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 find the dimension of \( \frac{ABC}{D} \), we start by analyzing the given equation for the position of a particle: \( x(t) = A \sin t + B \cos^2 t + Ct^2 + D \). The position \( x(t) \) has the dimension of length, denoted as \( [x] = L \).

Each term in the equation must also have the dimension of length.

• The term \( A \sin t \) has the dimension of \( L \), so \( [A] = L \).
• The term \( B \cos^2 t \) also has the dimension of \( L \), hence, \( [B] = L \).
• The term \( Ct^2 \) must have the dimension \( L \). Since \( [t] = T \), we have \( [Ct^2] = [C][T^2] = L \). Therefore, \( [C] = \frac{L}{T^2} \).
• The term \( D \) is a constant with the dimension of length, so \( [D] = L \).

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

\( \left[\frac{ABC}{D}\right] = \frac{L \times L \times \frac{L}{T^2}}{L} = \frac{L^3}{L \times T^2} = L^2 T^{-2} \).

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