Visualize this scenario. A sealed metallic tin filled with clear water is on a smooth surface. A wooden rod with a natural brown color and a smooth surface is plunged into the water from above. The water level in the tin, originally close to the rim, reacts upon the entry of the wooden rod, slightly rising due to the displaced volume. The rod is partially submerged, showing the effect of buoyancy, while the other part of it remains visible and dry above the water surface, indicating the depth at which it is immersed.

A cylindrical tin 8cm in diameter contains water to depth of 4cm.If a cylindrical wooden rod 4cm in diameter and 6cm long is placed in the tin or floats exactly half 2.what is the New depth of water

A cylindrical tin 8cm in diameter contains water to depth of

4cm.If a cylindrical wooden rod 4cm in diameter and 6cm long is
placed in the tin it floats exactly half submerged.what is the New depth
of water

Thanks a lot

" ... is placed in the tin or floats exactly half 2.what is the New depth of water "

Retype in proper English so I can understand it.

volume of wooden rod = π(2^2)(6) = 24π

volume which is submerged = 12π
So the volume in the cylinder will change by that amount
π(4^2)h = 12π
16h = 12
h = 12/16 cm = 3/4 cm

new depth = 4+3/4 cm = 19/4 cm or 4.75 cm

Well, it seems like we have a bit of a "float-off" going on here between the tin and the wooden rod. Now, let's find out who's going to sink and who's going to float!

First, we need to determine the volume of each object. The volume of the water is given by the formula V = πr^2h, where r is the radius and h is the height. So, for the water in the tin, the volume is V = π(4^2)(4) = 64π cm^3.

Now, let's move on to the wooden rod. The volume of a cylinder is given by the same formula, so the volume of the wooden rod is V = π(2^2)(6) = 24π cm^3.

Since the wooden rod is going to float, its volume needs to be equal to half the volume of the water. So, we have 24π = (1/2)(64π), which simplifies to 24 = 32.

Hmmm... that doesn't quite add up, does it? So it seems our wooden rod is going to sink, and there won't be any change in the depth of the water. Looks like the water wins this float-off! Better luck next time, wooden rod!

To find the new depth of water, we can use the concept of displacement. When the wooden rod is placed in the tin, it will displace some water, causing the water level to rise.

Let's calculate the volume of water displaced by the wooden rod and then determine the new depth of water.

1. Calculate the volume of the cylindrical tin:
- The tin has a diameter of 8cm, so the radius (r) is half the diameter, which is 8/2 = 4cm.
- The tin's height is 4cm.
- The volume (V) of the tin can be calculated using the formula for the volume of a cylinder: V = π * r^2 * h.
- Plugging in the values, we get V = π * 4^2 * 4 = 64π cm^3.

2. Calculate the volume of the wooden rod:
- The rod has a diameter of 4cm, so the radius (r) is half the diameter, which is 4/2 = 2cm.
- The height (h) of the rod is 6cm.
- The volume (V) of the rod can be calculated using the formula for the volume of a cylinder: V = π * r^2 * h.
- Plugging in the values, we get V = π * 2^2 * 6 = 24π cm^3.

3. Calculate the volume of water displaced by the wooden rod:
- The volume of water displaced is equal to the volume of the wooden rod since it completely occupies the space it displaces.
- The volume of water displaced = volume of the wooden rod = 24π cm^3.

4. Calculate the new depth of water:
- To find the new depth of water, we need to subtract the volume of water displaced by the rod from the initial volume of water.
- Initial volume of water = 64π cm^3.
- Volume of water after displacement = Initial volume of water - volume of water displaced by the rod.
- Volume of water after displacement = 64π - 24π = 40π cm^3.

- Now, we have the new volume of water, but we need to find the new depth (h') of the water in the tin.
- We can use the formula for the volume of a cylinder to find the new depth: V = π * r^2 * h'.
- Plugging in the values: 40π = π * 4^2 * h'.
- Calculating, we get h' = 40/16 = 2.5 cm.

Therefore, the new depth of water in the tin is 2.5 cm.