A) \[\frac{2\pi }{\sqrt{10}}\]
B) \[\frac{2\pi }{\sqrt{5}}\]
C) \[\frac{\sqrt{10}}{2\pi }\]
D) \[\frac{\sqrt{5}}{2\pi }\]
Correct Answer: C
Solution :
As radius is doubled hence, force will be halved. \[{{v}_{\min }}=r{{\omega }_{\min }}\] \[{{v}_{\min }}=r\times 2\pi {{n}_{\min }}\] \[{{n}_{\min }}\frac{{{v}_{\min }}}{2\pi r}=\frac{\sqrt{rg}}{2\pi r}\] \[\therefore \] \[{{n}_{\min }}=\frac{\sqrt{1\times 10}}{2\pi \times 1}=\frac{\sqrt{10}}{2\pi }\]
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