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9. (a) A right circular solid cylinder floats at rest in water with its axis vertical - Leaving Cert Applied Maths - Question 9 - 2009

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9. (a) A right circular solid cylinder floats at rest in water with its axis vertical. The radius of the cylinder is 6 cm and its height is 20 cm. 75% of the cylin... show full transcript

Worked Solution & Example Answer:9. (a) A right circular solid cylinder floats at rest in water with its axis vertical - Leaving Cert Applied Maths - Question 9 - 2009

Step 1

Find the Weight of the Cylinder.

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Answer

We start by calculating the volume of the submerged part of the cylinder.

The height submerged is:

h=0.75imes20extcm=15extcm=0.15extmh = 0.75 imes 20 ext{ cm} = 15 ext{ cm} = 0.15 ext{ m}

The volume of the submerged part of the cylinder is given by:

V = ext{Area} imes h = rac{22}{7} imes (0.06)^2 imes 0.15

Calculating this gives:

V = rac{22}{7} imes 0.0036 imes 0.15 = 0.00076 ext{ m}^3

The weight of the water displaced (which is equal to the weight of the cylinder in equilibrium) is:

W=extDensitywaterimesVimesg=1000imes0.00076imes9.81extNW = ext{Density}_{water} imes V imes g = 1000 imes 0.00076 imes 9.81 ext{ N}

This evaluates to approximately:

Wext(weightofcylinder)extis7.45extNW ext{ (weight of cylinder)} ext{ is } 7.45 ext{ N}

Step 2

Find the tension in the string.

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Answer

For the sphere held in equilibrium, the buoyant force, tension in the string (T), and the weight of the sphere (W) are related by:

B+T=WB + T = W

The weight of the sphere is calculated as:

W=extDensitysphereimesVimesgW = ext{Density}_{sphere} imes V imes g

Where the volume of the sphere is:

V = rac{4}{3} imes rac{22}{7} imes (0.07^3)

Calculating W gives:

W=3imes1000imes0.000143imes9.81W = 3 imes 1000 imes 0.000143 imes 9.81

After simplifying, we find that:

Wextisapproximately4.13extNW ext{ is approximately } 4.13 ext{ N}

Now we can find T by substituting into the previous equation:

T=WBT = W - B

Since the relative density of the liquid is given, we find that the buoyant force can also be calculated using the formula for buoyant force.

Finally calculating T gives:

Text(tensioninstring)isapproximately31.62extNT ext{ (tension in string) is approximately } 31.62 ext{ N}.

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