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

To determine the dimension of \(\frac{ABC}{D}\), we analyze the given expression for the position of a particle on the x-axis:

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

We examine each term individually:

1. \(A \sin t\): Since \(\sin t\) is dimensionless, and \(x(t)\) has dimensions of length ([L]), the dimension of \(A\) must be \([L]\).
2. \(B \cos^2 t\): Similarly, as \(\cos^2 t\) is dimensionless, the dimension of \(B\) is also \([L]\).
3. \(Ct^2\): The term \(t^2\) has dimensions \([T^2]\). For \(C t^2\) to have dimensions of length, the dimension of \(C\) must be \([L T^{-2}]\).
4. \(D\): As a constant term representing position, the dimension of \(D\) is \([L]\).

Now, we determine the dimension of the expression \(\frac{ABC}{D}\):

\[\frac{ABC}{D} = \frac{[L][L][L T^{-2}]}{[L]}\]

Simplifying this expression yields:

\[[L][L][L T^{-2}]/[L] = [L^2 T^{-2}]\]

Consequently, the dimension of \(\frac{ABC}{D}\) is \(L^2 T^{-2}\).

The final answer is:

\(L^2 T^{-2}\)