Steel ruptures when a shear of $$3.5 \times 10^8$$ N m$$^{-2}$$ is applied. The force needed to punch a 1 cm diameter hole in a steel sheet 0.3 cm thick is nearly:

The steel ruptures when the shear stress reaches $$3.5 \times 10^8$$ N m$$^{-2}$$. To punch a hole, we need to apply a force that causes this shear stress over the area being sheared.

When punching a hole of diameter 1 cm in a sheet of thickness 0.3 cm, the sheared area is the lateral surface area of the cylindrical hole. This area is calculated as the circumference of the hole multiplied by the thickness of the sheet.

First, convert all measurements to meters for consistency. The diameter is 1 cm, which is 0.01 m, so the radius $$r$$ is half of that: $$r = \frac{0.01}{2} = 0.005$$ m. The thickness $$t$$ is 0.3 cm, which is 0.003 m.

The circumference of the hole is $$2 \pi r$$. Therefore, the sheared area $$A$$ is:

$$A = 2 \pi r t$$

Substituting the values:

$$A = 2 \times \pi \times 0.005 \times 0.003$$

First, multiply the numerical parts:

$$0.005 \times 0.003 = 0.000015$$

Then:

$$A = 2 \times \pi \times 0.000015 = 0.00003 \pi$$

Using $$\pi \approx 3.1416$$,

$$A \approx 0.00003 \times 3.1416 = 0.000094248 \text{ m}^2$$

In scientific notation, $$A \approx 9.4248 \times 10^{-5}$$ m$$^2$$.

The shear stress $$\tau$$ is related to the force $$F$$ and area $$A$$ by:

$$\tau = \frac{F}{A}$$

At rupture, $$\tau = 3.5 \times 10^8$$ N m$$^{-2}$$. Solving for $$F$$:

$$F = \tau \times A$$

Substituting the values:

$$F = (3.5 \times 10^8) \times (9.4248 \times 10^{-5})$$

First, multiply the coefficients:

$$3.5 \times 9.4248 = 32.9868$$

Then multiply the powers of 10:

$$10^8 \times 10^{-5} = 10^{3}$$

So:

$$F \approx 32.9868 \times 10^3 = 32986.8 \text{ N}$$

Which is $$3.29868 \times 10^4$$ N. Rounding to two significant figures, as in the options, gives approximately $$3.3 \times 10^4$$ N.

Alternatively, using exact values:

$$A = 2 \pi r t = 2 \times \pi \times 0.005 \times 0.003 = 2 \times \pi \times 1.5 \times 10^{-5} = 3\pi \times 10^{-5} \text{ m}^2$$

Then:

$$F = (3.5 \times 10^8) \times (3\pi \times 10^{-5}) = 3.5 \times 3 \times \pi \times 10^{8-5} = 10.5 \pi \times 10^3 = 10500\pi \text{ N}$$

With $$\pi \approx 3.1416$$,

$$F \approx 10500 \times 3.1416 = 32986.8 \text{ N} \approx 3.3 \times 10^4 \text{ N}$$

Comparing with the options, $$3.3 \times 10^4$$ N corresponds to option C.

Hence, the correct answer is Option C.