Q. A wheel has a speed of $1200$ revolutions per minute and is made to slow down at a rate of $4rads^{2}$ . The number of revoluitions it makes before coming to rest is :-

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A

$143$

B

$272$

C

$314$

D

$722$

Solution:

$n=1200rpm$
$=\frac{1200}{60}rps$
$=20rps$
$w^{2}=w_{0}^{2}+2\alpha \theta $
$0=w_{0}^{2}-2\alpha \theta $
$\theta =\frac{w_{0}^{2}}{2 \alpha }$
No. of revolutions $=\frac{\theta }{2 \pi }$
$=\frac{1}{2 \pi }\frac{\left(\right. 2 \pi n \left(\left.\right)^{2}}{2 \alpha }=\pi \cdot \frac{n^{2}}{\alpha }=\pi \times \frac{20 \times 20}{4}=100\pi =314$

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