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A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of $θ$ , where $θ$ is the angle by which it has rotated, is given as ${kθ}^{2}$ . If its moment of inertia is I then the angular acceleration of the disc is

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$\frac{2 k}{I} θ$

$\frac{k}{2 I} θ$

$\frac{k}{4 I} θ$

$\frac{k}{I} θ$

answer is A.

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Detailed Solution

We know that,

Kinetic energy

$= \frac{1}{2} {Iω}^{2} = {kθ}^{2}$

$ω^{2} = \frac{2 {kθ}^{2}}{I}$

$ω \sqrt{\frac{2 k}{I}} θ______(1)$

differentiating the equation 1 with respect to time t

$\frac{dω}{dt} = α = \sqrt{\frac{2 k}{I}} (\frac{dθ}{dt})$

$α = \sqrt{\frac{2 k}{I}} \sqrt{\frac{2 k}{I}} θ$

$α = \frac{2 k}{I} θ$

Hence the correct answer is $\frac{2 k}{I} θ$

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