Class 11 Exam > Class 11 Questions > A uniform steel rod of length 1m and cross se...
Start Learning for Free
A uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ?
Most Upvoted Answer
A uniform steel rod of length 1m and cross sectional 20cm^2 is hanging...
Whats the unit for the ans.Is the increase in length in cm .?
power exactly doesnt match but this is the ans. the power may match after u check the unit for ans
now we know , F = YAl/L = mg
YAl/L= density × area ×g /2
we divide by 2 bcoz the rod is hanging and its COM lies at centre
so putting the values l =1.923 × 10^-2 metres .
hope it helps
power exactly doesnt match but this is the ans. the power may match after u check the unit for ans
now we know , F = YAl/L = mg
YAl/L= density × area ×g /2
we divide by 2 bcoz the rod is hanging and its COM lies at centre
so putting the values l =1.923 × 10^-2 metres .
hope it helps
Community Answer
A uniform steel rod of length 1m and cross sectional 20cm^2 is hanging...
Problem:
A uniform steel rod of length 1m and cross-sectional area 20cm^2 is hanging from a fixed support. Find the increase in the length of the rod. (Given: Young's modulus Y = 2×10^11 N/m^2, density = 7.85 × 10^3 kg/m^3)
Solution:
Step 1: Identify the given information
We are given:
- Length of the rod (L) = 1m
- Cross-sectional area of the rod (A) = 20cm^2
- Young's modulus (Y) = 2×10^11 N/m^2
- Density of steel (ρ) = 7.85 × 10^3 kg/m^3
Step 2: Calculate the volume and mass of the rod
The volume of the rod can be calculated using the formula:
Volume = Length × Cross-sectional area
Converting the cross-sectional area to square meters:
A = 20cm^2 = 20 × 10^-4 m^2
Volume = 1m × 20 × 10^-4 m^2 = 2 × 10^-4 m^3
The mass of the rod can be calculated using the formula:
Mass = Density × Volume
Mass = 7.85 × 10^3 kg/m^3 × 2 × 10^-4 m^3 = 1.57 kg
Step 3: Calculate the force acting on the rod
The force acting on the rod is the weight of the rod, given by:
Force = Mass × Acceleration due to gravity
Using the value of acceleration due to gravity (g) as 9.8 m/s^2:
Force = 1.57 kg × 9.8 m/s^2 = 15.386 N
Step 4: Calculate the stress on the rod
The stress on the rod can be calculated using the formula:
Stress = Force / Area
Converting the cross-sectional area to square meters:
Area = 20 × 10^-4 m^2
Stress = 15.386 N / 20 × 10^-4 m^2 = 7.693 × 10^4 N/m^2
Step 5: Calculate the strain on the rod
The strain on the rod can be calculated using the formula:
Strain = Stress / Young's modulus
Strain = 7.693 × 10^4 N/m^2 / 2 × 10^11 N/m^2 = 3.8465 × 10^-7
Step 6: Calculate the increase in length
The increase in length can be calculated using the formula:
Increase in length = Original length × Strain
Increase in length = 1m × 3.8465 × 10^-7 = 3.8465 × 10^-7 m
Step 7: Answer
The increase in length of the rod is 3.8465 × 10^-7 m, which can be approximated to 1.923 × 10^-5 m or 0.01923 mm.
A uniform steel rod of length 1m and cross-sectional area 20cm^2 is hanging from a fixed support. Find the increase in the length of the rod. (Given: Young's modulus Y = 2×10^11 N/m^2, density = 7.85 × 10^3 kg/m^3)
Solution:
Step 1: Identify the given information
We are given:
- Length of the rod (L) = 1m
- Cross-sectional area of the rod (A) = 20cm^2
- Young's modulus (Y) = 2×10^11 N/m^2
- Density of steel (ρ) = 7.85 × 10^3 kg/m^3
Step 2: Calculate the volume and mass of the rod
The volume of the rod can be calculated using the formula:
Volume = Length × Cross-sectional area
Converting the cross-sectional area to square meters:
A = 20cm^2 = 20 × 10^-4 m^2
Volume = 1m × 20 × 10^-4 m^2 = 2 × 10^-4 m^3
The mass of the rod can be calculated using the formula:
Mass = Density × Volume
Mass = 7.85 × 10^3 kg/m^3 × 2 × 10^-4 m^3 = 1.57 kg
Step 3: Calculate the force acting on the rod
The force acting on the rod is the weight of the rod, given by:
Force = Mass × Acceleration due to gravity
Using the value of acceleration due to gravity (g) as 9.8 m/s^2:
Force = 1.57 kg × 9.8 m/s^2 = 15.386 N
Step 4: Calculate the stress on the rod
The stress on the rod can be calculated using the formula:
Stress = Force / Area
Converting the cross-sectional area to square meters:
Area = 20 × 10^-4 m^2
Stress = 15.386 N / 20 × 10^-4 m^2 = 7.693 × 10^4 N/m^2
Step 5: Calculate the strain on the rod
The strain on the rod can be calculated using the formula:
Strain = Stress / Young's modulus
Strain = 7.693 × 10^4 N/m^2 / 2 × 10^11 N/m^2 = 3.8465 × 10^-7
Step 6: Calculate the increase in length
The increase in length can be calculated using the formula:
Increase in length = Original length × Strain
Increase in length = 1m × 3.8465 × 10^-7 = 3.8465 × 10^-7 m
Step 7: Answer
The increase in length of the rod is 3.8465 × 10^-7 m, which can be approximated to 1.923 × 10^-5 m or 0.01923 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
|
|
Similar Class 11 Doubts
Top Courses for Class 11View all
A uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ?
Question Description
A uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ? 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 uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ?.
A uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ? 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 uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ?.
Solutions for A uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ? 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 uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ? defined & explained in the simplest way possible. Besides giving the explanation of
A uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ?, a detailed solution for A uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ? has been provided alongside types of A uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ? theory, EduRev gives you an
ample number of questions to practice A uniform steel rod of length 1m and cross sectional 20cm^2 is hanging from a fixed support . find the increase in the length of the rod ?(Y= 2×10^11 N m ^-2, density = 7.85 ×10^8 kg m^-3) . The answer for this is 1.923×10^-5. Can u explain this ? 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