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# The formula of Young’s modulus $Y$ of a wire, is given by $Y = \dfrac{F}{A} \times \dfrac{L}{{\Delta L}}$, where $F$ is the force with which the wire is stretched, $\Delta L$ is the change in length of wire and $A$ is the area of cross-section. The conversion factor to change it from CGS to MKS system is(A) $1$(B) $0.1$(C) $0.001$(D) $0.01$

Last updated date: 13th Jun 2024
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Hint To solve this question, we need to consider the units of each of the quantities present in the given equation corresponding to both the CGS and the MKS systems. Then by substituting the values in the equation, we will get the answer.

The formula given in the question for the Young’s modulus of a wire is
$\Rightarrow Y = \dfrac{F}{A} \times \dfrac{L}{{\Delta L}}$ (i)
Now, we consider the units of each of the quantities present in (i), corresponding to both the systems. For that, we first need to determine the dimensions of each quantity.
Length – We know that length is a fundamental dimension. The whole dimension is $\left[ {{M^0}{L^1}{T^0}} \right]$. So, we have
$\Rightarrow \left[ L \right] = \left[ {{M^0}{L^1}{T^0}} \right]$ (ii)
Change in length – As the change in length is the difference between two lengths, so its dimension is equal to that of the length. This gives us
$\Rightarrow \left[ {\Delta L} \right] = \left[ {{M^0}{L^1}{T^0}} \right]$ (iii)
Area – As we know that the dimensions of the area are obtained by taking the square of the dimensions of the length, so its dimensions are
$\Rightarrow \left[ A \right] = \left[ {{M^0}{L^2}{T^0}} \right]$ (iv)
Force – We know from Newton’s second law that the force is equal to the mass times the acceleration, i.e.
$\Rightarrow F = ma$
Now, the dimensions of the mass is $\left[ M \right]$ and that of the acceleration is $\left[ {L{T^{ - 2}}} \right]$. So, the dimension of the force becomes
$\left[ F \right] = \left[ M \right]\left[ {L{T^{ - 2}}} \right]$
$\Rightarrow \left[ F \right] = \left[ {{M^1}{L^1}{T^{ - 2}}} \right]$ (v)
For CGS system:
We know that the units of the mass, length and time in this system are gram, centimetre and second respectively. So from (ii) unit of length, $L$ is $cm$, from (iii) unit of change in length, $\Delta L$ is $cm$, from (iv) the unit of area, $A$ is $c{m^2}$ and from (v) the unit of the force $F$ is $gcm{s^{ - 2}}$. Therefore, from (i) the unit of the force in CGS system is
$\Rightarrow {u_{cgs}} = \dfrac{{gcm{s^{ - 2}}}}{{c{m^2}}} \times \dfrac{{cm}}{{cm}}$
$\Rightarrow {u_{cgs}} = gc{m^{ - 1}}{s^{ - 2}}$ (vi)
For MKS system:
The units of the mass, length and time in this system are kilogram, metre and second respectively. So from (ii) unit of length, $L$ is $m$, from (iii) unit of change in length, $\Delta L$ is $m$, from (iv) the unit of area, $A$ is ${m^2}$ and from (v) the unit of the force $F$ is $kgm{s^{ - 2}}$. From (i) the unit of the force in MKS system is
$\Rightarrow {u_{mks}} = \dfrac{{kgm{s^{ - 2}}}}{{{m^2}}} \times \dfrac{m}{m}$
$\Rightarrow {u_{mks}} = kg{m^{ - 1}}{s^{ - 2}}$ (vii)
Dividing (vi) by (vii) we get
$\Rightarrow \dfrac{{{u_{cgs}}}}{{{u_{mks}}}} = \dfrac{{gc{m^{ - 1}}{s^{ - 2}}}}{{kg{m^{ - 1}}{s^{ - 2}}}}$
$\Rightarrow \dfrac{{{u_{cgs}}}}{{{u_{mks}}}} = \dfrac{g}{{kg}} \times \dfrac{m}{{cm}}$
We know that $1kg = 1000g$ and $1m = 100cm$. Therefore we have
$\Rightarrow \dfrac{{{u_{cgs}}}}{{{u_{mks}}}} = \dfrac{g}{{1000g}} \times \dfrac{{100cm}}{{cm}}$
$\Rightarrow \dfrac{{{u_{cgs}}}}{{{u_{mks}}}} = \dfrac{{100}}{{1000}}$
Finally, we get
$\Rightarrow {u_{cgs}} = 0.1{u_{mks}}$
Thus, the conversion factor to change the Young’s modulus from CGS to MKS system is $0.1$.
Hence, the correct answer is option B.

Note
For solving such types of questions, it is required to obtain each physical quantity given in the problem in terms of the fundamental or base quantities. So we must know the physical formula which can relate the quantity with the fundamental quantities. It must be noted that there can be multiple such formulas and any one of them can be used for this purpose.