Question

A bead P sliding on a frictionless semi-circular string... bead Q ejected... relation between $t_P$ and $t_Q$ is

• A. $t_P>t_Q$
• B. $t_P>1.25 t_Q$
• C. $t_P = t_Q$
• D. $t_P<t_Q$

Hint:

Brachistochrone principle: Curved path under gravity is faster.

Answer 1

The Correct Option is D

Let's analyze the motion of both beads, P and Q, on the semi-circular and straight paths respectively.

1. Bead P on the semi-circular path: 

• As bead P slides down the semi-circular string without friction, it undergoes circular motion under the force of gravity.
• The path of P is a semi-circle with radius \( R \). The length of the path is \( \pi R / 2 \).
• The acceleration of bead P is due to the gravitational component along the arc, \( g \sin \theta \), and the time taken is given by:
  \(t_P = \frac{\pi R}{2v_P}\), where \( v_P \) is the average velocity.

2. Bead Q on the straight path:

• Bead Q is ejected horizontally at 45° with initial speed. Therefore, bead Q follows a projectile motion.
• The time, \( t_Q \), for bead Q to reach point A is determined by the straight-line distance and the horizontal component of its initial velocity.

Comparison of \( t_P \) and \( t_Q \):

• The path travelled by P is longer than that of Q due to semi-circular motion versus straight-line travel.
• The horizontal component of Q’s velocity is greater than the average velocity of P along the arc due to lack of any component counteracting this motion.

Conclusion:

• Considering both the path length and velocity components, \(t_P \lt t_Q\) holds true.

Thus, the correct answer is: \(t_P \lt t_Q\)