JEE Physics 2023 Question with Solution

Given: Two forces have magnitudes $A$ and $\frac{A}{2}$, and they are perpendicular to each other.

Find: The magnitude of the resultant force.

For two perpendicular forces, use the Pythagorean relation:

$$F_r = \sqrt{F_1^2 + F_2^2}$$

Substituting $F_1 = A$ and $F_2 = \frac{A}{2}$,

$$F_r = \sqrt{A^2 + \left(\frac{A}{2}\right)^2}$$ $$F_r = \sqrt{A^2 + \frac{A^2}{4}}$$ $$F_r = \sqrt{\frac{4A^2 + A^2}{4}}$$ $$F_r = \sqrt{\frac{5A^2}{4}}$$ $$F_r = \frac{\sqrt{5A^2}}{2} = \frac{\sqrt{5}A}{2}$$

Therefore, the magnitude of the resultant force is $\frac{\sqrt{5}A}{2}$. The correct option is B.

Given: Two perpendicular forces are $A$ and $\frac{A}{2}$.

Find: Their resultant magnitude.

When two forces are perpendicular, their resultant is obtained directly by squaring each, adding, and taking the square root:

$$R = \sqrt{A^2 + \left(\frac{A}{2}\right)^2} = \sqrt{\frac{5A^2}{4}} = \frac{\sqrt{5A^2}}{2}$$

So the resultant is $\frac{\sqrt{5}A}{2}$, which matches option B.