Q141 / 160 · TS EAMCET 2014 · Physics
A wheel which is initially at rest is subjected to a constant angular acceleration about its axis. It rotates through an angle of $15^{\circ}$ in time $t$ sec. The increase in angle through which it rotates in the next $2 t \mathrm{sec}$ is
- A$90^{\circ}$
- Correct answerB$120^{\circ}$
- C$30^{\circ}$
- D$45^{\circ}$
Solution
If angular acceleration is constant, we have
$$
\theta=\omega_0 t+\frac{1}{2} \alpha t^2
$$
The given $\theta=15^{\circ}$
$$
\omega_0=0
$$
For the first conditions (time $=t \mathrm{sec}$ )
$$
\begin{aligned}
15^{\circ} & =0+\frac{1}{2} \alpha t^2 \\
\Rightarrow \quad 15^{\circ} & =\frac{1}{2} \alpha t^2
\end{aligned}
$$
For the second conditions (time $=3 t \mathrm{sec}$ )
$$
\theta_1=\frac{1}{2} \alpha(3 t)^2=\frac{1}{2}(\alpha) 9 t^2
$$
So,
$$
\begin{aligned}
\Delta \theta & =\theta_1-\frac{1}{2} \alpha t^2 \\
\Delta \theta & =9 \times \frac{1}{2} \alpha t^2-\frac{1}{2} \alpha t^2 \\
& =8 \frac{1}{2} \alpha t^2 \\
& =8 \times 15^{\circ}=120^{\circ}
\end{aligned}
$$
(from Eq. (i))
Hence, the correct answer is option: 'B'.