The position x of a particle at time t is given by x - v0a1 - e-at where v0 is a constant and a - 0- The dimensional formula of v0 and a are

The correct option is A [M^0L^1T^ 1] and [M^0L^0T^ 1] For finding v_0=ax1 e^a/t =m/s× m/s^21 e^ m/s^2× s =m^2s^2×sm =m/s[M^0LT^ 1] ...

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Byju's Answer

Standard XII

Physics

Dimensional Analysis

The position ...

Question

The position x of a particle at time t is given by $x = \frac{v_{0}}{a} (1 - e^{- a t})$ where $v_{0}$ is a constant and a > 0. The dimensional formula of $v_{0}$ and a are

[M^{0} L^{1} T^{- 1}] a n d [M^{0} L^{0} T^{- 1}]

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[M^{0} L^{1} T^{0}] a n d [M^{0} L^{0} T^{- 1}]

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[M^{0} L T^{- 1}] a n d [M^{0} L^{1} T^{- 2}]

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[M^{0} L^{1} T^{- 1}] a n d [M^{0} L^{0} T^{1}]

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Solution

The correct option is A $[M^{0} L^{1} T^{- 1}] a n d [M^{0} L^{0} T^{- 1}]$
For finding $v_{0} = \frac{a x}{1 - e^{a / t}} = \frac{m / s \times m / s^{2}}{1 - e^{- m / s^{2} \times s}} = \frac{m^{2}}{s^{2}} \times \frac{s}{m} = m / s [M^{0} L T^{- 1}]$
For finding,
$v_{0} = \frac{x a}{1 - e^{- a t}} \frac{v_{0}}{x} = \frac{a}{1 - e^{- a t}} \frac{x}{v_{0}} = \frac{1 - e^{- a t}}{a} a = \frac{v_{0}}{x} = \frac{m / s}{m} a = \frac{1}{s} a = s^{- 1} = [M^{0} L^{0} T^{- 1}]$

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The position x of a particle at time t is given by x=v0a(1−e−at) where v0 is a constant and a > 0. The dimensional formula of v0 and a are

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