$\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}=10^{12}$ where $\mathrm{m}=\mathrm{mass}$ of one atom
Mass of one atom of silver, $=\frac{108}{\left(6.02 \times 10^{23}\right)} \times 10^{-3} \mathrm{kg}$
$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{108 \times 10^{-3}} \times 6.02 \times 10^{23}}=10^{12}$
Solving we get, spring constant, $\mathrm{K}=7.1 \mathrm{N} / \mathrm{m}$
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| List-$I$ | List-$II$ |
| $P$ $\dot{r}(t)=\alpha t \hat{t}+\beta t \hat{j}$ | $1$ $\overrightarrow{ p }$ |
| $Q$ $\dot{r}(t)=\alpha \cos \omega t \hat{i}+\beta \sin \omega t \hat{j}$ | $2$ $\overrightarrow{ L }$ |
| $R$ $\dot{r}(t)=\alpha(\cos \omega t \hat{i}+\sin \omega t \hat{j})$ | $3$ $K$ |
| $S$ $\dot{r}(t)=\alpha t \hat{i}+\frac{\beta}{2} t^2 \hat{j}$ | $4$ $U$ |
| $5$ $E$ |
