A cylindrical block of wood (density = 650 kg m$$^{-3}$$), of base area 30 cm$$^2$$ and height 54 cm, floats in a liquid of density 900 kg m$$^{-3}$$. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly):

The block is cylindrical, so its volume is equal to the base area multiplied by the height. We first convert all the given quantities to SI units so that we can use them directly in the standard formulas.

The base area is given as $$30\ \text{cm}^2$$. Since $$1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$$, we have $$A = 30 \times 10^{-4}\ \text{m}^2 = 3.0 \times 10^{-3}\ \text{m}^2.$$

The height is given as $$54\ \text{cm}$$. Using $$1\ \text{cm} = 10^{-2}\ \text{m}$$, we write $$h = 54 \times 10^{-2}\ \text{m} = 0.54\ \text{m}.$$

Now the volume of the cylinder is $$V = A \times h = (3.0 \times 10^{-3})\ \text{m}^2 \times 0.54\ \text{m} = 1.62 \times 10^{-3}\ \text{m}^3.$$

The density of the wooden block is $$\rho_b = 650\ \text{kg m}^{-3}$$, so its mass is $$m = \rho_b V = 650\ \text{kg m}^{-3} \times 1.62 \times 10^{-3}\ \text{m}^3 = 1.053\ \text{kg}.$$

The liquid in which the block floats has density $$\rho_l = 900\ \text{kg m}^{-3}$$. When the block is pushed down by a small vertical displacement $$x$$ and released, the extra volume of liquid displaced is $$A x$$, and therefore the additional buoyant force is

$$F_b = \rho_l g (A x).$$

This force acts upward and tends to restore the block to equilibrium. Using Newton’s second law, the equation of motion for the block becomes

$$m \frac{d^2x}{dt^2} + \rho_l g A x = 0.$$

This is the standard form of a simple harmonic motion equation, $$\frac{d^2x}{dt^2} + \omega^2 x = 0,$$ where the angular frequency $$\omega$$ is given by

$$\omega^2 = \frac{\rho_l g A}{m}.$$

Hence $$\omega = \sqrt{\frac{\rho_l g A}{m}}.$$

The time period of oscillation is related to the angular frequency by the formula $$T = \frac{2\pi}{\omega}.$$ Substituting $$\omega$$ from above we get

$$T = 2\pi \sqrt{\frac{m}{\rho_l g A}}.$$

Now we substitute the numerical values: $$m = 1.053\ \text{kg},\quad \rho_l = 900\ \text{kg m}^{-3},\quad g = 9.8\ \text{m s}^{-2},\quad A = 3.0 \times 10^{-3}\ \text{m}^2.$$

First calculate the denominator inside the square root: $$\rho_l g A = 900 \times 9.8 \times 3.0 \times 10^{-3} = 8820 \times 3.0 \times 10^{-3} = 26.46.$$

Next compute the ratio inside the square root: $$\frac{m}{\rho_l g A} = \frac{1.053}{26.46} \approx 0.0398.$$

Taking the square root gives $$\sqrt{0.0398} \approx 0.1995.$$

Therefore the time period is $$T = 2\pi \times 0.1995 \approx 6.283 \times 0.1995 \approx 1.25\ \text{s}.$$

We are asked to find the length $$L$$ of a simple pendulum that has the same time period. For a simple pendulum, the time period is given by the well-known formula

$$T = 2\pi \sqrt{\frac{L}{g}}.$$

Equating the two expressions for the period, we write $$1.25 = 2\pi \sqrt{\frac{L}{g}}.$$

First isolate the square root term: $$\sqrt{\frac{L}{g}} = \frac{1.25}{2\pi} = \frac{1.25}{6.283} \approx 0.199.$$

Now square both sides: $$\frac{L}{g} = (0.199)^2 \approx 0.0396.$$

Finally multiply by $$g = 9.8\ \text{m s}^{-2}$$ to obtain $$L$$: $$L = 0.0396 \times 9.8 \approx 0.388\ \text{m}.$$

Converting metres to centimetres, $$L = 0.388\ \text{m} \times 100\ \text{cm m}^{-1} \approx 38.8\ \text{cm}.$$

The value is very close to $$39\ \text{cm}$$, which matches option C.

Hence, the correct answer is Option C.