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Physics · Newton's Laws of Motion

JEE Main 2026 — 24 January, Morning Shift — Question 34

A spring of force constant 15 N/m15 \mathrm{~N} / \mathrm{m} is cut into two pieces. If the ratio of their length is 1:31: 3, then the force constant of smaller piece is ____\_\_\_\_ N/m\mathrm{N} / \mathrm{m}

  1. Option A:

    15

  2. Option B:

    20

  3. Option C:

    60

    Correct
  4. Option D:

    45

Answer: C

Step-by-step solution

Kℓ=\mathrm{K} \ell= constant Kℓ=K′(ℓ4)\mathrm{K} \ell=\mathrm{K}^{\prime}\left(\frac{\ell}{4}\right) K′=4 K\mathrm{K}^{\prime}=4 \mathrm{~K} K′=60 N/m\mathrm{K}^{\prime}=60 \mathrm{~N} / \mathrm{m}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Newton's Laws of Motion
Topic
Spring Force and Combination of Springs
A spring of force constant 15 N / m is cut into two pieces. If the… | JEE Main 2026 PYQ with Solution · DhiX AI