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JIPMER 2013
Exam
JIPMER
Year
2013
Subject
Physics
Chapter
Units and Dimensions

JIPMER 2013 · Physics · Units and Dimensions

A force F is given F=a t+b t^2 , where, t is time. What are the...

A force FF is given F=at+bt2F=a t+b t^2, where, tt is time. What are the dimensions of aa and bb ?

Options

  1. A[MLT−1]\left[\mathrm{MLT}^{-1}\right] and [MLT0]\left[\mathrm{MLT}^0\right]
  2. B[MLT−3]\left[\mathrm{MLT}^{-3}\right] and [ML2 T4]\left[\mathrm{ML}^2 \mathrm{~T}^4\right]
  3. C[MLT−4]\left[\mathrm{MLT}^{-4}\right] and [MLT1]\left[\mathrm{MLT}^1\right]
  4. D[MLT−3]\left[\mathrm{MLT}^{-3}\right] and [MLT−4]\left[\mathrm{MLT}^{-4}\right]

Correct answer

D. [MLT−3]\left[\mathrm{MLT}^{-3}\right] and [MLT−4]\left[\mathrm{MLT}^{-4}\right]

Step-by-step solution

Force F=at+bt2F=a t+b t^2 From principle of homogeneity, Dimension of at =[F][t]=[MLT−2][T]=\frac{[F]}{[t]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{T}]} =[MLT−3]=\left[\mathrm{MLT}^{-3}\right] Similarly, dimensions of $b=\frac{[F]}{\left[t^2\right]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{T}^2\right]}\left[\mathrm{MLT}^{-4}\right]
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