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  2. Try this! - Rotational Motion - JEE Main-3
# Engineering

Moment of inertia of a body about a given axis is 1.5\:kg\:m^{2}. Initially the body is ate 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 for a duration of :

  • Option 1)

    2.5\:s

  • Option 2)

    2\:s

  • Option 3)

    5\:s

  • Option 4)

    3\:s

Answers (1)
A Aadil.Khan

Given 

K.E.= 1200 J

\Rightarrow \frac{1}{2}I\:w^{2}=K.E.

\Rightarrow w=\sqrt{\frac{2K.E.}{I}}

\tau\:t=I\omega

\Rightarrow \tau\: t=I \sqrt{ \frac{2K.E.}{I}}

\Rightarrow \: t=\frac{I \sqrt{ \frac{2K.E.}{I}}}{I \:\alpha }

\Rightarrow \: t= \sqrt{\frac{2K.E.}{I}}\times \frac{1}{\alpha }

\Rightarrow \: t= \sqrt{\frac{2\times 1200}{1.5}}\times \frac{1}{20 }=2\:\:seconds  

 


Option 1)

2.5\:s

Option 2)

2\:s

Option 3)

5\:s

Option 4)

3\:s

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