Discussion Forum



1)

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 


A) $5l_{1} - 4l_{2}$

B) $5l_{2} - 4l_{1}$

C) $9l_{1} - 8l_{2}$

D) $9l_{2} - 8l_{1}$

Answer:

Option B

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