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MHT CET · Physics · Units and Dimensions

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. A \([\mathrm{A}]=\left[\mathrm{L}^{1} \mathrm{M}^{0} \mathrm{~T}^{-1}\right], \quad[\mathrm{D}]=\left[\mathrm{T}^{1}\right],[\mathrm{E}]=\left[\mathrm{T}^{1}\right]\)
  2. B \([\mathrm{A}]=\left[\mathrm{L}^{0} \mathrm{M}^{0} \mathrm{~T}^{-1}\right], \quad[\mathrm{D}]=\left[\mathrm{T}^{1}\right],[\mathrm{E}]=\left[\mathrm{L}^{1} \mathrm{~T}^{1}\right]\)
  3. C \([\mathrm{A}]=\left[\mathrm{L}^{1} \mathrm{M}^{1} \mathrm{~T}^{0}\right], \quad[\mathrm{D}]=\left[\mathrm{T}^{2}\right],[\mathrm{E}]=\left[\mathrm{L}^{1} \mathrm{~T}^{2}\right]\)
  4. D \([\mathrm{A}]=\left[\mathrm{L}^{1} \mathrm{M}^{0} \mathrm{~T}^{-1}\right], \quad[\mathrm{D}]=\left[\mathrm{M}^{1} \mathrm{~T}^{1}\right],[\mathrm{E}]=\left[\mathrm{M}^{1} \mathrm{~T}^{1}\right]\)
Verified Solution

Answer & Solution

Correct Answer

(A) \([\mathrm{A}]=\left[\mathrm{L}^{1} \mathrm{M}^{0} \mathrm{~T}^{-1}\right], \quad[\mathrm{D}]=\left[\mathrm{T}^{1}\right],[\mathrm{E}]=\left[\mathrm{T}^{1}\right]\)

Step-by-step Solution

Detailed explanation

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 \([T]\)