A block and tackle system of pulleys consisting of 6 pulleys is used to raise a load of 600 N through a height of 15 cm. The work done against friction in the pulley is 900 J. Calculate the: (i) total work done by the effort; (ii) efficiency of the system; (iii) effort applied.

The work done on the load is the product of the load and the height it is raised.

$$W_L = \text{Load} \times \text{height}$$ $$W_L = 600 \text{ N} \times 0.15 \text{ m}$$ $$W_L = 90 \text{ J}$$

The total work done by the effort is the sum of the useful work done on the load and the work done against friction.

$$W_E = W_L + W_f$$ $$W_E = 90 \text{ J} + 900 \text{ J}$$ $$W_E = 990 \text{ J}$$

The total work done by the effort is $\boxed{\text{990 J}}$.

Efficiency ($\eta$) is the ratio of work output ($W_L$) to work input ($W_E$), expressed as a percentage.

$$\eta = \frac{W_L}{W_E} \times 100\%$$ $$\eta = \frac{90 \text{ J}}{990 \text{ J}} \times 100\%$$

$$\eta = \frac{1}{11} \times 100\%$$ $$\eta \approx 9.09\%$$

The efficiency of the system is approximately $\boxed{\text{9.09\%}}$.

For a block and tackle system, the velocity ratio is equal to the number of pulleys.

$$\text{VR} = \text{Number of pulleys} = 6$$

The velocity ratio is also the ratio of the distance moved by the effort to the distance moved by the load.

$$\text{VR} = \frac{h_E}{h_L}$$ $$h_E = \text{VR} \times h_L$$ $$h_E = 6 \times 0.15 \text{ m}$$ $$h_E = 0.9 \text{ m}$$

The total work done by the effort is the product of the effort and the distance it moves.

$$W_E = E \times h_E$$ $$E = \frac{W_E}{h_E}$$ $$E = \frac{990 \text{ J}}{0.9 \text{ m}}$$ $$E = 1100 \text{ N}$$

The effort applied is $\boxed{\text{1100 N}}$.