Question:
Position of a body with acceleration \(\alpha\) is given by \(x = k \alpha^m t^n\), here \(t\) is time. Find the dimension of \(m\) and \(n\)
Position of a body with acceleration \(\alpha\) is given by \(x = k \alpha^m t^n\), here \(t\) is time. Find the dimension of \(m\) and \(n\)
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For constant acceleration, \(x = \frac{1}{2}\alpha t^2\). So \(m = 1, n = 2\) matches.
Updated On: Apr 23, 2026
- \(m = 1, n = 1\)
- \(m = 1, n = 2\)
- \(m = 2, n = 1\)
- \(m = 2, n = 2\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Dimensions: \([x] = L\), \([\alpha] = LT^{-2}\), \([t] = T\).
Step 2: Detailed Explanation:
\([x] = [\alpha]^m [t]^n \Rightarrow L = (LT^{-2})^m (T)^n = L^m T^{-2m + n}\).
Equating powers: For L: \(1 = m \Rightarrow m = 1\).
For T: \(0 = -2m + n = -2(1) + n \Rightarrow n = 2\).
Step 3: Final Answer:
Thus, \(m = 1, n = 2\).
Dimensions: \([x] = L\), \([\alpha] = LT^{-2}\), \([t] = T\).
Step 2: Detailed Explanation:
\([x] = [\alpha]^m [t]^n \Rightarrow L = (LT^{-2})^m (T)^n = L^m T^{-2m + n}\).
Equating powers: For L: \(1 = m \Rightarrow m = 1\).
For T: \(0 = -2m + n = -2(1) + n \Rightarrow n = 2\).
Step 3: Final Answer:
Thus, \(m = 1, n = 2\).
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