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Q.
Force $F$ is given in terms of time $t$ and distance $x$ by $F=A \sin C t+B \cos D x .$ Then dimensions of $\frac{A}{B}$ and $\frac{C}{D}$ are
Physical World, Units and Measurements
Solution:
As $[A]=[F]$ and $[B]=[F]$
$\therefore \left[\frac{A}{B}\right]=\frac{[F]}{[F]}=\left[ M ^{0} L ^{0} T ^{0}\right]$
As $\sin \theta$ is a dimensionless,
$\therefore \quad[C t]=1$
or $[C]=\frac{1}{[t]}=\frac{1}{[ T ]}=\left[ T ^{-1}\right]$
$[D x]=1$ or $[D]=\frac{1}{[x]}=\frac{1}{[ L ]}=\left[ L ^{-1}\right]$
$\therefore \left[\frac{C}{D}\right]=\frac{\left[ T ^{-1}\right]}{\left[ L ^{-1}\right]}$
$=\left[ M ^{0} L T ^{-1}\right]$