Question

The dimention of(μ₀ε₀)⁻¹/² is

Answer 1

Answer:

The answer is $ML^{2} T^{-1}$

Explanation:

1/√ μ₀ε₀ =   $\sqrt{L^{2}T^{-2}}$     =    $LT^{-1}$

speed =$LT^{-1}$

stress, young's modulus =  $\frac{MLT^{-2}}{L^{2}}$ = $ML^{-1}T^{-2}$

Therefore, momentum   =$MLT^{-1}$

and   Torque work is     = $ML^{2}T^{-2}$

So, Plank's constant     =>   $\frac{ML^{2}T^{-2}}{LT^{-1}}* L$

                                     =>$ML^{2} T^{-1}$

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