Question

In the equation \( P = \frac{C+S}{D} \), \( P \) and \( S \) represent pressure and distance respectively. The dimensions of \( \frac{P}{D} \) are

• A. \( [L^1 M^1 T^2] \)
• B. \( [L^1 M^{-1} T^2] \)
• C. \( [L^1 M^0 T^{-1}] \)
• D. \( [L^0 M^1 T^1] \)

Hint:

To find the dimensions of a derived quantity, divide the dimensions of the numerator by the dimensions of the denominator and simplify.

Answer 1

The Correct Option is B

Step 1: Understanding the equation.
The equation \( P = \frac{C+S}{D} \) involves pressure \( P \), distance \( D \), and other variables \( C \) and \( S \). The dimensions of \( P \) (pressure) and \( D \) (distance) are: \[ [P] = [M L^{-1} T^{-2}], \quad [D] = [L] \] Step 2: Finding the dimensions of \( \frac{P}{D} \).
The dimensions of \( \frac{P}{D} \) are: \[ \left[\frac{P}{D}\right] = \frac{[M L^{-1} T^{-2}]}{[L]} = [M L^{-2} T^{-2}] \] Step 3: Conclusion.
Thus, the dimensions of \( \frac{P}{D} \) are \( [L^1 M^{-1} T^2] \), corresponding to option (B).