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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'. They are connected in parallel and the force constant is k". Then k':k"is:


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Solution

Step1: Given data

  1. The spring constant of the spring is k.
  2. The length of the spring is divided into three pieces of ratio 1:2:3.
  3. The spring constant of the system of spring when they are in series is k'.
  4. The spring constant of the system of spring when they are in parallel is k" .

Step2: Spring constant

  1. The spring constant of a spring is the force exerted on the spring of unit length.
  2. Spring constant is defined by the form, K=Fx, where, F is the force on the spring due to extension, and x is the amount of extension.
  3. So, springconstant1length.

Step3: Diagram

A spring of force constant k is cut into lengths of class 11 physics CBSE

Step4: Finding the spring constants

Let L be the length of the spring and it is cut into three segments of ratio 1:2:3. So, the three segments are L6, L3, L2.

As we know springconstant1length.So, the spring constant of the spring segments are

K1=6K, K2=3K, K3=2K

So, the spring constant of the spring system when they are connected in series is

K'=K1+K2+K3=6K+3K+2KorK'=11K.......................(1)

Again, the spring constant of the spring system when connected parallel is

1K''=1K1+1K2+1K3=16K+13K+12Kor1K''=1+2+36K=1Kor1K''=1K.....................(2)

Step5: Finding the ratio

Now, the ratio of the two given spring systems is

K'K''=11KK=11From1and2orK':K''=11:1

Therefore, the ratio of spring constant is K':K''=11:1.


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