Statement I-Two forcesP-Q and P-Q where P-Q- when act at an angle -1 to each other- the magnitude of their resultant is -3P2-Q2- when they act an angle -2- the magnitude of their resultant becomes -2P2-Q2- This is possible only when -1-2-Statement II-In the situtation given above- -1-60o and -2-90o In the light of the above statements choose the most appropriate answer from the options given below-

Given that P is perpendicular to Q, henceP . Q=0 Using paralellogram law, Resultant formula is given by nbsp; R=√(| nbsp; A|^2+| nbsp; B|^2+2| nbsp; A| ...

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Byju's Answer

Standard XII

Physics

Vector Addition

Statement I:T...

Question

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

$Statement II:$
In the situtation given above. $θ_{1} = 60^{o} and θ_{2} = 90^{o}$ In the light of the above statements choose the most appropriate answer from the options given below:

Both statement I and Statement II are false

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Statement I is true but Statement II is false.

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Statement i is false but Statement II is true.

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Both

Statement I

and

Statement II

are true.

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Solution

The correct option is D Both $Statement I$ and $Statement II$ are true.
Given that $\to P$ is perpendicular to $\to Q$ , hence $\to P . \to Q = 0$

Using paralellogram law, Resultant formula is given by
$R = \sqrt{| \to A |^{2} + | \to B |^{2} + 2 | \to A | | \to B} | cos θ$

Here $\to A = (\to P + \to Q) ⟹ | \to A | = \sqrt{P^{2} + Q^{2}}, (∵ \to P . \to Q = 0)$
$\to B = (\to P - \to Q) ⟹ | \to B | = \sqrt{P^{2} + Q^{2}} (∵ \to P . \to Q = 0)$

Hence $R = \sqrt{P^{2} + Q^{2} + P^{2} + Q^{2} + 2 (P^{2} + Q^{2})} cos θ$
$R = \sqrt{2 (P^{2} + Q^{2}) (1 + cos θ)}$

$(i)$ If $R = \sqrt{3 (P^{2} + Q^{2})}, θ = θ_{1}$

$\sqrt{3 (P^{2} + Q^{2})} = \sqrt{2 (P^{2} + Q^{2}) (1 + cos θ_{1})}$

$3 = 2 (1 + cos θ_{1})$

$cos θ_{1} = 0.5$

$θ_{1} = 60^{\circ}$

$(i i)$ If $R = \sqrt{2 (P^{2} + Q^{2})}, θ = θ_{2}$

$\sqrt{2 (P^{2} + Q^{2})} = \sqrt{2 (P^{2} + Q^{2}) (1 + cos θ_{2})}$

$2 = 2 (1 + cos θ_{2})$

$cos θ_{2} = 0$

$θ_{2} = 90^{\circ}$

Both $Statement I$ and $Statement II$ are true.

Suggest Corrections

Similar questions

Q. The resultant of two forces P and Q acting at an angle a is equal to

(2 m + 1) \sqrt{P^{2} + Q^{2}} .

When the forces act at an angle

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(2 m - 1) \sqrt{P^{2} + Q^{2}} .

Then value of

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Q. If

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and

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