Question:
Consider the Euler variables of polyhedral objects namely, P1, P2, P3, and P4 as given in the table below.
Polyhedral objects Faces (F) Edges (E) Vertices (V) Faces' inner loops (L) Bodies (B) Genus (G) P1 6 12 8 0 1 0 P2 5 8 5 0 1 0 P3 5 12 8 0 1 0 P4 10 24 16 0 1 0
Which one of the following options is NOT a topologically valid closed polyhedral object as per Euler's law?
Consider the Euler variables of polyhedral objects namely, P1, P2, P3, and P4 as given in the table below.
Which one of the following options is NOT a topologically valid closed polyhedral object as per Euler's law?
| Polyhedral objects | Faces (F) | Edges (E) | Vertices (V) | Faces' inner loops (L) | Bodies (B) | Genus (G) |
|---|---|---|---|---|---|---|
| P1 | 6 | 12 | 8 | 0 | 1 | 0 |
| P2 | 5 | 8 | 5 | 0 | 1 | 0 |
| P3 | 5 | 12 | 8 | 0 | 1 | 0 |
| P4 | 10 | 24 | 16 | 0 | 1 | 0 |
Which one of the following options is NOT a topologically valid closed polyhedral object as per Euler's law?
Show Hint
Check each object against the Euler-Poincare formula \(F-E+V=2(B-G)+L\) before picking the odd one out.
Updated On: Aug 14, 2026
- P3
- P1
- P4
- P2
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Write down Euler's law for solids.
For a closed polyhedral solid, the general Euler-Poincare formula is \(F - E + V = 2(B - G) + L\), where \(F\), \(E\), \(V\) are the face, edge, and vertex counts, \(B\) is the number of separate bodies, \(G\) is the genus (through holes), and \(L\) is the number of inner loops on the faces. Here every object has \(B = 1\), \(G = 0\), and \(L = 0\), so a valid solid must satisfy the simple form \(F - E + V = 2\).
Step 2: Check P1.
\(F - E + V = 6 - 12 + 8 = 2\). This matches, so P1 is a valid solid (it has the numbers of a cube).
Step 3: Check P2.
\(F - E + V = 5 - 8 + 5 = 2\). This also matches, consistent with a square pyramid.
Step 4: Check P3.
\(F - E + V = 5 - 12 + 8 = 1\). This does not equal 2, so P3 breaks Euler's law even though its \(B\) and \(G\) values claim it is a single, hole-free solid.
Step 5: Check P4.
\(F - E + V = 10 - 24 + 16 = 2\). This matches too.
Final Answer:
P1, P2, and P4 all satisfy \(F-E+V=2\), but P3 gives 1, so P3 is the one that cannot be a valid closed polyhedron. \[ \boxed{P3} \]
For a closed polyhedral solid, the general Euler-Poincare formula is \(F - E + V = 2(B - G) + L\), where \(F\), \(E\), \(V\) are the face, edge, and vertex counts, \(B\) is the number of separate bodies, \(G\) is the genus (through holes), and \(L\) is the number of inner loops on the faces. Here every object has \(B = 1\), \(G = 0\), and \(L = 0\), so a valid solid must satisfy the simple form \(F - E + V = 2\).
Step 2: Check P1.
\(F - E + V = 6 - 12 + 8 = 2\). This matches, so P1 is a valid solid (it has the numbers of a cube).
Step 3: Check P2.
\(F - E + V = 5 - 8 + 5 = 2\). This also matches, consistent with a square pyramid.
Step 4: Check P3.
\(F - E + V = 5 - 12 + 8 = 1\). This does not equal 2, so P3 breaks Euler's law even though its \(B\) and \(G\) values claim it is a single, hole-free solid.
Step 5: Check P4.
\(F - E + V = 10 - 24 + 16 = 2\). This matches too.
Final Answer:
P1, P2, and P4 all satisfy \(F-E+V=2\), but P3 gives 1, so P3 is the one that cannot be a valid closed polyhedron. \[ \boxed{P3} \]
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