MCQ
Equal torques act on the discs $A$ and $B$ of the previous problem, initially both being at rest. At a later instant, the linear speeds of a point on the rim of $A$ and another point on the rim of $B$ are $v_A$ and $v_B$ respectively. We have:
  • $v_A> v_B$
  • B
    $v_A= v_B$
  • C
    $v_A < v_B$
  • D
    The relation depends on the actual magnitude of the torques.

Answer

Correct option: A.
$v_A> v_B$
$\tau=\text{I}\alpha ($Magnitude$)$
For equal torque, we have:
$\text{I}_{\text{A}\propto\text{A}}=\text{I}_{\text{B}\propto\text{B}}$
$\text{I}_{\text{A}}<\text{I}_{\text{B}}$
$\Rightarrow\alpha_{\text{A}}>\alpha_{\text{B}}\ \dots(\text{i})$
Now, $\omega=\alpha\text{t}$
Or, $\frac{\text{v}}{\text{r}}=\alpha\text{t}$
$\text{v}_{\text{A}}>\text{v}_{\text{B}} [$Using $\text{(i)}]$

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