Question

Young's modulus of steel is $1.9 \times 10^{11}\, N \,m^{-2}$. When expressed in $CGS$ units of dynes $cm^{-2}$, it will be equal to
($1 \,N = 10^5$ dyne, $1\, m^2 = 10^4\, cm^2$)

• A. $1.9 \times 10^{10}$
• B. $1.9 \times 10^{11}$
• C. $1.9 \times 10^{12}$
• D. $1.9 \times 10^{13}$

Answer 1

Answer: (C)

Dyne and Newton, both are units of force. While Dyne is specified in $C-G-S$(Centimeter - Gram - Second) unit system, Newton is specified in modern $SI$ unit system which gives the relationship Young's Modulus between stress and strain. It is used to measure hardness of material.
Here stress is force applied per unit area, strain is deformation happened on object.
$E =\frac{6}{\epsilon}=\frac{\frac{F}{ A }}{\frac{\Delta L }{ L _{0}}}=\frac{ F l _{0}}{ A \Delta L }$
Where, $\sigma=$ Stress in Pascal
$\epsilon=$ Strain or deformation
$F =$ Force
$A=$ Cross sectioned area
$\Delta L =$ Change in Length
$L _{ o }=$ Actual Length
$E =$ Young's Modulus
Here $E = 1 . 9 \times 1 0 \,N / M ^{2}$
Let's convert it into CGS unit system.
$=1.9 \times 10^{11}$ Newtons $/$ meter $^{2}$
$=\frac{1.9 \times 10^{11} \times 10^{5}}{10^{4} cm ^{2}}$
$=1.9 \times 10^{12}\, dyne / cm ^{2}$
$E =1.9 \times 10^{12}\, dyne / cm ^{2}$