A mould cavity of 1200 cm³ volume has to be filled through a sprue of 10 cm length feeding a horizontal runner. Cross-sectional area at the base of the sprue is 2 cm². Consider acceleration due to gravity as 9.81 m/s². Neglecting frictional losses due to molten metal flow, the time taken to fill the mould cavity is _______ seconds (round off to 2 decimal places).

The correct answer is 4.28.

Step-by-step Explanation:

Let's list all the given values from the problem description:
Volume of the mould cavity, V = 1200 cm³
Length of the sprue, h_t = 10 cm
Cross-sectional area at the base of the sprue, A = 2 cm²
Acceleration due to gravity, g = 9.81 m/s²

To keep the units consistent, we convert the acceleration due to gravity from meters per second squared (m/s²) to centimeters per second squared (cm/s²):
g = 9.81 m/s² = 9.81 × 100 cm/s² = 981 cm/s²

Using Torricelli's theorem (neglecting frictional losses), the velocity of the molten metal at the base of the sprue is given by:
$$v=\sqrt{2\cdot g\cdot h_t}$$
Substituting the values:
$$v=\sqrt{2\times 981\times 10}=\sqrt{19620}\approx 140.07\text{ cm/s}$$

The volumetric flow rate (Q) of the molten metal entering the mould cavity is:
Q = A × v
Substituting the cross-sectional area and the velocity:
$$Q=2\times 140.07\approx 280.14{\text{ cm}}^3\text{/s}$$

Finally, the total time (t) required to completely fill the mould cavity of volume V is calculated as:
$$t=\frac{V}{Q}=\frac{1200}{2\times \sqrt{2\times 981\times 10}}$$
$$t=\frac{1200}{280.14}\approx 4.283\text{ seconds}$$
Rounding off to two decimal places, we get:
t = 4.28 seconds