JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 16)
A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance 'h', the square of angular velocity of wheel will be :
$${{2mgh} \over {I + 2m{r^2}}}$$
$${{2mgh} \over {I + m{r^2}}}$$
2gh
$${{2gh} \over {I + m{r^2}}}$$
Explanation
_26th_February_Evening_Shift_en_16_1.png)
Using energy conservation between A and B point
$$mgh = {1 \over 2}m{(wR)^2} + {1 \over 2}I{\omega ^2}$$
$$2mgh = (M{R^2} + I){\omega ^2}$$
$${\omega ^2} = {{2mgh} \over {I + M{R^2}}}$$
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