A small sphere of radius R is held against the inner surface of a larger sphere of radius 6R as shown in figure. The masses of the large and small sphere are 4M and M respectively. This arrangement is fixed on a horizontal table. No friction exists between any two surfaces in contact. The small sphere is now released. The (x, y) coordinates of the centre of the larger sphere when the smaller sphere reaches the other extreme position are:

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(L+2R, 0)

(L+R, 0)

$(L - 2 R, 0)$

$(L - R, 0)$

answer is B.

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Detailed Solution

${(X_{COM})}_{i} = {(X_{COM})}_{f}$

$(4 M) L + M (L + 5 R) = M (X - 5 R) + 4 M (X)$

$5 M (L + 2 R) = 5 MX \Rightarrow x = L + 2 R$

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