A spring of force constant is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is . Then they are connected in parallel and the force constant is . Then is [2017]

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Solution

Q.
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 the force constant is $k''$ . Then $k' : k''$ is [2017]

1 1 : 9

2 1 : 11

3 1 : 14

4 1 : 6

Ans.
(2)

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,

$k_{1} = \frac{k l}{l_{1}} = \frac{k l}{l / 6} = 6 k, k_{2} = \frac{k l}{l_{2}} = \frac{k l}{2 l / 6} = 3 k$ and $k_{3} = \frac{k l}{l_{3}} = \frac{k l}{3 l / 6} = 2 k$

When connected in series, $\frac{1}{k'} = \frac{1}{6 k} + \frac{1}{3 k} + \frac{1}{2 k} = \frac{1 + 2 + 3}{6 k} = \frac{1}{k}$

$∴$ $k' = k$

When connected in parallel, $k'' = 6 k + 3 k + 2 k = 11 k$

$∴$ $\frac{k'}{k''} = \frac{k}{11 k} = \frac{1}{11}$

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