A) 600
B) 1500
C) 1000
D) 3000
E) 2000
Correct Answer: E
Solution :
Angular displacement during time \[\theta =({{\omega }_{2}}-{{\omega }_{1}})t\] \[=(2\pi {{n}_{2}}-2\pi {{n}_{1}})t\] \[=(600\pi -200\pi )\times 10\] \[=4000\text{ }\pi \text{ }rad\] Therefore, number of revolutions made during this time \[=\frac{4000\pi }{2\pi }=2000\]
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