The velocity ratio of an inclined plane where angle of inclination in θ is
A.
Sinθ
B.
Cosθ
C.
Tanθ
D.
\(\frac{1}{\sin \theta}\)
Correct answer is D
Consider an inclined plane shown below, as the effort moves along OB, the load moves is lifted up through a vertical height AB
V.R = \(\frac{\text{distance moved by effort}}{\text{distance moved by load}}\) = \(\frac{OB}{AB}\)
But Sinθ = \(\frac{opp}{hyp}\) = \(\frac{AB}{OB}\)
therefore; = \(\frac{OB}{AB}\) = \(\frac{1}{\sin \theta}\)