9. (a) A U-tube whose limbs are vertical and of equal length has mercury poured in until the level is 26.2 cm from the top in each limb - Leaving Cert Applied Maths - Question 9 - 2007

The mass of the water and the mass of the added liquid can be computed using the formula:

$\text{mass of water} = 1000 \times 500 \times 10^{-6} = 0.5 \text{ kg}$ $\text{mass of liquid} = 1000 \times 300 \times 10^{-6} \times 1.2 = 0.36 \text{ kg}$

The total mass of the mixture is:

$m_{\text{mixture}} = 0.5 + 0.36 = 0.86 \text{ kg}$

The relative density of the mixture is:

$s = \frac{\text{mass of mixture}}{\text{volume of mixture}} = \frac{0.86}{0.0008}\approx 1.075.$

Using the tension relationships:

For the water, we have:

$T = \text{weight of water} - \text{upthrust from water}$

$0.0784 = (1000 \text{ g} \cdot m_{\text{sphere}} g) / s$

For the mixture, we have:

$B = T + mg$

Now substituting, we get:

$\text{mixture: } B = 0.1078, ext{ where } s=1.075.$

Setting up the mass equation, we find:

$m \times (g_{\text{mixture}}) = 0.1078.$

This leads us to find:

$m \approx 0.032 kg.$