Given information:
A circular waveguide carries TE11 mode whose radial electric-field is given by Er = E0J1(r)sinφ V/m, where 'r' is the radial distance in cm, from the waveguide axis.

Objective:
To determine the cut-off wavelength of the mode.

Solution:

1. Radial electric field equation:
The radial electric field of the TE11 mode in a circular waveguide is given by Er = E0J1(r)sinφ, where Er is the radial electric field in V/m, E0 is the amplitude of the electric field, J1(r) is the first-order Bessel function of the first kind, and φ is the azimuthal angle.

2. Definition of cut-off wavelength:
The cut-off wavelength of a mode in a waveguide is the wavelength below which the mode cannot propagate through the waveguide.

3. TE11 mode:
The TE11 mode is the fundamental mode in a circular waveguide, where both the electric field and magnetic field have only one variation across the cross-section of the waveguide. In the TE11 mode, the electric field has a radial variation and no variation in the azimuthal direction.

4. Bessel function:
The first-order Bessel function of the first kind, J1(r), is a mathematical function that describes oscillatory phenomena. It has zeros at certain values of r, which correspond to the points where the electric field goes to zero. These zeros determine the cut-off wavelengths of the mode in the waveguide.

5. Determining the cut-off wavelength:
To determine the cut-off wavelength of the TE11 mode, we need to find the values of r where J1(r) = 0. These values of r correspond to the points where the electric field goes to zero, indicating that the mode cannot propagate beyond those points.

6. Cut-off condition:
For the TE11 mode, the cut-off condition is given by J1(ρ) = 0, where ρ = kρa, and a is the radius of the waveguide.

7. Bessel function zeros:
The zeros of the Bessel function J1(r) occur at certain values of r. The first zero of J1(r) is approximately 3.832, the second zero is approximately 7.016, the third zero is approximately 10.173, and so on.

8. Cut-off wavelength:
Since the cut-off condition for the TE11 mode is J1(ρ) = 0, the first zero of J1(ρ) corresponds to the cut-off wavelength. Therefore, the cut-off wavelength of the TE11 mode is given by λcutoff = 2πa/ρ1, where ρ1 is the first zero of J1(r).

9. Answer:
The correct answer is option (c) 2πa/ρ1, which represents the cut-off wavelength of the TE11 mode in the circular waveguide.