Question

Young's modulus of steel is $1.9\times {10}^{11}N{m}^{-2}$. Expressed in CGS
(Given $1N={10}^5$ dyne, $1m^2={10}^{4}c{m}^2$)

• 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)

$1.9\times {10}^{12}$

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.

$E=\frac{6}{ϵ}=\frac{\frac{F}{A}}{\frac{\Delta L}{L_0}}=\frac{F{l}_0}{A\Delta L}$

Where, $\sigma =$ Stress in Pascal

$ϵ=$ Strain or deformation

$F=$ Force

$A=$ Cross sectioned area

$\Delta L=$ Change in Length

$L_o=$ Actual Length

$E=$ Young’s Modulus

Here $E=1.9\times 10N/M^2$

Let’s convert it into CGS unit system.

$=1.9\times {10}^{11}Newtons/mete{r}^2$

$=\frac{1.9\times {10}^{11}\times {10}^5}{{10}^{4}c{m}^2}$

$=1.9\times {10}^{12}dyne/c{m}^2$

$E=1.9\times {10}^{12}dyne/c{m}^2$