Question

If displacement of a particle is x=a^2t^3+bt^2+c then find dimensions of abc

Answer 1

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x = a²t³ + bt² + c

[x] = [a²t³] = [bt²] = [c]

Therefore,

[x] = [a²t³]

L = [a²] T³

[a²] = LT⁻³

[a] = L⁰°⁵ T⁻¹°⁵

[x] = [bt²]

[b] = LT⁻²

[x] = [c]

[c] = L

Therefore,

[abc] = L⁰°⁵ T⁻¹°⁵ * LT⁻² * L

         = L⁰°⁵ ⁺ ¹ ⁺ ¹  T⁻¹°⁵ ⁻ ²

         = L²°⁵ T ⁻³°⁵

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Answer 2

See the image. Hope it helps. Please mark as brainliest pleaseeeee