A uniform disc of diameter R/2 is put over another uniform disc of diameter R of same thickness and density. The peripheries of the two discs touch each other. The centre of mass of the system from centre of big disc is

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R/10

R/20

9R/10

11R/10

answer is B.

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Detailed Solution

For the disc we know that,

$m \propto R^{2} m_{1} \propto {(\frac{R}{2})}^{2} m_{1} = \frac{R^{2}}{4} k m_{2} \propto {(\frac{R}{4})}^{2} m_{2} = \frac{R^{2}}{16} k d = \frac{R}{2} - \frac{R}{4} d = \frac{R}{4}$

Distance of Cm from big disc =

$\frac{m_{2} d}{m_{1} + m_{2}} = \frac{(\frac{R^{2}}{16}) (\frac{R}{4})}{\frac{R^{2}}{4} + \frac{R^{2}}{16}} = \frac{R}{20}$

Hence the correct answer is R/20.

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