Question

In trees at a height of 75 meters, the magnitude of gravitational component of water potential in leaves is:

• A. - 0.25 MPa
• B. - 0.50 MPa
• C. - 0.75 MPa
• D. - 1.00 MPa

Hint:

To quickly calculate gravitational water potential, remember the direct conversion factor:
$\Psi_g$ changes by $0.01 \text{ MPa}$ for every $1 \text{ meter}$ of height.
Simply multiply the height in meters by $0.01$ to get the value in MPa.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The gravitational component of water potential ($\Psi_g$) represents the influence of gravity on water potential.
Gravity causes water to flow downward unless opposed by an equal and opposite force.
As water moves upward in a tall tree, work must be done against gravity.
This requirement decreases the overall water potential of water at the top of the tree relative to the ground.
Key Formula or Approach:
The gravitational component of water potential is calculated using the formula: \[ \Psi_g = \rho_w \cdot g \cdot h \] where:
$\rho_w$ is the density of water ($1000 \text{ kg/m}^3$).
$g$ is the acceleration due to gravity ($9.8 \text{ m/s}^2$).
$h$ is the height above the reference level (m).
This translates to a change of approximately $0.01 \text{ MPa}$ per meter of height.

Step 2: Detailed Explanation:
The height of the tree is given as $h = 75 \text{ meters}$.
Using the conversion rate of $0.01 \text{ MPa}$ per meter of vertical height: \[ \Psi_g = 75 \text{ m} \times 0.01 \text{ MPa/m} = 0.75 \text{ MPa} \] To pull water up to a height of 75 meters, the plant must generate a tension (negative pressure) that overcomes this gravitational head.
Thus, the effective potential drop experienced in the leaf cells due to height is represented as a negative value in the water potential budget.
Therefore, the magnitude of the gravitational component needed to be overcome is $-0.75 \text{ MPa}$.

Step 3: Final Answer:
The magnitude of the gravitational component of water potential at 75 meters is $-0.75 \text{ MPa}$.