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9. (a) A load of mass M acts on a light circular piston of diameter d - Leaving Cert Applied Maths - Question 9 - 2016
Question 9
9. (a) A load of mass M acts on a light circular piston of diameter d. The piston sits on a reservoir of oil. The density of the oil is ρ. The reservoir is connected... show full transcript
Worked Solution & Example Answer:9. (a) A load of mass M acts on a light circular piston of diameter d - Leaving Cert Applied Maths - Question 9 - 2016
Step 1
Find h in terms of M, ρ and d
Answer
To solve for h, we can start by applying the principle of hydrostatic pressure balance. The force due to the load M is given by:
The pressure exerted by the oil at the base of the piston is:
P = rac{F}{A} = rac{Mg}{A}
The area A of the circular piston can be represented in terms of its diameter d:
u d^2}{4}$$ Thus, the pressure at the base can also be expressed as: $$P = ρgh$$ Equating both expressions for pressure: $$ rac{Mg}{ rac{ u d^2}{4}} = ρgh$$ By simplifying, we get: $$h = rac{4Mg}{ρgd^2} = rac{4M}{ρ rac{ u d^2}{4}}$$ Hence, the relationship of h in terms of M, ρ, and d is: $$h = rac{4M}{ρ rac{ u d^2}{4}}$$Step 2
Find in terms of s the fraction of the length of the rod that is immersed in the water
Answer
Let the length of rod AB be ℓ, and the length of the part immersed in water be x. The weight of the water displaced by the rod must equal the weight of the rod:
ho g rac{x}{ u}$$ Using the geometry of the inclined rod, the buoyant force B acting on the rod is: $$B = rac{1}{2} W (1 - s)$$ Substituting for B: $$B imes rac{x}{ u} imes sin( heta) = W imes rac{1}{2} (sin( heta)$$ From this, solve for x: $$x = ℓ (1 - sqrt{1 - s})$$ Therefore, the fraction of the rod immersed in water is: $$ rac{x}{ℓ} = 1 - sqrt{1 - s}$$