A bead $P$ sliding on a frictionless semi-circular string $(A C B)$ and it is at point $S$ at $\mathrm{t}=0$ and at this instant the horizontal component of its velocity is $v$. Another bead $Q$ of the same mass as $P$ is ejected from point $A$ at $\mathrm{t}=0$ along the horizontal string $A B$, with the speed $v$, friction between the beads and the respective strings may be neglected in both cases. Let $t_{p}$ and $t_{Q}$ be the respective times taken by beads $P$ and $Q$ to reach the point $B$, then the relation between $t_{p}$ and $t_{Q}$ is

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Kinematics

A bead

P

sliding on a frictionless semi-circular string

(A C B)

and it is at point

S

t = 0

and at this instant the horizontal component of its velocity is

v

. Another bead

Q

of the same mass as

P

is ejected from point

A

t = 0

along the horizontal string

A B

, with the speed

v

, friction between the beads and the respective strings may be neglected in both cases. Let

t_{p}

and

t_{Q}

be the respective times taken by beads

P

and

Q

to reach the point

B

, then the relation between

t_{p}

and

t_{Q}

Answer: (B)

S

the bead's velocity is tangent to the circle and makes

4 5^{\circ}

with the horizontal, so if its horizontal component is

v

the speed there is

v_{S} = \frac{v}{cos 4 5 ^{\circ}} = 2 v

. Bead

Q

runs along the straight horizontal wire

A B

with no horizontal force, covering the diameter

2 R

at the constant speed

v

, so

t_{Q} = \frac{2 R}{v}

. Bead

P

travels the arc from

S

B

, which subtends

13 5^{\circ}

and has length

\frac{3 π}{4} R \approx 2.36 R

; as it first dips below

A B

gravity speeds it up, so its speed stays of order

2 v

or more along the path. Covering a path only a little longer than

2 R

at an average speed appreciably greater than

v

, bead

P

arrives before bead

Q

t_{P} < t_{Q}

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option