**Solution:**

Given,

Load (L) = 50 kg which means the actual load is 50 kg × 10 m/s^{2} = 500 N **[∵ F = mg ],
**Effort (E) = 120 N,

Effort distance (ED) = 20 cm,

- Load distance (LD) = ?

We have, from the principle of a simple machine,

L × LD = E × ED

Or, 500 × LD = 120 × 20

Or, \(LD=\frac{2400}{500}\)

**∴ LD = 4.8 cm**

- Mechanical advantage (MA) =?

We have,

Mechanical advantage, \(MA=\frac{L}{E}\)

Or, \(MA=\frac{500}{120}\)

**∴ MA = 4.17**

- Velocity ratio, VR =?

We have,

velocity ratio, \(VR=\frac{ED}{LD}\)

Or, \(VR=\frac{20}{4.8}\)

**∴ VR = 4.17**

- Efficiency (η) =?

Since MA = VR the efficiency of the given lever is 100%

Also using the formula for efficiency of a simple machine. we have,

Efficiency, \(η=\frac{MA}{VR}\times 100\%\)

Or, \(η=\frac{4.17}{4.17}\times 100\%\)

∴ Efficiency, η = 100 %

The given lever is 100% efficient. Which means that the lever is an ideal lever.