To solve this problem, we need to calculate the initial volume of water in the vessel, and then determine how many lead shots need to be dropped to make one-fourth of the water flow out.
The volume of an inverted cone can be calculated using the formula:
V_cone = (1/3) * π * r^2 * h
where r is the radius of the top of the cone and h is the height of the cone.
In this case, the height of the inverted cone is 8 cm, and the radius of its top is 5 cm. Let's calculate the initial volume of water in the vessel:
V_initial = (1/3) * π * 5^2 * 8
= (1/3) * 3.14 * 25 * 8
= 209.333 cm^3 (approximately)
Now, when lead shots are dropped into the vessel, one-fourth of the water flows out. This means the final volume of water in the vessel will be three-fourths (3/4) of the initial volume.
V_final = (3/4) * V_initial
= (3/4) * 209.333
= 156.99975 cm^3 (approximately)
To find the number of lead shots dropped, we need to determine their combined volume. The volume of a sphere can be calculated using the formula:
V_sphere = (4/3) * π * r^3
In this case, the radius of each lead shot is 0.5 cm. Let's calculate the volume of a single lead shot:
V_sphere = (4/3) * 3.14 * 0.5^3
= (4/3) * 3.14 * 0.125
= 0.523333 cm^3 (approximately)
To find the number of lead shots, we can divide the final volume of water by the volume of a single lead shot:
Number of shots = V_final / V_sphere
= 156.99975 / 0.523333
≈ 299.865 (approximately)
Therefore, the number of lead shots dropped in the vessel is approximately 300.
