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JEE Main 20231 FebruaryEvening Shift
Exam
JEE Main
Year
2023
Date
1 February
Shift
Evening Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2023 · 1 February · Evening Shift · Physics · Center of Mass, Momentum and Collision

Figures (a), (b), (c) and (d) show variation of force with time....

Figures (a), (b), (c) and (d) show variation of force with time.

Figures (a), (b), (c) and (d) show variation of force with time. The impulse is highest in figure.Figures (a), (b), (c) and (d) show variation of force with time. The impulse is highest in figure.Figures (a), (b), (c) and (d) show variation of force with time. The impulse is highest in figure.Figures (a), (b), (c) and (d) show variation of force with time. The impulse is highest in figure.

The impulse is highest in figure.

Options

  1. AFig (c)
  2. BFig (b)
  3. CFig (a)
  4. DFig (d)

Correct answer

B. Fig (b)

Step-by-step solution

As we know, impulse is the area under the curve of the force vs. time graph. So, analysing each figure area under curve.

(a)A1=12×1.0×0.5=14N‐s\left(a\right){A}_{1} = \dfrac{1}{2} \times 1 . 0 \times 0 . 5 = \dfrac{1}{4}N ‐ s

\left(b\right){A}_{2} = 0 . 5 \times 2 = 1N ‐ s</p><p>(c)A3=12×1×0.75=38N‐s</p><p>\left(c\right){A}_{3} = \dfrac{1}{2} \times 1 \times 0 . 75 = \dfrac{3}{8}N ‐ s

\left(d\right){A}_{4} = \dfrac{1}{2} \times 2 \times 0 . 5 = \dfrac{1}{2}N ‐ s

From above calculations, area under curve in figure (b) is maximum. Thus, fig (b) impulse is highest.

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