Concept:
When a conducting rod rotates in a uniform magnetic field about one of its ends, the free charges present in the rod experience a magnetic Lorentz force. This causes charge separation along the length of the rod and an emf is induced between its ends.
For a rod of length $L$ rotating with angular velocity $\omega$ in a magnetic field $B$ perpendicular to the plane of rotation, the induced emf is given by
\[
\varepsilon
=
\frac{1}{2}B\omega L^2
\]
Since angular velocity and frequency are related by
\[
\omega = 2\pi f
\]
the expression for induced emf can also be written as
\[
\varepsilon
=
\frac{1}{2}B(2\pi f)L^2
=
\pi BfL^2
\]
Step 1: Convert the given quantities into SI units
Length of the rod:
\[
L = 100\text{ cm} = 1\text{ m}
\]
Frequency of rotation:
\[
f = 14\text{ revolutions per second}
\]
Magnetic field:
\[
B = 2\text{ T}
\]
Step 2: Calculate the angular velocity
Using
\[
\omega = 2\pi f
\]
\[
\omega
=
2\pi(14)
=
28\pi\ \text{rad s}^{-1}
\]
Step 3: Apply the formula for induced emf
\[
\varepsilon
=
\frac{1}{2}B\omega L^2
\]
Substituting the values,
\[
\varepsilon
=
\frac{1}{2}\times 2\times 28\pi \times (1)^2
\]
\[
\varepsilon
=
28\pi
\]
Using
\[
\pi \approx \frac{22}{7}
\]
\[
\varepsilon
=
28\times\frac{22}{7}
\]
\[
\varepsilon
=
4\times22
\]
\[
\varepsilon
=
88\text{ V}
\]
Therefore, the induced emf between the two ends of the rod is
\[
\boxed{88\text{ V}}
\]