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The Quantity $X=\frac{\varepsilon_0 L V}{t}$ is the permittivity of free space, is length, is potential difference and is time. The dimensions of are same as that of
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The correct answer is:
Current
$\left[\varepsilon_0 L\right]=[C] \therefore X=\frac{\varepsilon_0 L V}{t}=\frac{C \times V}{t}=\frac{Q}{t}=$ current
or
Dimensional formula for \(\epsilon_0=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\right]\)
Dimensional formula for \(\mathrm{L}=[\mathrm{L}]\)
Dimensional formula for \(\Delta \mathrm{V}=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-1}\right]\)
Dimensional formula for \(\Delta t=\left[\mathrm{T}^1\right]\)
Therefore, dimensional formula for X is:
\(\frac{\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\right][\mathrm{L}]\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-1}\right]}{\left[\mathrm{T}^1\right]}=[\mathrm{I}]\)
Therefore, its the same as that of current.
or
Dimensional formula for \(\epsilon_0=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\right]\)
Dimensional formula for \(\mathrm{L}=[\mathrm{L}]\)
Dimensional formula for \(\Delta \mathrm{V}=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-1}\right]\)
Dimensional formula for \(\Delta t=\left[\mathrm{T}^1\right]\)
Therefore, dimensional formula for X is:
\(\frac{\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\right][\mathrm{L}]\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-1}\right]}{\left[\mathrm{T}^1\right]}=[\mathrm{I}]\)
Therefore, its the same as that of current.
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