Class 11 Exam > Class 11 Questions > A 14.5 kg mass, fastened to the end of a stee...
Start Learning for Free
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.
- a)1.339 × 10−4 m
- b)1.439 × 10−4 m
- c)1.539 × 10−4 m
- d)1.239 × 10−4 m
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A 14.5 kg mass, fastened to the end of a steel wire of unstretched len...
Mass, m = 14.5 kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s
Cross-sectional area of the wire, a = 0.065 cm2 = 0.065 × 10-4 m2
Let Δl be the elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is:
F = mg + mlω2
= 14.5 × 9.8 + 14.5 × 1 × (12.56)2
= 2429.53 N
Young’s modulus = Strss / Strain
Y = (F/A) / (∆l/l)
∴ ∆l = Fl / AY
Young’s modulus for steel = 2 × 1011 Pa
∆l = 2429.53 × 1 / (0.065 × 10-4 × 2 × 1011) = 1.87 × 10-3 m
Hence, the elongation of the wire is 1.87 × 10–3 m
Hence 1.87 × 10–3 m
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s
Cross-sectional area of the wire, a = 0.065 cm2 = 0.065 × 10-4 m2
Let Δl be the elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is:
F = mg + mlω2
= 14.5 × 9.8 + 14.5 × 1 × (12.56)2
= 2429.53 N
Young’s modulus = Strss / Strain
Y = (F/A) / (∆l/l)
∴ ∆l = Fl / AY
Young’s modulus for steel = 2 × 1011 Pa
∆l = 2429.53 × 1 / (0.065 × 10-4 × 2 × 1011) = 1.87 × 10-3 m
Hence, the elongation of the wire is 1.87 × 10–3 m
Hence 1.87 × 10–3 m
Most Upvoted Answer
A 14.5 kg mass, fastened to the end of a steel wire of unstretched len...
Mass, m = 14.5 kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s
Cross-sectional area of the wire, a = 0.065 cm2 = 0.065 × 10-4 m2
Let Δl be the elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is:
F = mg + mlω2
= 14.5 × 9.8 + 14.5 × 1 × (12.56)2
= 2429.53 N
Young’s modulus = Strss / Strain
Y = (F/A) / (∆l/l)
∴ ∆l = Fl / AY
Young’s modulus for steel = 2 × 1011 Pa
∆l = 2429.53 × 1 / (0.065 × 10-4 × 2 × 1011) = 1.87 × 10-3 m
Hence, the elongation of the wire is 1.87 × 10–3 m.
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s
Cross-sectional area of the wire, a = 0.065 cm2 = 0.065 × 10-4 m2
Let Δl be the elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is:
F = mg + mlω2
= 14.5 × 9.8 + 14.5 × 1 × (12.56)2
= 2429.53 N
Young’s modulus = Strss / Strain
Y = (F/A) / (∆l/l)
∴ ∆l = Fl / AY
Young’s modulus for steel = 2 × 1011 Pa
∆l = 2429.53 × 1 / (0.065 × 10-4 × 2 × 1011) = 1.87 × 10-3 m
Hence, the elongation of the wire is 1.87 × 10–3 m.
Free Test
FREE
| Start Free Test |
Community Answer
A 14.5 kg mass, fastened to the end of a steel wire of unstretched len...
To solve this problem, we can use the formula for elongation of a wire under tension:
Elongation = (F * L) / (A * Y)
Where:
- F is the force applied on the wire
- L is the length of the wire
- A is the cross-sectional area of the wire
- Y is the Young's modulus of the material
First, let's find the force applied on the wire when the mass is at the lowest point of its path. At the lowest point, the tension in the wire is equal to the weight of the mass:
F = m * g
Where:
- m is the mass
- g is the acceleration due to gravity
F = 14.5 kg * 9.8 m/s^2
F = 142.1 N
Next, let's find the length of the wire when the mass is at the lowest point of its path. The unstretched length of the wire is given as 1.0 m, and the elongation is equal to the radius of the circular path:
L = R
Next, let's find the cross-sectional area of the wire. The given cross-sectional area is in cm^2, so we need to convert it to m^2:
A = 0.065 cm^2 * (1 m / 100 cm)^2
A = 0.065 * 10^-4 m^2
A = 6.5 * 10^-6 m^2
Finally, let's find the Young's modulus of steel. The Young's modulus of steel is typically around 200 GPa (2 * 10^11 Pa):
Y = 2 * 10^11 Pa
Now we can calculate the elongation of the wire:
Elongation = (F * L) / (A * Y)
Elongation = (142.1 N * 1.0 m) / (6.5 * 10^-6 m^2 * 2 * 10^11 Pa)
Elongation = 2.184 * 10^-4 m
Elongation = 0.2184 mm
Therefore, the elongation of the wire when the mass is at the lowest point of its path is approximately 0.2184 mm.
Elongation = (F * L) / (A * Y)
Where:
- F is the force applied on the wire
- L is the length of the wire
- A is the cross-sectional area of the wire
- Y is the Young's modulus of the material
First, let's find the force applied on the wire when the mass is at the lowest point of its path. At the lowest point, the tension in the wire is equal to the weight of the mass:
F = m * g
Where:
- m is the mass
- g is the acceleration due to gravity
F = 14.5 kg * 9.8 m/s^2
F = 142.1 N
Next, let's find the length of the wire when the mass is at the lowest point of its path. The unstretched length of the wire is given as 1.0 m, and the elongation is equal to the radius of the circular path:
L = R
Next, let's find the cross-sectional area of the wire. The given cross-sectional area is in cm^2, so we need to convert it to m^2:
A = 0.065 cm^2 * (1 m / 100 cm)^2
A = 0.065 * 10^-4 m^2
A = 6.5 * 10^-6 m^2
Finally, let's find the Young's modulus of steel. The Young's modulus of steel is typically around 200 GPa (2 * 10^11 Pa):
Y = 2 * 10^11 Pa
Now we can calculate the elongation of the wire:
Elongation = (F * L) / (A * Y)
Elongation = (142.1 N * 1.0 m) / (6.5 * 10^-6 m^2 * 2 * 10^11 Pa)
Elongation = 2.184 * 10^-4 m
Elongation = 0.2184 mm
Therefore, the elongation of the wire when the mass is at the lowest point of its path is approximately 0.2184 mm.
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Top Courses for Class 11View all
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.a)1.339 × 10−4 mb)1.439 × 10−4 mc)1.539 × 10−4 md)1.239 × 10−4 mCorrect answer is option 'C'. Can you explain this answer?
Question Description
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.a)1.339 × 10−4 mb)1.439 × 10−4 mc)1.539 × 10−4 md)1.239 × 10−4 mCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.a)1.339 × 10−4 mb)1.439 × 10−4 mc)1.539 × 10−4 md)1.239 × 10−4 mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.a)1.339 × 10−4 mb)1.439 × 10−4 mc)1.539 × 10−4 md)1.239 × 10−4 mCorrect answer is option 'C'. Can you explain this answer?.
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.a)1.339 × 10−4 mb)1.439 × 10−4 mc)1.539 × 10−4 md)1.239 × 10−4 mCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.a)1.339 × 10−4 mb)1.439 × 10−4 mc)1.539 × 10−4 md)1.239 × 10−4 mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.a)1.339 × 10−4 mb)1.439 × 10−4 mc)1.539 × 10−4 md)1.239 × 10−4 mCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.a)1.339 × 10−4 mb)1.439 × 10−4 mc)1.539 × 10−4 md)1.239 × 10−4 mCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.a)1.339 × 10−4 mb)1.439 × 10−4 mc)1.539 × 10−4 md)1.239 × 10−4 mCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.a)1.339 × 10−4 mb)1.439 × 10−4 mc)1.539 × 10−4 md)1.239 × 10−4 mCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.a)1.339 × 10−4 mb)1.439 × 10−4 mc)1.539 × 10−4 md)1.239 × 10−4 mCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.a)1.339 × 10−4 mb)1.439 × 10−4 mc)1.539 × 10−4 md)1.239 × 10−4 mCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.a)1.339 × 10−4 mb)1.439 × 10−4 mc)1.539 × 10−4 md)1.239 × 10−4 mCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 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
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup