Question

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

• A. $\left[L^{-\frac{1}{2}} M^{\frac{3}{2}} T^{-2}\right]$
• B. $\left[L^{-1} M^{\frac{1}{2}} T^{-2}\right]$
• C. $\left[L^{-1} M^{\frac{3}{2}} T^{-2}\right]$
• D. $\left[L^{-\frac{1}{2}} M^{\frac{1}{2}} T^{-2}\right]$

Answer 1

Answer: (A)

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