Question

position of a body with acceleration 'a' is given by x= Ka to the power m into t to the power n , here t is time. the values of m and n are: (K is dimensionless )

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{x}={\mathrm{ka}}^{\mathrm{m}}{\mathrm{t}}^{\mathrm{n}} \\ \mathrm{Dimesional}\mathrm{formula}\mathrm{of}\mathrm{x}=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right] \\ \mathrm{Dimesional}\mathrm{formula}\mathrm{of}\mathrm{a}=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-2}\right] \\ \mathrm{Dimesional}\mathrm{formula}\mathrm{of}\mathrm{t}=\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^1\right] \\ \left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]={\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-2}\right]}^{\mathrm{m}}\times {\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^1\right]}^{\mathrm{n}} \\ \left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]=\left[{\mathrm{M}}^0{\mathrm{L}}^{\mathrm{m}}{\mathrm{T}}^{-2\mathrm{m}}\right]\times \left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^{\mathrm{n}}\right] \\ \left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]=\left[{\mathrm{M}}^{0+0}{\mathrm{L}}^{\mathrm{m}+0}{\mathrm{T}}^{-2\mathrm{m}+\mathrm{n}}\right] \\ \mathrm{Compare}\mathrm{the}\mathrm{coefficients}, \\ \mathrm{m}=1 \\ -2\mathrm{m}+\mathrm{n}=0 \\ -2\left(1\right)+\mathrm{n}=0 \\ \mathrm{n}=2 \end{gathered}$$