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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 SHM  of time period T. If the mass is increased by ‘m’ the time period becomes 5T3 , then the ratio of mM  is 

a
35
b
259
c
169
d
53

detailed solution

Correct option is C

As time period  T=2πMK------(1)new time period when mass m is added,T1=2πM+mK5T3=2πM+mK---------(2)Dividing equation (1) by equation (2), we get35=MM+m squaring on both sides,925=MM+m9M+9m=25M16M=9mmM=169.

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