#### A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed in the x-y plane as shown in the figure : Which of the following statements is false for the angular momentum  about the origin? Option: 1 $\vec{L}= -\frac{mv}{\sqrt{2}}R\hat{k}$when the particle is moving from A to B. Option: 2 $\vec{L}= mv\left [ \frac{R}{\sqrt{2}}-a \right ]\hat{k}$when the particle is moving from C to D. Option: 3 $\vec{L}= mv\left [ \frac{R}{\sqrt{2}}+a \right ]\hat{k}$when the particle is moving from B to C. Option: 4 when the particle is moving from D to A.

$\\ In\ option\ (a), co-ordinates \ of\ A are \left(\frac{R}{\sqrt{2}}, \frac{R}{\sqrt{2}}\right) \\ \therefore \vec{r}=\left(\frac{R}{\sqrt{2}} \hat{i}+\frac{R}{\sqrt{2}} \hat{j}\right) and \vec{v}=v \hat{i}\\ \vec{L} m(\vec{r} \times \vec{v})=m\left(\frac{R}{\sqrt{2}} \hat{i}+\frac{R}{\sqrt{2}} \hat{j}\right) \times v \hat{i}\\ \vec{L}=-\frac{m R}{\sqrt{2}} v \hat{k}$

$\\ in\ option \ (b)\ it \ moves\ from \ C \ to\ D\\ L=\left(\frac{R}{\sqrt{2}}+a\right) m v(\hat{k})$

so option b is correct option

$\\ in \ option\ (c), \ For\ B \ to\ C, \\ we have L=\left(\frac{R}{\sqrt{2}}+a\right) m v(\hat{k})$

$\\ in \ option \ (d), \ When \ a \ particle\ is \ moving\ from \ D \ to\ A\\ L=\frac{R}{\sqrt{2}} m v(-k)$