Question 285: Units , Dimensions and Error Analysis

Question

A quantity X is given by $${\varepsilon _0}L{{\Delta V} \over {\Delta t}}$$ where $${\varepsilon _0}$$ is the permittivity of the free space, L is a length, $${\Delta V}$$ is a potential difference and $${\Delta t}$$ is a time interval. The dimentional formula for X is the same as that of

Accepted Answer

Given that $$X = {\varepsilon _0}L\left( {{{\Delta V} \over {\Delta t}}} \right)$$
Therefore, dimensional formula for X is
$$[{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}][L]{{[M{L^2}{T^{ - 3}}{I^{ - 1}}]} \over {[T]}} = [I]$$
Thus, dimensional formula for X is same as that of current.
Answer (D)

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Units , Dimensions and Error Analysis

A quantity X is given by ${\varepsilon _0}L{{\Delta V} \over {\Delta t}} where {\varepsilon _0} is the permittivity of the free space, L is a length, {\Delta V} is a potential difference and {\Delta t}$ is a time interval. The dimentional formula for X is the same as that of

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