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होम › भौतिकी › Hooke's Law and Springs

Hooke's Law and Springs — Physics Questions & Practice

354 Hooke's Law and Springs questions from Physics for competitive exams such as UPSC, Banking, SSC and Railway. Read the worked questions below, check every answer with an explanation, and track what you get wrong. का अभ्यास करें

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Hooke's Law and Springs पर अभ्यास प्रश्न — उत्तर सहित

Try each question first, then open the answer. Explanations are included.

1. A spring with a spring constant of 185 N/m is stretched by 0.3 m. What is the restoring force in the spring? (F = kx, Hooke's Law)

  1. A57.5
  2. B43.98
  3. C60.5
  4. D55.5
उत्तर दिखाएँ

उत्तर: D

By Hooke's Law, F = kx = 185 x 0.3 = 55.5 N.

2. A spring with a spring constant of 140 N/m is compressed by 0.15 m. What is the elastic potential energy stored in the spring? (PE = 1/2 kx^2)

  1. A1.575
  2. B1.943
  3. C-0.425
  4. D1.816
उत्तर दिखाएँ

उत्तर: A

Elastic PE = (1/2)kx^2 = 0.5 x 140 x 0.15^2 = 0.5 x 140 x 0.0225 = 1.575 J.

3. A spring with a spring constant of 190 N/m is stretched by 0.1 m. What is the restoring force in the spring? (F = kx, Hooke's Law)

  1. A29
  2. B15.221
  3. C20.072
  4. D19
उत्तर दिखाएँ

उत्तर: D

By Hooke's Law, F = kx = 190 x 0.1 = 19 N.

4. A spring with a spring constant of 125 N/m is stretched by 0.3 m. What is the restoring force in the spring? (F = kx, Hooke's Law)

  1. A37.5
  2. B37.232
  3. C33.262
  4. D36.464
उत्तर दिखाएँ

उत्तर: A

By Hooke's Law, F = kx = 125 x 0.3 = 37.5 N.

5. A spring with a spring constant of 20 N/m is compressed by 0.15 m. What is the elastic potential energy stored in the spring? (PE = 1/2 kx^2)

  1. A-1.775
  2. B10.225
  3. C-9.775
  4. D0.225
उत्तर दिखाएँ

उत्तर: D

Elastic PE = (1/2)kx^2 = 0.5 x 20 x 0.15^2 = 0.5 x 20 x 0.0225 = 0.225 J.

Hooke's Law and Springs और प्रश्न हल करें

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