The lower end of a 4m long uniform rod AB is pulled with constant speed v = 4m/s. Then speed of centre of mass of the rod at $\mathrm{\theta }={60}^0$ will be 

From the figure,

$\mathrm{OB}=\mathrm{x}=\mathrm{ABcos\theta }$

$\mathrm{x}=4\mathrm{cos\theta }$

$\frac{\mathrm{dx}}{\mathrm{dt}}=(-4\mathrm{sin\theta })\frac{\mathrm{d\theta }}{\mathrm{dt}}\mathrm{\_\_\_\_\_}\left(1\right)$

Velocity along x direction $\frac{\mathrm{dx}}{\mathrm{dt}}=4\mathrm{m}/\mathrm{s}$

from equation 1,

${\left(\frac{\mathrm{d\theta }}{\mathrm{dt}}\right)}_{\mathrm{at} \mathrm{\theta }={60}^{\mathrm{o}}}=-\mathrm{cosec}{60}^{\mathrm{o}}=-\frac{2}{\sqrt{3}}\mathrm{rad}/\mathrm{s}$

the rod's centre of mass's radius,

$\overrightarrow{{\mathrm{r}}_{\mathrm{c}}}=\left(2\mathrm{cos\theta }\right)\hat{\mathrm{i}}+\left(2\mathrm{sin\theta }\right)\hat{\mathrm{j}}$

velocity of the provided rod's centre of mass

$\overrightarrow{{\mathrm{V}}_{\mathrm{c}}}=\frac{\mathrm{d}\overrightarrow{{\mathrm{r}}_{\mathrm{c}}}}{\mathrm{dt}}=\left(-2\mathrm{sin\theta }\frac{\mathrm{d\theta }}{\mathrm{dt}}\right)\hat{\mathrm{i}}+\left(2\mathrm{cos\theta }\frac{\mathrm{d\theta }}{\mathrm{dt}}\right)\hat{\mathrm{j}}$

$\mathrm{at} \mathrm{\theta }={60}^{\mathrm{o}}$

$\overrightarrow{{\mathrm{V}}_{\mathrm{c}}}=\left(-2\times \frac{\sqrt{3}}{2}\times \frac{-2}{\sqrt{3}}\right)\hat{\mathrm{i}}+\left(2\times \frac{1}{2}\times \frac{-2}{\sqrt{3}}\right)\hat{\mathrm{j}}$

$\overrightarrow{{\mathrm{V}}_{\mathrm{c}}}=2\hat{\mathrm{i}}-\frac{2}{\sqrt{3}}\hat{\mathrm{j}}$

$\left|\overrightarrow{{\mathrm{V}}_{\mathrm{c}}}\right|=\sqrt{4+\frac{4}{3}}=\frac{4}{\sqrt{3}}\mathrm{m}/\mathrm{s}$

Hence the correct answer is $\frac{4}{\sqrt{3}}\mathrm{m}/\mathrm{s}.$