Question

To determine the Young's modulus of a wire, the formula is \[Y = \dfrac{{FL}}{{A\Delta L}}\] ​; where L=length, A= area of cross-section of the wire, \[\Delta L\]= Change in length of the wire when stretched with a force F. The conversion factor to change it from CGS to MKS system is
(A) 1
(B) 10
(C) 0.1
(D) 0.01

Answer 1

Hint: Young’s modulus is dependent on factors such as force, length and are inversely dependent on factors such as area. Now, we know the SI units of the same which are in the MKS system. To find the factor, equate the units in CGS system and MKS system and find the factor by which it varies.

Complete Step by Step Solution:
 Young’s modulus, also called as modulus of elasticity, is the measure of the tensile stiffness of the given material. It also measures the stiffness of the material when there is a force applied on it which causes the change in length due to compression or expansion. Young’s modulus directly depends upon the force applied on the body and inversely varies upon the area of the surface where the force is applied. This can be represented mathematically as,
\[Y = \dfrac{{FL}}{{A\Delta L}}\]
Where, F is the force applied and A is the area of the surface.
 Force in the CGS unit is measured in dyne and area is measured in centimeters. In the MKS system, the force is measured as Newton’s and area and change in length are measured in meters. Using this the units for young’s modulus is given as,
\[ \Rightarrow Y = dyne/c{m^2}\] in CGS unit and
\[ \Rightarrow Y = N/{m^2}\] in MKS units
 Now, taking the CGS equation, we know that \[1dyne = {10^{ - 5}}N\]. Substituting this in the equation we get,
\[ \Rightarrow Y = \dfrac{{{{10}^{ - 5}}N}}{{c{m^2}}}\]
We know that \[1cm = {10^{ - 2}}m\]. Substituting this , we get
\[ \Rightarrow Y = \dfrac{{{{10}^{ - 5}}N}}{{{{({{10}^{ - 2}}m)}^2}}}\]
On further simplification, we get,
\[ \Rightarrow Y = \dfrac{{{{10}^{ - 5}}N}}{{{{10}^{ - 4}}{m^2}}}\]
Taking the common term to the numerator ,we get
\[ \Rightarrow Y = \dfrac{{{{10}^{ - 5 + 4}}N}}{{{m^2}}}\]
\[ \Rightarrow Y = \dfrac{{{{10}^{ - 5 + 4}}N}}{{{m^2}}} = \dfrac{{{{10}^{ - 1}}N}}{{{m^2}}}\]
Hence, \[\dfrac{{1dyne}}{{c{m^2}}} = 0.1\dfrac{N}{{{m^2}}}\]

Therefore, option (c) is the right answer for the given question.

Note: The CGS system is abbreviated as Centimeter-Gram-Second system, which is a type of Standard Unit system that is quite useful in measuring small quantities in chemical fields. MKS system abbreviated as Meter-Kilogram-Second is used for standard measurements globally across various fields.