(a) State the Principle of Archimedes - Leaving Cert Applied Maths - Question 9 - 2008

To find the volume of the metal, we start by calculating the weight of the water displaced:

Weight of the water displaced = 28 N - 18 N = 10 N.

Using the density of water, we can find the volume:

egin{align*} ext{Weight} &= ext{mass} imes g \ 10 ext{ N} &= ho imes V imes g \ V &= rac{10}{1000} = 0.01 ext{ m}^3 ext{ therefore, the volume of the metal } V = 0.01 ext{ m}^3. \end{align*}

Relative density ( ho) is defined as the ratio of the density of the substance to the density of water. Using the volume calculated:

egin{align*}

ho &= rac{M}{V} \

ho &= rac{28 ext{ N}}{0.01 ext{ m}^3} imes rac{1}{g} \

ho &= rac{28}{0.01} = 2800 ext{ kg/m}^3 \ ext{thus, the relative density of the metal is }
ho_{relative} = 2.8. \end{align*}

First, determine the weight of the cone:

egin{align*} W_{ic} &= ext{Relative Density} imes ext{Volume} imes ext{Density of water} \ ext{Volume of cone} &= rac{1}{3} imes ext{Base Area} imes ext{Height} \ W_c &= 0.6 imes rac{1}{3} imes rac{22}{7} imes (0.06)^2 imes 0.15 \ ext{Calculating the tension: } \ B &= T + W \ 0.6 imes 1000 imes g &= T + W\ ext{Find } T \text{ such that: } \ T &= rac{1}{2} imes (0.6) imes (6^2 imes rac{22}{7} imes (0.15) imes 1000) \ T &= 1.7 N.\end{align*}