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Physics-

General

Easy

Question

A particle of mass $m$ is rotating in a horizontal circle of radius $R$ and is attached to a hanging mass $M$ as shown in the figure. The speed of rotation required by the mass $m$ keep $M$ steady is

$\sqrt{\frac{m g R}{M}}$
$\sqrt{\frac{m g R}{m}}$
$\sqrt{\frac{m g}{M R}}$
$\sqrt{\frac{m R}{M g}}$

The correct answer is: $square root of fraction numerator m g R over denominator m end fraction end root$

To keep the mass M steady, let T is the tension in the string joining the two. Then for particle $m comma$
$T equals fraction numerator m v to the power of 2 end exponent over denominator R end fraction open parentheses i close parentheses$
For mass $M comma$
$T equals M g left parenthesis i i right parenthesis$
From Eqs. (i) and (ii)
$fraction numerator m v to the power of 2 end exponent over denominator R end fraction equals M g ⟹ v equals square root of fraction numerator M g R over denominator m end fraction end root$

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