Question

The energy \(E\) of a system is a function of time \(t\) and is given by \[ E(t)=\alpha t-\beta t^3 \] where \(\alpha\) and \(\beta\) are constants. The dimensions of \(\alpha\) and \(\beta\) are

• A. \([ML^2T^{-1}]\) and \([ML^2T]\)
• B. \([LT^{-1}]\) and \([LT]\)
• C. \([ML^2T^{-3}]\) and \([ML^2T^{-5}]\)
• D. \([MLT^{-1}]\) and \([MLT]\)

Hint:

In dimensional analysis, every term added or subtracted in a physical equation must have the same dimensions.

Answer 1

The Correct Option is C

Step 1: Recall the dimensional formula of energy.
Energy has the dimensional formula $[E] = [ML^2T^{-2}]$. Since the equation $E(t) = \alpha t - \beta t^3$ must be dimensionally consistent, every term on the right side must also have the dimension of energy.
Step 2: Find the dimension of $\alpha$.
From the first term $\alpha t$: \[ [\alpha][T] = [ML^2T^{-2}] \] Dividing both sides by $[T]$: \[ [\alpha] = \frac{[ML^2T^{-2}]}{[T]} = [ML^2T^{-3}] \]
Step 3: Find the dimension of $\beta$.
From the second term $\beta t^3$: \[ [\beta][T^3] = [ML^2T^{-2}] \] Dividing both sides by $[T^3]$: \[ [\beta] = \frac{[ML^2T^{-2}]}{[T^3]} = [ML^2T^{-5}] \]
Step 4: Check the negative sign does not affect dimensions.
The subtraction in $\alpha t - \beta t^3$ does not change dimensional requirements. Both terms independently must equal $[ML^2T^{-2}]$. So the analysis for each constant is correct.
Step 5: Verify the answers.
We found $[\alpha] = [ML^2T^{-3}]$ and $[\beta] = [ML^2T^{-5}]$. Checking option 3: $[ML^2T^{-3}]$ and $[ML^2T^{-5}]$ - this matches exactly.
Step 6: State the final result.
The correct dimensions are $[\alpha] = [ML^2T^{-3}]$ and $[\beta] = [ML^2T^{-5}]$, corresponding to option 3. \[ \boxed{[ML^2T^{-3}] \text{ and } [ML^2T^{-5}]} \]