A) \[\frac{R}{\sqrt{2}}\]
B) \[\frac{R}{2}\]
C) \[\frac{3R}{4}\]
D) \[\frac{\sqrt{3}R}{4}\]
Correct Answer: C
Solution :
Case 1 \[mgh\,\,=\,\frac{1}{2}\,m{{\nu }^{2}}+\frac{1}{2}m{{k}^{2}}{{\omega }^{2}}\] Case 2 \[mgh\,\,=\,\frac{1}{2}\,m\times \,\,{{\left( \frac{5\nu }{4} \right)}^{2}}\] \[\frac{1}{2}m{{k}^{2}}\frac{{{v}^{2}}}{{{R}^{2}}}+\frac{1}{2}m{{\nu }^{2}}=\frac{1}{2}m\times \frac{25{{\nu }^{2}}}{16}\] \[\frac{{{k}^{2}}}{{{R}^{2}}}+1=\frac{25}{16}\,\,\,\,\,\,\,\Rightarrow \,\,\,k=\frac{3\,R}{4}\]You need to login to perform this action.
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