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(D) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.For mass $m_1 = 4 \ kg$,the time period is $T_1 = 2\pi \sqrt{\fr

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$100$

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(D) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.For mass $m_1 = 4 \ kg$,the time period is $T_1 = 2\pi \sqrt{\frac{4}{k}} = 2\pi \cdot \frac{2}{\sqrt{k}} = \frac{4\pi}{\sqrt{k}}$.For mass $m_2 = 9 \ kg$,the time period is $T_2 = 2\pi \sqrt{\frac{9}{k}} = 2\pi \cdot \frac{3}{\sqrt{k}} = \frac{6\pi}{\sqrt{k}}$.Given that the time period increases by $0.2\pi \ s$,we have $T_2 - T_1 = 0.2\pi$.Substituting the expressions for $T_1$ and $T_2$: $\frac{6\pi}{\sqrt{k}} - \frac{4\pi}{\sqrt{k}} = 0.2\pi$.$\frac{2\pi}{\sqrt{k}} = 0.2\pi$.Dividing both sides by $\pi$: $\frac{2}{\sqrt{k}} = 0.2$.$\sqrt{k} = \frac{2}{0.2} = 10$.Squaring both sides,we get $k = 100 \ N \ m^{-1}$.

When the mass attached to a spring is increased from $4 \ kg$ to $9 \ kg$,the time period of oscillation increases by $0.2 \pi \ s$. Then the spring constant of the spring is (in $N \ m^{-1}$)

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