Final answer:
The surface of the water in the drum will be raised by approximately 0.0826m.
Step-by-step explanation:
To find the amount by which the surface of the water in the drum will be raised, we need to calculate the volume of the solid iron sphere and then determine the increase in volume of the water when the sphere is submerged.
First, let's find the volume of the iron sphere. The radius of the sphere is half of its diameter, so the radius is 42cm/2 = 21cm = 0.21m. The volume of a sphere is given by (4/3)πr^3, so the volume is (4/3)π(0.21m)^3 = 0.40527m^3.
Now, let's determine the increase in volume of the water. The volume of a cylinder is given by πr^2h, where r is the radius and h is the height. The radius of the drum is 1.4m, so its volume is π(1.4m)^2h. The volume of the water after the iron sphere is dropped in is the volume of the drum plus the volume of the sphere, so π(1.4m)^2h + 0.40527m^3 = π(1.4m)^2(h+0.40527m^3/(π(1.4m)^2)).
The increase in volume of the water is equal to the volume of the sphere, so h+0.40527m^3/(π(1.4m)^2) = 0.40527m^3/(π(1.4m)^2). Solving for h, we find that h ≈ 0.0826m.