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Question

A large fluid star oscillates in shape under the influence of its own gravitational field. Using the method of dimensions, obtain a relation for the period of oscillation (T) in terms of a radius (R) of the star, its density (ρ) and universal gravitational constant (G) is :

A
T=KRpG
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B
T=KpG
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C
T=KpRG
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D
T=Kp2G2
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Solution

The correct option is D T=KpG
Let T=KRaρbGc, where k is a constant.
Thus, equating the dimensions for both sides, using dimensions for G as L3M1T2.
T=La(ML3)b(L3M1T2)c
Or, 2c=1, giving c=12
bc=0 or b=c=12
a3b+3c=0 or a=0
Thus we get, T=K(ρ.G)1/2=KρG

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