Two springs with spring constants $k_{1} = 1200 N / m$ and $k_{2} = 500 N / m$ are stretched by the same force. The ratio of potential energy in the springs will be

Option 1 - <math> <mfrac> <mrow> <mrow> <mn>25</mn> </mrow> </mrow> <mrow> <mrow> <mn>144</mn> </mrow> </mrow> </mfrac> </math>

Option 2 - <math> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>Q</mi> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> </mfrac> </math>

Option 3 - <math> <mfrac> <mrow> <mrow> <mi>Q</mi> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>k</mi> </mrow> </mrow> </mfrac> </math>

Option 4 - <math> <mfrac> <mrow> <mrow> <mn>3</mn> <mi>Q</mi> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>k</mi> </mrow> </mrow> </mfrac> </math>

1 Views|Posted 4 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Alok Kumar Singh | Contributor-Level 10

4 months ago

Correct Option - 3

Detailed Solution:

Elastic potential energy stored in the spring

$U = \frac{1}{2} k x^{2}$

$= \frac{1}{2} F x = \frac{F^{2}}{2 k}$

$H e n c e \frac{U_{1}}{U_{2}} = \frac{k_{2}}{k_{1}} [F i s s a m e f o r b o t h s p r i n g s]$

$= \frac{500}{1200} = \frac{5}{12}$

Upvote

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

1.2K

Exams

6.9L

Reviews

1.8M

Answers

Learn more about...

Physics Work, Energy and Power 2021

View Exam Details

4 Answered Questions

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

See what others like you are asking & answering

Two springs with spring constants k 1 = 1200 N / m and k 2 = 500 N / m are stretched by the same force. The ratio of potential energy in the springs will be

Similar Questions for you