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A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes S.H.M. of time period T. If the mass is increased by m, the time period becomes 5T/3. Then the ratio of m/M is 

a
53
b
35
c
259
d
169

detailed solution

Correct option is D

T∝m⇒T2T1=m2m1⇒53=M+mM⇒259=M+mM⇒mM=169

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