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: Concept
According to the principle of dimensional homogeneity, the argument of a trigonometric function must be dimensionless, and the dimensions of both sides of an equation must match.

Step 2: Meaning
$P$ represents power, which has the formula $\text{Work} / \text{Time}$. The term $\beta t$ must be dimensionless, meaning $[\beta][t] = [M^{0}L^{0}T^{0}]$.

Step 3: Analysis
Dimensions of Force $F = [MLT^{-2}]$ and Velocity $v = [LT^{-1}]$. Therefore, $P = [MLT^{-2}][LT^{-1}] = [ML^{2}T^{-3}]$. For $\beta t$, $[\beta][T] = 1 \implies [\beta] = [T^{-1}]$.

Step 4: Conclusion
The dimensions are $ML^{2}T^{-3}$ for $P$ and $T^{-1}$ for $\beta$.
Final Answer: (A)