The position of a particle at time t is given by the relation xt-v0-1-c-a t where v0 and - are constants - The dimensions of v0 and - are respectively

The correct option is B MoL1T 1 and T 1Dimension of α t = [M^0L^0 T^0] ∴ [α]=[T^ 1] Again [v_0α]= [L] so [v_0] =[LT^ 1] ...

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Standard VIII

Mathematics

Dimensional Analysis

The position ...

Question

The position of a particle at time t is given by the relation $x (t) = \frac{v_{o}}{α} (1 - c^{- α t})$
$v_{o}$ and $c$ are constants . The dimensions of $v_{o}$ and $α$ are respectively:

M^oL¹T^-1 and T^-1

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M^oL¹T^o and T^-1

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M^oL¹T^-1 and LT^-2

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M^oL¹T^-1 and T

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Solution

The correct option is: (B)
Given Data:
The position of a particle at time t is given by the relation; $x (t) = \frac{v_{o}}{α} (1 - c^{- α t})$
Dimension of length, $= [L]$
Since, $c$ is constant, so $α t$ should be unitless,
So, $α t = [M^{0} L^{0} T^{0}]$
$α [T] = [M^{0} L^{0} T^{0}]$
$α = [T^{- 1}]$

p
utting value of dimension on both side of equation:
$[L] = \frac{v_{o}}{α}$
$[L] = \frac{v_{o}}{[T^{- 1}]}$
$v_{o} = [L T^{- 1}]$
So, dimension of
$α = [T^{- 1}]$
, and
$v_{o} = [L T^{- 1}]$

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The position of a particle at time t is given by the relation x(t)=voα(1−c−αt) vo and c are constants . The dimensions of vo and α are respectively:

The position of a particle at time t is given by the relation $x (t) = \frac{v_{o}}{α} (1 - c^{- α t})$
$v_{o}$ and $c$ are constants . The dimensions of $v_{o}$ and $α$ are respectively: