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

The impulse is highest in figure.

Option: 1

Fig (c)


Option: 2

Fig (b)


Option: 3

Fig (d)


Option: 4

Fig (a)


Answers (1)

best_answer

As we know that impulse is given by
$$ \mathrm{I}=\Delta \mathrm{P}=\mathrm{F} \times \Delta \mathrm{t}
Or
$$ I=\text { Area of f-t graph }
For fig (a) \rightarrow \mathrm{I}=\frac{1}{2} \times base \times height
$$ =\frac{1}{2} \times 0.5 \times 1=0.25 \mathrm{~N}-\mathrm{sec} .
For fig (b) \mathrm{I}= length \times width
$$ =2 \times 0.5=1 \mathrm{~N}-\mathrm{sec}
For fig (c) I =\frac{1}{2} \times$ base $\times height
$$ =\frac{1}{2} \times 1 \times 0.75=0.375 \mathrm{~N}-\mathrm{sec} .
For fig (d) I=\frac{1}{2} \times$ base $\times height
$$ =\frac{1}{2} \times 2 \times 0.5=0.5 \mathrm{~N}-\mathrm{sec} .
Impulse is highest for that figure, whose area under F-t is maximum and i.e. figure(b) Option (2) is correct.

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Pankaj

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