JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 21)
The angular speed of truck wheel is increased from 900 rpm to 2460 rpm in 26 seconds. The number of revolutions by the truck engine during this time is _____________. (Assuming the acceleration to be uniform).
Answer
728
Explanation
$${\omega _f} = 2460 \times {{2\pi } \over {60}}$$
$$ = 82\pi $$
$${\omega _i} = {{900 \times 2\pi } \over {60}} = 30\pi $$
$$\alpha = {{{\omega _f} - {\omega _i}} \over t}$$
$$ = {{82\pi - 30\pi } \over {26}}$$
= 2 $$\pi$$ rad/sec2
$$\theta = {{\omega _f^2 - \omega _i^2} \over {2\alpha }}$$
$$ = {{(82\pi + 30\pi )(82\pi - 30\pi )} \over {2 \times 2\pi }}$$
$$ = {{(112 \times 52){\pi ^2}} \over {4\pi }}$$
No. of revolution $$ = {{(112 \times 13)\pi } \over {2\pi }}$$
= 728
$$ = 82\pi $$
$${\omega _i} = {{900 \times 2\pi } \over {60}} = 30\pi $$
$$\alpha = {{{\omega _f} - {\omega _i}} \over t}$$
$$ = {{82\pi - 30\pi } \over {26}}$$
= 2 $$\pi$$ rad/sec2
$$\theta = {{\omega _f^2 - \omega _i^2} \over {2\alpha }}$$
$$ = {{(82\pi + 30\pi )(82\pi - 30\pi )} \over {2 \times 2\pi }}$$
$$ = {{(112 \times 52){\pi ^2}} \over {4\pi }}$$
No. of revolution $$ = {{(112 \times 13)\pi } \over {2\pi }}$$
= 728
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