A particle moves with uniform acceleration along a straight line from rest. The percentage increase in displacement during 6^th compared to that in 5^th is about

Solution (By Examveda Team)

The question asks about the percentage increase in displacement of a particle moving with uniform acceleration.
It compares the distance covered in the 6th second to the distance covered in the 5th second.

Understanding the Concept:
Since the particle starts from rest and has uniform acceleration, it means its velocity increases at a constant rate.
The distance covered in each successive second will keep increasing.

Formula to Use:
The distance traveled in the *n*th second (Sn) can be calculated using the formula: Sn = u + (a/2)(2n - 1)
Where:
* u = initial velocity (0 in this case, since it starts from rest)
* a = acceleration
* n = the second you are interested in

Calculations:
Distance in 5th second (S5) = 0 + (a/2)(2*5 - 1) = (a/2) * 9 = 4.5a
Distance in 6th second (S6) = 0 + (a/2)(2*6 - 1) = (a/2) * 11 = 5.5a

Percentage Increase:
Percentage increase = [(S6 - S5) / S5] * 100
Percentage increase = [(5.5a - 4.5a) / 4.5a] * 100
Percentage increase = (a / 4.5a) * 100
Percentage increase = (1 / 4.5) * 100
Percentage increase ≈ 22.22%

Answer:
Therefore, the percentage increase is approximately 22%. So the correct answer is Option B: 22%