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The velocity vv of a particle at time tt is given by v=at+bt+cv = at + \frac{b}{t+c}, where aa, bb and cc are constants. The dimensions of aa, bb and cc are respectively:

A

[LT2],[L][L T^{-2}], [L] and [T][T]

B

[L2],[T][L^2], [T] and [LT2][LT^2]

C

[LT2],[LT][LT^2], [LT] and [L][L]

D

[L],[LT][L], [LT] and [T2][T^2]

Step-by-Step Solution

According to the principle of homogeneity of dimensions, only physical quantities with the same dimensions can be added or subtracted .

  1. Dimension of cc: In the equation, cc is added to tt (time) in the denominator of the second term. Therefore, the dimensions of cc must be the same as that of time tt. So, [c]=[T][c] = [T].
  2. Dimension of aa: Each additive term in the equation must have the same dimensions as the quantity on the left-hand side, which is velocity vv [LT1][L T^{-1}]. For the term atat: [a][t]=[v][a][T]=[LT1][a]=[LT1][T]=[LT2][a][t] = [v] \Rightarrow [a][T] = [L T^{-1}] \Rightarrow [a] = \frac{[L T^{-1}]}{[T]} = [L T^{-2}].
  3. Dimension of bb: For the term bt+c\frac{b}{t+c}, the entire term must have the dimensions of velocity. So, [b][t+c]=[v]\frac{[b]}{[t+c]} = [v]. Since [t+c]=[T][t+c] = [T], we have [b][T]=[LT1][b]=[LT1]×[T]=[L]\frac{[b]}{[T]} = [L T^{-1}] \Rightarrow [b] = [L T^{-1}] \times [T] = [L].

Thus, the dimensions of aa, bb, and cc are [LT2],[L][L T^{-2}], [L], and [T][T] respectively.

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