Question

An elastic spring has a length $l_{1}$ when it is stretched with a force of $2\, N$ and a length of $l_{2}$ when it is stretched with a force of $3\, N$. What will be the length of the spring if it is stretched with force of $5\, N$ ?

• A. $\left(l_{1}+l_{2}\right)$
• B. $\frac{1}{2}\left(l_{1}+l_{2}\right)$
• C. $\left(3l_{1}-2l_{1}\right)$
• D. $\left(3l_{1}-2l_{2}\right)$

Answer 1

Answer: (C)

In equilibrium position for a spring reaction.
$k y_{0}=m g$
where $y_{0}$ is the extension in spring.
Let $l$ be the length of the spring.
Then for $l_{1}$
Extension $=\left(l_{1}-l\right)$
$\therefore k\left(l_{1}-l\right)=2$ ...(i)
Similarly for $l_{2}$
$k\left(l_{2}-l\right)=3$
Eliminating $k$ from Eqs. (i) and (ii), we get
$l=3 l_{1}-2 l_{2}$
Moreover adding Eqs. (i) and (ii), we get
Substituting the value of $l$ in above equation we get
$k\left[5 l_{2}-5 l_{1}\right]=5$
$5 l_{2}-5 l_{1}$ is the extension for $5\, N$ force
$\therefore $ Total length $=5 l_{2}-5 l_{1}+l$
$=3 l_{2}-2 l_{1}$