A solid sphere, of radius 14 cm, floats at rest in water - Leaving Cert Applied Maths - Question 9 - 2011

The weight of the cone can be determined using its relative density. Given the relative density of the cone and the given height and radius, the weight can be calculated as follows:

1. Calculate the volume of the cone:

$V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (0.1)^2 (0.12) \approx 0.0039 \text{ m}^3$

2. Determine the weight of the cone using its density:

$W = \text{relative density} \times V_{cone} \times \rho_{water} = 7 \times 0.0039 \times 1000 \approx 27.3 \text{ N}$

3. The buoyant force when the cone is submerged in the liquid:

$B = \rho_{liquid} \cdot g \cdot V_{cone} = 0.9 \cdot 1000 \cdot 9.81 imes 0.0039 \approx 3.46 \text{ N}$

4. Now apply the equilibrium of forces:

$T + B = W$

where:

• T is the tension in the string
• B is the buoyant force
• W is the weight of the cone

Substituting values:

$T + 3.46 = 27.3$

Thus, the tension in the string is calculated as:

$T = 27.3 - 3.46 \\ = 23.84 \approx 24 \text{ N}$

Therefore, the tension in the string is approximately 24 N, correct to the nearest Newton.