- Home
- Standard 11
- Physics
A spring of force constant $k$ is cut into lengths of ratio $1:2:3$ . They are connected in series and the new force constant is $k'$ . Then they are connected in parallel and force constant is $k''$ . Then $k':k''$ is
$1:11$
$1:14$
$1:16$
$1:9$
Solution
Let us assume, the length of spring be $l$.
When we cut the spring into ratio of length $1: 2: 3,$ we
get three springs of lengths $\frac{l}{6}, \frac{2 l}{6}$ and $\frac{3 l}{6}$ with force
constant,
$\therefore k_{1}=\frac{k l}{l_{1}}=\frac{k l}{l / 6}=6 k$
${k_{2}=\frac{k l}{l_{2}}=\frac{k l}{2 l / 6}=3 k}$
${k_{3}=\frac{k l}{l_{3}}=\frac{k l}{3 l / 6}=2 k}$
When connected in series,
$\frac{1}{k^{\prime}}=\frac{1}{6 k}+\frac{1}{3 k}+\frac{1}{2 k}=\frac{1+2+3}{6 k}=\frac{1}{k}$
$\therefore \quad \overline{k^{\prime}}=k$
When connected in parallel,
${k^{\prime \prime}=6 k+3 k+2 k=11 k}$
${\frac{k^{\prime}}{k^{\prime \prime}}=\frac{k}{11 k}=\frac{1}{11}}$
