JEE Exam > JEE Questions > An uncalibrated spring balance is found to ha...
Start Learning for Free
An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Take π = 3.14
- a)2.45 cm
- b)2.40 cm
- c)2.30 cm
- d)2.10 cm
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
An uncalibrated spring balance is found to have a period of oscillatio...
Here. T = 0.314 s ; m = 1 kg
or
When spring is loaded with a weight 1 kg:
m g = k I
or
or
or
When spring is loaded with a weight 1 kg:
m g = k I
or
or
Most Upvoted Answer
An uncalibrated spring balance is found to have a period of oscillatio...
To find the elongation of the spring, we can use Hooke's Law which states that the force exerted by a spring is directly proportional to the displacement or elongation of the spring from its equilibrium position.
Let's assume the elongation of the spring is x meters.
The restoring force exerted by the spring is given by F = kx, where k is the spring constant.
In this case, the weight of the 1 kg mass is equal to the force exerted by the spring. So we have F = mg, where m is the mass of the weight and g is the acceleration due to gravity.
Therefore, mg = kx.
We know that the period of oscillation, T, is given by T = 2π√(m/k).
Given that T = 0.314 s and m = 1 kg, we can rearrange the equation to solve for k:
T = 2π√(m/k)
0.314 = 2π√(1/k)
0.314/2π = √(1/k)
(0.314/2π)^2 = 1/k
(0.314^2)/(4π^2) = 1/k
0.0313/π^2 = 1/k
k = π^2/0.0313
Now we can substitute this value of k into the equation mg = kx:
1 * 9.8 = (π^2/0.0313) * x
9.8 = (π^2/0.0313) * x
x = (9.8 * 0.0313) / π^2
Using a calculator, we find that x ≈ 0.097 meters.
Therefore, the spring elongates by approximately 0.097 meters when a 1 kg weight is suspended from it.
Let's assume the elongation of the spring is x meters.
The restoring force exerted by the spring is given by F = kx, where k is the spring constant.
In this case, the weight of the 1 kg mass is equal to the force exerted by the spring. So we have F = mg, where m is the mass of the weight and g is the acceleration due to gravity.
Therefore, mg = kx.
We know that the period of oscillation, T, is given by T = 2π√(m/k).
Given that T = 0.314 s and m = 1 kg, we can rearrange the equation to solve for k:
T = 2π√(m/k)
0.314 = 2π√(1/k)
0.314/2π = √(1/k)
(0.314/2π)^2 = 1/k
(0.314^2)/(4π^2) = 1/k
0.0313/π^2 = 1/k
k = π^2/0.0313
Now we can substitute this value of k into the equation mg = kx:
1 * 9.8 = (π^2/0.0313) * x
9.8 = (π^2/0.0313) * x
x = (9.8 * 0.0313) / π^2
Using a calculator, we find that x ≈ 0.097 meters.
Therefore, the spring elongates by approximately 0.097 meters when a 1 kg weight is suspended from it.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Takeπ = 3.14a)2.45 cmb)2.40 cmc)2.30 cmd)2.10 cmCorrect answer is option 'A'. Can you explain this answer?
Question Description
An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Takeπ = 3.14a)2.45 cmb)2.40 cmc)2.30 cmd)2.10 cmCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Takeπ = 3.14a)2.45 cmb)2.40 cmc)2.30 cmd)2.10 cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Takeπ = 3.14a)2.45 cmb)2.40 cmc)2.30 cmd)2.10 cmCorrect answer is option 'A'. Can you explain this answer?.
An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Takeπ = 3.14a)2.45 cmb)2.40 cmc)2.30 cmd)2.10 cmCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Takeπ = 3.14a)2.45 cmb)2.40 cmc)2.30 cmd)2.10 cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Takeπ = 3.14a)2.45 cmb)2.40 cmc)2.30 cmd)2.10 cmCorrect answer is option 'A'. Can you explain this answer?.
Solutions for An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Takeπ = 3.14a)2.45 cmb)2.40 cmc)2.30 cmd)2.10 cmCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Takeπ = 3.14a)2.45 cmb)2.40 cmc)2.30 cmd)2.10 cmCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Takeπ = 3.14a)2.45 cmb)2.40 cmc)2.30 cmd)2.10 cmCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Takeπ = 3.14a)2.45 cmb)2.40 cmc)2.30 cmd)2.10 cmCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Takeπ = 3.14a)2.45 cmb)2.40 cmc)2.30 cmd)2.10 cmCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Takeπ = 3.14a)2.45 cmb)2.40 cmc)2.30 cmd)2.10 cmCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup