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A torque of $30\; N-m$ is applied on a 5 kg wheel whose moment of inertia is $2 kg\;m^2$ for $10 \;seconds$. The angle covered by the wheel in $10\; s$ will be

\[\begin {array} {1 1} (a)\;750\;rad & \quad (b)\;1500\;rad \\ (c)\;3000\;rad & \quad  (d)\;6000\;rad \end {array}\]
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$\alpha= \large\frac{z}{I}$
$\quad = \large\frac{30}{2}$
$\quad= 15 \;rad/s^2$
$\theta= \large\frac{1}{2}$$ \alpha\; t^2$
$\quad= \large\frac{1}{2} $$\times 15 \times 100$
$\quad =750 \;rad$


answered Nov 25, 2013 by meena.p

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