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NEET PHYSICSOscillationsMedium

Question

A spring of force constant kk is cut into lengths of ratio 1:2:31:2:3. They are connected in series and the new force constant is k′k'. Then they are connected in parallel and the force constant is k′′k''. Then k′:k′′k':k'' is:

A

1:9

B

1:11

C

1:14

D

1:6

Step-by-Step Solution

The spring constant kk is inversely proportional to the length of the spring (k∝1Lk \propto \frac{1}{L}).

  1. Cutting the Spring: Let the original length be LL and the original spring constant be kk. The spring is cut in the ratio 1:2:31:2:3. The lengths of the segments are: L1=16L,L2=26L=L3,L3=36L=L2L_1 = \frac{1}{6}L, \quad L_2 = \frac{2}{6}L = \frac{L}{3}, \quad L_3 = \frac{3}{6}L = \frac{L}{2} Since kL=constantkL = \text{constant}, the spring constants of the segments are: k1=6k,k2=3k,k3=2kk_1 = 6k, \quad k_2 = 3k, \quad k_3 = 2k

  2. Series Connection (k′k'): When connected in series, the reciprocal of the equivalent spring constant is the sum of the reciprocals of individual constants: 1k′=1k1+1k2+1k3\frac{1}{k'} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} 1k′=16k+13k+12k=1+2+36k=66k=1k\frac{1}{k'} = \frac{1}{6k} + \frac{1}{3k} + \frac{1}{2k} = \frac{1 + 2 + 3}{6k} = \frac{6}{6k} = \frac{1}{k} k′=kk' = k (Note: Connecting parts back in series restores the original spring).

  3. Parallel Connection (k′′k''): When connected in parallel, the equivalent spring constant is the sum of individual constants: k′′=k1+k2+k3k'' = k_1 + k_2 + k_3 k′′=6k+3k+2k=11kk'' = 6k + 3k + 2k = 11k

  4. Ratio: k′k′′=k11k=111\frac{k'}{k''} = \frac{k}{11k} = \frac{1}{11}

Thus, the ratio is 1:111:11.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Oscillations. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSOscillationsspringconstantlengthsconnectedseries

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