Statement-I : Two forces ( → P + → Q ) and ( → P - → Q ) where → P ⊥ → Q , when act at an angle θ _ 1 each other, the magnitude of their resultant is 3( P ^ 2 + Q ^ 2 ) , when they act at an angle θ _ 2 , the magnitude

Correct answer: B. Both Statement I and Statement II are true. → F 1 and → F 2 at θ _ 1 F _ net1 = P ^ 2 + Q ^ 2 + P ^ 2 + Q ^ 2 + 2( P ^ 2 + Q ^ 2 )cos θ _ 1 F _ net2 = P ^ 2 + Q ^ 2 + P ^ 2 + Q ^ 2 + 2( P ^ 2 + Q ^ 2 )cos θ _ 2 If F _ net1 = 3( P ^ 2 + Q ^ 2 ) = 2( P ^ 2 + Q ^ 2 ) + 2( P ^ 2 + Q ^ 2 )cos θ _ 1 ⇒ cos θ _ 1 = P ^ 2 + Q ^ 2 2( P ^ 2 + Q ^ 2 ) ⇒ θ _ 1 = 60 ° F _ net2 = 2( P ^ 2 + Q ^ 2 ) = 2( P ^ 2 + Q ^ 2 ) + 2( P ^ 2 + Q ^ 2 )cos θ _ 2 ⇒ cos θ _ 2 = 0 ⇒ θ _ 2 = 90 °

NEET

Exam: NEET
Subject: Physics
Chapter: Mathematics in Physics

NEET · Physics · Mathematics in Physics

Statement-I : Two forces ( → P + → Q ) and ( → P - → Q ) where → P...

Statement-I : Two forces

\left(\overset{\to}{P} + \overset{\to}{Q}\right)

and \left(\overset{\to}{P} - \overset{\to}{Q}\right)

where

\overset{\to}{P} ⊥ \overset{\to}{Q},

when act at an angle

{\theta}_{1}

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.<br/>{\theta}_{1} = 60 ° and {\theta}_{2} = 90 °

In the light of the above statement, choose the most appropriate answer from the options given below : Statement-I : Two forces ( → P + → Q ) and ( → P - → Q ) where → P ⊥ → Q , when act at an angle θ _ 1 each other, the

Statement-I : Two forces ( → P + → Q ) and ( → P - → Q ) where → P ⊥ → Q , when act at an angle θ _ 1 each other, the

Options

AStatement I is false but Statement II is true.
BBoth Statement I and Statement II are true.
CBoth Statement I and Statement II are false.
DStatement I is true but Statement II is false.

Correct answer

B. Both Statement I and Statement II are true.

Step-by-step solution

\overset{\to}{F}1

and

\overset{\to}{F}2

{\theta}_{1}

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

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

{F}_{net1} = \sqrt{3\left({P}^{2} + {Q}^{2}\right)} = \sqrt{2\left({P}^{2} + {Q}^{2}\right) + 2\left({P}^{2} + {Q}^{2}\right)cos{\theta}_{1}}

⇒ cos{\theta}_{1} = \dfrac{{P}^{2} + {Q}^{2}}{2\left({P}^{2} + {Q}^{2}\right)}

⇒ {\theta}_{1} = 60 °

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

⇒ cos{\theta}_{2} = 0 ⇒ {\theta}_{2} = 90 °

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