The correct answer is $\boxed{\text{C. 55\%}}$.
The efficiency of a machine is defined as the ratio of the work done by the machine to the work done on the machine. In this case, the work done by the machine is the lifting of the 500 kg load through a distance of 13 cm. The work done on the machine is the effort of 25 kg moving through a distance of 650 cm. Therefore, the efficiency of the machine is:
$$\text{efficiency} = \frac{\text{work done by machine}}{\text{work done on machine}} = \frac{500 \times 0.13}{25 \times 0.65} = 0.55 = 55\%$$
Option A is incorrect because it is the percentage of the load
213.5V175.2l142.7 81.2-142.7 81.2z"/> Subscribe on YouTube