The force 'F' acting on a body of density 'd' are related by the relation F = (y/√ d) The dimensions of 'y ' are

Question

The force 'F' acting on a body of density 'd' are related by the relation F = (y/√ d) The dimensions of 'y ' are (A) [L-(1/2) M(3/2) T-2] (B) [L-1 M(1/2) T-2] (C) [L-1 M(3/2) T-2] (D) [L-(1/2) M(1/2) T-2]

Tardigrade Admin · Accepted Answer

The dimensions of force [ F ]=[M L T-2] and density [d]=[M L-3 T0] From the given relation, F=(y/√d) ⇒ y=F √d Substituting the above dimensions, we get [y] =[F][d]1 / 2=[M L T-2][M L-3 T0]1 / 2 =[M3 / 2 L-1 / 2 T-2]

Q. The force $^{'} F^{'}$ acting on a body of density $^{'} d^{'}$ are related by the relation $F = \frac{y}{d}$ The dimensions of $^{'} y^{'}$ are

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

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

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

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

The dimensions of force $[F] = [M L T^{- 2}]$ and density $[d] = [M L^{- 3} T^{0}]$
From the given relation, $F = \frac{y}{d}$
$\Rightarrow y = F d$
Substituting the above dimensions, we get
$[y] = [F] [d]^{1/2} = [M L T^{- 2}] [M L^{- 3} T^{0}]^{1/2}$
$= [M^{3/2} L^{- 1/2} T^{- 2}]$

Q. The force ′F′ acting on a body of density ′d′ are related by the relation F=d​y​ The dimensions of ′y′ are

[L−21​M23​T−2]

[L−1M21​T−2]

[L−1M23​T−2]

[L−21​M21​T−2]

The dimensions of force [F]=[MLT−2] and density [d]=[ML−3T0] From the given relation, F=d​y​ ⇒y=Fd​ Substituting the above dimensions, we get [y]=[F][d]1/2=[MLT−2][ML−3T0]1/2 =[M3/2L−1/2T−2]