Given: vecF=( veca/t)+ vecbt2 where symbols have their usual meaning. The dimensions of a and b are, respectively,

Question

Given: vecF=( veca/t)+ vecbt2 where symbols have their usual meaning. The dimensions of a and b are, respectively, (A) [ML-1T-1] and [MLT-4] (B) [MLT-1] and [MLT-4] (C) [MLT-1] and [ML] (D) [MLT-1] and [ML-1T-4]

Tardigrade Admin · Accepted Answer

[a]=[ veca]=[ vecFt]=[MLT-2]×[T]=[MLT-1] [b]=[ vecb]=[( vecF/t2)]=[MlT-2] [(1/T2)]=[MLT-4]

Q. Given: $F = \frac{a}{t} + b t^{2}$ where symbols have their usual meaning. The dimensions of $a$ and $b$ are, respectively,

$[M L^{- 1} T^{- 1}]$ and $[M L T^{- 4}]$

$[M L T^{- 1}]$ and $[M L T^{- 4}]$

$[M L T^{- 1}]$ and $[M L]$

$[M L T^{- 1}]$ and $[M L^{- 1} T^{- 4}]$

$[a] = [a] = [F t] = [M L T^{- 2}] \times [T] = [M L T^{- 1}]$
$[b] = [b] = [\frac{F}{t ^{2}}] = [Ml T^{- 2}] [\frac{1}{T ^{2}}] = [M L T^{- 4}]$

Q. Given: F=ta​+bt2 where symbols have their usual meaning. The dimensions of a and b are, respectively,

[ML−1T−1] and [MLT−4]

[MLT−1] and [MLT−4]

[MLT−1] and [ML]

[MLT−1] and [ML−1T−4]