A certain clock's pendulum has a time period of 2.04 s . How fast...

A standard seconds pendulum has a time period $T = 2\text{ s}$.

The given pendulum has a time period $T' = 2.04\text{ s}$. Since $T' > T$, the pendulum takes more time to complete one oscillation, meaning the clock will run slow.

In a true time of $24\text{ hours}$ ($24 \times 3600 = 86400\text{ s}$), the number of oscillations made by the pendulum is:

$n = \dfrac{86400}{2.04}$

Since the clock registers $2\text{ s}$ for every oscillation, the time shown by the clock in $24\text{ hours}$ is:

$t_{\text{clock}} = n \times 2 = \dfrac{86400}{2.04} \times 2 = \dfrac{172800}{2.04}\text{ s}$

The time lost by the clock in $24\text{ hours}$ is the difference between the true time and the clock time:

$\Delta t = 86400 - \dfrac{172800}{2.04}$

\Delta t = 86400 \left(1 -\dfrac{2}{2.04}\right) =$86400 \left(\dfrac{0.04}{2.04}\right)<br><br>$\Delta t = $\dfrac{3456}{2.04} \approx$1694.12\text{ s}$<br><br>Converting the time lost into minutes:<br><br>$\Delta t = $\dfrac{1694.12}{60} \approx$28.235\text{ minutes}$<br><br>Therefore, the clock will run approximately$28.2\text{ minutes}$slow.<br><br>Answer:$28.2\text{ minutes}$slow$