Question

A rod $AB$ is shown in figure. End $A$ of the rod is fixed on the ground. Block is moving with velocity $\sqrt{3}\text{m/s}$ towards right. The velocity of end $B$ of rod when rod makes an angle of ${60}^{\circ }$ with the ground is

• A. $\sqrt{3}\text{m/s}$
• B. $2\text{m/s}$
• C. $2\sqrt{3}\text{m/s}$
• D. $3\text{m/s}$

Answer 1

The correct option is B $2\text{m/s}$

As we take point B

From diagram we can say that
velocity of block can be given as
$v=v_t\cos {30}^{\circ }$
where $v_t=$ velocity of point B as rod moves in circular trajectory
So
$\sqrt{3}=v_t\times \frac{\sqrt{3}}{2}\Rightarrow v_t=2\text{m/s}$