Question

Each situation in column I gives graph of a particle moving in circular path. The variables and t represent angular speed (at any time t ), angular displacement (in time t) and time respectively. Column II gives certain resulting interpretation. Match the graphs in column I with statements in column II.

• A. P $\to$ 3, Q $\to$ 2, R $\to$ 3, S $\to$ 4
• B. P $\to$ 2, Q $\to$ 3, R $\to$ 4, S $\to$ 1
• C. P $\to$ 2, Q $\to$ 1, R $\to$ 1, S $\to$ 2
• D. P $\to$ 4, Q $\to$ 3, R $\to$ 3, S $\to$ 2

Answer 1

The correct option is C

P $\to$ 2, Q $\to$ 1, R $\to$ 1, S $\to$ 2

From graph (a) $\Rightarrow \omega =k\theta$ where k is positive constant

angular acceleration $=\omega \frac{d\omega }{d\theta }=k\theta \times k=k^2\theta$

angular acceleration is non uniform and directly proportional to $\theta$.

From graph (b) $\Rightarrow {\omega }^2=k\theta$. Differentiating both sides with respect to $\theta$.

$2\omega \frac{d\omega }{d\theta }=k$ or $\omega \frac{d\omega }{d\theta }=\frac{k}{2}$ Hence angular acceleration is uniform.

From graph (c) $\Rightarrow \omega =kt$

angular acceleration $=\frac{d\omega }{dt}=k$ Hence angular acceleration is unifomr

Form graph (d) $\Rightarrow \omega =k{t}^2$

angular acceleration $=\frac{d\omega }{dt}=2kt$

Hence angular acceleration is non uniform and directly proportional to t.