Question

A simple machine displaces a load of 125 kgf through a distance of 0.30 m, when an effort of 12.5 kgf acts through a distance of 4.0 m. Answer the following question.

The percentage efficiency of the machine is:

पर्याय

• 60%

• 65%

• 75%

• 70%

Answer 1

75%

Explanation:

$$\mathrm{MA} = \frac{L}{E}$$

$$ = \frac{125\ \mathrm{kgf}}{12.5\ \mathrm{kgf}}$$

= 10

\[ \text{VR} = \frac{\text{Distance moved by effort}}{\text{Distance moved by load}} \]

\[= \frac{4.0}{0.30} \]

= 13.33

\[ \eta = \dfrac{\text{MA}}{\text{VR}} \times 100\% \]

\[ \eta = \dfrac{10}{13.33} \times 100\% \]

= 0.7501 × 100%

= 75%