When a long glass capillary tube of radius 0.015 cm is dipped in a liquid, the liquid rises to a height of 15 cm within it. If the contact angle between the liquid and glass is close to 0°, the surface tension of the liquid, in milliNewton m$$^{-1}$$, is $$\left[\rho_{\text{(liquid)}} = 900 \text{ kg m}^{-3},\; g = 10 \text{ m s}^{-2}\right]$$ (Given answer in closed integer)

We first collect the data given in the question. The radius of the capillary is $$r = 0.015 \text{ cm}$$ and the height to which the liquid rises is $$h = 15 \text{ cm}$$. We convert these centimetre values into metre because SI units must be used throughout. Since $$1 \text{ cm} = 10^{-2} \text{ m}$$, we have $$r = 0.015 \times 10^{-2} \text{ m} = 0.00015 \text{ m}$$ and $$h = 15 \times 10^{-2} \text{ m} = 0.15 \text{ m}$$. The density of the liquid is $$\rho = 900 \text{ kg m}^{-3}$$ and the acceleration due to gravity is $$g = 10 \text{ m s}^{-2}$$. The contact angle is nearly $$0^{\circ}$$, so $$\cos \theta \approx 1$$.

Now we recall the capillary rise formula. For a liquid that wets the walls of a capillary (contact angle $$\theta$$) the equilibrium height $$h$$ to which the liquid rises is related to the surface tension $$T$$ by the expression $$h = \frac{2T \cos\theta}{\rho g r}.$$

Because $$\theta \approx 0^{\circ}$$, we can put $$\cos\theta = 1$$, and we rearrange the formula to make $$T$$ the subject: $$T = \frac{h \, \rho \, g \, r}{2}.$$

Substituting each numerical value step by step, we write $$T = \frac{0.15 \times 900 \times 10 \times 0.00015}{2} \; \text{N m}^{-1}.$$

We now carry out the multiplication in the numerator: $$0.15 \times 900 = 135,$$ so the expression becomes $$T = \frac{135 \times 10 \times 0.00015}{2}.$$ Next, $$135 \times 10 = 1350,$$ giving $$T = \frac{1350 \times 0.00015}{2}.$$ Multiplying further, $$1350 \times 0.00015 = 0.2025,$$ and therefore $$T = \frac{0.2025}{2} \; \text{N m}^{-1}.$$ Dividing by $$2$$ we obtain $$T = 0.10125 \; \text{N m}^{-1}.$$

We are asked for the answer in milliNewton per metre. Remembering that $$1 \text{ N} = 1000 \text{ mN}$$, we multiply by $$1000$$: $$T = 0.10125 \times 1000 = 101.25 \; \text{mN m}^{-1}.$$

Finally, rounding to the nearest integer as demanded in the closed-integer answer format, we get $$T \approx 101 \text{ mN m}^{-1}.$$

So, the answer is $$101$$.