Mathematics in Physics

JEE Main level, single correct, medium

14 of 121

Question 14 of 121

Two forces having magnitude AA and A2\dfrac{A}{2} are perpendicular to each other. The magnitude of their resultant is:

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Answer: (B) 5A2\dfrac{\sqrt{5}A}{2}

Using parallelogram law of vector addition, the formula can be written as
Anet=A12+A22+2A1A2cos⁡θ{A}_{net}=\sqrt{{{A}_{1}}^{2}+{{A}^{2}}_{2}+2{A}_{1}{A}_{2}\cos \theta }
The data given is
∣A1⃗∣=A,∣A2⃗∣=A2\left|\vec{{A}_{1}}\right|=A,\left|\vec{{A}_{2}}\right|=\dfrac{A}{2}
Since the vectors are perpendicular cos⁡90∘=0\cos 90^{\circ}=0
⇒Fnet=A2+A24\Rightarrow {F}_{net}=\sqrt{{A}^{2}+\dfrac{{A}^{2}}{4}}=5A2=\dfrac{\sqrt{5}A}{2}

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