Q130 / 150 · MHT CET 2020 (12 Oct Shift 1) · Physics

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 period $\mathrm{T}$. If the mass is increased by 'm', the time period becomes $\frac{5 \mathrm{~T}^{\prime}}{3}$. What is the ratio $\left(\frac{\mathrm{M}}{\mathrm{m}}\right)$ ?

  • A
    $\frac{25}{9}$
  • B
    $\frac{16}{9}$
  • C
    $\frac{9}{25}$
  • Correct answer
    D
    $\frac{9}{16}$

Solution

$T=2 \pi \sqrt{\frac{M}{k}} ; \quad \frac{5 T}{3}=2 \pi \sqrt{\frac{M+m}{k}}$ $\therefore \frac{5}{3} \times 2 \pi \sqrt{\frac{M}{k}}=2 \pi \sqrt{\frac{M+m}{k}}$ $\therefore \frac{25}{9} \cdot \frac{M}{k}=\frac{M+m}{k}$ $\therefore \frac{25}{9} M=M+m$ Dividing by $M, \frac{25}{9}=1+\frac{m}{M} \quad \therefore \frac{m}{M}=\frac{25}{9}-1=\frac{16}{9}$ $\frac{M}{m}=\frac{9}{16}$

Hence, the correct answer is option: 'D'.

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