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

1. Basics of 3D Mensuration and Volume (basic)

In our journey toward mastering Quantitative Aptitude, the first step is understanding what Volume actually represents. At its core, volume is the measure of three-dimensional space occupied by an object. While a flat sheet of paper has an area, a 3D object like a notebook or a shoe box occupies a specific capacity in the physical world. As we learn in Science, Class VIII, NCERT (Revised ed 2025), The Amazing World of Solutes, Solvents, and Solutions, p.143, matter is defined as anything that possesses mass and occupies space.

To quantify this space, we use specific units. The Standard International (SI) unit of volume is the cubic metre (m³), which is the space occupied by a cube where every side is exactly one metre long. However, for everyday objects, we use smaller units like cubic centimetres (cm³ or cc) or Litres (L). A key conversion factor to remember for your exams is that 1 Litre is equivalent to 1 cubic decimetre (dm³), and 1,000 cubic centimetres make up exactly 1 Litre Science, Class VIII, NCERT (Revised ed 2025), The Amazing World of Solutes, Solvents, and Solutions, p.143.

When dealing with regular shapes, like a cuboid, we calculate volume by multiplying three dimensions: Length × Width × Height Science, Class VIII, NCERT (Revised ed 2025), The Amazing World of Solutes, Solvents, and Solutions, p.145. For irregular objects, like a stone, we can determine volume through displacement—measuring how much the water level rises in a cylinder when the object is submerged Science, Class VIII, NCERT (Revised ed 2025), The Amazing World of Solutes, Solvents, and Solutions, p.146. Understanding this "displacement principle" is vital because it links the physical geometry of a container to the liquid it holds.

Unit	Equivalent to...	Common Use Case
1 cm³ (1 cc)	1 millilitre (mL)	Small doses, medicine, or small stones.
1 dm³	1 Litre (L)	Beverage cartons (e.g., buttermilk or juice).
1 m³	1,000 Litres (L)	Large overhead water tanks.

Finally, we must distinguish volume from Density. Density is the mass present in a unit volume of a substance (Density = Mass / Volume). While the volume changes if you resize an object, the density remains a fundamental property of the material itself Science, Class VIII, NCERT (Revised ed 2025), The Amazing World of Solutes, Solvents, and Solutions, p.140.

Sources: Science, Class VIII, NCERT (Revised ed 2025), The Amazing World of Solutes, Solvents, and Solutions, p.140, 143, 145, 146

2. Properties of Spheres and Hemispheres (basic)

To master quantitative aptitude, we must first visualize a sphere as a three-dimensional object where every point on its surface is at an equal distance (the radius, r) from a central point. This perfect symmetry means the sphere has no edges or vertices. In physical geography, we often study the Earth as a sphere divided into two equal halves known as hemispheres Exploring Society: India and Beyond, Oceans and Continents, p.29. While our planet's Northern Hemisphere is dominated by landmass and the Southern by water Physical Geography by PMF IAS, Horizontal Distribution of Temperature, p.287, in mathematical terms, a hemisphere is simply a sphere cut exactly in half through its center.

Understanding the measurement of these shapes requires two main formulas: Volume and Surface Area. Just as we measure the volume of a cuboid by multiplying its dimensions Science, Class VIII, The Amazing World of Solutes, Solvents, and Solutions, p.145, the volume of a sphere is derived from its radius as (4/3)πr³. Consequently, the volume of a hemisphere is exactly half of that, or (2/3)πr³. This relationship is vital when calculating the capacity of containers or the amount of material needed to create a hemispherical bowl.

However, you must be particularly cautious with Surface Area. A sphere has only one continuous curved surface with an area of 4πr². When you cut a solid sphere to create a hemisphere, you expose a new, flat circular base. Therefore, a solid hemisphere has a Curved Surface Area (CSA) of 2πr², but its Total Surface Area (TSA) includes that flat base (πr²), totaling 3πr².

Property	Sphere	Solid Hemisphere
Volume	(4/3)πr³	(2/3)πr³
Curved Surface Area	4πr²	2πr²
Total Surface Area	4πr²	3πr²

Sources: Exploring Society: India and Beyond, Oceans and Continents, p.29; Physical Geography by PMF IAS, Horizontal Distribution of Temperature, p.287; Science, Class VIII, The Amazing World of Solutes, Solvents, and Solutions, p.145

3. Geometry of the Right Circular Cylinder (basic)

Let’s dive into the geometry of the Right Circular Cylinder, one of the most common shapes you will encounter in both daily life and competitive exams. Imagine taking a circle and stacking identical circles directly on top of it until you reach a certain height. This 'stack' forms a cylinder. It is called a Right cylinder because the axis (the line connecting the centers of the top and bottom circles) is exactly perpendicular to the base, and Circular because its base is a circle. To define any cylinder, we need two primary dimensions: the Radius (r) of its base and its Height (h). Just as you might use a measuring cylinder in a lab to find the volume of a liquid Science, Class VIII, The Amazing World of Solutes, Solvents, and Solutions, p.143, the mathematical volume represents the total space inside the container. The formula for the Volume (V) of a cylinder is derived by taking the area of the circular base (πr²) and multiplying it by the height (h), giving us: V = πr²h. This volume tells us exactly how much 'juice' or 'water' a cylindrical vessel can hold. In practical applications, the relationship between radius and height is crucial. For instance, in scientific experiments, the accuracy of measuring volume can change based on the capacity and dimensions of the cylinder used Science, Class VIII, The Amazing World of Solutes, Solvents, and Solutions, p.144. Similarly, in optics, we often relate the radius of a curved surface to other dimensions like focal length (R = 2f) Science, class X, Light – Reflection and Refraction, p.137. In geometry, we apply similar logic: if you know the ratio between the radius and the height, you can express the entire volume in terms of just one variable, which is a common trick used to solve complex aptitude problems.

Sources: Science, Class VIII, The Amazing World of Solutes, Solvents, and Solutions, p.143; Science, Class VIII, The Amazing World of Solutes, Solvents, and Solutions, p.144; Science, class X, Light – Reflection and Refraction, p.137

4. Ratios and Percentage Increases in Dimensions (intermediate)

Concept: Ratios and Percentage Increases in Dimensions

5. Liquid Displacement and Volume Conservation (intermediate)

At its core, the study of liquids in quantitative aptitude rests on the Law of Conservation of Volume. Unlike gases, liquids are incompressible; therefore, the total volume remains constant regardless of the shape of the container it occupies. As observed in basic science experiments, when you transfer 200 mL of water between differently shaped containers, the liquid level might change, but the quantity of matter does not Science, Class VIII NCERT, Particulate Nature of Matter, p.104. This principle is vital for solving problems where 'recasting' occurs—such as melting a metal sphere to form a wire or pouring juice from a bowl into a glass. To master these problems, you must look beyond the visual height of the liquid and focus on the geometric volume formulas. For example, a hemisphere and a cylinder might look vastly different, but if their dimensions are mathematically related (like a cylinder's height being exactly two-thirds of its radius), their capacities can be identical. Another critical application is the Displacement Method. When a solid object is submerged in a liquid, it displaces a volume of liquid exactly equal to its own volume Science, Class VIII NCERT, The Amazing World of Solutes, Solvents, and Solutions, p.146. This 'Final Volume minus Initial Volume' calculation is the standard way to find the volume of irregular solids. Finally, it is helpful to understand the behavior of liquids in communicating vessels. If multiple containers of different shapes are connected at the base, the liquid level will remain equal across all of them due to atmospheric pressure acting uniformly Science, Class VIII NCERT, Pressure, Winds, Storms, and Cyclones, p.94.

Concept	Mathematical Logic	Physical Result
Conservation	Volume₁ = Volume₂	100% of liquid fills the new space if capacities match.
Displacement	V_object = V_final - V_initial	Used to measure irregular solids.
Equilibrium	Pressure is equal at the base	Liquid levels stay the same in connected vessels.

Sources: Science, Class VIII NCERT, Particulate Nature of Matter, p.104; Science, Class VIII NCERT, The Amazing World of Solutes, Solvents, and Solutions, p.146; Science, Class VIII NCERT, Pressure, Winds, Storms, and Cyclones, p.94

6. Algebraic Substitution in Mensuration (exam-level)

In competitive examinations, mensuration problems rarely give you direct numerical values for every dimension. Instead, they provide relationships between different parts of a solid—for instance, stating that the radius is a certain percentage of the height. Algebraic substitution is the technique of expressing all variables in terms of a single 'pivot' variable (usually the radius r) to simplify complex formulas. This approach is similar to how we use the lens formula to relate object-distance (u) and image-distance (v) through a single focal length (f), as discussed in Science, Class X NCERT (2025 ed.), Light – Reflection and Refraction, p.155.

To master this, you must first translate verbal descriptions into algebraic equations. If a problem states that the radius (r) of a cylinder is 50% more than its height (h), you write this as r = 1.5h or r = (3/2)h. To make the volume calculation easier, it is often smarter to rearrange this to h = (2/3)r. By substituting this expression for h back into the standard volume formula (V = πr²h), you eliminate one variable entirely. This reduction allows you to compare different shapes—like a cylinder and a hemisphere—on a 'level playing field' using only the variable r.

The beauty of this method lies in its precision. While we often measure objects with scales or measuring cylinders to find volume in mL or cm³, as seen in Science, Class VIII NCERT (Revised ed 2025), The Amazing World of Solutes, Solvents, and Solutions, p.145-146, algebraic substitution allows us to prove mathematical equalities (like two containers having identical capacities) without knowing a single physical measurement. When you substitute the relationship into the formulas, terms like π and r often cancel out or combine, revealing the underlying ratio immediately.

Step	Action	Example (r = 2h)
1	Identify the Relationship	r = 2h
2	Isolate the secondary variable	h = r/2
3	Substitute into Formula	V = πr²(r/2) = (1/2)πr³

Sources: Science, Class VIII NCERT (Revised ed 2025), The Amazing World of Solutes, Solvents, and Solutions, p.145-146; Science, Class X NCERT (2025 ed.), Light – Reflection and Refraction, p.155

7. Solving the Original PYQ (exam-level)

Now that you have mastered the geometric volume formulas and percentage-to-ratio conversions, this question serves as the perfect synthesis of those building blocks. To solve this, you must recall the Volume of a Hemisphere ($2/3 \pi r^3$) and the Volume of a Cylinder ($\pi r^2 h$). The constraint that the diameter is the same for both tells you immediately that they share a common radius, r. The critical step is translating the phrase "radius is 50% more than its height" into the equation $r = 1.5h$, which simplifies to $h = (2/3)r$.

By substituting this expression for height into the cylinder's volume formula, you get $\pi \times r^2 \times (2/3)r$, which results in $(2/3) \pi r^3$. Notice how the two volumes are mathematically identical. This means the beverage from the bowl will occupy the cylindrical vessel exactly, leading us to the correct answer (C) 100%. This problem exemplifies how UPSC tests your ability to manipulate variables rather than just performing rote calculations.

UPSC often includes distractor options like (A) and (D) to catch students who rush or rely on visual intuition. Option (A) is a common trap for those who might invert the ratio (mistakenly setting $h = 1.5r$), while Option (D) targets students who assume a physical transfer always involves some loss or overflow. By sticking to the algebraic equivalence derived from the formulas in Standard UPSC Quantitative Aptitude Guides, you avoid these conceptual pitfalls and ensure accuracy.