The dimensions of dfrac - alpha - beta - in the equation F-dfrac - alpha - t - 2 - - beta - - 2 - - - where F is the force- is velocity and t is time- isleft- ML- T - -1 - right- left- ML- -1 - T - -2 - right- left- ML-3 - T - -4 - right- left- ML- 2 - T - -4 - right-

Question

The dimensions of dfrac - alpha - beta -  in the equation F-dfrac - alpha - t - 2 - - beta - - 2 - - - where F is the force- is velocity and t is time- isleft- ML- T - -1 - right- left- ML- -1 - T - -2 - right- left- ML-3 - T - -4 - right- left- ML- 2 - T - -4 - right-

Toppr Admin · Accepted Answer

FV2 - -x3B1-x3B2- -t2-x3B2-xA0-or -x3B1-x3B2- has dim- of-xA0-FV2-xA0- MLT-x2212-2L2T-x2212-2 - ML3T-x2212-4Hence C is correct

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
$[M L T^{- 1}]$
$[M L^{- 1} T^{- 2}]$
$[M L^{3} T^{- 4}]$
$[M L^{2} T^{- 4}]$

$F V^{2}$ = $\frac{α}{β}$ - $\frac{t^{2}}{β}$

or $\frac{α}{β}$ has dim. of $F V^{2}$

= $M L T^{- 2} L^{2} T^{- 2}$ = $M L^{3} T^{- 4}$
Hence C is correct

The dimensions of αβ in the equation F=α−t2βv2, where F is the force, v is velocity and t is time, is[MLT−1][ML−1T−2][ML3T−4][ML2T−4]

FV2 = αβ -t2β or αβ has dim. of FV2= MLT−2L2T−2 = ML3T−4Hence C is correct

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
$[M L T^{- 1}]$
$[M L^{- 1} T^{- 2}]$
$[M L^{3} T^{- 4}]$
$[M L^{2} T^{- 4}]$

$F V^{2}$ = $\frac{α}{β}$ - $\frac{t^{2}}{β}$

or $\frac{α}{β}$ has dim. of $F V^{2}$

= $M L T^{- 2} L^{2} T^{- 2}$ = $M L^{3} T^{- 4}$
Hence C is correct