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A particle of mass $m$ is moving along the side of a square of side '$a$', with a uniform speed $v$ in the $x-y$ plane as shown in the figure
Which of the following statement is false for the angular momentum $\vec L$ about the origin ?
$\;\left( a \right)\overrightarrow {\;L} $ $ = mv\left[ {\frac{R}{{\sqrt 2 }} + a} \right]\hat k$ ,when the particle is moving from $B$ to $C$
$\vec L$ $ = \frac{{mv}}{{\sqrt 2 }}\;R\;\hat k$ , when the particle is moving from $D$ to $A$
$\left( d \right)\overrightarrow {\;L} $ $ = mv\left[ {\frac{R}{{\sqrt 2 }} - a} \right]\hat k$ ,when the particle is moving from $C$ to $D$
Both (b) and (c)
Solution
$\overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{P}}$ or $\overrightarrow{\mathrm{L}}=\mathrm{rpsin} \theta \hat{\mathrm{n}}$
$\overrightarrow{\mathrm{L}}=\mathrm{r}_{\perp}(\mathrm{P}) \hat{\mathrm{n}}$
For $D$ to $A$
$\overrightarrow{\mathrm{L}}=\frac{\mathrm{R}}{\sqrt{2}} \mathrm{mV}(-\hat{\mathrm{k}})$
For $A$ to $B$
$\overrightarrow{\mathrm{L}}=\frac{\mathrm{R}}{\sqrt{2}} \mathrm{mV}(-\hat{\mathrm{k}})$
For $\mathrm{C}$ to $\mathrm{D}$
$\overrightarrow{\mathrm{L}}=\left(\frac{\mathrm{R}}{\sqrt{2}}+\mathrm{a}\right) \mathrm{mV}(\hat{\mathrm{k}})$
For $B$ to $C$
$\overrightarrow{\mathrm{L}}=\left(\frac{\mathrm{R}}{\sqrt{2}}+\mathrm{a}\right) \mathrm{mV}(\hat{\mathrm{k}})$
