Question

A force of $250\text{ N}$ is required to lift a mass of $75\text{ kg}$ through a pulley system. In order to lift this mass through $3\text{ m}$, the rope has to be pulled through $12\text{ m}$. The efficiency of the system is:

• A. $50\%$
• B. $75\%$
• C. $33\%$
• D. $90\%$

Hint:

Efficiency can also be written as:
$\eta = \frac{\text{Mechanical Advantage (MA)}}{\text{Velocity Ratio (VR)}} \times 100\%$.
Here, $\text{MA} = \frac{mg}{F} = \frac{750}{250} = 3$, and $\text{VR} = \frac{d}{h} = \frac{12}{3} = 4$.
Thus, $\eta = \frac{3}{4} \times 100\% = 75\%$.

Answer 1

The Correct Option is B

Step 1: Use mechanical advantage and velocity ratio instead of raw work values.
Mechanical advantage is the ratio of load to effort, and velocity ratio is the ratio of the distance the effort moves to the distance the load moves.
Step 2: Compute both ratios.
The load is $mg = 75\times10 = 750\text{ N}$ (taking $g=10\text{ ms}^{-2}$) and the effort is $250\text{ N}$, so $\text{MA} = \frac{750}{250} = 3$. The rope moves $12\text{ m}$ while the load rises $3\text{ m}$, so $\text{VR} = \frac{12}{3} = 4$.
Step 3: Combine them for efficiency. \[ \eta = \frac{\text{MA}}{\text{VR}}\times100\% = \frac{3}{4}\times100\% \]
\[ \boxed{\eta = 75\%} \]