- AStationary
- ✓Not stationary
- CCentralized
- DNone of these
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$\mathop {\begin{array}{*{20}{c}}
{O\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{||\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C - C{H_2} - Cl}
\end{array}}\limits_{\left( I \right)} $ $\underset{(II)}{\mathop{\begin{matrix}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||\,\,\, \\
Cl-C{{H}_{2}}-C-H \\
\end{matrix}}}\,$
$\mathop {\begin{array}{*{20}{c}}
O \\
{||} \\
{H - C - H}
\end{array}}\limits_{\left( {III} \right)} $ $\mathop {\begin{array}{*{20}{c}}
O \\
{||} \\
{{H_3}C - C - C{H_3}}
\end{array}}\limits_{\left( {IV} \right)} $
${i}=\left\{\sqrt{42} \sin \left(\frac{2 \pi}{{T}} {t}\right)+10\right\} {A}$
The $r.m.s.$ value of this current is ${A}$
$(A)$ $|\vec{\tau}|=\frac{1}{3} \mathrm{Nm}$
$(B)$ The torque $\vec{\tau}$ is in the direction of the unit vector $+\hat{\mathrm{k}}$
$(C)$ The velocity of the body at $\mathrm{t}=1$ is $\overrightarrow{\mathrm{v}}=\frac{1}{2}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}) \mathrm{m} \mathrm{s}^{-1}$
$(D)$ The magnitude of displacement of the body at $\mathrm{t}=1 \mathrm{~s}$ is $\frac{1}{6} \mathrm{~m}$
| Column $I$ | Column $II$ |
| $(A)$ Horizontal component of velocity | $(p)$ $5$ SI unit |
| $(B)$ Vertical component of velocity | $(q)$ $10$ SI unit |
| $(C)$ Horizontal displacement | $(r)$ $15$ SI unit |
| $(D)$ Vertical displacement | $(s)$ $20$ SI unit |