A cylindrical vessel of radius 6 cm is partially filled with water. If a solid sphere of radius 5 cm is completely immersed in the vessel, by how much will the water level rise? (Take π = 22/7.)

Introduction / Context:
This question concerns volume displacement. When a solid object such as a sphere is fully immersed in a liquid, it displaces a volume of liquid equal to its own volume. In a cylindrical vessel, this displaced volume causes the water level to rise. By equating the volume of the sphere to the volume increase of water in the cylinder, we can compute the height by which the water level rises. This principle is widely used in physics and engineering, especially in fluid mechanics and measurement methods based on displacement.

Given Data / Assumptions:
• The radius of the cylindrical vessel, R, is 6 cm.
• The radius of the solid sphere, r, is 5 cm.
• The sphere is completely immersed in the water.
• π is to be taken as 22/7.
• The water level rise is h cm, which we must find.
• The sphere is fully below the water surface and does not touch the bottom in a way that changes the effective volume relationship.

Concept / Approach:
The volume of a sphere is V_sphere = (4/3)πr³. When this sphere is immersed in the cylindrical vessel, it displaces an equal volume of water. The cylinder volume associated with a height increase h is V_cylinder = πR²h. Setting these equal gives (4/3)πr³ = πR²h. After cancelling π, we can solve for h. Finally, substitute the numeric values and simplify to obtain the rise in water level. The question options are given in centimetres rounded to two decimal places.

Step-by-Step Solution:
Step 1: Write the displaced volume equality: (4/3)πr³ = πR²h. Step 2: Cancel π from both sides to get (4/3)r³ = R²h. Step 3: Substitute r = 5 cm and R = 6 cm: (4/3) * 5³ = 6² * h. Step 4: Compute 5³ = 125 and 6² = 36, so (4/3) * 125 = 36h. Step 5: Evaluate (4/3) * 125 = 500/3. Step 6: So we have 500/3 = 36h, giving h = (500/3) / 36 = 500 / 108. Step 7: Simplify the fraction: divide numerator and denominator by 4 to get h = 125 / 27. Step 8: Evaluate 125 / 27 ≈ 4.63 cm when rounded to two decimal places.

Verification / Alternative check:
We can check the reasonableness of 4.63 cm by approximating. The sphere volume is roughly (4/3) * 3.14 * 5³ ≈ 4.19 * 125 ≈ 523 cubic centimetres. The cylinder cross section area is πR² ≈ 3.14 * 36 ≈ 113 cubic centimetres. Dividing 523 by 113 gives approximately 4.6 to 4.7 cm, which agrees closely with 4.63 cm. This consistency between detailed fraction based computation and rough estimation confirms the correctness of the result.

Why Other Options Are Wrong:
4.50 cm is slightly smaller than the correct value and would underestimate the displaced volume when multiplied by the cylinder's cross sectional area.
5.56 cm, 6.67 cm, and 6.94 cm are all larger and would imply displaced volumes greater than that of the sphere. None of these match the exact fraction 125 / 27 derived from equating the volumes.

Common Pitfalls:
Students sometimes forget to cancel π from both sides and carry it incorrectly through the equation. Another frequent mistake is to use the diameter instead of the radius in the sphere volume or cylinder area formulas. Errors in cube and square calculations, such as miscomputing 5³ or 6², also lead to wrong answers. Finally, some may round too early and get inaccurate final values. Keeping exact fractions and only rounding at the final step helps maintain accuracy.

Final Answer:
The water level in the cylindrical vessel rises by approximately 4.63 cm.