Gujarati
Hindi
3-1.Vectors
hard

Two forces $P + Q$ and $P -Q$ make angle $2 \alpha$ with each other and their  resultant make $\theta$ angle with bisector of angle between them. Then :

A

$P\, tan\, \theta = Q\, tan \, \alpha$

B

$P\, sin\, \theta = Q\, sin\, \alpha$

C

$P\, cos\, \alpha = Q\, sin\, \theta$

D

$P\, sin\, \alpha = Q\, sin\, \theta$

Solution

As shown in figure resolve vector of magnitude $(P+Q)$ and $(P-Q)$ along line of resultant and perpendicular to it then sine components cancel each other.

$(P+Q) \sin (\alpha-\theta)=(P-Q) \sin (\alpha+\theta)$

$\mathrm{P}[\sin (\alpha+\theta)-\sin (\alpha-\theta)]=Q[\sin (\alpha-\theta)+\sin (\alpha+\theta)]$

$\mathrm{P} .2 \cos \alpha \sin \theta=\mathrm{Q} \cdot 2 \sin \alpha \cos \theta$

$\mathrm{P} \tan \theta=\theta \tan \alpha$

Standard 11
Physics

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