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)= $(\frac{v_{0}}{α}) (1 - c^{- a t})$ where $v_{0}$ and $α$ are constants . The dimensions of $v_{0}$ 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 equation is $x (t) = (\frac{v_{0}}{α}) (1 - C^{- α t})$
According to the question $v_{0}$ is constant and $α > 0$
Since C is constant so $α t$ should be unit less
So unit of $α = {s e c}^{- 1} = [T^{- 1}]$
x is length, so unit of x is meter.
$v_{0} = x α$
$v_{0} = M {s e c}^{- 1}$
So now we can say

$v_{0} = [L T^{- 1}]$
$α = [T^{- 1}]$
So the dimensions are as above.

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The position of a particle at time t is given by the relation x(t)= (v0α)(1−c−at) where v0 and α are constants . The dimensions of v0 and α are respectively

The position of a particle at time t is given by the relation x(t)= $(\frac{v_{0}}{α}) (1 - c^{- a t})$ where $v_{0}$ and $α$ are constants . The dimensions of $v_{0}$ and $α$ are respectively