A plane is inclined at an angle \(\theta\) to the horizontal. Its velocity ratio is
\(sin\theta\)
\(tan\theta\)
\(\frac{1}{sin\theta}\)
\(\frac{1}{tan\theta}\)
Correct answer is C
Velocity ratio = \(\frac{Effort distance}{Load distance}\)
On an inclined plane, it is the inverse of \(sin\theta\) = \(\frac{1}{sin\theta}\)