1. A right circular cylindrical container of base diameter 1 cm contains some water. Six identical metal spheres are totally immersed in water and the water level rises by 0.25 cm . Find the radius of each metal sphere correct to 3 significant figures. 2. A right circular cylindrical container of base radius 10 cm contains some water. Some identical metal spheres of radii 1.5 cm are totally immersed in water and the water level rises by 0.9 cm . Find the number of metal spheres put in the container.
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To solve these problems, let's dive into the world of cylindrical containers and spherical volumes! For the first scenario, the volume of water displaced when six spheres are immersed in the container can be calculated using the formula for the volume of a cylinder (V = πr²h). The base diameter is 1 cm, which gives us a radius of 0.5 cm. The height increase of water is 0.25 cm, so the volume of the water displaced is V = π(0.5)²(0.25) = π(0.0625)(0.25) = π(0.015625) ≈ 0.0491 cm³. This volume must equal the total volume of the six spheres, which can be calculated using the formula for the volume of a sphere (V = (4/3)πr³). Hence, V_sphere = V_displaced/6. In the second scenario, we have a cylinder with a radius of 10 cm. The volume of water displaced is again derived from the height increase and the base area of the cylinder (V = π(10)²(0.9) = π(100)(0.9) = 90π cm³). Each sphere has a radius of 1.5 cm, so the volume of one sphere is V = (4/3)π(1.5)³ ≈ 14.137 cm³. Now, to find the number of spheres, divide the displaced volume by the volume of one sphere, N = (90π) / (14.137) ≈ 20.1. Hence, you can fit about 20 spheres into the container! Behold, the beauty of geometry and water dynamics in action! Engaging with these shapes unlocks wondrous applications, from engineering to simple problem-solving in everyday life. Sphere-tastic, right?