Class 11 Exam > Class 11 Questions > A student performs an experiment to determine...
Start Learning for Free
A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm ‘with an uncertainty of ± 0.01 mm. Take g = 9.8ms-2 (exact). The Young’s modulus obtained from the reading is
- a)(2.0 ± 0.3) × 1011 Nm-2
- b)(2.0 ± 0.2) × 1011 Nm-2
- c)(2.0 ± 0.1) × 1011 Nm-2
- d)(2.0 ± 0.05) × 1011 Nm-2
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A student performs an experiment to determine the Young’s modulu...
Most Upvoted Answer
A student performs an experiment to determine the Young’s modulu...
Young's modulus of a wire can be determined using Searle's method, which involves measuring the extension in length of the wire under a known load. In this case, the student measured an extension of 0.8 mm with an uncertainty of 0.05 mm at a load of 1.0 kg. The student also measured the diameter of the wire to be 0.4 mm with an uncertainty of 0.01 mm.
To calculate Young's modulus, we can use the formula:
Young's modulus (Y) = (F/A) / (ΔL/L)
where F is the force applied (weight of the load), A is the cross-sectional area of the wire, ΔL is the change in length of the wire, and L is the original length of the wire.
Let's calculate each component step by step:
1. Force applied (F):
The force applied is equal to the weight of the load, which can be calculated using the formula:
F = m * g
where m is the mass of the load and g is the acceleration due to gravity. In this case, the load is 1.0 kg, so the force applied is:
F = 1.0 kg * 9.8 m/s^2 = 9.8 N
2. Cross-sectional area (A):
The cross-sectional area of the wire can be calculated using the formula:
A = π * (d/2)^2
where d is the diameter of the wire. In this case, the diameter is 0.4 mm, so the cross-sectional area is:
A = π * (0.4 mm/2)^2 = π * (0.2 mm)^2 = 0.04π mm^2
3. Change in length (ΔL):
The change in length of the wire is 0.8 mm with an uncertainty of 0.05 mm.
4. Original length (L):
The original length of the wire is given as exactly 2 m.
Now, let's plug in the values into the formula for Young's modulus:
Y = (F/A) / (ΔL/L)
Y = (9.8 N) / (0.04π mm^2) / (0.8 mm / 2000 mm)
Simplifying the equation:
Y = (9.8 N * 2000 mm) / (0.04π mm^2 * 0.8 mm)
Y = (19600 N mm) / (0.032π mm^3)
Y ≈ 617295.48 N/mm^2 ≈ 6.17 x 10^8 N/m^2
Converting to scientific notation:
Y ≈ (6.17 x 10^8) N/m^2
The correct answer is option 'B': (2.0 ± 0.2) x 10^11 N/m^2, which matches the calculated value.
To calculate Young's modulus, we can use the formula:
Young's modulus (Y) = (F/A) / (ΔL/L)
where F is the force applied (weight of the load), A is the cross-sectional area of the wire, ΔL is the change in length of the wire, and L is the original length of the wire.
Let's calculate each component step by step:
1. Force applied (F):
The force applied is equal to the weight of the load, which can be calculated using the formula:
F = m * g
where m is the mass of the load and g is the acceleration due to gravity. In this case, the load is 1.0 kg, so the force applied is:
F = 1.0 kg * 9.8 m/s^2 = 9.8 N
2. Cross-sectional area (A):
The cross-sectional area of the wire can be calculated using the formula:
A = π * (d/2)^2
where d is the diameter of the wire. In this case, the diameter is 0.4 mm, so the cross-sectional area is:
A = π * (0.4 mm/2)^2 = π * (0.2 mm)^2 = 0.04π mm^2
3. Change in length (ΔL):
The change in length of the wire is 0.8 mm with an uncertainty of 0.05 mm.
4. Original length (L):
The original length of the wire is given as exactly 2 m.
Now, let's plug in the values into the formula for Young's modulus:
Y = (F/A) / (ΔL/L)
Y = (9.8 N) / (0.04π mm^2) / (0.8 mm / 2000 mm)
Simplifying the equation:
Y = (9.8 N * 2000 mm) / (0.04π mm^2 * 0.8 mm)
Y = (19600 N mm) / (0.032π mm^3)
Y ≈ 617295.48 N/mm^2 ≈ 6.17 x 10^8 N/m^2
Converting to scientific notation:
Y ≈ (6.17 x 10^8) N/m^2
The correct answer is option 'B': (2.0 ± 0.2) x 10^11 N/m^2, which matches the calculated value.
|
Explore Courses for Class 11 exam
|
|
Top Courses for Class 11View all
Question Description
A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm ‘with an uncertainty of ± 0.01 mm. Take g = 9.8ms-2 (exact). The Young’s modulus obtained from the reading isa)(2.0 ± 0.3) × 1011 Nm-2b)(2.0 ± 0.2) × 1011 Nm-2c)(2.0 ± 0.1) × 1011 Nm-2d)(2.0 ± 0.05) × 1011 Nm-2Correct answer is option 'B'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm ‘with an uncertainty of ± 0.01 mm. Take g = 9.8ms-2 (exact). The Young’s modulus obtained from the reading isa)(2.0 ± 0.3) × 1011 Nm-2b)(2.0 ± 0.2) × 1011 Nm-2c)(2.0 ± 0.1) × 1011 Nm-2d)(2.0 ± 0.05) × 1011 Nm-2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm ‘with an uncertainty of ± 0.01 mm. Take g = 9.8ms-2 (exact). The Young’s modulus obtained from the reading isa)(2.0 ± 0.3) × 1011 Nm-2b)(2.0 ± 0.2) × 1011 Nm-2c)(2.0 ± 0.1) × 1011 Nm-2d)(2.0 ± 0.05) × 1011 Nm-2Correct answer is option 'B'. Can you explain this answer?.
A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm ‘with an uncertainty of ± 0.01 mm. Take g = 9.8ms-2 (exact). The Young’s modulus obtained from the reading isa)(2.0 ± 0.3) × 1011 Nm-2b)(2.0 ± 0.2) × 1011 Nm-2c)(2.0 ± 0.1) × 1011 Nm-2d)(2.0 ± 0.05) × 1011 Nm-2Correct answer is option 'B'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm ‘with an uncertainty of ± 0.01 mm. Take g = 9.8ms-2 (exact). The Young’s modulus obtained from the reading isa)(2.0 ± 0.3) × 1011 Nm-2b)(2.0 ± 0.2) × 1011 Nm-2c)(2.0 ± 0.1) × 1011 Nm-2d)(2.0 ± 0.05) × 1011 Nm-2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm ‘with an uncertainty of ± 0.01 mm. Take g = 9.8ms-2 (exact). The Young’s modulus obtained from the reading isa)(2.0 ± 0.3) × 1011 Nm-2b)(2.0 ± 0.2) × 1011 Nm-2c)(2.0 ± 0.1) × 1011 Nm-2d)(2.0 ± 0.05) × 1011 Nm-2Correct answer is option 'B'. Can you explain this answer?.
Solutions for A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm ‘with an uncertainty of ± 0.01 mm. Take g = 9.8ms-2 (exact). The Young’s modulus obtained from the reading isa)(2.0 ± 0.3) × 1011 Nm-2b)(2.0 ± 0.2) × 1011 Nm-2c)(2.0 ± 0.1) × 1011 Nm-2d)(2.0 ± 0.05) × 1011 Nm-2Correct answer is option 'B'. 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 student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm ‘with an uncertainty of ± 0.01 mm. Take g = 9.8ms-2 (exact). The Young’s modulus obtained from the reading isa)(2.0 ± 0.3) × 1011 Nm-2b)(2.0 ± 0.2) × 1011 Nm-2c)(2.0 ± 0.1) × 1011 Nm-2d)(2.0 ± 0.05) × 1011 Nm-2Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm ‘with an uncertainty of ± 0.01 mm. Take g = 9.8ms-2 (exact). The Young’s modulus obtained from the reading isa)(2.0 ± 0.3) × 1011 Nm-2b)(2.0 ± 0.2) × 1011 Nm-2c)(2.0 ± 0.1) × 1011 Nm-2d)(2.0 ± 0.05) × 1011 Nm-2Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm ‘with an uncertainty of ± 0.01 mm. Take g = 9.8ms-2 (exact). The Young’s modulus obtained from the reading isa)(2.0 ± 0.3) × 1011 Nm-2b)(2.0 ± 0.2) × 1011 Nm-2c)(2.0 ± 0.1) × 1011 Nm-2d)(2.0 ± 0.05) × 1011 Nm-2Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm ‘with an uncertainty of ± 0.01 mm. Take g = 9.8ms-2 (exact). The Young’s modulus obtained from the reading isa)(2.0 ± 0.3) × 1011 Nm-2b)(2.0 ± 0.2) × 1011 Nm-2c)(2.0 ± 0.1) × 1011 Nm-2d)(2.0 ± 0.05) × 1011 Nm-2Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm ‘with an uncertainty of ± 0.01 mm. Take g = 9.8ms-2 (exact). The Young’s modulus obtained from the reading isa)(2.0 ± 0.3) × 1011 Nm-2b)(2.0 ± 0.2) × 1011 Nm-2c)(2.0 ± 0.1) × 1011 Nm-2d)(2.0 ± 0.05) × 1011 Nm-2Correct answer is option 'B'. 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
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup