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 (A/B) and (C/D) are

Question

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 (A/B) and (C/D) are (A) [ M 0 L 0 T 0],[ M 0 L 0 T -1] (B) [ MLT -2],[ M 0 L -1 T 0] (C) [ M 0 L 0 T 0],[ M 0 LT -1] (D) [ M 0 LT -1],[ M 0 L 0 T 0]

Tardigrade Admin · Accepted Answer

As [A]=[F] and [B]=[F] ∴ [(A/B)]=([F]/[F])=[ M 0 L 0 T 0] As sin θ is a dimensionless, ∴ [C t]=1 or [C]=(1/[t])=(1/[ T ])=[ T -1] [D x]=1 or [D]=(1/[x])=(1/[ L ])=[ L -1] ∴ [(C/D)]=([ T -1]/[ L -1]) =[ M 0 L T -1]

Q. Force $F$ is given in terms of time $t$ and distance $x$ by $F = A sin Ct + B cos D x .$ Then dimensions of $\frac{A}{B}$ and $\frac{C}{D}$ are

$[M^{0} L^{0} T^{0}], [M^{0} L^{0} T^{- 1}]$

$[M L T^{- 2}], [M^{0} L^{- 1} T^{0}]$

$[M^{0} L^{0} T^{0}], [M^{0} L T^{- 1}]$

$[M^{0} L T^{- 1}], [M^{0} L^{0} T^{0}]$

Q. Force F is given in terms of time t and distance x by F=AsinCt+BcosDx. Then dimensions of BA​ and DC​ are

[M0L0T0],[M0L0T−1]

[MLT−2],[M0L−1T0]

[M0L0T0],[M0LT−1]

[M0LT−1],[M0L0T0]

As [A]=[F] and [B]=[F] ∴[BA​]=[F][F]​=[M0L0T0] As sinθ is a dimensionless, ∴[Ct]=1 or [C]=[t]1​=[T]1​=[T−1] [Dx]=1 or [D]=[x]1​=[L]1​=[L−1] ∴[DC​]=[L−1][T−1]​ =[M0LT−1]

Q. Force $F$ is given in terms of time $t$ and distance $x$ by $F = A sin Ct + B cos D x .$ Then dimensions of $\frac{A}{B}$ and $\frac{C}{D}$ are

$[M^{0} L^{0} T^{0}], [M^{0} L^{0} T^{- 1}]$

$[M L T^{- 2}], [M^{0} L^{- 1} T^{0}]$

$[M^{0} L^{0} T^{0}], [M^{0} L T^{- 1}]$

$[M^{0} L T^{- 1}], [M^{0} L^{0} T^{0}]$