Question

A particle is rotating in a circle of radius $1\text{m}$ with constant speed $4\text{m/s}$. In time $1\text{s}$
$\begin{array}{cccc}& \text{Column - I}& & \text{Column - II}\\ & \text{(Quantity)}& & \text{(Value in SI units)}\\ \text{(a)}& \text{Displacement}& \text{(p)}& 8\sin 2\\ \text{(b)}& \text{Distance}& \text{(q)}& 4\\ \text{(c)}& \text{Average velocity}& \text{(r)}& 2\sin 2\\ \text{(d)}& \text{Average acceleration}& \text{(s)}& 4\sin 2\end{array}$

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

Answer 1

The correct option is D a - r; b - q; c - r; d - p

(a) → (r); (b) → (q); (c) → (r); (d) → (p)
Given, $v=4\text{m/s},r=1\text{m}$
(a) In $1\text{s}$,



$\theta =\frac{l}{r}=\frac{4}{1}=4$
Now in $\Delta OMB$
$\sin \frac{\theta }{2}=\frac{BM}{BO}\Rightarrow \sin \left(\frac{4}{2}\right)=\frac{BM}{r}$
$\therefore BM=r\sin 2$
$\therefore$ required displacement
$=2BM$
$=2\sin 2$
(b) In one second distance $=v\times t$
$=4\times 1=4\text{m}$
(c) $v_{\text{average}}=\frac{2\sin 2}{1}=2\sin 2$
(d) Change in velocity $=\overrightarrow{v_f}-\overrightarrow{v_i}=\overrightarrow{v_f}+(-\overrightarrow{v_i})$
$=\sqrt{v^2+v^2+2\times v\times v\cos (\pi -4)}$
$=\sqrt{2v^2-2v^2\cos 4}$
$=\sqrt{2v^2(1-\cos 4)}=2v\sin 2$
$a_{\text{average}}=\frac{\text{change in velocity}}{t}=\frac{2v\sin 2}{t}$
$=2\times v\sin 2=8\sin 2$