- Home
- Standard 11
- Physics
3-1.Vectors
hard
Two forces of magnitude $P$ & $Q$ acting at a point have resultant $R$. The resolved part of $R$ in the direction of $P$ is of magnitude $Q$. Angle between the forces is :
A
$2{\sin ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)$
B
$2{\sin ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)^{\frac{1}{2}}$
C
$2{\cos ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)$
D
$2{\cos ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)^{\frac{1}{2}}$
Solution
$\cos \theta=\frac{Q-P}{Q}$
$\cos \theta=1-\frac{P}{Q}$
$1-2 \sin ^{2} \frac{\theta}{2}=1-\frac{P}{Q}$
$\therefore 2 \sin ^{2} \frac{\theta}{2}=\frac{P}{Q}$
$\sin ^{2} \frac{\theta}{2}=\frac{P}{2 Q}$
$\Rightarrow \sin \frac{\theta}{2}=\left(\frac{P}{2 Q}\right)^{1 / 2}$
$\Rightarrow \theta=2 \sin ^{-1}\left(\frac{P}{2 Q}\right)^{1 / 2}$
Standard 11
Physics
