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(B) Let the unit vector towards East be $\hat{i}$ and towards North be $\hat{j}$.$s_1 = \sqrt{2} \cos 45^{\circ} \hat{i} + \sqrt{2} \sin 45^{\circ} \h

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$\sqrt{14 - 4\sqrt{3}} \ m$

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(B) Let the unit vector towards East be $\hat{i}$ and towards North be $\hat{j}$.$s_1 = \sqrt{2} \cos 45^{\circ} \hat{i} + \sqrt{2} \sin 45^{\circ} \hat{j} = \hat{i} + \hat{j}$$s_2 = -2 \hat{j}$ (since it is due south)$s_3 = 4 \cos 150^{\circ} \hat{i} + 4 \sin 150^{\circ} \hat{j} = 4(-\frac{\sqrt{3}}{2}) \hat{i} + 4(\frac{1}{2}) \hat{j} = -2\sqrt{3} \hat{i} + 2 \hat{j}$Net displacement $\vec{s} = s_1 + s_2 + s_3 = (1 - 2\sqrt{3}) \hat{i} + (1 - 2 + 2) \hat{j} = (1 - 2\sqrt{3}) \hat{i} + \hat{j}$Magnitude $|\vec{s}| = \sqrt{(1 - 2\sqrt{3})^2 + (1)^2} = \sqrt{1 + 12 - 4\sqrt{3} + 1} = \sqrt{14 - 4\sqrt{3}} \ m$.

$A$ particle undergoes three successive displacements given by $s_1 = \sqrt{2} \ m$ north-east,$s_2 = 2 \ m$ due south,and $s_3 = 4 \ m$ at $30^{\circ}$ north of west. The magnitude of the net displacement is:

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