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If the displacement of a particle is represented by $y=A T+B T^{2}$ where $y$ is in metre and $t$ is in second, then calculate the dimension of $\mathrm{A}$ and B.

Solution : $\left.\left[\mathrm{LT}^{-1}\right] ; \mathrm{LT}^{-2}\right]$

L.H.S.; $y=[\mathrm{L}]$

R. H. S.; $A T=[L] . \quad \therefore \quad A=\frac{L}{T}=\left[L^{-1}\right]$

and, $\left.\quad \mathrm{BT}^{2}=[\mathrm{L}] \quad \therefore \mathrm{B}=\mathbf{L T}^{-2}\right]$

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