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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 5T3. Then the ratio of mM is

a
53
b
35
c
259
d
169

detailed solution

Correct option is

T ∝ m ⇒ T2T1 = m2m1   ⇒53= M+mM⇒ 259 = M+mM ⇒mM = 169

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