JEE Main 2023 — Laws Of Motion Question with Solution

As we know that impulse is given by

$$I=\Delta P=F\times \Delta t$$

or $I=$ Area of $f-t$ graph

For fig (a)

$\to \mathrm{I}=\frac{1}{2}\times$ base $\times$ height

$$=\frac{1}{2}\times 0.5\times 1=0.25\mathrm{N}-\sec .$$

For fig (b),

$I=$ length $\times$ width

$$=2\times 0.5=1\mathrm{N}\text{-sec }$$

For fig (c),

I $=\frac{1}{2}\times$ base $\times$ height

$$=\frac{1}{2}\times 1\times 0.75=0.375\mathrm{N}\text{-sec. }$$

For fig (d),

$I=\frac{1}{2}\times$ base $\times$ height

$$=\frac{1}{2}\times 2\times 0.5=0.5\mathrm{N}-\sec \text{. }$$

Impulse is highest for that figure, whose area under $F$-t is maximum and i.e. figure(b)

Option (D) is correct.