Question

Acceleration $a$ in case of circular motion is given by $a=\sqrt{a_r^2+a_t^2}$, where $a_r$ is radial component of acceleration and $a_t$ the tangential component of acceleration. $a_t$ governs the magnitude of velocity $\overrightarrow{v}$ and $a_r$ its direction of motion.

$\begin{array}{cccc}& \text{Column - I}& & \text{Column-II}\\ \text{(a)}& \text{If}{a}_r=0,a_t=0& \text{(p)}& \text{motion is non-uniform circular}\\ \text{(b)}& \text{If}{a}_r=0,a_t\ne 0& \text{(q)}& \text{motion is uniform circular}\\ \text{(c)}& \text{If}{a}_r\ne 0,a_t=0& \text{(r)}& \text{motion is accelerated translatory}\\ \text{(d)}& \text{If}{a}_r\ne 0,a_t\ne 0& \text{(s)}& \text{motion is uniform translatory}\\ & & \text{(t)}& \text{no conclusion can be drawn regarding motion}\end{array}$

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

Answer 1

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

$\left(a\right)\to \left(s\right);\left(b\right)\to \left(r\right);\left(c\right)\to \left(q\right);\left(d\right)\to \left(p\right)$
$a=\sqrt{a_c^2+a_t^2}$
$a_t=\frac{dv}{dt}=0$ (if $v=\text{constant}$)
$a_c=\frac{v^2}{R}$

(a) $a_r=0$ and $a_t=0$
$a=0\to$ uniform and translatory
(b) $a_r=0$ and $a_t\ne 0$
$a=a_t$
i.e., $v=$ variable and $a_r\ne 0$
$\therefore a_r=\frac{v^2}{R}\to$ must variable
but it is given, $a_r=0$
motion is accelerated translatory.
(c) $a_r\ne 0,a_t=0$
$a_t=0\to v=\text{constant}$
so, $a_r=\frac{v^2}{r}\ne 0$
so motion is uniform circular motion
(d) $a_r\ne 0$ and $a_t\ne 0$
$a_t\ne 0\Rightarrow v\ne \text{constant}$
and so $a_r=\frac{v^2}{r}\ne \text{constant}\ne 0$
so motion is non-uniform circular motion.