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
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
Show Hint
In dimensional analysis, every term added or subtracted in a physical equation must have the same dimensions.
Updated On: Jun 22, 2026
- \([ML^2T^{-1}]\) and \([ML^2T]\)
- \([LT^{-1}]\) and \([LT]\)
- \([ML^2T^{-3}]\) and \([ML^2T^{-5}]\)
- \([MLT^{-1}]\) and \([MLT]\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Write the dimensional formula of energy.
The dimensional formula of energy is \[ [E]=[ML^2T^{-2}] \] Given, \[ E(t)=\alpha t-\beta t^3 \] Since both terms on the right side represent energy, each term must have the same dimension as energy.
Step 2: Find the dimension of \(\alpha\).
From the term \[ \alpha t \] we have \[ [\alpha t]=[E] \] So, \[ [\alpha][T]=[ML^2T^{-2}] \] Therefore, \[ [\alpha]=\frac{[ML^2T^{-2}]}{[T]} \] \[ [\alpha]=[ML^2T^{-3}] \]
Step 3: Find the dimension of \(\beta\).
From the term \[ \beta t^3 \] we have \[ [\beta t^3]=[E] \] So, \[ [\beta][T^3]=[ML^2T^{-2}] \] Therefore, \[ [\beta]=\frac{[ML^2T^{-2}]}{[T^3]} \] \[ [\beta]=[ML^2T^{-5}] \]
Step 4: Final conclusion.
Hence, the dimensions of \(\alpha\) and \(\beta\) are respectively \[ \boxed{[ML^2T^{-3}]\ \text{and}\ [ML^2T^{-5}]} \]
The dimensional formula of energy is \[ [E]=[ML^2T^{-2}] \] Given, \[ E(t)=\alpha t-\beta t^3 \] Since both terms on the right side represent energy, each term must have the same dimension as energy.
Step 2: Find the dimension of \(\alpha\).
From the term \[ \alpha t \] we have \[ [\alpha t]=[E] \] So, \[ [\alpha][T]=[ML^2T^{-2}] \] Therefore, \[ [\alpha]=\frac{[ML^2T^{-2}]}{[T]} \] \[ [\alpha]=[ML^2T^{-3}] \]
Step 3: Find the dimension of \(\beta\).
From the term \[ \beta t^3 \] we have \[ [\beta t^3]=[E] \] So, \[ [\beta][T^3]=[ML^2T^{-2}] \] Therefore, \[ [\beta]=\frac{[ML^2T^{-2}]}{[T^3]} \] \[ [\beta]=[ML^2T^{-5}] \]
Step 4: Final conclusion.
Hence, the dimensions of \(\alpha\) and \(\beta\) are respectively \[ \boxed{[ML^2T^{-3}]\ \text{and}\ [ML^2T^{-5}]} \]
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