Find Dimensions of ABC/D JEE Main Question

 

❓ Question

The position of a particle moving along the $x$-axis is given by

$$x(t)=A\sin t+B\cos^2 t+Ct^2+D$$

where $t$ is time.

Find the dimensions of

$$\frac{ABC}{D}$$

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✍️ Solution

According to the principle of dimensional homogeneity, every term in the equation must have the same dimensions as position $x$.

$$[x]=[A\sin t]=[B\cos^2 t]=[Ct^2]=[D]$$

Since position has dimension

$$[x]=[L]$$

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Step 1: Dimension of $A$

The trigonometric function is dimensionless:

$$[\sin t]=1$$

Therefore,

$$[A\sin t]=[L]$$
$$\boxed{[A]=[L]}$$

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Step 2: Dimension of $B$

Similarly,

$$[\cos^2 t]=1$$

Therefore,

$$[B\cos^2 t]=[L]$$
$$\boxed{[B]=[L]}$$

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Step 3: Dimension of $C$

Since

$$[Ct^2]=[L]$$

and

$$[t^2]=[T^2]$$

we get

$$[C][T^2]=[L]$$

Therefore,

$$\boxed{[C]=[LT^{-2}]}$$

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Step 4: Dimension of $D$

Since $D$ is added directly to position,

$$\boxed{[D]=[L]}$$

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Step 5: Dimension of $\frac{ABC}{D}$

$$\left[\frac{ABC}{D}\right] = \frac{[A][B][C]}{[D]}$$

Substituting:

$$= \frac{[L][L][LT^{-2}]}{[L]}$$
$$= [L^2T^{-2}]$$

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✅ Final Answer

$$\boxed{\left[\frac{ABC}{D}\right]=[L^2T^{-2}]}$$

Key Idea

Trigonometric functions such as $\sin$ and $\cos$ are dimensionless, so the coefficients $A$ and $B$ must carry the dimension of position.