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-A- M O L 1 T -1 and T -1B- M O L 1 T - and T -1C- M - L 1 T -1 and LT -2D- M o L 1 T -1 and T

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

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Standard IX

Mathematics

Factor Theorem

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 A M^oL¹T^-1 and T^-1
Dimension of $α t$ = $[M^{0} L^{0} T^{0}] t h e r e f o r e [α] = [T^{- 1}]$
Again $[\frac{v_{0}}{α}] = [L]$
$s o [v_{0}] = [L T^{- 1}]$

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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 correct option is A MoL1T-1 and T-1Dimension of αt = [M0L0T0] therefore[α]=[T−1] Again [v0α]=[L] so [v0]=[LT−1]

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:

The correct option is A M^oL¹T^-1 and T^-1
Dimension of $α t$ = $[M^{0} L^{0} T^{0}] t h e r e f o r e [α] = [T^{- 1}]$
Again $[\frac{v_{0}}{α}] = [L]$
$s o [v_{0}] = [L T^{- 1}]$