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 A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed \nu in the x-y plane as shown in the figure : Which of the following statements is false for the angular momentum  \vec{L} about the
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 L=\frac{R}{\sqrt{2}} m v(-k)when the particle is moving from D to A.

Answers (1)


\\ 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)

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Ritika Jonwal

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