The displacement of a particle as a function of time is given by:$x(t)=A\sin ⁡(t)+B{\cos ⁡}^{2}(t)+C{t}^{2}+D$.

Find the dimension of $\frac{ABC}{D}$

(Shift I Memory Based)

Question

The displacement of a particle as a function of time is given by:$x(t)=A\sin ⁡(t)+B{\cos ⁡}^{2}(t)+C{t}^{2}+D$. Find the dimension of $\frac{ABC}{D}$ (Shift I Memory Based)

YESPYQ · Accepted Answer

Correct answer: b) $\left[{\mathrm{L}}^{2}{\mathrm{T}}^{−2}\right]$. The displacement $x(t)$ has the dimension $[L]$. Each term in the equation must also have the same dimension $[L]$: $A\sin ⁡(t)$: $\sin ⁡(t)$ is dimensionless, so $A$ has $[L]$. $B{\cos ⁡}^{2}(t)$: ${\cos ⁡}^{2}(t)$ is dimensionless, so $B$ has $[L]$. $C{t}^{2}$: ${t}^{2}$ has dimensions $[{T}^{2}]$, so $C$ must have $[L{T}^{−2}]$. $D$: As part of the displacement, $D$ has $[L]$. Thus, the dimensions of $\frac{ABC}{D}$​ are: $\frac{[L]⋅[L]⋅[L{T}^{−2}]}{[L]}=[{L}^{2}{T}^{−2}]$ Therefore, the correct answer is $(b)$.

The displacement of a particle as a function of time is given by:\(x(t)=A\sin ⁡(t)+B{\cos ⁡}^{2}(t)+C{t}^{2}+D\). Find t…

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