A cylindrical jar with base radius 15 cm is filled with water to a height of 20 cm. A solid iron sphere of radius 10 cm is fully submerged in the jar. By how many centimetres does the water level rise?

Introduction / Context:
Submerging a solid displaces an equal volume of water. The rise in water height equals displaced volume divided by the jar’s base area. The initial 20 cm level is irrelevant to the rise amount as long as there is capacity.

Given Data / Assumptions:

• Jar radius R = 15 cm.
• Sphere radius r = 10 cm.
• Sphere volume V_s = (4/3)πr^3.
• Jar base area A = πR^2.

Concept / Approach:
Rise Δh = V_displaced / A = V_s / (πR^2) since the sphere is fully submerged.

Step-by-Step Solution:

Verification / Alternative check:
Cancel π; reduce the fraction 4000/675 by 25 to get 160/27 exactly.

Why Other Options Are Wrong:
Other fractions correspond to incorrect radius usage or arithmetic; 160/27 is the exact simplified value.

Common Pitfalls:
Using diameter instead of radius; forgetting to square R in the base area; including the initial 20 cm in the rise computation.

Final Answer:
160/27 cm