Answer

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Hint: Recall the values of \[{{\mu }_{_{0}}}\]and \[{{\varepsilon }_{0}}\], thereafter by rearranging terms and performing basic mathematical calculations, students will get the required formula easily.

Complete step by step answer:

We know that the values of \[\dfrac{{{\mu }_{_{0}}}}{4\pi }\]and \[\dfrac{1}{4\pi {{\varepsilon }_{0}}}\]are as follows:

\[\dfrac{{{\mu }_{_{0}}}}{4\pi }={{10}^{-7}}H/m\]--------(1)

\[\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N-{{m}^{2}}/{{C}^{2}}\]--------(2)

On rearranging terms, the value of \[{{\mu }_{_{0}}}\]and \[{{\varepsilon }_{0}}\]is given by,

\[{{\mu }_{_{0}}}=4\pi \times {{10}^{-7}}H/m\]--------(3)

\[{{\varepsilon }_{0}}=4\pi \times 9\times {{10}^{9}}N-{{m}^{2}}/{{C}^{2}}\]--------(4)

Now on multiplying equations (3) and (4), we get

\[\Rightarrow {{\mu }_{_{0}}}{{\varepsilon }_{0}}=4\pi \times {{10}^{-7}}\times \dfrac{1}{4\pi \times 9\times {{10}^{9}}}\]

On solving,

\[\Rightarrow {{\mu }_{_{0}}}{{\varepsilon }_{0}}=\dfrac{1}{9\times {{10}^{16}}}\]

This may be written as,

\[\Rightarrow {{\mu }_{_{0}}}{{\varepsilon }_{0}}=\dfrac{1}{{{\left( 3\times {{10}^{8}} \right)}^{2}}}\]

We know that the speed of light in vacuum is, \[\text{c=3 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{8}}}\text{m/s}\]

\[\Rightarrow {{\mu }_{_{0}}}{{\varepsilon }_{0}}=\dfrac{1}{{{\left( c \right)}^{2}}}\]

\[\Rightarrow {{\left( c \right)}^{2}}=\dfrac{1}{{{\mu }_{_{0}}}{{\varepsilon }_{0}}}\]

On taking square root,

\[\Rightarrow c=\dfrac{1}{\sqrt{{{\mu }_{_{0}}}{{\varepsilon }_{0}}}}\]

Hence, the correct option is B, i.e., \[c=\dfrac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}\]

Additional Information:

Here, \[{{\varepsilon }_{0}}\] is the permittivity of free space. Basically, this is a mathematical quantity that represents how much electric field is permitted (penetrated) in free space or vacuum.

In this solution, \[\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N-{{m}^{2}}/{{C}^{2}}\]this number tells us that \[9\times {{10}^{9}}\]field lines are crossing by a charge in a vacuum but for any medium, this number may change and the number of field lines penetrating is also changed.

Also \[{{\mu }_{_{0}}}\]is the permeability of free space (vacuum) is a physical constant equal to \[1.257\times {{10}^{-6}}H/m\](approximately). Permeability in general is symbolized by µ, and is a constant of proportionality that exists between magnetic flux density and magnetic field strength in a given medium. In certain metals, notably iron and nickel and alloys containing them, µ is greater than \[{{\mu }_{_{0}}}\].

Note: Students should try to memorize this kind of very usual relationship. If they do not remember then it is a better option to write basic rememberable values of \[{{\varepsilon }_{0}}\] and \[{{\mu }_{_{0}}}\] thereafter performing some calculation and rearranging of terms they will get final relation easily.

Complete step by step answer:

We know that the values of \[\dfrac{{{\mu }_{_{0}}}}{4\pi }\]and \[\dfrac{1}{4\pi {{\varepsilon }_{0}}}\]are as follows:

\[\dfrac{{{\mu }_{_{0}}}}{4\pi }={{10}^{-7}}H/m\]--------(1)

\[\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N-{{m}^{2}}/{{C}^{2}}\]--------(2)

On rearranging terms, the value of \[{{\mu }_{_{0}}}\]and \[{{\varepsilon }_{0}}\]is given by,

\[{{\mu }_{_{0}}}=4\pi \times {{10}^{-7}}H/m\]--------(3)

\[{{\varepsilon }_{0}}=4\pi \times 9\times {{10}^{9}}N-{{m}^{2}}/{{C}^{2}}\]--------(4)

Now on multiplying equations (3) and (4), we get

\[\Rightarrow {{\mu }_{_{0}}}{{\varepsilon }_{0}}=4\pi \times {{10}^{-7}}\times \dfrac{1}{4\pi \times 9\times {{10}^{9}}}\]

On solving,

\[\Rightarrow {{\mu }_{_{0}}}{{\varepsilon }_{0}}=\dfrac{1}{9\times {{10}^{16}}}\]

This may be written as,

\[\Rightarrow {{\mu }_{_{0}}}{{\varepsilon }_{0}}=\dfrac{1}{{{\left( 3\times {{10}^{8}} \right)}^{2}}}\]

We know that the speed of light in vacuum is, \[\text{c=3 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{8}}}\text{m/s}\]

\[\Rightarrow {{\mu }_{_{0}}}{{\varepsilon }_{0}}=\dfrac{1}{{{\left( c \right)}^{2}}}\]

\[\Rightarrow {{\left( c \right)}^{2}}=\dfrac{1}{{{\mu }_{_{0}}}{{\varepsilon }_{0}}}\]

On taking square root,

\[\Rightarrow c=\dfrac{1}{\sqrt{{{\mu }_{_{0}}}{{\varepsilon }_{0}}}}\]

Hence, the correct option is B, i.e., \[c=\dfrac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}\]

Additional Information:

Here, \[{{\varepsilon }_{0}}\] is the permittivity of free space. Basically, this is a mathematical quantity that represents how much electric field is permitted (penetrated) in free space or vacuum.

In this solution, \[\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N-{{m}^{2}}/{{C}^{2}}\]this number tells us that \[9\times {{10}^{9}}\]field lines are crossing by a charge in a vacuum but for any medium, this number may change and the number of field lines penetrating is also changed.

Also \[{{\mu }_{_{0}}}\]is the permeability of free space (vacuum) is a physical constant equal to \[1.257\times {{10}^{-6}}H/m\](approximately). Permeability in general is symbolized by µ, and is a constant of proportionality that exists between magnetic flux density and magnetic field strength in a given medium. In certain metals, notably iron and nickel and alloys containing them, µ is greater than \[{{\mu }_{_{0}}}\].

Note: Students should try to memorize this kind of very usual relationship. If they do not remember then it is a better option to write basic rememberable values of \[{{\varepsilon }_{0}}\] and \[{{\mu }_{_{0}}}\] thereafter performing some calculation and rearranging of terms they will get final relation easily.

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