Statement I: Two forces (P+Q) and (P−Q)... | Laws of Motion

JEE MAIN 2021`

31-08-2021 S1

Question

Statement I: Two forces $\left(\stackrel{\Delta \stackrel{\Delta}{\mathrm{P}}+\stackrel{\Delta}{\mathrm{Q}})}{(\stackrel{\Delta}{\mathrm{P}}-\stackrel{\Delta}{\mathrm{Q}})}\right.$ and where $^{(\stackrel{\Delta}{\mathrm{P}} \perp \stackrel{\Delta}{\mathrm{Q}})}$, when act at an angle $\theta_1$ to each other, the magnitude of their resultant is $\sqrt{3\left(P^2+Q^2\right)}$ when they act at an angle $\theta 2$, the magnitude of their resultant becomes $\sqrt{2\left(P^2+Q^2\right)}$. This is possible only when $\theta_1<\theta_2$. Statement II : In the situation given above. $$ \theta_1=60^{\circ} \text { and } \theta_2=90^{\circ} $$ In the light of the above statements, choose the most appropriate answer from the options given below :-

Select the correct option:

Statement-I is false but Statement-II is true

Both Statement-I and Statement-II are true

Statement-I is true but Statement-II is false

Both Statement-I and Statement-II are false

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$$ \begin{aligned} & \stackrel{\varpi 口}{\mathrm{~A}}=\stackrel{\varpi \oplus}{\mathrm{P}}+\stackrel{\varpi}{\mathrm{Q}}, \stackrel{\Delta n}{\mathrm{~B}}=\stackrel{\varpi}{\mathrm{P}}-\stackrel{\varpi 口}{\mathrm{Q}} \\ & \stackrel{\square}{\mathrm{P}} \perp \stackrel{\mathbb{Q}}{\mathrm{Q}} \\ & |\stackrel{N}{A}|=|\stackrel{W}{B}|=\sqrt{P^2+Q^2} \\ & |\stackrel{\theta}{\mathrm{~A}}+\stackrel{\Omega}{\mathrm{B}}|=\sqrt{2\left(\mathrm{P}^2+\mathrm{Q}^2\right)(1+\cos \theta)} \\ & \text { For }|\stackrel{\otimes}{\mathrm{A}}+\stackrel{\otimes}{\mathrm{B}}|=\sqrt{3\left(\mathrm{P}^2+\mathrm{Q}^2\right)} \\ & \theta_1=60^{\circ} \\ & \text { For }|\hat{A}+B|=\sqrt{2\left(P^2+Q^2\right)} \\ & \theta_2=90^{\circ} \end{aligned} $$

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JEE Main

Physics

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