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JIPMER 2019
Exam
JIPMER
Year
2019
Subject
Physics
Chapter
Units and Dimensions

JIPMER 2019 · Physics · Units and Dimensions

What is dimensions of energy in terms of linear momentum (P), area...

What is dimensions of energy in terms of linear momentum (P), area (A) and Time (T)

Options

  1. A[P1 A1 T1]\left[\mathrm{P}^1 \mathrm{~A}^1 \mathrm{~T}^1\right]
  2. B[P2 A2 T−1]\left[\mathrm{P}^2 \mathrm{~A}^2 \mathrm{~T}^{-1}\right]
  3. C[P1 A1/2 T−1]\left[\mathrm{P}^1 \mathrm{~A}^{1 / 2} \mathrm{~T}^{-1}\right]
  4. D[P1/2A1/2T−1]\left[\mathrm{P}^{1 / 2} A^{1 / 2} T^{-1}\right]

Correct answer

C. [P1 A1/2 T−1]\left[\mathrm{P}^1 \mathrm{~A}^{1 / 2} \mathrm{~T}^{-1}\right]

Step-by-step solution

Dimensions of energy in terms of momentum (P). Area (A) and Time (T), is related of E=PaAbTC…(i)\mathrm{E}=\mathrm{P}^{\mathrm{a}} \mathrm{A}^{\mathrm{b}} \mathrm{T}^{\mathrm{C}}\ldots(i) Writing dimensional formula of both side, [ML2 T−2]=[MLT−1]a[L2]b[ T]c[ML2 T−2]]=[MaLa+T−a+c]\begin{aligned} & {\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]=\left[\mathrm{MLT}^{-1}\right]^a\left[L^2\right]^b[\mathrm{~T}]^{\mathrm{c}}} \\ & {\left[\mathrm{ML}^2 \mathrm{~T}^{-2]}\right]=\left[\mathrm{M}^{\mathrm{a}} \mathrm{L}^{\mathrm{a}+\mathfrak{}} \mathrm{T}^{-\mathrm{a}+\mathrm{c}}\right]} \\ & \end{aligned} Comparing the exponents, a=1,a+2 b=22 b=1 b=12−a+c=−2−1+c=−2⇒C=−1∴\mathrm{a}=1, \mathrm{a}+2 \mathrm{~b}=2 \begin{gathered} 2 \mathrm{~b}=1 \\ \mathrm{~b}=\frac{1}{2} \\ -\mathrm{a}+\mathrm{c}=-2-1+\mathrm{c}=-2 \\ \Rightarrow \quad \mathrm{C}=-1 \end{gathered} \therefore Form Eq. (i), we have \Rightarrow \quad \mathrm{E}=\left[\mathrm{PA}^{1 / 2} \mathrm{~T}^{-1}\right]
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