Question

In circular motion, suppose $\overrightarrow{a}$ is the acceleration of particle,
$\overrightarrow{v}$ its linear velocity and $\overrightarrow{\omega }$ its angular velocity.
$$\begin{array}{cc}& \\ \text{Column-I}& \text{Column-II}\\ \text{(a)}|\overrightarrow{a}\times \overrightarrow{v}|& \text{(p) positive}\\ \text{(b)}\overrightarrow{a}.\overrightarrow{v}& \text{(q) Negative}\\ \text{(c)}\overrightarrow{\omega }.\overrightarrow{v}& \text{(r) zero}\\ \text{(d)}|\overrightarrow{\omega }\times \overrightarrow{a}|& \text{(s) Undefined}\end{array}$$

• A. $a-p;b-p,q;c-r;d-p$
• B. $a-p;b-p,q,r;c-r;d-p$
• C. $a-p,q;b-p,q;c-r;d-p$
• D. $a-p;b-p,q;c-q,r;d-p$

Answer 1

The correct option is B $a-p;b-p,q,r;c-r;d-p$

$\left(a\right)\to \left(p\right);\left(b\right)\to \left(p\right),\left(q\right),\left(r\right)$
$\left(c\right)\to \left(r\right);\left(d\right)\to \left(p\right)$
In circular motion
(a) Here $\overrightarrow{a}$ is net acceleration
$|\overrightarrow{a}\times \overrightarrow{v}|=av\sin \theta$ also $0\le \theta \le \pi ,\sin \theta \to +ve$
so $|\overrightarrow{a}\times \overrightarrow{v}|=+ve$
(b) $\overrightarrow{a}.\overrightarrow{v}=av\cos \theta$
For $0<\theta <\pi /2\Rightarrow \cos \theta \to +ve$
and for $\theta =\pi /2\Rightarrow \cos \theta =0$
$\therefore \pi /2<\theta <\pi \Rightarrow \cos \theta =-ve$
$\overrightarrow{a}.\overrightarrow{v}\to$ may be +ve , 0 and -ve
(c) $\overrightarrow{\omega }\perp \overrightarrow{v}$ i.e., $\overrightarrow{\omega }.\overrightarrow{v}=0$
(d) $\because \overrightarrow{\omega }\perp \overrightarrow{a}$
$\Rightarrow |\overrightarrow{\omega }\times \overrightarrow{a}|=\omega a\sin {90}^{\circ }=\omega a\to +ve$