Question:
$\beta=\frac{F}{v^{2}}\cos(\alpha t)$, if $F$ is force, $v$ is velocity, $t$ is time, then the dimensional formulae of $\alpha$, $\beta$ are respectively
$\beta=\frac{F}{v^{2}}\cos(\alpha t)$, if $F$ is force, $v$ is velocity, $t$ is time, then the dimensional formulae of $\alpha$, $\beta$ are respectively
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Always remember that transcendental arguments (inside cosines, exponents, or logarithms) have no dimensions, letting you isolate unknown values instantly.
Updated On: Jun 3, 2026
- $M^{0}L^{0}T^{0}$, $ML^{-1}T^{0}$
- $M^{0}L^{0}T^{-1}$, $MLT^{0}$
- $M^{0}L^{0}T^{-1}$, $ML^{-1}T^{0}$
- $ML^{0}T^{-1}$, $ML^{-1}T$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
The argument of a trigonometric function must be completely dimensionless, meaning $[\alpha t] = M^0L^0T^0$. Furthermore, the overall trigonometric function component $\cos(\alpha t)$ itself is dimensionless.
Step 2: Meaning
Since the argument is dimensionless, the dimension of $\alpha$ must be the reciprocal of time: $[\alpha] = \frac{1}{[t]} = T^{-1} = M^0L^0T^{-1}$.
Step 3: Analysis
For the coefficient $\beta$, we equate its dimensions to the remaining terms: $[\beta] = \frac{[F]}{[v]^2}$. Substituting the standard dimensional formulas for force ($[F] = MLT^{-2}$) and velocity ($[v] = LT^{-1}$): $[\beta] = \frac{MLT^{-2}}{(LT^{-1})^2} = \frac{MLT^{-2}}{L^2T^{-2}} = ML^{-1}T^0$.
Step 4: Conclusion
Combining both results gives $[\alpha] = M^0L^0T^{-1}$ and $[\beta] = ML^{-1}T^0$, which precisely matches option (C).
Final Answer: (C)
The argument of a trigonometric function must be completely dimensionless, meaning $[\alpha t] = M^0L^0T^0$. Furthermore, the overall trigonometric function component $\cos(\alpha t)$ itself is dimensionless.
Step 2: Meaning
Since the argument is dimensionless, the dimension of $\alpha$ must be the reciprocal of time: $[\alpha] = \frac{1}{[t]} = T^{-1} = M^0L^0T^{-1}$.
Step 3: Analysis
For the coefficient $\beta$, we equate its dimensions to the remaining terms: $[\beta] = \frac{[F]}{[v]^2}$. Substituting the standard dimensional formulas for force ($[F] = MLT^{-2}$) and velocity ($[v] = LT^{-1}$): $[\beta] = \frac{MLT^{-2}}{(LT^{-1})^2} = \frac{MLT^{-2}}{L^2T^{-2}} = ML^{-1}T^0$.
Step 4: Conclusion
Combining both results gives $[\alpha] = M^0L^0T^{-1}$ and $[\beta] = ML^{-1}T^0$, which precisely matches option (C).
Final Answer: (C)
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