Machines — Question 4

(i) Given,

Effort = 25 kgf

Load = 75 kgf

As we know that,

$M.A. = \dfrac{\text{Load}}{\text{Effort}} \\[0.5em]$

Substituting the values in the formula we get,

$M.A. = \dfrac{75}{25} \\[0.5em] M.A. = \dfrac{3}{1} \\[0.5em] \Rightarrow M.A. = 3 \\[0.5em]$

(ii) As we know,

V.R. = n = 3

(iii) Also, $\eta = \dfrac{M.A.}{V.R.} \\[0.5em] \eta = \dfrac{3}{3} \\[0.5em] \Rightarrow\eta = 1 \\[0.5em]$

Therefore, efficiency = 100%