Question

An elastic spring has a length $l_1$ when it is stretched with a force of $2N$ and a length of $l_2$ when it is stretched with a force of $3N$. What will be the length of the spring if it is stretched with force of $5N$ ?

• 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.
$k{y}_0=mg$
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
$l=3l_1-2l_2$
Moreover adding Eqs. (i) and (ii), we get
Substituting the value of $l$ in above equation we get
$k\left[5l_2-5l_1\right]=5$
$5l_2-5l_1$ is the extension for $5N$ force
$\therefore$ Total length $=5l_2-5l_1+l$
$=3l_2-2l_1$