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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 - 2013
Question 9
9. (a) A right circular solid cylinder floats at rest in water with its axis vertical. The radius of the cylinder is 5 cm and its height is 12 cm. 95% of the cyli... 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 - 2013
Step 1
Find the weight of the cylinder.
Answer
To find the weight of the cylinder, we first calculate the volume of the part submerged in water. Since 95% of the cylinder lies below the water surface:
- Height submerged = 0.95 × 12 cm = 11.4 cm.
- Radius = 5 cm.
The volume of the submerged part, V, can be calculated using the formula for the volume of a cylinder:
V = ext{Base Area} imes ext{Height} = rac{22}{7} imes (5^2) imes 11.4
Then we can find the buoyant force (B) which equals the weight of the water displaced:
Thus, the weight of the cylinder (W) is equal to the buoyant force:
$$W = B = 895 ext{ N}.$
Step 2
Find the tension in the string.
Answer
To find the tension in the string, we will first calculate the buoyant force acting on the sphere.
The volume of the sphere (B) is:
B = rac{4}{3} imes rac{22}{7} imes (0.02)^3 imes 10.
Substituting in the values, we get:
Next, we need to calculate the weight of the sphere:
W = ext{Density of sphere} imes V imes g = 700 imes rac{4}{3} imes rac{22}{7} imes (0.02)^3 imes 10 = 0.075 ext{ N}.
Since the sphere is at rest, the total upward force (T + W) equals the buoyant force:
T = B - W = 0.128 - 0.075 = 0.053 ext{ N}.$$ Therefore, the tension in the string is: $$T = 0.017 ext{ N}.$