Statement-I : Two forces ( P+ Q) and ( P- Q) where Pperp Q, when act at an angle θ _1 each other, the magnitude… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Mechanics

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Statement-I : Two forces $(\vec{P}+\vec{Q})$ and $(\vec{P}-\vec{Q})$ where $\vec{P}\perp \vec{Q},$ when act at an angle ${\theta }_{1}$ each other, the magnitude of their resultant is $\sqrt{3({P}^{2}+{Q}^{2})},$ when they act at an angle ${\theta }_{2},$ the magnitude of their resultant becomes $\sqrt{2({P}^{2}+{Q}^{2})}.$ This is possible only when ${\theta }_{1}<{\theta }_{2}.$

Statement-II : In the situation given above.

${\theta }_{1}=60^{\circ}$ and ${\theta }_{2}=90^{\circ}$

In the light of the above statement, choose the most appropriate answer from the options given below :

Official previous-year question

Held on 31 Aug 2021 · Verified 6 Jul 2026.

Options

A
Statement I is false but Statement II is true.
B
Both Statement I and Statement II are true.
C
Both Statement I and Statement II are false.
D
Statement I is true but Statement II is false.

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Solution

${\vec{F}}_{1}$ and ${\vec{F}}_{2}$ at ${\theta }_{1}$

${F}_{net1}=\sqrt{{P}^{2}+{Q}^{2}+{P}^{2}+{Q}^{2}+2({P}^{2}+{Q}^{2})\mathrm{cos}{\theta }_{1}}$

${F}_{net2}=\sqrt{{P}^{2}+{Q}^{2}+{P}^{2}+{Q}^{2}+2({P}^{2}+{Q}^{2})\mathrm{cos}{\theta }_{2}}$

If ${F}_{net1}=\sqrt{3({P}^{2}+{Q}^{2})}=\sqrt{2({P}^{2}+{Q}^{2})+2({P}^{2}+{Q}^{2})\mathrm{cos}{\theta }_{1}}$

$\Rightarrow \mathrm{cos}{\theta }_{1}=\frac{{P}^{2}+{Q}^{2}}{2({P}^{2}+{Q}^{2})}$

$\Rightarrow {\theta }_{1}=60^{\circ}$

${F}_{net2}=\sqrt{2({P}^{2}+{Q}^{2})}=\sqrt{2({P}^{2}+{Q}^{2})+2({P}^{2}+{Q}^{2})\mathrm{cos}{\theta }_{2}}$

$\Rightarrow \mathrm{cos}{\theta }_{2}=0\Rightarrow {\theta }_{2}=90^{\circ}$

JEE Main 2021 Physics, Mathematics in Physics — solution figure

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Solution

${\vec{F}}_{1}$ and ${\vec{F}}_{2}$ at ${\theta }_{1}$

${F}_{net1}=\sqrt{{P}^{2}+{Q}^{2}+{P}^{2}+{Q}^{2}+2({P}^{2}+{Q}^{2})\mathrm{cos}{\theta }_{1}}$

${F}_{net2}=\sqrt{{P}^{2}+{Q}^{2}+{P}^{2}+{Q}^{2}+2({P}^{2}+{Q}^{2})\mathrm{cos}{\theta }_{2}}$

If ${F}_{net1}=\sqrt{3({P}^{2}+{Q}^{2})}=\sqrt{2({P}^{2}+{Q}^{2})+2({P}^{2}+{Q}^{2})\mathrm{cos}{\theta }_{1}}$

$\Rightarrow \mathrm{cos}{\theta }_{1}=\frac{{P}^{2}+{Q}^{2}}{2({P}^{2}+{Q}^{2})}$

$\Rightarrow {\theta }_{1}=60^{\circ}$

${F}_{net2}=\sqrt{2({P}^{2}+{Q}^{2})}=\sqrt{2({P}^{2}+{Q}^{2})+2({P}^{2}+{Q}^{2})\mathrm{cos}{\theta }_{2}}$

$\Rightarrow \mathrm{cos}{\theta }_{2}=0\Rightarrow {\theta }_{2}=90^{\circ}$

JEE Main 2021 Physics, Mathematics in Physics — solution figure

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Solution

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JEE Main Mechanics in other years

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Other Physics topics for JEE Main

Electromagnetism

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Optics

Thermodynamics

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