Question

If $P = F \cdot v \sin \beta t$ where $F$ is force and $v$ is velocity then the dimensions of $P$ and $\beta$ are

• A. $ML^{2}T^{-3}$, $T^{-1}$
• B. $MLT^{-2}$, $T^{-2}$
• C. $ML^{2}T^{-1}$, $T^{-1}$
• D. $ML^{2}T^{3}$, $T^{-2}$

Hint:

Always remember: Angles and arguments of trig, log, or exponential functions are always dimensionless ($M^{0}L^{0}T^{0}$).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In any physical equation, the dimensions on both sides must be equal. Furthermore, the argument of a trigonometric function (like sine) must be dimensionless.
Step 2: Key Formula or Approach:
1. Dimensions of Force (\(F\)): \( [MLT^{-2}] \).
2. Dimensions of Velocity (\(v\)): \( [LT^{-1}] \).
3. Dimensions of Time (\(t\)): \( [T] \).
Step 3: Detailed Explanation:
For \( P = F \cdot v \sin \beta t \): The term \( \sin \beta t \) is dimensionless. Therefore, the dimensions of \( P \) are simply the dimensions of \( F \times v \): \[ [P] = [MLT^{-2}] \times [LT^{-1}] = [ML^2T^{-3}] \] For the argument \( \beta t \) to be dimensionless: \[ [\beta t] = [M^0L^0T^0] \] \[ [\beta][T] = [1] \implies [\beta] = [T^{-1}] \]
Step 4: Final Answer:
The dimensions are \( [P] = ML^2T^{-3} \) and \( [\beta] = T^{-1} \).