Question

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 $\frac{AD}{B}$ is:

• A. $\left[M^1{L}^1{T}^{-2}\right]$
• B. $\left[M^2{L}^2{T}^{-3}\right]$
• C. $\left[M{L}^2{T}^{-3}\right]$
• D. $\left[M^{0}L{T}^{-1}\right]$

Answer 1

The correct option is C $\left[M{L}^2{T}^{-3}\right]$

The argument of the trigonometric functions should be dimensionless.

$\therefore$ The dimensions of

$\left[B\right]=\left[L^{-1}\right]$

$\left[D\right]=\left[T^{-1}\right]$

As the trigonometric functions are dimensionless,

$A\text{and}C$ should have the dimension of force.

$\therefore$ The dimension of

$\left[A\right]=\left[ML{T}^{-2}\right]$

$\Rightarrow \left[\frac{AD}{B}\right]=\frac{\left[ML{T}^{-2}\right]\left[T^{-1}\right]}{\left[L^{-1}\right]}$

$\Rightarrow \left[\frac{AD}{B}\right]=\left[M{L}^2{T}^{-3}\right]$

Hence, option $\left(B\right)$ is correct.