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# A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string is wrapped over its rim and a mass of 4 kg is hung. An angular acceleration of $8\;rad-s^{-2}$ is produced in it due to the torque. Then moment of inertia of the wheel is :

$\begin {array} {1 1} (1)\;2\;kg-m^{2} & \quad (2)\;1\;kg-m^{2} \\ (3)\;4\;kg-m^{2} & \quad (4)\;8\;kg-m^{-2} \end {array}$

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A)
We know that Torque=r×f=I alpha Given., R=0.4m ,alpha(angular acceleration)=8 radian/s^2 ,M=4kg So..By equation. Of torque I×alpha=R×F I×8=0.4×mg I×8=0.4×4×10 By solving we get I=2kg-m^2