AIIMS2019Physics-Mechanics

AIIMS 2019 Physics Impulse and Momentum MCQ Question

Type: MCQ-numerical-Medium-Class 11

Consider the given expression. F = (100 - 0.5 × 10⁵ t) N Given that, F = 0 (100 - 0.5 × 10⁵ t) = 0 t = 100 / (0.5 × 10⁵) = 2 × 10⁻³ sec Calculate the impulse as, I = ∫ F dt = ∫ (100 - 0.5 × 10⁵ t) dt = [100t - (10⁵ t² / 2)] = [100(2 × 10⁻³) - (10⁵ (2 × 10⁻³)² / 2)] = 0.1 Ns

A

0.1 Ns

B

0.2 Ns

C

0.3 Ns

D

0.4 Ns

Correct Answer

Option C

Detailed Explanation

The impulse II is calculated by integrating the force F=(1000.5×105t)F = (100 - 0.5 \times 10^5 t) over time, which yields I=Fdt=[100t(0.5×105t2)]I = \int F \, dt = [100t - (0.5 \times 10^5 t^2)]. Evaluating this from t=0t = 0 to t=2×103t = 2 \times 10^{-3} seconds gives I=0.1NsI = 0.1 \, \text{Ns}, confirming that option A is correct. Options B, C, and D are incorrect as they do not match the calculated impulse value, demonstrating the importance of careful integration and evaluation in determining impulse from varying forces.

Found an issue with this question?