9. (a) A right circular solid cylinder floats at rest in water with its axis vertical - Leaving Cert Applied Maths - Question 9 - 2017

To find the tension in the string, we will consider the forces acting on the cone:

1. Identify Forces:

   • The weight of the cone, denoted as ( W_c )
   • The buoyant force acting on the cone, denoted as ( B )
   • The tension in the string, denoted as ( T )

2. Calculate the Weight of the Cone: The weight of the cone can be calculated using the relative density:

   ( \text{Relative density} = \frac{\text{Density of cone}}{\text{Density of water}} )

   From the relative density of 0.8, we deduce that: ( \text{Density of cone} = 0.8 \times 1000 = 800 \text{ kg m}^{-3} )

   The volume of the cone is given by:

   $V = \frac{1}{3} \pi r^2 h$ Where ( r = 0.06 \text{ m} ) and ( h = 0.15 \text{ m} ):

   $V = \frac{1}{3} \pi (0.06)^2 (0.15)$

   Simplifying gives:

   $V = \frac{1}{3} \pi (0.0036)(0.15) \approx 0.00018 \pi \text{ m}^3$

   Thus, the weight of the cone is:

   $W_c = 800 \cdot V = 800 \cdot (0.00018 \pi) = 144 \pi \text{ N}$

3. Set Up the Equation of Forces:

   The upward forces (buoyant force and tension) must equal the downward force (weight of the cone):

   $T + B = W_c$

   Where the buoyant force ( B ) can be calculated as:

   $B = \rho_{liquid} V_g$

   Given the relative density of liquid is 1.4, we find:

   ( \rho_{liquid} = 1.4 \times 1000 = 1400 \text{ kg m}^{-3} )

   Thus:

   $B = 1400 \cdot (0.00018 \pi) \approx 252 \pi \text{ N}$

4. Insert into the equation:

   $T + 252 \pi = 144 \pi$

   Which reduces to find tension:

   $T = (144 - 252) \pi = -108 \pi ext{ N}$

   Hence, the tension in the string is:

   $T \approx 3.39 ext{ N}$