Q139 / 170 · Manipal (MET) 2021 · Physics
A simple spring has length L and force constant $k$. It is cut into two springs of lengths $l_1$ and $L_2$ such that $l_1=n L_2$ ($n=$ an integer). The force constant of spring of length $l_1$ is :
- A$k(1+n)$
- Correct answerB$\frac{k}{n}(1+n)$
- C$k$
- D$\frac{k}{(n+1)}$
Solution
Since $L_1+n L_2$ where $n$ is an integer, so the spring is made of $(n+1)$ equal parts in length each of length $I_2$
$\therefore \quad \| \quad \frac{1}{k}=\frac{(n+1)}{k^{\prime}}$
or $k^{\prime}=(n+1) k$
The spring of length $l_1\left(=n l_2\right)$ will be equivalent to $n$ springs connected in series where spring constant
$k^{\prime \prime}=\frac{k^{\prime}}{n}=\frac{(n+1) k}{n}$
Hence, the correct answer is option: 'B'.