The position x of a particle at time t is given by x - V0a1 - e-at- where V0 is constant and a - 0- The dimensions of V0 and a are

The correct option is A M^0LT^ 1 and T^ 1 x= V _ 0 a ( 1 e ^ at ) at should be dimensions #160; ⇒ Dim( a ) = 1 Dim( T ) #160; = T ...

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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 constant and $a > 0$ . The dimensions of $V_{0}$ and $a$ are

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

and

T^{- 1}

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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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Solution

The correct option is A. $M^{0} L T^{- 1}$ and $T^{- 1}$
Given,
$x = \frac{V_{0}}{a} (1 - e^{- a t})$
'at' should be dimensionless $\Rightarrow D i m (a) = \frac{1}{D i m (T)} = T^{- 1}$
$x$ is position $\Rightarrow D i m (x) = M^{0} L T^{0} = D i m (\frac{V_{0}}{a})$
$∴$ $D i m (V_{0}) = D i m (x) \times D i m (a) = M^{0} L T^{- 1}$ .

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The position x of a particle at time t is given by x=V0a(1−e−at), where V0 is constant and a>0. The dimensions of V0 and a are

The correct option is A. M0LT−1 and T−1Given,x=V0a(1−e−at)'at' should be dimensionless ⇒Dim(a)=1Dim(T)=T−1x is position ⇒Dim(x)=M0LT0=Dim(V0a)∴ Dim(V0)=Dim(x)×Dim(a)=M0LT−1.

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 constant and $a > 0$ . The dimensions of $V_{0}$ and $a$ are