A 5m plank of weight 50N is supported by two strings A and B. A block of weight 100N is suspended 2.5m from the left end. Rope A is attached 2m from the left end and Rope B is attached 1.5m from the r

$$
\text{Total volume} = \text{Final reading} - \text{Initial reading}
$$
$$
\text{Total volume} = 56 \text{ cm}^3 - 22 \text{ cm}^3 = 34 \text{ cm}^3
$$

$$
\text{Volume per drop} = \frac{\text{Total volume}}{\text{Number of drops}}
$$
$$
\text{Volume per drop} = \frac{34 \text{ cm}^3}{50}
$$
$$
\text{Volume per drop} = 0.68 \text{ cm}^3
$$
The volume of each drop is $\boxed{\text{0.68 cm}^3}$.

*   **Plank details:**
    *   Length of plank = $5 \text{ m}$
    *   Weight of plank ($W_p$) = $50 \text{ N}$ (acts at the center of the plank, $2.5 \text{ m}$ from either end).
    *   Block weight ($W_b$) = $100 \text{ N}$ (suspended from the center of the plank, $2.5 \text{ m}$ from either end).
*   **Positions from the left end:**
    *   Rope A: $2 \text{ m}$
    *   Center of plank (where $W_p$ and $W_b$ act): $2.5 \text{ m}$
    *   Rope B: $5 \text{ m} - 1.5 \text{ m} = 3.5 \text{ m}$

*   Tension $T_A$ (upwards): Distance from B = $3.5 \text{ m} - 2 \text{ m} = 1.5 \text{ m}$. This creates an anti-clockwise moment.
*   Weight of plank $W_p = 50 \text{ N}$ (downwards): Distance from B = $3.5 \text{ m} - 2.5 \text{ m} = 1.0 \text{ m}$. This creates a clockwise moment.
*   Weight of block $W_b = 100 \text{ N}$ (downwards): Distance from B = $3.5 \text{ m} - 2.5 \text{ m} = 1.0 \text{ m}$. This creates a clockwise moment.

$$
T_A \times 1.5 \text{ m} = (W_p \times 1.0 \text{ m}) + (W_b \times 1.0 \text{ m})
$$
$$
T_A \times 1.5 = (50 \text{ N} \times 1.0 \text{ m}) + (100 \text{ N} \times 1.0 \text{ m})
$$
$$
1.5 T_A = 50 + 100
$$
$$
1.5 T_A = 150
$$

$$
T_A = \frac{150}{1.5}
$$
$$
T_A = 100 \text{ N}
$$
The tension $T_A$ on the string A is $\boxed{\text{100 N}}$.