The position of a particle at time t is given by the equation x t - V 0 - A 1- e AT where V 0 is a constant and A - 0- Dimensions of V 0 and A respectively are-

The correct option is C M ^ 0 LT ^ 1 and T ^ 1 x(t)=v_0A(1 e^AT) ⇒ x(t) has the dimension of M^0L^1T^0 V_0A should have the ...

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Byju's Answer

Standard XII

Physics

Dimensional Analysis

The position ...

Question

The position of a particle at time t is given by the equation $x (t) = \frac{V_{0}}{A} (1 - e^{A T})$ where $V_{0}$ is a constant and $A > 0$ . Dimensions of $V_{0}$ and $A$ respectively are:

M^{0} {L T}^{0}

and

T^{- 1}

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M^{0} {L T}^{- 1}

and

{L T}^{- 2}

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M^{0} {L T}^{- 1}

and

T

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M^{0} {L T}^{- 1}

and

T^{- 1}

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Solution

The correct option is C $M^{0} {L T}^{- 1}$ and $T^{- 1}$
$x (t) = \frac{v_{0}}{A} (1 - e^{A T})$
$\Rightarrow x (t)$ has the dimension of $M^{0} L^{1} T^{0}$
$\frac{V_{0}}{A}$ should have the dimension of $M^{0} L^{1} T^{0}$
$A t$ should have the dimension $M^{0} L^{0} T^{0}$
$\Rightarrow A T = M^{0} L^{0} T^{0}$
$\Rightarrow A = T^{- 1}$
$\frac{v_{0}}{A} = L^{1} \Rightarrow V_{0} = L^{1} A = L^{1} T^{- 1} M^{0}$
Hence, option D is correct.

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The position of a particle at time t is given by the equation x(t)=V0A(1−eAT) where V0 is a constant and A>0. Dimensions of V0 and A respectively are:

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