Question

Two beads P & Q move along two wires straight and semi circle. At some instant both are shown in figure having same horizontal component of velocity V. Find relation of time taken to reach A by beads :

• A. \(t_p>t_Q\)
• B. \(t_p<t_Q\)
• C. \(t_p = t_Q\)
• D. None of these

Hint:

In comparative motion problems, remember that the straight-line path between two points is the shortest distance.
A particle moving along this straight path will have the highest possible average velocity component in that direction compared to any other particle moving along a longer, curved path between the same two points, assuming comparable speeds.

Answer 1

The Correct Option is B

In this problem, we need to find the relation of the time taken by two beads, \(P\) and \(Q\), to reach point \(A\). Bead \(P\) moves along a straight path, and bead \(Q\) moves along a semicircular path, both having the same horizontal component of velocity \(V\). 

To determine the relation between the times \(t_P\) and \(t_Q\), let's consider the motion of both beads:

1. Horizontal Velocity Component:
   • Both beads have the same horizontal component of velocity, \(V\). Since they start moving horizontally from the same line, the time taken to cover the same horizontal distance will primarily depend on the total path length.
2. Path Length:
   • Bead \(P\) is traveling along a straight path, and its horizontal distance traveled is simply the straight-line distance directly to point \(A\).
   • Bead \(Q\) is traveling along a semicircular path, which is longer than the straight path taken by bead \(P\).
3. Time Relation:
   • The time taken (\(t = \frac{\text{distance}}{\text{velocity}}\)) will be greater for a longer path when the horizontal velocity components are equal.
   • This implies that bead \(Q\) will take a longer time to reach point \(A\) as compared to bead \(P\).
4. Conclusion:
   • Thus, \(t_P < t_Q\).
   • The correct answer is \(t_p<t_Q\).