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AIIMS 201827 May
Exam
AIIMS
Year
2018
Date
27 May
Subject
Physics
Chapter
Units and Dimensions

AIIMS 2018 · 27 May · Physics · Units and Dimensions

What are the dimensions of (A)/(B) in the relation F=A x +B t^2 ,...

What are the dimensions of AB\frac{A}{B} in the relation F=Ax+Bt2F=A \sqrt{x}+B t^2, where FF is the force, xx is distance and tt is time ?

Options

  1. A[ML2 T−2]\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]
  2. B[L−1/2 T2]\left[\mathrm{L}^{-1 / 2} \mathrm{~T}^2\right]
  3. C[L−1/2 T−1]\left[\mathrm{L}^{-1 / 2} \mathrm{~T}^{-1}\right]
  4. D[LT−2]\left[\mathrm{LT}^{-2}\right]

Correct answer

B. [L−1/2 T2]\left[\mathrm{L}^{-1 / 2} \mathrm{~T}^2\right]

Step-by-step solution

Given, F=Ax+Bt2F=A \sqrt{x}+B t^2 Here, F=F= force, x=x= distance and t=t= time Dimensionally, F=Ax=Bt2F=A \sqrt{x}=B t^2 \Rightarrow \quad \frac{A}{B}=\frac{t^2}{\sqrt{x}} \Rightarrow \frac{A}{B}=\frac{\left[\mathrm{T}^2\right]}{\left[\mathrm{L}^{1 / 2}\right]}=\left[\mathrm{L}^{-1 / 2} \mathrm{~T}^2\right]
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