Back to Practice
AIIMS2019Physics-Mechanics

AIIMS 2019 Physics Impulse and Momentum MCQ Question

Type: MCQ-numerical-Medium-Class 11

A gun applies a force F\text{F} on a bullet which is given by F=(100−0.5×105t)N\text{F} = \left(100 - 0.5 \times 10^5\text{t}\right)\text{N}. The bullet emerges out with speed 400 m/s400\text{ m/s}. Then find out the impulse exerted till force on bullet becomes zero.

A

0.2 N−s0.2\text{ N}-\text{s}

B

0.3 N−s0.3\text{ N}-\text{s}

C

0.1 N−s0.1\text{ N}-\text{s}

D

0.4 N−s0.4\text{ N}-\text{s}

Correct Answer

Option C

Detailed Explanation

The impulse II is calculated by integrating the force F=(100−0.5×105t)F = (100 - 0.5 \times 10^5 t) over time, which yields I=∫F dt=[100t−(0.5×105t2)]I = \int F \, dt = [100t - (0.5 \times 10^5 t^2)]. Evaluating this from t=0t = 0 to t=2×10−3t = 2 \times 10^{-3} seconds gives I=0.1 NsI = 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?

Related Questions

Explore More Questions

All Physics QuestionsMechanical Properties of Fluids QuestionsAIIMS 2019 PYQ