As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be: |
[JEE MAIN Held on 07-01-2020 Morning] |
A) \[r\sqrt{\frac{3}{2gh}}\]
B) \[r\sqrt{\frac{3}{4gh}}\]
C) \[\frac{1}{r}\sqrt{\frac{4gh}{3}}\]
D) \[\frac{1}{r}\sqrt{\frac{2gh}{3}}\]
Correct Answer: C
Solution :
[c] |
\[mgh=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}l{{\omega }^{2}}\] |
\[=\frac{1}{2}m{{(\omega r)}^{2}}+\frac{1}{2}\times \frac{m{{r}^{2}}}{2}\times {{\omega }^{2}}\] |
\[\Rightarrow mgh=\frac{3}{4}m{{\omega }^{2}}{{r}^{2}}\] |
\[\Rightarrow \omega =\sqrt{\frac{4gh}{3{{r}^{2}}}}=\frac{1}{r}\sqrt{\frac{4gh}{3}}\] |
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