The elongation of a wire on the surface of the earth is 10⁻⁴ m. The same wire of same dimensions is elongated by 6 × 10⁻⁵ m on another planet. The acceleration due to gravity on the planet will be ____________ ms⁻². (Take acceleration due to gravity on the surface of earth = 10 ms⁻²)

Solution

Physics sir · 3 steps

Same wire, same mass hanging from it, so the stretch is just proportional to g. Take the ratio.

1. Elongation formula. A mass m hung from a wire pulls with force mg. From Y = FL/(A Δ L), the elongation is Δ L = mgL/AY.
2. What stays fixed. The wire, its length, its area and the hanging mass are the same on both planets. Only g changes, so Δ L ∝ g.

   gₚ/gₑ = (Δ Lₚ)/(Δ Lₑ) = (6 × 10⁻⁵)/10⁻⁴ = 0.6

3. So. gₚ = 0.6 × 10 = 6 m s⁻². The answer is 6.

The common slip: 16.7. Turning the ratio upside down, as gₚ = 10 × 10⁻⁴/(6 × 10⁻⁵), gives about 16.7, not 6. A smaller stretch means a smaller pull, so g must be smaller too.

The idea to keep. When the same load hangs from the same wire in two places, only g changes. The stretch scales directly with g, so a simple ratio does the job.

Answer from the NTA final answer key. Solution by PrepWiser.