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QNEET-2017-Ph-990: NEET 2017 Physics - Class 11 NEET-2017-Ph-990 Solution: A spring of force constant k is cut into lengths of ratio…
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 1:9 1:11 1:14 1:6
Multiple Choice Options
- A 1:09
- B 1:11
- C 1:14
- D 1:06
Step-by-Step Verified 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 l 6 , 2 l 6 and 3 l 6 with force constant,
_ k 1 = k l l 1 = k l l / 6 = 6 k
k 2 = k l l 2 = k l 2 l / 6 = 3 k
k 3 = k l l 3 = k l 3 l / 6 = 2 k
When connected in series,
1 k _ = 1 6 k + 1 3 k + 1 2 k = 1 + 2 + 3 6 k = 1 k
_ k _ = k
When connected in parallel,
k'' = 6k + 3k + 2k = 11k
k _ k _ = k 11 k = 1 11