If a disc of mass M and radius R rotates with an angular acceleration a_, of the torque acting on the disc is

Correct answer: (B) (M R^2 a)/2. ∴ Moment of inertia of disc, Y=1/2 M R^2 Angular acceleration, α=a (given) So, torque τ=Y α , τ= (1/2 M R^2 )⟨ a⟩ ⇒ τ=(M R^2 a)/2

Q131 / 160 · TS EAMCET 2018 (07 May Shift 1) · Physics

If a disc of mass $M$ and radius $R$ rotates with an angular acceleration $a_{\text {, }}$ of the torque acting on the disc is

A
$M R^2 a$
Correct answer
B
$\frac{M R^2 a}{2}$
C
$\frac{2 M R^2 a}{5}$
D
$\frac{M R^2 a}{12}$

Solution

$\therefore$ Moment of inertia of disc, $Y=\frac{1}{2} M R^2$ Angular acceleration, $\alpha=a$ (given) So, torque $\tau=Y \alpha$, $$ \tau=\left(\frac{1}{2} M R^2\right)\langle a\rangle \Rightarrow \tau=\frac{M R^2 a}{2} $$

Hence, the correct answer is option: 'B'.

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