A metallic rod of length 1.0 ~m is pivoted at one end and rotated...

First, calculate the induced electromotive force (emf) across the ends of the rotating rod. The formula for the induced emf $\varepsilon$ is:
$\varepsilon = \dfrac{1}{2} B \omega L^2$

Given values:
$B = 0.5 \mathrm{~T}$
$\omega = 20 \mathrm{~rad/s}$
$L = 1.0 \mathrm{~m}$

Substituting the values:
$\varepsilon = \dfrac{1}{2} \times 0.5 \times 20 \times (1.0)^2$
$\varepsilon = 0.5 \times 10 \times 1 = 5 \mathrm{~V}$

Next, calculate the electrical power $P$ dissipated in the resistor using the formula:
$P = \dfrac{\varepsilon^2}{R}$

Given $R = 5 \Omega$:
$P = \dfrac{(5)^2}{5} = \dfrac{25}{5} = 5 \mathrm{~W}$

If the $\dfrac{1}{2}$ factor in the emf formula is omitted, the emf would be calculated as $10 \mathrm{~V}$, leading to an incorrect power of $20 \mathrm{~W}$. If the formula $P = \varepsilon \times R$ is mistakenly used, the result would be $25 \mathrm{~W}$.

Answer: $5 \mathrm{~W}$