentheses i close parentheses$
Also, the horizontal component of force is equal centrifugal force.
$therefore blank m g cos invisible function application theta equals fraction numerator m v to the power of 2 end exponent over denominator R end fraction open parentheses i i close parentheses$
From Eq. (i)
$therefore blank m g cos invisible function application theta equals fraction numerator 2 m g h over denominator R end fraction blank open parentheses i i i close parentheses$
From $increment A O B comma$
$cos invisible function application theta equals fraction numerator left parenthesis R minus h over denominator R end fraction$
$⟹ m g open parentheses fraction numerator open parentheses R minus h close parentheses over denominator R end fraction close parentheses equals fraction numerator 2 m g h over denominator R end fraction$
$⟹ blank 3 h equals R$
$⟹ h equals fraction numerator R over denominator 3 end fraction$

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