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A pulley of radius $2 \mathrm{~m}$ is rotated about its axis by a force $\mathrm{F}=\left(20 \mathrm{t}-5 \mathrm{t}^2\right)$ Newton (where $\mathrm{t}$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation made by the pulley before its direction of motion if reversed, is :
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The correct answer is:
more than 3 but less than 6
more than 3 but less than 6
$\mathrm{r} \times \mathrm{F}=\mathrm{l} \times \alpha$
$2\left(20 \mathrm{t}-5 \mathrm{t}^2\right)=10 \alpha \Rightarrow \alpha=4 \mathrm{t}-\mathrm{t}^2$
$\frac{\mathrm{d} \omega}{\mathrm{dt}}=4 \mathrm{t}-\mathrm{t}^2$
$\mathrm{~d} \omega=\left(4 \mathrm{t}^2-\mathrm{t}^2\right) \mathrm{dt}$
$\omega=2 \mathrm{t}^2-\frac{\mathrm{t}^3}{3}$ (on integration)
$\omega=0 \Rightarrow \mathrm{t}=6 \mathrm{~s}$
$2\left(20 \mathrm{t}-5 \mathrm{t}^2\right)=10 \alpha \Rightarrow \alpha=4 \mathrm{t}-\mathrm{t}^2$
$\frac{\mathrm{d} \omega}{\mathrm{dt}}=4 \mathrm{t}-\mathrm{t}^2$
$\mathrm{~d} \omega=\left(4 \mathrm{t}^2-\mathrm{t}^2\right) \mathrm{dt}$
$\omega=2 \mathrm{t}^2-\frac{\mathrm{t}^3}{3}$ (on integration)
$\omega=0 \Rightarrow \mathrm{t}=6 \mathrm{~s}$
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