A rectangular block has dimensions 5 cm by 6 cm by 7 cm and is gently lowered into a cylindrical vessel of radius 6 cm containing a liquid. As a result, the level of the liquid in the cylinder rises by 4 cm. Assuming no overflow, what is the volume of liquid displaced in cubic centimetres?

Introduction / Context:
This problem connects volume displacement with a rise in liquid level inside a cylindrical vessel. When a solid block is lowered into the liquid, it displaces some liquid, causing the liquid level to rise. The extra liquid level, combined with the cross sectional area of the cylinder, directly gives the volume of liquid displaced, regardless of the detailed shape of the block.

Given Data / Assumptions:

• Cylindrical vessel radius R = 6 cm.
• Initial liquid level is not needed explicitly.
• Rise in liquid level after block is inserted = 4 cm.
• No overflow of liquid occurs.
• We are asked for the volume of liquid displaced in cubic centimetres.

Concept / Approach:
In a right circular cylinder, the volume corresponding to a small change in height is equal to the base area multiplied by that change in height. The base area of a cylinder is:
Base area = π * R^2. If the height increases by h, then the additional volume is:
Volume displaced = π * R^2 * h. Here, R and h are given, so we can compute the displaced volume directly.

Step-by-Step Solution:
Step 1: Compute the base area of the cylindrical vessel. Base area = π * R^2 = π * 6^2 = π * 36. Step 2: Use the rise in height of the liquid. Rise in level h = 4 cm. Step 3: Compute the volume displaced. Volume displaced = base area * rise = π * 36 * 4. Volume displaced = 144π cubic centimetres.

Verification / Alternative check:
If we approximate π as 22 / 7, the volume becomes 144 * 22 / 7 = 3168 / 7 ≈ 452.6 cm^3. This is a reasonable volume for a modest rise of 4 cm in a vessel with radius 6 cm. The presence of the rectangular block in the problem statement simply provides the physical cause of the displacement but does not change the computation based on the cylinder geometry and the observed rise in liquid level.

Why Other Options Are Wrong:
72π corresponds to only half the correct rise in liquid level and would match a 2 cm increase instead of 4 cm.
108π corresponds to three quarters of the correct displaced volume and would result from using an incorrect height or radius.
216π and 288π are larger volumes that would require a greater rise in liquid level or a larger radius than given in the problem.

Common Pitfalls:
A common mistake is to attempt to use the dimensions of the rectangular block directly, rather than focusing on the observed rise in the cylinder. While Archimedes principle states that the volume of displaced liquid equals the volume of the submerged part of the block, the easiest route here is to work with the cylinder geometry. Another pitfall is forgetting to square the radius when computing the base area or miscalculating 6^2 as 12 rather than 36.

Final Answer:
The volume of liquid displaced in the cylindrical vessel is 144π cubic centimetres.