Verified Solution • 2025–26 Session moderate

QJEEM21Phdaa1: JEE Main 2021 Physics - Class 11 JEEM21Phdaa1 Solution: Statement I : Two forces and where , when act at an angle 1 to…

Statement I :

Two forces (P→+Q→)\left( {\overrightarrow P + \overrightarrow Q } \right) and (P→−Q→)\left( {\overrightarrow P - \overrightarrow Q } \right) where P→⊥Q→\overrightarrow P \bot \overrightarrow Q , when act at an angle θ\theta 1 to each other, the magnitude of their resultant is 3(P2+Q2)\sqrt {3({P^2} + {Q^2})} , when they act at an angle θ\theta 2, the magnitude of their resultant becomes 2(P2+Q2)\sqrt {2({P^2} + {Q^2})} . This is possible only when {\theta _1} < {\theta _2} .

Statement II :

In the situation given above.

θ\theta 1 = 60 ∘^\circ and θ\theta 2 = 90 ∘^\circ

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

Multiple Choice Options

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

Step-by-Step Verified Solution

A→=P→+Q→\overrightarrow A = \overrightarrow P + \overrightarrow Q

B→=P→−Q→\overrightarrow B = \overrightarrow P - \overrightarrow Q

P→⊥Q→\overrightarrow P \bot \overrightarrow Q

∣A→∣=∣B→∣=2(P2+Q2)(1+cos⁡θ)\left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})(1 + \cos \theta )}

For ∣A→+B→∣=3(P2+Q2)\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {3({P^2} + {Q^2})}

θ1=60∘{\theta _1} = 60^\circ

For ∣A→+B→∣=2(P2+Q2)\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})}

θ2=90∘{\theta _2} = 90^\circ

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