JEE Exam > JEE Questions > One end of a string of length L is tied to th...
Start Learning for Free
One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.
- a)Wave speed is increasing as it travels from bottom to top.
- b)Acceleration of the wave on the string is uniform.
- c)Time taken by the pulse to reach from bottom to top will be √2L/g .
- d)All of the above
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
One end of a string of length L is tied to the celling of a lift accel...
Ans.
Method to Solve :
This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
One end of a string of length L is tied to the celling of a lift accel...
In order to solve this problem, we need to consider the forces acting on the string.
First, let's consider the tension in the string. At the free end of the string, there is no tension as it is free to move. However, at the other end of the string (tied to the ceiling of the lift), there is a tension force acting downwards. Let's call this tension force T.
Next, let's consider the weight of the string. The weight of the string is given by the formula w = mg, where m is the linear mass density and g is the acceleration due to gravity.
Since the linear mass density of the string varies linearly from 0 to 𝜇, we can express the linear mass density as a function of the length of the string x. Let's call this function 𝜇(x).
The weight of the string at any point x can be calculated as w(x) = 𝜇(x)gx.
Now, let's consider the acceleration of the string. The entire string is accelerating upwards with an acceleration of 3g. This means that the acceleration of any point on the string is also 3g.
Using Newton's second law, we can write the equation of motion for the string as:
T - w(x) = m(x)a
where T is the tension force, w(x) is the weight of the string at point x, m(x) is the linear mass density at point x, and a is the acceleration of the string.
Substituting the values we have:
T - 𝜇(x)gx = m(x)(3g)
Since the linear mass density varies linearly from 0 to 𝜇, we can express it as 𝜇(x) = (𝜇/L)x, where L is the length of the string.
Substituting this into the equation of motion:
T - (𝜇/L)gx = m(x)(3g)
T - (𝜇/L)gx = (𝜇/L)x(3g)
T - (𝜇/L)gx = (𝜇/L)3gx
T = (𝜇/L)4gx
So, the tension in the string is given by T = (𝜇/L)4gx.
First, let's consider the tension in the string. At the free end of the string, there is no tension as it is free to move. However, at the other end of the string (tied to the ceiling of the lift), there is a tension force acting downwards. Let's call this tension force T.
Next, let's consider the weight of the string. The weight of the string is given by the formula w = mg, where m is the linear mass density and g is the acceleration due to gravity.
Since the linear mass density of the string varies linearly from 0 to 𝜇, we can express the linear mass density as a function of the length of the string x. Let's call this function 𝜇(x).
The weight of the string at any point x can be calculated as w(x) = 𝜇(x)gx.
Now, let's consider the acceleration of the string. The entire string is accelerating upwards with an acceleration of 3g. This means that the acceleration of any point on the string is also 3g.
Using Newton's second law, we can write the equation of motion for the string as:
T - w(x) = m(x)a
where T is the tension force, w(x) is the weight of the string at point x, m(x) is the linear mass density at point x, and a is the acceleration of the string.
Substituting the values we have:
T - 𝜇(x)gx = m(x)(3g)
Since the linear mass density varies linearly from 0 to 𝜇, we can express it as 𝜇(x) = (𝜇/L)x, where L is the length of the string.
Substituting this into the equation of motion:
T - (𝜇/L)gx = m(x)(3g)
T - (𝜇/L)gx = (𝜇/L)x(3g)
T - (𝜇/L)gx = (𝜇/L)3gx
T = (𝜇/L)4gx
So, the tension in the string is given by T = (𝜇/L)4gx.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.a)Wave speed is increasing as it travels from bottom to top.b)Acceleration of the wave on the string is uniform.c)Time taken by the pulse to reach from bottom to top will be √2L/g .d)All of the aboveCorrect answer is option 'D'. Can you explain this answer?
Question Description
One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.a)Wave speed is increasing as it travels from bottom to top.b)Acceleration of the wave on the string is uniform.c)Time taken by the pulse to reach from bottom to top will be √2L/g .d)All of the aboveCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.a)Wave speed is increasing as it travels from bottom to top.b)Acceleration of the wave on the string is uniform.c)Time taken by the pulse to reach from bottom to top will be √2L/g .d)All of the aboveCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.a)Wave speed is increasing as it travels from bottom to top.b)Acceleration of the wave on the string is uniform.c)Time taken by the pulse to reach from bottom to top will be √2L/g .d)All of the aboveCorrect answer is option 'D'. Can you explain this answer?.
One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.a)Wave speed is increasing as it travels from bottom to top.b)Acceleration of the wave on the string is uniform.c)Time taken by the pulse to reach from bottom to top will be √2L/g .d)All of the aboveCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.a)Wave speed is increasing as it travels from bottom to top.b)Acceleration of the wave on the string is uniform.c)Time taken by the pulse to reach from bottom to top will be √2L/g .d)All of the aboveCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.a)Wave speed is increasing as it travels from bottom to top.b)Acceleration of the wave on the string is uniform.c)Time taken by the pulse to reach from bottom to top will be √2L/g .d)All of the aboveCorrect answer is option 'D'. Can you explain this answer?.
Solutions for One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.a)Wave speed is increasing as it travels from bottom to top.b)Acceleration of the wave on the string is uniform.c)Time taken by the pulse to reach from bottom to top will be √2L/g .d)All of the aboveCorrect answer is option 'D'. 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 One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.a)Wave speed is increasing as it travels from bottom to top.b)Acceleration of the wave on the string is uniform.c)Time taken by the pulse to reach from bottom to top will be √2L/g .d)All of the aboveCorrect answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.a)Wave speed is increasing as it travels from bottom to top.b)Acceleration of the wave on the string is uniform.c)Time taken by the pulse to reach from bottom to top will be √2L/g .d)All of the aboveCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.a)Wave speed is increasing as it travels from bottom to top.b)Acceleration of the wave on the string is uniform.c)Time taken by the pulse to reach from bottom to top will be √2L/g .d)All of the aboveCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.a)Wave speed is increasing as it travels from bottom to top.b)Acceleration of the wave on the string is uniform.c)Time taken by the pulse to reach from bottom to top will be √2L/g .d)All of the aboveCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice One end of a string of length L is tied to the celling of a lift accelerating upwards with an acceleration 3g. The other end of the string is free. The linear mass density of the string varies linearly from 0 to λ - from bottom to top. Mark out the correct statement for a wave pulse travelling on the string.a)Wave speed is increasing as it travels from bottom to top.b)Acceleration of the wave on the string is uniform.c)Time taken by the pulse to reach from bottom to top will be √2L/g .d)All of the aboveCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
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