The dimensions of (α/β) in the equation F=(α-t2/β v2), where F is the force, v is velocity and t is time, is

Question

The dimensions of (α/β) in the equation F=(α-t2/β v2), where F is the force, v is velocity and t is time, is (A) [ MLT -1] (B) [ ML -1 T -2] (C) [ ML 3 T -4] (D) [ ML 2 T -4]

Tardigrade Admin · Accepted Answer

F=(α-t2/β v2) Dimensionally, α=[T2] [ MLT -2]=([ T 2]/β[ L 2 T -2]) β=(T2/[M L T-2 ⋅ L2 T-2]) ⇒ β=[ M -1 L-3 T 6] Dimensions of (α/β)=( T 2/ M -1 L-3 T 6) =[ ML 3 T -4]

Q. The dimensions of $\frac{α}{β}$ in the equation $F = \frac{α - t ^{2}}{β v ^{2}}$ , where $F$ is the force, $v$ is velocity and $t$ is time, is

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

$[M L^{- 1} T^{- 2}]$

$[M L^{3} T^{- 4}]$

$[M L^{2} T^{- 4}]$

Q. The dimensions of βα​ in the equation F=βv2α−t2​, where F is the force, v is velocity and t is time, is

[MLT−1]

[ML−1T−2]

[ML3T−4]

[ML2T−4]

F=βv2α−t2​ Dimensionally, α=[T2] [MLT−2]=β[L2T−2][T2]​ β=[MLT−2⋅L2T−2]T2​ ⇒β=[M−1L−3T6] Dimensions of βα​=M−1L−3T6T2​ =[ML3T−4]

Q. The dimensions of $\frac{α}{β}$ in the equation $F = \frac{α - t ^{2}}{β v ^{2}}$ , where $F$ is the force, $v$ is velocity and $t$ is time, is

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

$[M L^{- 1} T^{- 2}]$

$[M L^{3} T^{- 4}]$

$[M L^{2} T^{- 4}]$