A horizontal rod AB is hinged at end A and carries three point loads of 3.0 kg, 7.0 kg, and 10.0 kg at distances 2.0 cm, 9.0 cm, and 15.0 cm respectively from A. Neglecting the rod weight, at what distance from A is the balance point (centre of gravity) located?

Introduction / Context:
Locating the centre of gravity of a system of discrete masses along a straight line is a standard statics problem. For a hinged rod with point loads, the rod will balance (no net moment) at the weighted-average position of the loads.

Given Data / Assumptions:

• Masses: 3.0 kg at 2.0 cm, 7.0 kg at 9.0 cm, 10.0 kg at 15.0 cm from A.
• Rod is weightless; hinge at A is the reference.
• Acceleration due to gravity is common and cancels in ratios.

Concept / Approach:
The balance point x̄ from A equals the sum of moments divided by the sum of weights.
x̄ = (Σ W_i * x_i) / (Σ W_i)

Step-by-Step Solution:
Compute total weight: W_total = 3 + 7 + 10 = 20 kg.Compute moment about A: M_A = 3*2 + 7*9 + 10*15 = 6 + 63 + 150 = 219 kg·cm.Find x̄: x̄ = 219 / 20 = 10.95 cm.Therefore, the rod balances 10.95 cm from A.

Verification / Alternative check:
Check reasonableness: the largest load (10 kg) at 15 cm pulls the CG toward the far end, so x̄ should be closer to the 15 cm side than to 9 cm, matching 10.95 cm.

Why Other Options Are Wrong:

• 11.25, 12.55, 13.25 cm overestimate the CG shift; they would require larger far-end moments.
• 9.90 cm underestimates the influence of the 10 kg load at 15 cm.

Common Pitfalls:
Forgetting to multiply each weight by its lever arm; mixing centimetres with metres; rounding too early.

Final Answer:
10.95 cm from A