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A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2. It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the elongation. Take Young's modulus of steel as 20 × 1010 Pa
- a)2.0 mm
- b)1.6 mm
- c)2.2 mm
- d)1.8 mm
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2. It is h...
σ=Stress and ε=strain
σ=F/A= (550kg) × (9.81m/s2)3×10-5m2/=0.18GPA
ε=Δl/l0=σ/Υ=0.18×109/200×109=9×10-4
Δl=εl0= (9×10-4) (2m) = 0.0018m=1.8mm
σ=F/A= (550kg) × (9.81m/s2)3×10-5m2/=0.18GPA
ε=Δl/l0=σ/Υ=0.18×109/200×109=9×10-4
Δl=εl0= (9×10-4) (2m) = 0.0018m=1.8mm
Most Upvoted Answer
A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2. It is h...
Young's modulus= stress/strain
stress= F/A=mg/A
=550Kg×10m/s2÷3×10^-5m2
=18.3×10^7N
strain=X/L=X/2m
Young's modulus=20×10^10Pa
let Y~ Young's modulus
Y=18.3×10^7N÷(X÷2m)
20×10^10Pa=18.3×10^7N÷(X÷2m)
X=2m×18.3×10^7N÷20×10^10Pa
X=1.83×10^-3m=1.83mm~1.8mm
stress= F/A=mg/A
=550Kg×10m/s2÷3×10^-5m2
=18.3×10^7N
strain=X/L=X/2m
Young's modulus=20×10^10Pa
let Y~ Young's modulus
Y=18.3×10^7N÷(X÷2m)
20×10^10Pa=18.3×10^7N÷(X÷2m)
X=2m×18.3×10^7N÷20×10^10Pa
X=1.83×10^-3m=1.83mm~1.8mm
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Community Answer
A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2. It is h...
To determine the elongation of the steel rod, we can use Hooke's Law, which states that the elongation of an object is directly proportional to the force applied to it and inversely proportional to its cross-sectional area and Young's modulus.
The formula for elongation (ΔL) is given by:
ΔL = (F × L) / (A × Y)
Where:
ΔL = elongation
F = force applied
L = original length of the rod
A = cross-sectional area of the rod
Y = Young's modulus
Given data:
Length of the rod (L) = 2.0 m
Cross-sectional area (A) = 0.30 cm^2 = 0.30 × 10^-4 m^2
Force applied (F) = weight of the milling machine = mass × acceleration due to gravity = 550 kg × 9.8 m/s^2 = 5390 N
Young's modulus (Y) = 20 × 10^10 Pa
Substituting the values into the formula, we have:
ΔL = (5390 N × 2.0 m) / (0.30 × 10^-4 m^2 × 20 × 10^10 Pa)
Simplifying the expression:
ΔL = (10780 N·m) / (6 × 10^-6 N·m^2)
ΔL = 1.796 × 10^6 m^-1
Converting the elongation from meters per meter to millimeters per meter:
ΔL = 1.796 × 10^6 mm/m
ΔL = 1.796 mm
Therefore, the elongation of the steel rod is 1.8 mm (rounded to one decimal place).
Hence, option D (1.8 mm) is the correct answer.
The formula for elongation (ΔL) is given by:
ΔL = (F × L) / (A × Y)
Where:
ΔL = elongation
F = force applied
L = original length of the rod
A = cross-sectional area of the rod
Y = Young's modulus
Given data:
Length of the rod (L) = 2.0 m
Cross-sectional area (A) = 0.30 cm^2 = 0.30 × 10^-4 m^2
Force applied (F) = weight of the milling machine = mass × acceleration due to gravity = 550 kg × 9.8 m/s^2 = 5390 N
Young's modulus (Y) = 20 × 10^10 Pa
Substituting the values into the formula, we have:
ΔL = (5390 N × 2.0 m) / (0.30 × 10^-4 m^2 × 20 × 10^10 Pa)
Simplifying the expression:
ΔL = (10780 N·m) / (6 × 10^-6 N·m^2)
ΔL = 1.796 × 10^6 m^-1
Converting the elongation from meters per meter to millimeters per meter:
ΔL = 1.796 × 10^6 mm/m
ΔL = 1.796 mm
Therefore, the elongation of the steel rod is 1.8 mm (rounded to one decimal place).
Hence, option D (1.8 mm) is the correct answer.
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A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2. It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the elongation. Take Young's modulus of steel as 20 × 1010 Paa)2.0 mmb)1.6 mmc)2.2 mmd)1.8 mmCorrect answer is option 'D'. Can you explain this answer?
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A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2. It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the elongation. Take Young's modulus of steel as 20 × 1010 Paa)2.0 mmb)1.6 mmc)2.2 mmd)1.8 mmCorrect answer is option 'D'. 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 steel rod 2.0 m long has a cross-sectional area of 0.30 cm2. It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the elongation. Take Young's modulus of steel as 20 × 1010 Paa)2.0 mmb)1.6 mmc)2.2 mmd)1.8 mmCorrect answer is option 'D'. 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 steel rod 2.0 m long has a cross-sectional area of 0.30 cm2. It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the elongation. Take Young's modulus of steel as 20 × 1010 Paa)2.0 mmb)1.6 mmc)2.2 mmd)1.8 mmCorrect answer is option 'D'. Can you explain this answer?.
A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2. It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the elongation. Take Young's modulus of steel as 20 × 1010 Paa)2.0 mmb)1.6 mmc)2.2 mmd)1.8 mmCorrect answer is option 'D'. 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 steel rod 2.0 m long has a cross-sectional area of 0.30 cm2. It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the elongation. Take Young's modulus of steel as 20 × 1010 Paa)2.0 mmb)1.6 mmc)2.2 mmd)1.8 mmCorrect answer is option 'D'. 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 steel rod 2.0 m long has a cross-sectional area of 0.30 cm2. It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the elongation. Take Young's modulus of steel as 20 × 1010 Paa)2.0 mmb)1.6 mmc)2.2 mmd)1.8 mmCorrect answer is option 'D'. Can you explain this answer?.
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