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Two solid hemispheres of radii R and R/2 with centers $O a n d O'$ respectively as shown in figure. The density of bigger hemisphere is $ρ$ and that of smaller hemisphere is $2ρ$ .Taking center of bigger hemisphere is at origin and the distance between centers of two hemispheres $OO'$ is R/10 find co-ordinates of center of mass of the system.

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$(\frac{- R}{50}, \frac{21 R}{80})$

$(\frac{- R}{30}, \frac{21 R}{80})$

$(\frac{- R}{50}, \frac{7 R}{16})$

$(\frac{- R}{30}, \frac{7 R}{16})$

answer is A.

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Detailed Solution

$(x_{1}, y_{1}) = (0, \frac{3 R}{8}) & (x_{2}, y_{2}) = (\frac{- R}{10}, \frac{- 3}{8} (\frac{R}{2})) = (\frac{- R}{10}, \frac{- 3 R}{16})$

$M = \frac{4}{3} {πr}^{3} ρ$

$\frac{M_{1}}{M_{2}} = {(\frac{r_{1}}{r_{2}})}^{3} \frac{ρ_{1}}{ρ_{2}} = {(\frac{R}{\frac{R}{2}})}^{3} \frac{ρ}{2 ρ} = 4$

$X_{c m} = \frac{M_{1} (0) - M_{2} (\frac{R}{10})}{M_{1} + M_{2}} = \frac{- M_{2} (\frac{R}{10})}{5 M_{2}} = \frac{- R}{50}$

$Y_{c m} = \frac{M_{1} (\frac{3 R}{8}) - M_{2} (\frac{3 R}{16})}{M_{1} + M_{2}} = \frac{4 M_{2} (\frac{3 R}{8}) - M_{2} (\frac{3 R}{16})}{4 M_{2} + M_{2}} = \frac{21 R}{80}$

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