A) more than 3 but less than 6
B) more than 6 but less than 9
C) more than 9
D) less than 3
Correct Answer: A
Solution :
[a] \[F=20t-5{{t}^{2}}\] \[\therefore \]\[\alpha =\frac{FR}{I}=4t-{{t}^{2}}\] \[\Rightarrow \]\[\frac{d\omega }{dt}=4t-{{t}^{2}}\] \[\Rightarrow \]\[\int\limits_{0}^{\omega }{\,\,d\omega =\int\limits_{0}^{t}{{}}}\left( 4t-{{t}^{2}}dt \right)\] \[\Rightarrow \]\[\omega =2{{t}^{2}}-\frac{{{t}^{3}}}{3}\](as \[\omega =0\]at \[t=0,\]6s) \[\int\limits_{0}^{\theta }{{}}d\theta =\int\limits_{0}^{6}{{}}\left( 2{{t}^{2}}-\frac{{{t}^{3}}}{3} \right)dt\] \[\Rightarrow \]\[\theta =36\,\text{rad}\] \[\Rightarrow \]\[n=\frac{36}{2\pi }<6\]
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