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(D) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{M}{k}}$.Since $2\pi$ and $k$ are constant,we have $T \propto \sqrt{M}$.I

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$16/9$

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(D) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{M}{k}}$.Since $2\pi$ and $k$ are constant,we have $T \propto \sqrt{M}$.Initially,the time period is $T_1 = T$ for mass $M_1 = M$.When the mass is increased by $m$,the new mass is $M_2 = M + m$ and the new time period is $T_2 = \frac{5T}{3}$.Using the proportionality $T \propto \sqrt{M}$,we get $\frac{T_2}{T_1} = \sqrt{\frac{M_2}{M_1}}$.Substituting the values: $\frac{5T/3}{T} = \sqrt{\frac{M + m}{M}}$.$\frac{5}{3} = \sqrt{1 + \frac{m}{M}}$.Squaring both sides: $\frac{25}{9} = 1 + \frac{m}{M}$.Therefore,$\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}$.

$A$ 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

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