JEE Main 2019 — Physics solution

JEE Main2019Physics

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 k $θ^{2}$ . If its moment of inertia is l then the angular acceleration of the disc is:

A
$\frac{2 k}{l} θ$
B
$\frac{k}{l} θ$
C
$\frac{k}{2 l} θ$
D
$\frac{k}{4 l} θ$

Solution

$τ = \frac{d E}{d θ} 2 k θ = l α α = \frac{2 k θ}{l}$

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