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The linear velocity of a rotating body is given by $\vec{v}=\vec{\omega} \times \vec{r}$, where $\vec{\omega}$ is the angular velocity and $\vec{r}$ is the radius vector. The angular velocity of a body is $\vec{\omega}=\hat{i}-2 \hat{j}+2 \hat{k} \quad$ and the radius vector $\vec{r}=4 \hat{j}-3 \hat{k}$, then $|\vec{v}|$ is
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Verified Answer
The correct answer is:
$\sqrt{29}$ units
The correct option is $\mathrm{1} \sqrt{29}$ units
We know that, $\vec{v}=\vec{\omega} \times \vec{r}=(\hat{i}-2 \hat{j}+2 \hat{k}) \times(4 \hat{j}-3 \hat{k})$
$\begin{aligned} \vec{v} & =\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 0 & 4 & -3\end{array}\right| \\ \vec{v} & =-2 \hat{i}+3 \hat{j}+4 \hat{k}\end{aligned}$
Thus, the magnitude of $\vec{v}$ is, $|\vec{v}|=\sqrt{(-2)^2+3^2+4^2}=\sqrt{29}$ units
Hence option 1 is the correct answer
We know that, $\vec{v}=\vec{\omega} \times \vec{r}=(\hat{i}-2 \hat{j}+2 \hat{k}) \times(4 \hat{j}-3 \hat{k})$
$\begin{aligned} \vec{v} & =\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 0 & 4 & -3\end{array}\right| \\ \vec{v} & =-2 \hat{i}+3 \hat{j}+4 \hat{k}\end{aligned}$
Thus, the magnitude of $\vec{v}$ is, $|\vec{v}|=\sqrt{(-2)^2+3^2+4^2}=\sqrt{29}$ units
Hence option 1 is the correct answer
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