JEE MAIN - Physics (2023 - 15th April Morning Shift - No. 11)
A vector in $x-y$ plane makes an angle of $30^{\circ}$ with $y$-axis. The magnitude of $\mathrm{y}$-component of vector is $2 \sqrt{3}$. The magnitude of $x$-component of the vector will be :
$\sqrt{3}$
2
6
$\frac{1}{\sqrt{3}}$
Explanation
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$$ \begin{aligned} & \mathrm{A}_{\mathrm{y}}=\mathrm{A} \cos 30^{\circ}=2 \sqrt{3} \\\\ & \Rightarrow \mathrm{A} \frac{\sqrt{3}}{2}=2 \sqrt{3} \\\\ & \Rightarrow \mathrm{A}=4 \end{aligned} $$
Now $A_x=A \sin 30^{\circ}=4 \times \frac{1}{2}=2$
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