Calculate the capillary rise of water in a glass tube with a radius of 0.5 mm, given surface tension = 75 dyne/cm, density of water = 1000 kg/m³ and g = 9.8 m/s².

Given:

• Radius, $r = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m}$
• Surface tension, $\gamma = 75 \text{ dyne cm}^{-1}$
• Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$
• Density of water, $\rho = 1000 \text{ kg/m}^3$

Convert surface tension from dyne/cm to N/m:
We know that $1 \text{ dyne} = 10^{-5} \text{ N}$ and $1 \text{ cm} = 10^{-2} \text{ m}$.

$$\gamma = 75 \frac{\text{dyne}}{\text{cm}} \times \frac{10^{-5} \text{ N}}{1 \text{ dyne}} \times \frac{1 \text{ cm}}{10^{-2} \text{ m}}$$ $$\gamma = 75 \times \frac{10^{-5}}{10^{-2}} \frac{\text{N}}{\text{m}} = 75 \times 10^{-3} \frac{\text{N}}{\text{m}}$$

Assuming the contact angle $\theta = 0^\circ$, so $\cos\theta = 1$.

$$h = \frac{2 \times (75 \times 10^{-3} \text{ N/m}) \times 1}{(1000 \text{ kg/m}^3) \times (9.8 \text{ m/s}^2) \times (0.5 \times 10^{-3} \text{ m})}$$

$$h = \frac{0.15 \text{ N/m}}{1000 \times 9.8 \times 0.0005 \text{ kg/(m}^2\text{s}^2) \cdot \text{m}}$$ $$h = \frac{0.15 \text{ N/m}}{4.9 \text{ N/m}}$$ $$h \approx 0.030612 \text{ m}$$

$$h = 0.030612 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}}$$ $$h \approx 3.06 \text{ cm}$$

Comparing this result with the given options:

(a) 3.06 cm

(b) 6.12 cm

(c) 1.53 cm

(d) 9.42 cm

(e) None of the above

The calculated height matches option (a).

The final answer is $\boxed{\text{3.06 cm}}$.