1. Interpret the geometry

A rod of length $\ell$ is tied to a string of length $2\ell$, and the free end of the string is fixed.

So the rod rotates in the horizontal plane about the fixed end of the string with angular speed $\omega$.

The rod lies along the same radial line as the string, hence:

• inner end of rod is at distance $r_1 = 2\ell$ from the axis,
• outer end of rod is at distance $r_2 = 2\ell + \ell = 3\ell$ from the axis.

A uniform magnetic field $\vec B$ is vertical, i.e. perpendicular to the plane of rotation.

2. Use motional emf formula for a rotating rod segment

For a small element of rod at distance $r$ from the axis, its speed is $v = \omega r.$

Since $\vec v \perp \vec B$, the motional electric field magnitude is $E = vB = B\omega r.$

For a small rod element of length $dr$ along the radial direction, the small emf is $d\mathcal{E} = B\omega r\,dr.$

Hence total emf across the rod is $\mathcal{E} = \int_{r_1}^{r_2} B\omega r\,dr.$

Substitute limits $r_1=2\ell$, $r_2=3\ell$: $\mathcal{E} = B\omega \int_{2\ell}^{3\ell} r\,dr.$

3. Evaluate the integral

$\mathcal{E} = B\omega \left[\frac{r^2}{2}\right]_{2\ell}^{3\ell}$

$\mathcal{E} = \frac{B\omega}{2}\left((3\ell)^2 - (2\ell)^2\right)$

$\mathcal{E} = \frac{B\omega}{2}(9\ell^2 - 4\ell^2)$

$\mathcal{E} = \frac{5B\omega \ell^2}{2}.$

4. Match with options

The induced emf is $\boxed{\frac{5B\omega \ell^2}{2}}.$

This corresponds to Option D.

5. Comparison with stored correct answer

Stored correct answer: D

Derived answer: D

So they agree.