Solved Numerical Problem No. 17.1 ( Unit-17: Physics of Solids); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

A 1.25 cm diameter cylinder is subjected to a load of 2500 kg. Calculate the stress on the bar in mega Pascal. (Ans: 200 MPa) 

Given Data:

Diameter of the cylinder = D =  1.25 cm = 0.0125 m

Radius of the cylinder = r = `\frac {D}{2}` = `\frac {0.0125 m }{2}` = 0.00625 m

Mass of load = m = 2500 kg

value of gravitation acceleration = g = 9.81 m s⁻²

To Find:

Tensile Stress = σ = ?

Solution:

Tensile stress is defined as:

Tensile Stress = σ = `\frac {F}{A}`

where [ A = ㄫ r²]  and F = Weight = mg So

σ = `\frac {mg}{ㄫ r²}`

by putting values

σ = `\frac {2500 kgｘ9.81 m s⁻²}{3.1416 ｘ(0.00625 m)²}`

σ = `\frac {24525 kg m s⁻²}{3.1416 ｘ0.0000390625 m²}`

σ = `\frac {24525 kg m s⁻²}{0.00012271875 m²}`

σ = 199847211.611 N m⁻²

or

 σ = 199.8 ｘ 10⁶ Pa 

or

σ = 200 MPa-----------Ans.

Similar Questions:

A cylindrical steel rod 0.50 m long and 1 cm in radius is subjected to a tensile force of 1 ｘ 10⁴ N. (a) What is the tensile stress? (b) What is tensile strain? (c) By what amount does the rod stretch? Answer ((a) 0.31 ｘ 10⁸ N m⁻² (b) 1.6 ｘ 10⁻⁴ (c) 0.8 ｘ 10⁻⁴ m)

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