The modulus of elasticity is dimensionally equivalent to
A. Surface tension
B. Stress
C. Strain
D. None of these
Answer
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Hint: The modulus of elasticity can be defined as a measure of a material's ability to regain its original dimensions after the removal of a load or force. The modulus is the slope of the straight line portion of stress versus strain graph up to the proportional limit.
Complete step-by-step answer:
We have studied that the dimension of mass is represented by $M$ , length is represented by L and time is represented by $T$. So we have to find the dimensional formula of the modulus of elasticity in terms of M,L and T. The modulus of elasticity is dimensionally equivalent to stress because the modulus of elasticity $ = \dfrac{{stress}}{{strain}}$ and the unit of stain is nothing as it is unitless quantity. We know that the unit of stress is \[Newton/mete{r^2}\] or Pascal. We have studied that the dimension of Newton is equivalent to $ML{T^{ - 2}}$ because Newton is unit of force , $F = ma$ where $F$ is force, $m$ is mass and $a$ is acceleration and dimension of meter is $L$. Dimension of modulus of elasticity $ = \dfrac{{ML{T^{ - 2}}}}{{{L^2}}} = M{L^{ - 1}}{T^{ - 2}}$. Hence option B is the correct answer to this problem because the modulus of elasticity is dimensionally equivalent to stress.
Note: We know that the ratio of tangential stress to the shearing strain in the range of elastic limit. Since strain it is a dimensionless quantity because it is the ratio of change in length to the initial length of object. So the calculation of dimension or unit of The modulus of elasticity becomes more easy as it will be equal to the dimension or unit of stress.
Complete step-by-step answer:
We have studied that the dimension of mass is represented by $M$ , length is represented by L and time is represented by $T$. So we have to find the dimensional formula of the modulus of elasticity in terms of M,L and T. The modulus of elasticity is dimensionally equivalent to stress because the modulus of elasticity $ = \dfrac{{stress}}{{strain}}$ and the unit of stain is nothing as it is unitless quantity. We know that the unit of stress is \[Newton/mete{r^2}\] or Pascal. We have studied that the dimension of Newton is equivalent to $ML{T^{ - 2}}$ because Newton is unit of force , $F = ma$ where $F$ is force, $m$ is mass and $a$ is acceleration and dimension of meter is $L$. Dimension of modulus of elasticity $ = \dfrac{{ML{T^{ - 2}}}}{{{L^2}}} = M{L^{ - 1}}{T^{ - 2}}$. Hence option B is the correct answer to this problem because the modulus of elasticity is dimensionally equivalent to stress.
Note: We know that the ratio of tangential stress to the shearing strain in the range of elastic limit. Since strain it is a dimensionless quantity because it is the ratio of change in length to the initial length of object. So the calculation of dimension or unit of The modulus of elasticity becomes more easy as it will be equal to the dimension or unit of stress.
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