$(A)$ $v=\sqrt{\frac{k}{2 m}} R$
$(B)$ $v=\sqrt{\frac{k}{m}} R$
$(C)$ $\mathrm{L}=\sqrt{\mathrm{mk}} \mathrm{R}^2$
$(D)$ $\mathrm{L}=\sqrt{\frac{\mathrm{mk}}{2}} \mathrm{R}^2$
- A$A,C$
- ✓$B,C$
- C$A,D$
- D$A,C,D$
$(A)$ $v=\sqrt{\frac{k}{2 m}} R$
$(B)$ $v=\sqrt{\frac{k}{m}} R$
$(C)$ $\mathrm{L}=\sqrt{\mathrm{mk}} \mathrm{R}^2$
$(D)$ $\mathrm{L}=\sqrt{\frac{\mathrm{mk}}{2}} \mathrm{R}^2$
$\mathrm{V}=\frac{\mathrm{kr}^2}{2}$
$\because \mathrm{F}=-\frac{\mathrm{dv}}{\mathrm{dr}} \quad \therefore \mathrm{F}=-\mathrm{kr}$
(imag)
$\text { At } \mathrm{r}=\mathrm{R}, \mathrm{F}=-\mathrm{kR}$
To keep the particle in circular motion, this force must be equal to centripetal force. So, $k R=\frac{\mathrm{mv}^2}{\mathrm{R}} \quad$ or $\quad v=\sqrt{\frac{k R^2}{m}}=\sqrt{\frac{k}{m}} \cdot R$
The angular momentum of the particle is, $\mathrm{L}=\mathrm{mvR}=\mathrm{m} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}} \cdot \mathrm{R}^2$
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