Solution:

Given data:

Increase in normal depth = 20%

Now, we need to find out the percentage increase in discharge.

To find out the percentage increase in discharge, we can use the Manning's equation which relates the discharge (Q) with various parameters like channel slope (S), hydraulic radius (R), Manning's roughness coefficient (n) and normal depth (y).

The Manning's equation is given as:

Q = (1/n) * A * R^(2/3) * S^(1/2)

where,

Q = Discharge
n = Manning's roughness coefficient
A = Cross-sectional area of flow
R = Hydraulic radius
S = Channel slope
y = Normal depth

Now, let's assume that the channel width (B) is constant.

Therefore, the cross-sectional area (A) can be written as:

A = B * y

Now, we can write the Manning's equation in terms of normal depth (y) as:

Q = (1/n) * B * y * R^(2/3) * S^(1/2)

Now, let's assume that the slope (S) and roughness coefficient (n) are constant.

Therefore, the Manning's equation can be written as:

Q ∝ y * R^(2/3)

This means that the discharge (Q) is directly proportional to the normal depth (y) and the hydraulic radius (R) raised to the power of 2/3.

Now, let's consider two cases:

Case 1: Normal depth = y

Case 2: Normal depth = 1.2y (i.e. increase in normal depth by 20%)

Let's assume that the hydraulic radius (R) remains constant in both cases.

Therefore, we can write:

Q1 = k * y * R^(2/3)

Q2 = k * 1.2y * R^(2/3)

where k is a constant of proportionality.

Dividing Q2 by Q1, we get:

Q2/Q1 = (1.2y * R^(2/3)) / (y * R^(2/3))

Q2/Q1 = 1.2

Therefore, the percentage increase in discharge is:

(1.2 - 1) * 100 = 20%

Hence, the answer is not correct.

However, if we assume that the channel width (B) increases in proportion to the normal depth (y), then the cross-sectional area (A) will increase as the square of the normal depth (y^2).

Therefore, the Manning's equation can be written as:

Q ∝ y^2 * R^(2/3)

Now, let's consider two cases:

Case 1: Normal depth = y

Case 2: Normal depth = 1.2y (i.e. increase in normal depth by 20%)

Let's assume that the hydraulic radius (R) remains constant in both cases.

Therefore, we can write:

Q1 = k * y^2 * R^(2/3)

Q2 = k * (1.2y)^2 * R^(2/3)

where k is a constant of proportionality.

Dividing Q2 by Q1, we get:

Q2/Q1 = (1.2y)^2 / y^2

Q2/Q1 = 1.44

Therefore, the percentage