The emf induced in a rotating conductor in a magnetic field is given by \[\mathcal{E} = \frac{1}{2} B \omega l^2 = \frac{1}{2} B (2 \pi f) l^2 = \pi B f l^2,\]where $B$ is the magnetic field strength, $\omega$ is the angular velocity, $f$ is the frequency, and $l$ is the length of the conductor.
In this case, we are given that $l = 4 \, {m}$, $f = 3 \, {rev/s}$, and $B = 40 \, \mu {T} = 40 \times 10^{-6} \, {T}$, so \[\mathcal{E} = \pi B f l^2 = \pi \cdot 40 \times 10^{-6} \cdot 3 \cdot 4^2 \approx \boxed{6 \, {mV}}.\]
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Electromagnetic waves carry energy but not momentum.
Reason (R): Mass of a photon is zero. In the light of the above statements.
choose the most appropriate answer from the options given below:
The dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ is equal to that of: (Where $ \mu_0 $ is the vacuum permeability and $ \epsilon_0 $ is the vacuum permittivity)
If the roots of $\sqrt{\frac{1 - y}{y}} + \sqrt{\frac{y}{1 - y}} = \frac{5}{2}$ are $\alpha$ and $\beta$ ($\beta > \alpha$) and the equation $(\alpha + \beta)x^4 - 25\alpha \beta x^2 + (\gamma + \beta - \alpha) = 0$ has real roots, then a possible value of $y$ is: