The relation between force F and density d is F - dfrac-x-sqrt-d- The dimensions of x areleft-L-1-2-M-1-2-T-2-right-left-L-1-2-M-3-2-T-2-right-left-L-1-M-3-2-T-2-right-left-L-1-M-1-2-T-2-right-

Question

Toppr Admin · Accepted Answer

Given- F-x-x221A-dSubstituting dimensionsMLT-x2212-2-x-x221A-ML-x2212-3-x21D2-x-M3-2L-x2212-1-2T-x2212-2-

The relation between force $F$ and density $d$ is $F = \frac{x}{\sqrt{d}}$ . The dimensions of $x$ are
$[L^{- 1 / 2} M^{1 / 2} T^{- 2}]$
$[L^{- 1 / 2} M^{3 / 2} T^{- 2}]$
$[L^{- 1} M^{3 / 2} T^{- 2}]$
$[L^{- 1} M^{1 / 2} T^{- 2}]$

Given, $F = \frac{x}{\sqrt{d}}$
Substituting dimensions
$M L T^{- 2} = \frac{x}{\sqrt{M L^{- 3}}}$
$\Rightarrow x = [M^{3 / 2} L^{- 1 / 2} T^{- 2}]$

The relation between force F and density d is F=x√d. The dimensions of x are[L−1/2M1/2T−2][L−1/2M3/2T−2][L−1M3/2T−2][L−1M1/2T−2]

Given, F=x√dSubstituting dimensionsMLT−2=x√ML−3⇒x=[M3/2L−1/2T−2]

The relation between force $F$ and density $d$ is $F = \frac{x}{\sqrt{d}}$ . The dimensions of $x$ are
$[L^{- 1 / 2} M^{1 / 2} T^{- 2}]$
$[L^{- 1 / 2} M^{3 / 2} T^{- 2}]$
$[L^{- 1} M^{3 / 2} T^{- 2}]$
$[L^{- 1} M^{1 / 2} T^{- 2}]$

Given, $F = \frac{x}{\sqrt{d}}$
Substituting dimensions
$M L T^{- 2} = \frac{x}{\sqrt{M L^{- 3}}}$
$\Rightarrow x = [M^{3 / 2} L^{- 1 / 2} T^{- 2}]$