If a load of 1 kg stretches a cord by 1.2cm, what is the force constant of the cord? {g=10$ms^{-2}$}

Analyze the question: A 1 kg load stretches a cord by 1.2 cm. We need the force constant (spring constant) k, using Hooke's law.

Concept: Hooke's law states that F = ke, where F is the applied force, e is the extension, and k is the force constant.

Solution:

1. Force (weight of load): F = mg = 1 × 10 = 10 N
2. Extension: e = 1.2 cm = 1.2 × 10⁻² m = 0.012 m
3. Force constant:

k = F / e = 10 / 0.012 = 833.3 N/m

Answer: k ≈ 833 N/m, which matches option B.

Tip: Always convert centimetres to metres before substituting, because the unit of k is N/m. Watch for the g value given in the question (here 10, not 9.8), since using the wrong one will lead you to a wrong option. A quick check is 10 ÷ 0.012 = 10000 ÷ 12 ≈ 833.