The force is given in terms of time $t$ and displacement $x$ by the equation $\mathrm{F}=\mathrm{A} \cos \mathrm{Bx}+\mathrm{C} \sin \mathrm{Dt}$ The dimensional formula of $\frac{A D}{B}$ is :

(C) $\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-3}\right]$

(A) $\left[\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-3}\right]$

(B) $\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right]$

(D) $\left[\mathrm{M}^{0} \mathrm{LT}^{-1}\right]$

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Units & Measurement

The force is given in terms of time

t

and displacement

x

by the equation

F = A cos Bx + C sin Dt

The dimensional formula of

\frac{A D}{B}

is :

Answer: (C)

Since

cos B x

is dimensionless,

[B] = [L^{- 1}]

. Since

sin D t

is dimensionless,

[D] = [T^{- 1}]

. Force has dimensions

[A] = [M L T^{- 2}]

. Therefore,

[\frac{A D}{B}] = \frac{[ M L T ^{- 2} ] [ T ^{- 1} ]}{[ L ^{- 1} ]} = [M L^{2} T^{- 3}]

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option