Question

y=25 × 10^10 N/m^2. determine the value of young's modulus in cgs system

Answer 1

y = 25×10^10N/m²

N/m² = kgms-²/m² = kgm-¹s-²

25×10^10kgm-¹s-² = ? gcm-¹s-²

n1 = n2(M2/M1)^a(L2/L1)^b(T2/T1)^c

n1 = 25×10^10(1kg/g)(m/cm)-¹(s/s)-²

n1 = 25×10^10(1000g/g)(100cm/cm)-¹(1)-²

n1 = 25×10^10 ×10³ ×(1/100)

n1 = 25×10^10 × 10³ × 10-²

n1 = 25×10¹³-²

n1 = 25×10¹¹

So,

y = 25×10^10N/m² in cgs system is 25×10¹¹dyne/cm²

Cheers!

Answer 2

The value of young's modulus in the CGS system is $\\ \mathrm{E}=25\times 10^{11} \mathrm{dyne} / \mathrm{cm}^{2}$

Given:

$Y=25$×$10^{10} N/m^2$

To find:

The value of young's modulus in the CGS system

Solution:

Two very different Dyne & Newton are units of force. Although Dyne is described in the C-G-S (Centimeter - Gram - Second) unit system, Newton is specified in the current SI unit system, which provides the Young's Modulus link between stress and strain. It is used to determine the hardness of a substance.

Stress is defined as the force exerted per unit area, while strain is defined as object deformation.

As we know the formula,

$\mathrm{E}=\frac{6}{\epsilon}=\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{\Delta \mathrm{L}}{\mathrm{L}_{0}}}=\frac{\mathrm{Fl}_{0}}{\mathrm{~A} \Delta \mathrm{L}}$

Where,

$\sigma=$ Stress in Pascal

$\epsilon=$Strain or deformation

$\mathrm{A}=$Cross sectioned area

$\Delta \mathrm{L}=$ Change in Length

$\mathrm{L}_{\circ}=$ Actual Length

$\mathrm{E}=$ Young's Modulus

Now, $\\ \mathrm{E}=25 \times 10^{10} \mathrm{N} / \mathrm{m}^{2}$

Let us now convert that to the CGS unit system.

$=25 \times 10^{10} N /{\text {m }^{2}}$

$=\frac{25 \times 10^{10} \times 10^{5}}{10^{4} \mathrm{~cm}^{2}}$

$=25 \times 10^{11} \mathrm{dyne} / \mathrm{cm}^{2}$

$\\ \mathrm{E}=25\times 10^{11} \mathrm{dyne} / \mathrm{cm}^{2}$

Therefore, The value of young's modulus in the CGS system is $\\ \mathrm{E}=25\times 10^{11} \mathrm{dyne} / \mathrm{cm}^{2}$

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