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BITSAT 2012
Exam
BITSAT
Year
2012
Subject
Physics
Chapter
Units and Dimensions

BITSAT 2012 · Physics · Units and Dimensions

If the force is given by F=at+bt^ 2 with t as time. The dimensions...

If the force is given by F=at+bt2\mathrm{F}=\mathrm{at}+\mathrm{bt}^{2} with t\mathrm{t} as time. The dimensions of a and b are

Options

  1. A[MLT−4]\left[\mathrm{MLT}^{-4}\right] and [MLT−2]\left[\mathrm{MLT}^{-2}\right]
  2. B[MLT−3]\left[\mathrm{MLT}^{-3}\right] and [MLT−4]\left[\mathrm{MLT}^{-4}\right]
  3. C[ML2 T−3]\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right] and [ML2 T−2]\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]
  4. D[ML2 T−3]\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right] and [ML3 T−4]\left[\mathrm{ML}^{3} \mathrm{~T}^{-4}\right]

Correct answer

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

Step-by-step solution

Dimension of at = Dimension of F [at]=[F]⇒[a]=[Ft][a t]=[F] \Rightarrow[a]=\left[\frac{F}{t}\right] [b]=[MLT−2T]⇒[a]=[MLT−3][b]=\left[\frac{M L T^{-2}}{T}\right] \Rightarrow[a]=\left[M L T^{-3}\right] Dimension of bt2=\mathrm{bt}^{2}= Dimension of F\mathrm{F} [bt2]=[F]⇒[b]=[Ft2]\left[\mathrm{bt}^{2}\right]=[\mathrm{F}] \Rightarrow[\mathrm{b}]=\left[\frac{\mathrm{F}}{\mathrm{t}^{2}}\right] [b]=[MLT−4 T2]⇒[b]=[MLT−4][\mathrm{b}]=\left[\frac{\mathrm{ML} \mathrm{T}^{-4}}{\mathrm{~T}^{2}}\right] \Rightarrow[\mathrm{b}]=\left[\mathrm{MLT}^{-4}\right]
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