A hemispherical bowl is filled to the brim with a beverage. The contents of a bowl are transferred into a cylindrical vessel whose radius is 50% more than its height. If the diameter is same for both bowl and cylinder, then the volume of the beverage in t

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Q: 144 (IAS/1999)
A hemispherical bowl is filled to the brim with a beverage. The contents of a bowl are transferred into a cylindrical vessel whose radius is 50% more than its height. If the diameter is same for both bowl and cylinder, then the volume of the beverage in the, cylindrical vessel will be

question_subject: 

Maths

question_exam: 

IAS

stats: 

0,0,4,0,2,0,2

keywords: 

{'hemispherical bowl': [0, 1, 0, 0], 'cylindrical vessel': [0, 0, 0, 1], 'volume': [0, 0, 1, 0], 'bowl': [0, 1, 0, 1], 'beverage': [0, 1, 0, 0], 'cylinder': [0, 2, 1, 2], 'liquid': [0, 0, 0, 1], 'diameter': [0, 3, 3, 4], 'radius': [0, 0, 2, 2], 'brim': [0, 1, 0, 0], 'height': [0, 0, 1, 2]}

Let`s assume that the diameter of the hemispherical bowl and the cylindrical vessel is "d" units.

The radius of the bowl is d/2, and the height of the cylindrical vessel is h units.

Given that the radius of the cylindrical vessel is 50% more than its height, we can express it as:

Radius of cylindrical vessel = (h + 0.5h) = 1.5h

The volume of the hemispherical bowl is given by:

V1 = (2/3)?(d/2)^3 = (1/6)?d^3

The volume of the cylindrical vessel is given by:

V2 = ?(1.5h)^2h = 2.25?h^3

Now, we need to compare the volume of the beverage in the cylindrical vessel (V2) to the volume of the bowl (V1).

Since the bowl is filled to the brim, the volume of the beverage in the cylindrical vessel will be equal to or less than the volume of the bowl.

Let`s calculate the percentage of the volume in the cylindrical vessel compared to the volume in the bowl:

Percentage = (V2 / V1) * 100

Substituting the values of V1 and V2, we have:

Percentage = (2.25?h^3) / ((1/6)?d^3) * 100

Canceling out ? from the numerator and denominator, we get:

Percentage = (2.25h^3) / ((1/6)d^3) * 100

Since the diameter is the same for both the bowl and the cylindrical vessel (d is constant), the percentage will depend only on the ratio of heights:

Percentage = (2.25h^3) / ((1/6)d^3) * 100

Percentage = (2.25h^3) / ((1/6)d^3) * 100

Percentage = 2.25 * 6 * (h^3 / d^3) * 100

Percentage = 13.5 * (h^3 / d^3) * 100

Since the diameter is the same for both the bowl and the cylindrical vessel, d^3 is a constant. Therefore, the percentage will be directly proportional to the cube of the height:

Percentage ? h^3

This means that the percentage of the volume in the cylindrical vessel compared to the volume in the bowl will depend on the cube of the height.

Since we don`t have information about the specific values of h, we cannot determine the exact percentage. However, we can conclude that the percentage will be less than or equal to 100%.

Therefore, the correct answer is: The volume of the beverage in the cylindrical vessel will be 100% or less than 100%.

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