A uniform disc of radius R is put over another uniform disc of radius 2R of same thickness and density. The peripheries of the two discs touch each other. The position of their centre of mass is

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at $\frac{R}{3}$ from the centre of the bigger disc towards the centre of the smaller disc

at $\frac{R}{5}$ from the centre of the bigger disc towards the centre of the smaller disc

at $\frac{2 R}{5}$ from the centre of the bigger disc towards the centre of the smaller disc

none of the above

answer is B.

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Detailed Solution

Let the mass per unit area is $σ$

A disc with a radius of 2R $M_{1} = σ 4 {πR}^{2}$

A disc with a radius of R $M_{2} = σ \times {πR}^{2}$

Centre of mass,

$x_{cm} = \frac{4 {σπR}^{2} \times 0 + {σπR}^{2} \times R}{4 {σπR}^{22} + σπR} = \frac{R}{5}$

Hence the correct answer is at $\frac{R}{5}$ from the centre of the bigger disc towards the centre of the smaller disc.

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