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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.
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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 ?
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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 ?.
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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
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