- Chemical Engineering Basics - Section 1
- Chemical Engineering Basics - Section 2
- Chemical Engineering Basics - Section 3
- Chemical Engineering Basics - Section 4
- Chemical Engineering Basics - Section 5
- Chemical Engineering Basics - Section 6
- Chemical Engineering Basics - Section 7
- Chemical Engineering Basics - Section 8
- Chemical Engineering Basics - Section 9
- Chemical Engineering Basics - Section 10
- Chemical Engineering Basics - Section 11
- Chemical Engineering Basics - Section 12
- Chemical Engineering Basics - Section 13
- Chemical Engineering Basics - Section 14
- Chemical Engineering Basics - Section 15
- Chemical Engineering Basics - Section 16
- Chemical Engineering Basics - Section 17
- Chemical Engineering Basics - Section 18
- Chemical Engineering Basics - Section 19
- Chemical Engineering Basics - Section 20
- Chemical Engineering Basics - Section 21
- Chemical Engineering Basics - Section 22
- Chemical Engineering Basics - Section 23
- Chemical Engineering Basics - Section 24
- Chemical Engineering Basics - Section 25
- Chemical Engineering Basics - Section 26
- Chemical Engineering Basics - Section 27
- Chemical Engineering Basics - Section 28


Chemical Engineering Basics - Engineering
Q1: A metal having a Poisson's ratio = 0.3 is elastically deformed under uniaxial tension. If the longitudinal strain = 0.8, then the magnitude of thickness strain isA -0.8
B 0.8
C +0.08
D -0.24
ANS:D - -0.24 The Poisson's ratio (ν) relates the lateral strain (εlateral) to the longitudinal strain (εlongitudinal) in a material under uniaxial tension. It is given by the formula: ν=−εlateral/εlongitudina Given that the Poisson's ratio (ν) is 0.3 and the longitudinal strain (longitudinal) is 0.8, we can use the formula to find the lateral strain (εlateral): 0.3=−0.8εlateral εlateral=−0.3×0.8 εlateral=−0.24 Therefore, the magnitude of the thickness strain is -0.24. So, the correct option is: −0.24 |


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