Moment of inertia of a body about a given axis is 1.5, kg, m^2 Initially body is at rest. In order t

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6.System of Particles and Rotational Motion

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Moment of inertia of a body about a given axis is $1.5\, kg\, m^2$ Initially the body is at rest. In order to produce a rotational kinetic energy of $1200\, J$, the angular acceleration of $20\, rad/s^2$ must be applied about the axis of rotation for a duration of ......... $\sec$.

A

$2$

B

$5$

C

$2.5$

D

$3$

(JEE MAIN-2019)

Solution

$KE = \frac{1}{2}I{\omega ^2} = 1200\left( {given} \right)$

$ \Rightarrow \omega = 40\,rad/s$

$\,\,\,\,\,\,\,\omega = {\omega _0} + \alpha t$

$\,\,\,\,\,\,\,40 = 0 + \left( {20} \right)\,t$

$ \Rightarrow t = 2\,\sec .$

Standard 11

Physics

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