In a hydraulic lift, if the input power is 2000 W and the output power is 1800 W

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Question: In a hydraulic lift, if the input power is 2000 W and the output power is 1800 W, what is the efficiency of the lift?

Options:

Correct Answer: 80%

Solution:

Efficiency is calculated as (Output Power / Input Power) * 100%. Here, efficiency = (1800 W / 2000 W) * 100% = 90%.

In a hydraulic lift, if the input power is 2000 W and the output power is 1800 W, what is the efficiency of the lift?

In a hydraulic lift, if the input power is 2000 W and the output power is 1800 W, what is the efficiency of the lift?

Correct Answer: 90%

Step 1: Identify the input power of the hydraulic lift, which is given as 2000 W.
Step 2: Identify the output power of the hydraulic lift, which is given as 1800 W.
Step 3: Use the formula for efficiency, which is (Output Power / Input Power) * 100%.
Step 4: Substitute the values into the formula: (1800 W / 2000 W) * 100%.
Step 5: Calculate the division: 1800 W divided by 2000 W equals 0.9.
Step 6: Multiply the result by 100%: 0.9 * 100% equals 90%.
Step 7: Conclude that the efficiency of the hydraulic lift is 90%.

Efficiency – Efficiency is a measure of how much input energy is converted into useful output energy, expressed as a percentage.
Power – Power is the rate at which work is done or energy is transferred, measured in watts (W).