Question

Two bodies $A$ and $B$ rotate about an axis, such that angle ${\theta }_A$ (in radians) covered by the first body is proportional to square of time, and ${\theta }_B$ (in radians) covered by the second body varies linearly. At $t=0,{\theta }_A={\theta }_B=0$. If $A$ completes its first revolution in $\sqrt{\pi }\text{sec}$ and $B$ needs $4\pi \text{sec}$ to complete half a revolution, then angular velocity ${\omega }_A$ and ${\omega }_B$ at $t=5\text{sec}$ are in the ratio

• A. 4 : 1
• B. 20 : 1
• C. 80 : 1
• D. 20 : 4

Answer 1

The correct option is C 80 : 1

Given ${\theta }_A\propto t^2$ & ${\theta }_B\propto t$
Then, ${\theta }_A=k_1{t}^2$ & ${\theta }_B=k_{2}t$

From given condition, calculate $k_1$ and $k_2$
$2\pi =k_1\times \pi$ & $\pi =k_2\times 4\pi$
$\Rightarrow k_1=2$ & $k_2=\frac{1}{4}$
$\therefore {\theta }_A=2t^2$ & ${\theta }_B=\frac{t}{4}$
${\omega }_A=\frac{d{\theta }_A}{dt}=4t$ &
${\omega }_B=\frac{d{\theta }_B}{dt}=\frac{1}{4}$

At $t=5s$.
${\omega }_A=20rad/s$ & ${\omega }_B=\frac{1}{4}rad/s$
$\therefore {\omega }_A:{\omega }_B=80:1$