Question

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

• A. 1 : 6
• B. 1 : 9
• C. 1 : 11
• D. 1 : 14

Answer 1

The Correct Option is C

 To solve the problem, we need to find the ratio \(k': k''\) for a spring of force constant \(k\) that is cut into multiple segments and then connected in series and parallel. Here are the step-by-step computations and explanations:

1. The original spring is divided into segments with a ratio of 1:2:3. Let their lengths be \(L_1 = x\), \(L_2 = 2x\), and \(L_3 = 3x\).
2. The force constant of a spring is inversely proportional to its length when a spring is divided. Thus, the force constants for parts \(L_1\), \(L_2\), and \(L_3\) would be \(k_1 = \frac{k}{1}\), \(k_2 = \frac{k}{2}\), and \(k_3 = \frac{k}{3}\) respectively.
3. When springs are connected in series, the equivalent spring constant \(k'\) is given by: 

\[\frac{1}{k'} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3}\]

1. Substituting the values: 

\[\frac{1}{k'} = \frac{1}{k} + \frac{1}{\frac{k}{2}} + \frac{1}{\frac{k}{3}} = \frac{1}{k} + \frac{2}{k} + \frac{3}{k} = \frac{6}{k}\]

1. Solving for \(k'\): 

\[k' = \frac{k}{6}\]

1. When springs are connected in parallel, the equivalent spring constant \(k''\) is given by: 

\[k'' = k_1 + k_2 + k_3\]

1. Substituting the values: 

\[k'' = k + \frac{k}{2} + \frac{k}{3}\]

1. Combining these gives: 

\[k'' = k\left(1 + \frac{1}{2} + \frac{1}{3}\right) = k \left(\frac{11}{6}\right) = \frac{11k}{6}\]

1. Finally, the required ratio \(k':k''\) is: 

\[\frac{k'}{k''} = \frac{\frac{k}{6}}{\frac{11k}{6}} = \frac{1}{11}\]

Therefore, the ratio \(k':k''\) is 1:11, making the correct answer

1 : 11

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