The length of a wire becomes l₁ and l₂ when 100 N and 120 N tensions are applied respectively. If 10 l₂=11 l₁, the natural length of wire will be 1/x l₁. Here the value of x is _____________.

Solution

Physics sir · 4 steps

The stretch is proportional to tension, so compare the two extensions, not the two lengths.

1. Extension follows tension. Let the natural length be L. For the same wire, Hooke's law says the extension l - L is proportional to the tension, since area, length and Young's modulus do not change.

   (l₁ - L)/(l₂ - L) = 100/120 = 5/6

2. Solve for L. Cross-multiplying gives 6l₁ - 6L = 5l₂ - 5L, so L = 6l₁ - 5l₂.
3. Use the given relation. From 10 l₂ = 11 l₁, we get l₂ = 1.1 l₁. Then L = 6l₁ - 5.5 l₁ = 0.5 l₁ = 1/2 l₁.
4. So. The natural length is 1/2 l₁, so x = 2.

The common slip: 0.5. Reporting the ratio L/l₁ = 0.5 gives 0.5. The question defines L = 1/x l₁, so x is the reciprocal, 2.

The idea to keep. When one wire is loaded twice, write each extension as length minus natural length and take their ratio. The ratio equals the ratio of the tensions, and Y and A cancel out.

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