Plane pin-jointed frames — condition for imperfection: A framed structure (simple, plane, pin-jointed) is considered imperfect if the number of members m is __________ (2j − 3), where j is the number of joints.

Introduction / Context:
For plane pin-jointed trusses, the simple test m = 2j − 3 helps classify whether a frame is statically determinate (perfect), unstable (deficient), or redundant (excess). This criterion is widely used in structural analysis as a quick screening before detailed calculations.

Given Data / Assumptions:

• Plane, pin-jointed frame.
• No special geometric constraints (no internal mechanisms) beyond simple connections.
• m = number of members; j = number of joints.

Concept / Approach:
A simple (perfect) truss satisfies m = 2j − 3. If m < 2j − 3, the truss is deficient in members (mechanism/unstable). If m > 2j − 3, it is redundant and therefore imperfect from a purely statical-determinacy standpoint.

Step-by-Step Solution:
Perfect frame condition: m = 2j − 3.Deficient (unstable): m < 2j − 3.Redundant (statically indeterminate): m > 2j − 3.Imperfect = not perfect ⇒ m ≠ 2j − 3 ⇒ either less than or greater than 2j − 3.

Verification / Alternative check:
Classical examples: A triangular frame (j = 3) with m = 3 is perfect (3 = 2*3 − 3). Removing one member (m = 2) makes it a mechanism (deficient). Adding an extra member (m = 4) introduces redundancy (excess). Both non-equal cases are imperfect.

Why Other Options Are Wrong:

• equal to: That is the perfect condition, not imperfect.
• less than: Only one type of imperfection (deficient) but misses redundant case.
• greater than: Only redundant case; misses deficient case.

Common Pitfalls:

• Assuming m = 2j − 3 guarantees stability in all geometric layouts; certain special geometries can still be mechanisms.
• Confusing determinacy (solvability by statics alone) with structural safety or serviceability.

Final Answer:
either (b) or (c)