To find the weight of the water raised in the capillary tube, we can use the formula for weight, which is mass times acceleration due to gravity (W = mg). The mass of the water can be found by multi

To find the weight of the water raised in the capillary tube, we can use the formula for weight, which is mass times acceleration due to gravity ($W = mg$). The mass of the water can be found by multiplying its density by its volume ($m = \rho V$). The volume of the water in the capillary tube is the volume of a cylinder, $V = \pi r^2 h$.

From the previous question (Q11), we have the following values:

• Radius of the capillary tube, $r = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m}$
• Density of water, $\rho = 1000 \text{ kg/m}^3$
• Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$
• Height of water raised, $h = \frac{2\gamma \cos\theta}{\rho g r}$. From the previous calculation, we found $h = \frac{3}{98} \text{ m}$.

Step 1: Calculate the volume of the raised water.
The volume of the cylindrical column of water is $V = \pi r^2 h$.

$$V = \pi \times (0.5 \times 10^{-3} \text{ m})^2 \times \left(\frac{3}{98} \text{ m}\right)$$ $$V = \pi \times (0.25 \times 10^{-6} \text{ m}^2) \times \left(\frac{3}{98} \text{ m}\right)$$ $$V = \frac{0.75 \pi}{98} \times 10^{-6} \text{ m}^3$$

Step 2: Calculate the mass of the raised water.
The mass is $m = \rho V$.

$$m = (1000 \text{ kg/m}^3) \times \left(\frac{0.75 \pi}{98} \times 10^{-6} \text{ m}^3\right)$$ $$m = \frac{0.75 \pi}{98} \times 10^{-3} \text{ kg}$$

Step 3: Calculate the weight of the raised water.
The weight is $W = mg$.

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