(D) The argument of the sine function,$(Bx + Ct + D)$,must be dimensionless.Therefore,each term must be dimensionless.For $Bx$: $[B][x] = [M^0 L^0 T^0

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(D) The argument of the sine function,$(Bx + Ct + D)$,must be dimensionless.Therefore,each term must be dimensionless.For $Bx$: $[B][x] = [M^0 L^0 T^0] \implies [B] = [L^{-1}]$.For $Ct$: $[C][t] = [M^0 L^0 T^0] \implies [C] = [T^{-1}]$.For $D$: $[D] = [M^0 L^0 T^0]$ (dimensionless).For $A$: Since $y$ is displacement,$[A] = [L]$.Now,calculate the dimensions of $ABCD$:$[ABCD] = [A][B][C][D] = [L] \cdot [L^{-1}] \cdot [T^{-1}] \cdot [1] = [T^{-1}] = [M^0 L^0 T^{-1}]$.

Class 11 Physics Units and Measurements Dimensional Analysis, Uses and Limitations

Medium

Given that the displacement of an oscillating particle is given by $y = A \sin(Bx + Ct + D)$. The dimensional formula for $ABCD$ is: (Here $x$ is position,$t$ is time)

A
$[M^0 L^{-1} T^0]$
B
$[M^0 L^0 T^0]$
C
$[M^0 L^{-1} T^{-1}]$
D
$[M^0 L^0 T^{-1}]$

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Given that the displacement of an oscillating particle is given by $y = A \sin(Bx + Ct + D)$. The dimensional formula for $ABCD$ is: (Here $x$ is position,$t$ is time)

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The potential energy of a particle varies with distance $x$ from a fixed origin as $U = \frac{A \sqrt{x}}{x^2 + B}$,where $A$ and $B$ are dimensional constants. Then,the dimensional formula for $AB$ is:

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