Discussion Forum

The length of elastic string, obeying Hooke's law is l₁ metres when the tension 4N and l₂ metres when the tension is 5N. The length in metres when the tension is 9N is 

Answer:

Explanation:

Let l₀ be the unstretched length and l₃ be the length under a tension of 9N. Then

Y = $\frac{4l_{0}}{A(l_{1}-l_{0})}$ = $\frac{5l_{0}}{A(l_{2}-l_{0})}$

= $\frac{9l_{0}}{A(l_{3}-l_{0})}$ 

These give

$\frac{4}{(l_{1}-l_{0})}$ = $\frac{5}{(l_{2}-l_{0})}$ 

l₀ = 5l₁ - 4l₂

Further, $\frac{4}{(l_{1}-l_{0})}$ = $\frac{9}{(l_{2}-l_{0})}$

Substituting the value of l₀ and solving, and we get l₃ = 5l₂ - 4l₁