Question

Two beads A and B move along a semicircular wire frame as shown in figure. 
The beads are connected by an inelastic string which always remains tight. 
At an instant the speed of A is u, ∠BAC = 45° and ∠BOC = 75°, where O is the centre of the semicircular arc. The speed of bead B at the instant is 
(A) 2 u                (B) u                  (C) u/2(2)^1/2            (D) (2/3)^1/2 u 

Answer 1

Bead B is moving on the semicircle so its speed is always tangential to the path

Let say the speed of bead B is "v"

now it is given that the string is inelastic as well as it always remains tight

So here the component of velocity of two beads along the length of the string will always remain the same for both

Now the speed of bead A is given "u"

so its component of speed along the length of the string will be given as

$v_a = ucos45$

also the tangential speed of bead B is making an angle of 60 degree with the string so the component of speed of bead B along the length of the string is given as

$v_b = v cos60$

now as we know that both must be same

$vcos60 = ucos45$

$v = \frac{ucos45}{cos60}$

$v = \frac{u\frac{1}{\sqrt2}}{\frac{1}{2}}$

$v = \frac{2u}{\sqrt2}$

so above is the speed of the bead B