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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 resulting strain. Young's Modulus of Steel (Y) = 20×1010 Pa

  • a)
    8.0 × 10−4

  • b)
    10.0 × 10−4

  • c)
    7.0 × 10−4

  • d)
    9.0 × 10−4

Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2 . It is ...
from the Hooke’s law :
ε = σ/Y = (1.8 x 10^8 Pa)/(20 x 10^10 Pa) 
= 9.0 x 10^−4
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A steel rod 2.0 m long has a cross-sectional area of 0.30 cm2 . It is ...
To determine the resulting strain, we need to calculate the stress first.

The stress (σ) is given by the formula:

σ = F / A

Where:
σ = stress
F = force applied
A = cross-sectional area

In this case, the force applied is the weight of the milling machine, which is given by:

F = m * g

Where:
m = mass of the milling machine
g = acceleration due to gravity

In this case, the mass of the milling machine is 550 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. Plugging in these values, we can calculate the force:

F = 550 kg * 9.8 m/s^2
F = 5390 N

Now, we can calculate the stress:

σ = 5390 N / 0.30 cm^2

Since the cross-sectional area is given in cm^2, we need to convert it to m^2:

1 cm^2 = 0.0001 m^2

Therefore, the cross-sectional area in m^2 is:

0.30 cm^2 * 0.0001 m^2/cm^2
0.00003 m^2

Plugging in the values, we can calculate the stress:

σ = 5390 N / 0.00003 m^2
σ = 179,666,666.67 N/m^2

Now that we have the stress, we can determine the strain using Young's modulus (Y). The formula for strain (ε) is:

ε = σ / Y

Plugging in the values, we can calculate the strain:

ε = 179,666,666.67 N/m^2 / 20
ε = 8,983,333.33

Therefore, the resulting strain is approximately 8,983,333.33.
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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 resulting strain.Youngs Modulus of Steel (Y) = 20×1010 Paa)8.0 × 10−4b)10.0 × 10−4c)7.0 × 10−4d)9.0 × 10−4Correct 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 resulting strain.Youngs Modulus of Steel (Y) = 20×1010 Paa)8.0 × 10−4b)10.0 × 10−4c)7.0 × 10−4d)9.0 × 10−4Correct 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 resulting strain.Youngs Modulus of Steel (Y) = 20×1010 Paa)8.0 × 10−4b)10.0 × 10−4c)7.0 × 10−4d)9.0 × 10−4Correct 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 resulting strain.Youngs Modulus of Steel (Y) = 20×1010 Paa)8.0 × 10−4b)10.0 × 10−4c)7.0 × 10−4d)9.0 × 10−4Correct answer is option 'D'. Can you explain this answer?.
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