Question

The length of wire becomes l₁ and l₂ when 100N and 120 N tensions are applied respectively. If 10 l₂ = 11 l₁, the natural length of wire will be \(\frac{l}{x}l_1\). Here the value of x is ____.

Answer 1

Correct answer is 2

Let the original length be 'l'

When T₁ = 100 N, Extension = l₁ - l₀

When T₂ = 120 N, Extension = l₂ - l₀

Then 100 = K(l₁ - l₀) ....(1)

Then 120 = K(l₂ - l₀) ....(2)