Q115 / 150 · MHT CET 2024 (09 May Shift 2) · Physics
A disc at rest is subjected to a uniform angular acceleration about its axis. Let $\theta$ and $\theta^1$ be the angle made by the disc in $2^{\text {nd }}$ and $3^{\text {rd }}$ second of its motion. The ratio $\frac{\theta}{\theta^1}$ is
- A$2: 3$
- B$1: 2$
- C$2: 3$
- Correct answerD$4: 5$
Solution
Kinematic equation for rotational motion,
$\theta=\omega_0 t+\frac{1}{2} \alpha t^2$
Disk is initially at rest, $\omega_0=0$ $\Rightarrow \theta=\frac{1}{2} \alpha t^2...(i)$ Angle described in $2^{\text {nd }}$ second is, $\theta_1=\frac{1}{2} \alpha(2)^2=2 \alpha$
Angle described in first 3 seconds will be, $\theta_2=\frac{1}{2} \alpha(3)^2=4.5 \alpha$
Angle described in $3^{\text {rd }}$ second will be, $\theta^{\prime}=\theta_2-\theta_1$ $\begin{aligned} & =4.5 \alpha-2 \alpha=2.5 \alpha \\ \therefore \quad & \frac{\theta}{\theta^{\prime}} \end{aligned}=\frac{2}{2.5}=\frac{4}{5}$
Disk is initially at rest, $\omega_0=0$ $\Rightarrow \theta=\frac{1}{2} \alpha t^2...(i)$ Angle described in $2^{\text {nd }}$ second is, $\theta_1=\frac{1}{2} \alpha(2)^2=2 \alpha$
Angle described in first 3 seconds will be, $\theta_2=\frac{1}{2} \alpha(3)^2=4.5 \alpha$
Angle described in $3^{\text {rd }}$ second will be, $\theta^{\prime}=\theta_2-\theta_1$ $\begin{aligned} & =4.5 \alpha-2 \alpha=2.5 \alpha \\ \therefore \quad & \frac{\theta}{\theta^{\prime}} \end{aligned}=\frac{2}{2.5}=\frac{4}{5}$
Hence, the correct answer is option: 'D'.