# Two beads A and B move along a semicircular wire frame as shown. The beads are connected by an inelastic string which always remains tight. At an instant the speed of A is $u$ . $\angle BAC=45^{\circ}$ and angle $BOC=75 ^{\circ}$ where O is center of the semicircular arc. The speed of bead B at that instant is

$(a)\;\sqrt 2 u \quad (b)\; u \quad (c)\;\frac{u}{2 \sqrt 2} \quad(d)\;\sqrt {\frac{2}{3}}u$

Let V be the speed of bead B
Now since the string remains tight velocity component of A and B along the string AB must be same
$u \cos \theta=v \cos \alpha$
$\theta=45^{\circ}$ and by geometring we get
$\alpha=45^{\circ}$
$u \large\frac{1}{\sqrt 2}$$=v \times \large\frac{1}{\sqrt2}$
$v=\ u$
Hence b is the correct answer.
edited May 27, 2014
Well bro u were always wrong...don't post answers until u r 200% confident...it may confuse others ......
Correction: How can u take direction of velocity in the direction of string...
Direction of velocity is always tangential to the path....correct it ur self