A machine is operated by an effort of 25 N and the effort has a downward displacement of 2.5 m in raising a load of weight 100 N through 10 cm. What is the efficiency of machine (in %)?

Options

A. 250
B. 40
C. 25
D. 16

Answer

D

Explanation

Given: Effort (E) = 25 N
Load (L) = 100 N
`S_(E)` = displacement of effort = 0.25 m
= 25 cm
Displacement of load = 10 cm = `S_(L)`
`MA=L/(E)=100/(25)=4`
`V.R=(SE)/(SL)`
`=250/(10)=25`
`therefore` efficiency of machines is
`eta=(MA)/(VR)=4/(25)=0.16`
`eta="efficiency"=0.16`
% of efficiency `=0.16xx100=16%`
Hence, the correct option is (d).