A conoical funnel 12cm deep and 15cm in diameter is full of liquid.it is emptied into a cylindrical tin 10cm in diameter.calculate the height of the liquid in the tin
Volume in funnel = (1/3)π(7.5^2)(12) cm^3
Assume the height in the cylinder will be h cm
Volume in cylinder = π(5^2)h cm^3
so
π(5^2)h = (1/3)π(7.5^2)(12)
25h = 4(56.25)
h = 225/25 = 9 cm
just find h such that
π/3 (15/2)^2 * 12 = π*(10/2)^2 h
that is,
225π = 25π h
You find the area and not the volume
Well, aren't we in a "funnel" situation? Let's see if we can pour some humor into math!
First things first, let's find the volume of the liquid in the funnel. The volume of a cone can be calculated with the formula V = (1/3)πr²h, where V is the volume, r is the radius, and h is the height.
Now, the funnel has a diameter of 15cm, which means a radius of 15/2 = 7.5cm. Given that the height is 12cm, we can calculate the volume of the liquid in the funnel. Let me grab my calculator, and...
*beep boop beep*
Calculating... calculating... Voilà! The volume of the liquid in the funnel is approximately 883.03 cubic centimeters.
Now, we'll empty this liquid into a cylindrical tin with a diameter of 10cm. To calculate the height of the liquid in the tin, we'll rearrange the formula for the volume of a cylinder: V = πr²h.
Let's plug in the values we know: the radius is 10/2 = 5cm, and the volume is 883.03 cubic centimeters. Another round of calculations, and...
*beep boop beep*
Drumroll, please... According to my calculations, the height of the liquid in the tin is approximately 35.32 centimeters.
So, the height of the liquid in the tin is approximately 35.32cm. Now, that's a tall glass of humor, isn't it? Cheers!
To calculate the height of the liquid in the tin, we need to find the volume of the liquid in the funnel and then divide it by the cross-sectional area of the tin.
1. First, let's calculate the volume of the liquid in the funnel.
The volume of a cone can be calculated using the formula V = (1/3)πr²h, where r is the radius of the base of the cone and h is the height of the cone.
In this case, the radius of the cone is half the diameter, so r = 15/2 = 7.5 cm and h = 12 cm.
Substituting the values into the formula, we get V = (1/3) * π * (7.5)^2 * 12.
Evaluating the expression, we find that the volume of the liquid in the funnel is approximately 1767.15 cm³ (rounded to two decimal places).
2. Now, let's calculate the cross-sectional area of the tin.
The cross-sectional area of a cylinder can be calculated using the formula A = πr², where r is the radius of the cylinder.
In this case, the radius of the tin is 10/2 = 5 cm.
Substituting the value into the formula, we get A = π * 5².
Evaluating the expression, we find that the cross-sectional area of the tin is approximately 78.54 cm² (rounded to two decimal places).
3. Finally, let's find the height of the liquid in the tin.
We can use the formula V = Ah, where V is the volume, A is the cross-sectional area, and h is the height.
Rearranging the formula to solve for h, we have h = V / A.
Substituting the values we calculated earlier, we get h = 1767.15 / 78.54.
Evaluating the expression, we find that the height of the liquid in the tin is approximately 22.49 cm (rounded to two decimal places).
Therefore, the height of the liquid in the tin is approximately 22.49 cm.