Question

The position of a particle moving on x-axis is given by \( x(t) = A \sin t + B \cos^2 t + Ct^2 + D \), where \( t \) is time. The dimension of \( \frac{ABC}{D} \) is:

• A. \( L \)
• B. \( L^3 T^{-2} \)
• C. \( L^2 T^{-2} \)
• D. \( L^2 \)

Hint:

When dealing with equations involving time, make sure to carefully account for the powers of time, such as \( T^2 \) in the case of terms like \( Ct^2 \). Dimensionless trigonometric functions like \( \sin t \) or \( \cos^2 t \) do not affect the dimensional analysis.

Answer 1

The Correct Option is C

We are asked to find the dimension of \( \frac{ABC}{D} \). Let's first determine the dimensions of \( A \), \( B \), \( C \), and \( D \), based on the given position equation for the particle:
The equation is: \[ x(t) = A \sin t + B \cos^2 t + Ct^2 + D, \] where \( x(t) \) is the position of the particle at time \( t \), and \( A \), \( B \), \( C \), and \( D \) are constants to be determined. We know that the position of the particle \( x(t) \) has the dimension of length, i.e., \( [x(t)] = [L] \).
Step 1: Dimension of \( A \)
The term \( A \sin t \) is dimensionally consistent, so the dimension of \( A \) must be \( [L] \). This is because \( \sin t \) is dimensionless. \[ [A] = [L] \] Step 2: Dimension of \( B \)
Similarly, \( B \cos^2 t \) is dimensionally consistent, and since \( \cos^2 t \) is dimensionless, the dimension of \( B \) must be \( [L] \). \[ [B] = [L] \] Step 3: Dimension of \( C \)
The term \( C t^2 \) must have the dimension of length. Since \( t^2 \) has the dimension \( [T^2] \), the dimension of \( C \) must be \( [L T^{-2}] \) to make the term \( C t^2 \) have the dimension of length.
\[ [C] = [L T^{-2}] \] Step 4: Dimension of \( D \)
The term \( D \) is simply a constant term added to the equation. Since it is a length, the dimension of \( D \) is also \( [L] \). \[ [D] = [L] \] Step 5: Dimension of \( \frac{ABC}{D} \)
Now, we find the dimension of the expression \( \frac{ABC}{D} \): \[ \left[ \frac{ABC}{D} \right] = \frac{[A] \times [B] \times [C]}{[D]} = \frac{[L] \times [L] \times [L T^{-2}]}{[L]} = [L^2 T^{-2}]. \] Thus, the dimension of \( \frac{ABC}{D} \) is \( [L^2 T^{-2}] \). Therefore, the correct answer is: \[ \boxed{L^2 T^{-2}}. \]