As shown in the figure, five Carnot engines, each with efficiency $\eta$ and same number of cycles per unit time, are operating between six heat reservoirs. The amount of heat released per cycle by one engine is completely absorbed by the next engine. Consider $Q_0$ to be the amount of heat absorbed per cycle by the first engine and $W$ as the amount of total work done by all the engines per cycle, then the net efficiency of the system is found to be $\eta_{\text{net}} = \frac{W}{Q_0} = \frac{211}{243}$. The value of $\eta$ is:

Step 1: Express the heat rejected by the first engine.

The first engine absorbs heat $Q_0$ and operates with efficiency $\eta$.

$$Q_1 = Q_0(1-\eta)$$

Step 2: Express the heat rejected by the fifth engine.

Since the heat rejected by each engine is completely absorbed by the next engine in the series, the heat rejected by the fifth engine is determined sequentially.

$$Q_5 = Q_0(1-\eta)^5$$

Step 3: Calculate the total work done by all engines.

The total work $W$ is the difference between the heat absorbed by the first engine and the heat rejected by the last engine.

$$W = Q_0 - Q_5$$ $$W = Q_0\left[1 - (1-\eta)^5\right]$$

Step 4: Relate the total work to the net efficiency.

The net efficiency of the system is the ratio of the total work done to the initial heat absorbed.

$$\eta_{\text{net}} = \frac{W}{Q_0} = 1 - (1-\eta)^5$$

Step 5: Solve for the efficiency of each engine.

Substitute the given net efficiency $\eta_{\text{net}} = \frac{211}{243}$ into the equation.

$$1 - (1-\eta)^5 = \frac{211}{243}$$ $$(1-\eta)^5 = 1 - \frac{211}{243}$$ $$(1-\eta)^5 = \frac{32}{243}$$ $$(1-\eta)^5 = \left(\frac{2}{3}\right)^5$$ $$1 - \eta = \frac{2}{3}$$ $$\eta = \frac{1}{3}$$

Answer: The value of $\eta$ is $\dfrac{1}{3}$.