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The total angle turned through during this interval is

  1. 48 rad

  2. 24 rad

  3. 96 rad

  4. 72 rad

3rd option is correct

\omega _{0}=0[initial]

\omega = 24 rad/s[final]

t=8sec

 \alpha = \frac{\omega }{t} = \frac{24}{8} = 3[angular acceleration]

\theta = \frac{1}{2} \alpha t^{2} = \frac{1}{2}\times 3 \times 8^{2} = 96 is the answer..!

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