AP EAMCET202316 May 2023Morning ShiftPhysicsMechanical Properties of SolidsActual
A steel rod of radius 20 ~mm and length of 2 ~m is acted upon by a force of 400 kN along the length. The values of stress and strain are respectively (Y_ steel =2 10¹¹ Nm ⁻² )
Options
- A1.96 10^8 Nm ⁻², 0.16 %
- B3.18 10^8 Nm ⁻², 0.16 %
- C3.18 10^8 Nm ⁻², 0.32 %
- D4 10^8 Nm ⁻², 0.2 %
Correct answer
B. 3.18 10^8 Nm ⁻², 0.16 %
Step-by-step solution
Radius of rod, r =20 ~mm =20 10⁻³ ~m Length of rod, L=2 ~m Force, F =400 ~K ~N =400 10^3 ~N Y _ steel =2 10¹¹ ~N ~m ⁻² Stress = F A = F r^2 = 400 10^3 3.14 (20 10⁻³ )^2 =3.18 10^8 ~N ~m ⁻² Young modulus, Y= stress strain Strain = Stress Y = 3.18 10^8 2 10¹¹ =0.16 %