We are provided with the following values:

• Upward force on the top pulley: $F_1 = 110 \text{ N}$
• Downward force on the bottom pulley: $F_2 = 90 \text{ N}$
• Linear mass density of the rope: $\lambda = 0.25 \text{ kg/m}$
• Acceleration due to gravity: $g = 10 \text{ m/s}^2$

We need to find the total length of the rope, $L$.

Let us consider the entire system comprising the two pulleys and the rope as a single unit in vertical equilibrium. We analyze the forces acting along the vertical direction ($\hat{j}$).

Upward Forces:

• The force exerted on the axle of the top pulley lifts the rope system upwards. This force is $F_1$.

Downward Forces:

• The force exerted on the axle of the bottom pulley pulls the rope system downwards. This force is $F_2$.
• The total weight of the rope itself acts downwards. The weight $W_{rope}$ is given by: $$ W_{rope} = (\text{Total Mass}) \cdot g = (\lambda L) g $$

For the system to remain at rest, the net upward force exerted by the pulleys on the rope must balance the net downward forces, including the weight of the rope.

The tension at the level of dotted line is same, so the block’s weight is same as the force from right support.

Thus, we can write the force balance equation:

$$ F_{up} = F_{down} $$ $$ F_1 = F_2 + W_{rope} $$

Rearranging for the weight of the rope:

$$ F_1 – F_2 – \lambda g L = 0 $$

Substitute the given values into the equation:

$$ 110 – 90 – (0.25 \times 10 \times L) = 0 $$ $$ 20 – 2.5 L = 0 $$ $$ 2.5 L = 20 $$ $$ L = \frac{20}{2.5} $$ $$ L = 8 \text{ m} $$