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

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.

Ask Question

Tag:MathPhyChemBioOther

Take Test

...

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