In the expression $A=B+\frac{C}{D+E}$, the dimensions of physical quantities $B$ and $C$ are…

In the expression $A=B+\frac{C}{D+E}$, the dimensions of physical quantities $B$ and $C$ are $\left[\mathrm{L}^{1} \mathrm{M}^{0} \mathrm{~T}^{-1}\right]$ and $\left[\mathrm{L}^{1} \mathrm{M}^{0} \mathrm{~T}^{0}\right]$ respectively. The dimensions of quantities $\mathrm{A}, \mathrm{D}$ and $\mathrm{E}$ are
  1. $[\mathrm{A}]=\left[\mathrm{L}^{1} \mathrm{M}^{0} \mathrm{~T}^{-1}\right], \quad[\mathrm{D}]=\left[\mathrm{T}^{1}\right],[\mathrm{E}]=\left[\mathrm{T}^{1}\right]$
  2. $[\mathrm{A}]=\left[\mathrm{L}^{0} \mathrm{M}^{0} \mathrm{~T}^{-1}\right], \quad[\mathrm{D}]=\left[\mathrm{T}^{1}\right],[\mathrm{E}]=\left[\mathrm{L}^{1} \mathrm{~T}^{1}\right]$
  3. $[\mathrm{A}]=\left[\mathrm{L}^{1} \mathrm{M}^{1} \mathrm{~T}^{0}\right], \quad[\mathrm{D}]=\left[\mathrm{T}^{2}\right],[\mathrm{E}]=\left[\mathrm{L}^{1} \mathrm{~T}^{2}\right]$
  4. $[\mathrm{A}]=\left[\mathrm{L}^{1} \mathrm{M}^{0} \mathrm{~T}^{-1}\right], \quad[\mathrm{D}]=\left[\mathrm{M}^{1} \mathrm{~T}^{1}\right],[\mathrm{E}]=\left[\mathrm{M}^{1} \mathrm{~T}^{1}\right]$

Solution

Given, Dimension of $B$ is $\left[M^{0} L T^{-1}\right]$ \& $C$ is $\left[M^{0} L T^{0}\right]$ Now we know that, same dimensions can be added and the results we get are in same dimensions. So, dimension of $A=B=\left[M^{0} L T^{-1}\right]$ Let, the dimension of $D$ \& $E$ is $\left[M^{X} L^{Y} T^{Z}\right]$ $\therefore M^{0} L T^{-1}=\frac{M^{0} L T^{0}}{M^{X} L^{Y} T^{Z}}$ $\Rightarrow M^{X} L^{Y} T^{Z}=M^{0} L^{0} T^{1}$ $X=0, \quad Y=0 \quad \& \quad Z=1$ So, the dimension of $D$ \& $E$ is $[T]$

Asked in: MHT CET 2020 (20 Oct Shift 2)

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