Rotational Motion

JEE Main level, numerical answer, medium

207 of 280

Question 207 of 280

The position vector of 1kg1kg object is r⃗=(3i^−j^)m\vec{r}=\left(3\hat{i}-\hat{j}\right)m and its velocity v⃗=(3j^+k^)ms−1\vec{v}=\left(3\hat{j}+\hat{k}\right)m{s}^{-1}. The magnitude of its angular momentum is xNms\sqrt{x}Nms, where xx is

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Answer: 91

Angular momentum of a particle is given by,
L⃗=r⃗×P⃗=r⃗×mv⃗\vec{L}=\vec{r}\times \vec{P}=\vec{r}\times m\vec{v} also m=1kgm=1kg.
⇒L⃗=(3i^−j^)×(3j^+k^)=(9k^−3j^−i^)Nms\Rightarrow \vec{L}=\left(3\hat{i}-\hat{j}\right)\times \left(3\hat{j}+\hat{k}\right)=\left(9\hat{k}-3\hat{j}-\hat{i}\right)Nms
The magnitude of the angular momentum will be,
⇒∣L⃗∣=92+32+12=91Nms\Rightarrow \left|\vec{L}\right|=\sqrt{{9}^{2}+{3}^{2}+{1}^{2}}=\sqrt{91}Nms
Therefore, the value of x=91x=91.

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