Q19 / 214 · NTA Abhyas (NEET) 2020 · Physics · Oscillations

A uniform spring has an unstretched length L and a force constant k. The spring is cut into two parts of unstretched lengths l1 and l2 such that l2=ηl2, where η is an integer. The corresponding force constants are k1 and k2, then

  • A
    force has to be kept same to find k1 and k2
  • B
    k1=kη+1η and k2=kη-1
  • C
    k1=kη-1η and k2=kη+1
  • Correct answer
    D
    k1=kη+1η and k2=kη+1

Solution

In case of cutting of a spring, l×k=constant

k1l1=k2l2

k1ηl2=k2l2       ⇒ηk1=k2                        ...............(1)

Now both the springs when connected in series should give the same effect as that of the original spring. Hence, by applying the concept of springs connected in series, we obtain,

1k=1k1+1k2      ...............(2)

1k=1k1+1ηk1      ⇒k1k=η+1η            (using (1))

k1=η+1ηk             ............(3)

Using (3) again in (2), we obtain,

k2=η+1k

Hence, the correct answer is option: 'D'.

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