Free Young's Modulus and Stress-Strain MCQs with Answers

1. if both the length and radius of the rod are doubled, then modulus of elasticity will:

• A. inerease
• B. Deerease
• C. Remains the same
• D. Be doubled

Explanation: When the length ( L ) and radius ( r ) of a rod are doubled, the new length and radius are ( 2L ) and ( 2r ) respectively. The modulus of elasticity ( E ) is defined as the ratio of stress to strain, and is given by: E= Stress/StrainBy solving it, we find that modulus of elasticity become half.But it is an intrinsic property of a material, meaning it does not change with the dimensions of the material. Therefore, if both the length and radius of the rod are doubled, the modulus of elasticity will remain unchanged.

2. A force of 500 N is applied to one end of a cylindrical steel rod of diameter 50cm, the tensile stress is;

• A. 1.5 x 10^5Nm- 2
• B. 2.5 x 10^5Nm- 2
• C. 2.5 x 10^3Nm- 2
• D. 1 x 10^5Nm- 2

Explanation: We use the formula: σ= F/A Given:Force (F) = 500 NDiameter of the rod = 50 cm = 0.5 mThe cross-sectional area of a cylinder is calculated using the formula for the area of a circle ( A = πr2 ), where ( r ) is the radius of the circle.The radius is half of the diameter, so:r =0.5m/2 =0.25mNow, we can calculate the area:A = π(0.25)2 = π(0.0625 = 0.19635m² (rounded to 5 decimal places)Finally, we can calculate the tensile stress: σ = 500/0.19635 = 2546.48 Nm-2So, the tensile stress is approximately 2546.48 or 2.5×103Nm- 2.

3. Which of the following is the Young Modulus of steel?

• A. 20 x 10^9N/m2
• B. 3.9 x 10^-9N/m2
• C. 2 x 10^11N/m2
• D. 1.5 x 10^9N/m2

Explanation: The young modulus of steel is 20×1010 Pa which is 20×109N/m2.

4. A wire is stretched to four times of its length its strain is

• A. 0.5
• B. 4
• C. 3
• D. 1

Explanation: As, Strain= ΔL/Lwhere, ΔL=4L-L=3LSo, the strain is: Strain= ΔL/L =3L/L =3So, the strain is 3. This is a unitless quantity as it is a ratio of lengths.

5. If a material becomes more resistant to shear deformation, then shear modulus will

• A. increase.
• B. decrease.
• C. become zero.
• D. remain constant.

Explanation: A higher shear modulus indicates that a material is more resistant to shear deformation. Conversely, a material that is easily deformed under shear stress will have a lower shear modulus.Conclusion:Therefore, if a material becomes more resistant to shear deformation, its shear modulus will increase.

6. The dimension of Young's Modulus is:

• A. M2L-1T2
• B. ML-1T-1
• C. ML-1T-2
• D. ML-2T-2

Explanation: Young's modulus is the slope of the initial section of the curve (i.e. m in y = mx + b). When a material reaches a certain stress, the material will begin to deform. It is up to a point where the structure of the material is stretching and not deforming. Unit of Young's Modulus =N/m2=kgm/s2m2=ms2kg=ML−1T−2.

7. Volume stress divided by volume strain equal to:

• A. Young's modulus
• B. Bulk modulus
• C. Shear modulus
• D. Hyper modulus

Explanation: Bulk modulus is the ratio or hydrostatic stress to the volumetric strain. Within the elastic range is called bulk modulus. It is denoted by K or B. Sometimes referred to as the incompressibility, the bulk modulus is a measure of the ability of a substance to withstand changes in volume when under compression on all sides. It is equal to the quotient of the applied pressure divided by the relative deformation.