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Easy

Question

$A space r i n g space a n d space a space p a r t i c l e space o f space m a s s e s space 2 m space a n d space m space a r e space k e p t space a s space s h o w n space i n space t h e space f i g u r e comma space s u c h space t h a t space t h e space C M space o f space t h e space c o m b i n a t i o n space l i e s space a t space t h e space o r i g i n. space F i n d space x.$

$\frac{6 R}{π}$
$\frac{2 R}{π}$
$\frac{5 R}{π}$
$\frac{4 R}{π}$

The correct answer is: $\frac{4 R}{π}$

$T h e space r i n g space i s space s u b s t i t u t e d space b y space a space p a r t i c l e space o f space m a s s space m space a t space a space d i s tan c e space fraction numerator 2 R over denominator pi end fraction space f r o m space t h e space o r i g i n.$

$text Now, the CM of the two-particle system is at origin, hence end text 2 m open parentheses negative fraction numerator 2 R over denominator pi end fraction close parentheses plus m x equals 0 text or end text x equals fraction numerator 4 R over denominator pi end fraction$

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Centre of mass of three particles of masses 1kg, 2 kg and 3kg lies at the point (1,2,3) and centre of mass of another system of particles 3kg and 2kg lies at the point (-1,3,-2). Where should we put a particle of mass 5kg so that the centre of mass of entire system lies at the centre of mass of first system?

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