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Home » Young’s modulus of steel is . When expressed in CGS units of , it will be equal to:a) b) c) d)
Young’s modulus of steel is 1.9\times10^{11} N/m^{2}. When expressed in CGS units of dynes/cm^{2}, it will be equal to:
a) 1.9\times 10^{10}
b) 1.9\times 10^{11}
c) 1.9\times 10^{12}
d) 1.9\times 10^{13}
CBSE | Class 11 | NCERT Exemplar | Physics | Units and Measurements
Young’s modulus of steel is 1.9\times10^{11} N/m^{2}. When expressed in CGS units of dynes/cm^{2}, it will be equal to:
a) 1.9\times 10^{10}
b) 1.9\times 10^{11}
c) 1.9\times 10^{12}
d) 1.9\times 10^{13}

Correct answer is c) 1.9\times 10^{12}

Both Dyne and Newton are force units. While Dyne is measured in the C-G-S (Centimeter – Gram – Second) system, Newton is measured in the contemporary SI system, which yields the Young’s Modulus relationship between stress and strain. It is used to determine the material’s hardness.
Here, stress refers to the force applied per unit area, while strain refers to the object’s deformation.

\mathrm{E}=\frac{6}{\epsilon}=\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{\Delta \mathrm{L}}{\mathrm{L}_{0}}}=\frac{\mathrm{F} \mathrm{l}_{0}}{\mathrm{~A} \Delta \mathrm{L}} Where, \sigma= Stress in Pascal

\epsilon= Strain or deformation

\mathrm{F}= Force

\mathrm{A}= Cross sectioned area

\Delta \mathrm{L}= Change in Length

\mathrm{L}_{\mathrm{o}}= Actual Length

\begin{array}{l} \mathrm{E}=\text { Young's Modulus } \\ \text { Here } \mathrm{E}=1.9 \times 10 \mathrm{~N} / \mathrm{M}^{2} \end{array}

Converting it to CGS system, we get,

\begin{array}{l} =1.9 \times 10^{11} \text { Newtons } / \text { meter }^{2} \\ =\frac{1.9 \times 10^{11} \times 10^{5}}{10^{4} \mathrm{~cm}^{2}} \\ =1.9 \times 10^{12} \mathrm{dyne} / \mathrm{cm}^{2} \\ \mathrm{E}=1.9 \times 10^{12} \text { dyne } / \mathrm{cm}^{2} \end{array}

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