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

• A. $M^{0}L^{0}T^{0}$, $ML^{-1}T^{0}$
• B. $M^{0}L^{0}T^{-1}$, $MLT^{0}$
• C. $M^{0}L^{0}T^{-1}$, $ML^{-1}T^{0}$
• D. $ML^{0}T^{-1}$, $ML^{-1}T$

Hint:

Always remember that transcendental arguments (inside cosines, exponents, or logarithms) have no dimensions, letting you isolate unknown values instantly.

Answer 1

The Correct Option is C

Step 1: Look at the rule for angles.
Whatever sits inside a $\cos$, $\sin$ or $\tan$ must be a pure number with no units. So the part $\alpha t$ has to be dimensionless.

Step 2: Find the dimensions of $\alpha$.
If $\alpha t$ has no units, then $\alpha$ must cancel the unit of time. So $[\alpha] = \dfrac{1}{[t]} = T^{-1}$, which we write as $M^{0}L^{0}T^{-1}$.

Step 3: Understand $\beta$.
The $\cos$ part is just a number, so it adds no units. That means $\beta$ has the same dimensions as $\dfrac{F}{v^{2}}$.

Step 4: Write the known dimensions.
Force has $[F] = MLT^{-2}$ and velocity has $[v] = LT^{-1}$. So $[v^{2}] = L^{2}T^{-2}$.

Step 5: Divide to get $\beta$.
\[ [\beta] = \frac{MLT^{-2}}{L^{2}T^{-2}} = ML^{-1}T^{0} \]
Step 6: Match with the options.
We got $[\alpha] = M^{0}L^{0}T^{-1}$ and $[\beta] = ML^{-1}T^{0}$. This is option 3.
\[ \boxed{[\alpha]=M^{0}L^{0}T^{-1},\ [\beta]=ML^{-1}T^{0}} \]