S - A 1- e - Bxt - where S is speed and x is displacement- The unit of B is A- m -2 sB- s -2C- s -1D- m -1 s -1

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Byju's Answer

Standard XII

Physics

Dimensional Analysis

S = A 1- e - ...

Question

$S = A (1 - e^{- B x t})$ , where $S$ is speed and $x$ is displacement. The unit of $B$ is

m^{- 1} s^{- 1}

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m^{- 2} s

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s^{- 2}

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s^{- 1}

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Solution

The correct option is A $m^{- 1} s^{- 1}$
$S = A - A e^{- B x t}$

$[A] = [S] = [L T^{- 2}]$

(by principle of homogeneity)

$⟹ [e^{- B x t}] = [M^{0} L^{0} T^{0}]$

$⟹ [- B x t] = [M^{0} L^{0} T^{0}]$

$⟹ [B] [L T] = [M^{0} L^{0} T^{0}]$

$⟹ [B] = [L^{- 1} T^{- 1}]$

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S=A(1−e−Bxt), where S is speed and x is displacement. The unit of B is

The correct option is A m−1s−1S=A−Ae−Bxt [A]=[S]=[LT−2] (by principle of homogeneity) ⟹[e−Bxt]=[M0L0T0] ⟹[−Bxt]=[M0L0T0] ⟹[B][LT]=[M0L0T0] ⟹[B]=[L−1T−1]

$S = A (1 - e^{- B x t})$ , where $S$ is speed and $x$ is displacement. The unit of $B$ is

The correct option is A $m^{- 1} s^{- 1}$
$S = A - A e^{- B x t}$

$[A] = [S] = [L T^{- 2}]$

(by principle of homogeneity)

$⟹ [e^{- B x t}] = [M^{0} L^{0} T^{0}]$

$⟹ [- B x t] = [M^{0} L^{0} T^{0}]$

$⟹ [B] [L T] = [M^{0} L^{0} T^{0}]$

$⟹ [B] = [L^{- 1} T^{- 1}]$