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AIIMS 2004
Exam
AIIMS
Year
2004
Subject
Physics
Chapter
Units and Dimensions

AIIMS 2004 · Physics · Units and Dimensions

Using mass (M) , length (L) , time (T) and current (A) as...

Using mass (M)(M), length (L)(L), time (T)(T) and current (A)(A) as fundamental quantities, the dimension of permittivity is

Options

  1. A[ML−2 T2 A]\left[\mathrm{ML}^{-2} \mathrm{~T}^2 \mathrm{~A}\right]
  2. B[M−1 L−3 T4 A2]\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]
  3. C[MLT−2 A]\left[\mathrm{MLT}^{-2} \mathrm{~A}\right]
  4. D[ML2 T−1 A2]\left[\mathrm{ML}^2 \mathrm{~T}^{-1} \mathrm{~A}^2\right].

Correct answer

B. [M−1 L−3 T4 A2]\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]

Step-by-step solution

Force of attraction between two charges is given by F=14πε0q1q2r2⇒ε0=14π1Fq1q2r2=A2T2MLT−2L2=[M−1L−3T4A2].\begin{aligned}F & =\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}\\ \Rightarrow \quad \varepsilon_0 & =\frac{1}{4 \pi} \frac{1}{F} \frac{q_1 q_2}{r^2}=\frac{\mathrm{A}^2 \mathrm{ T}^2}{\mathrm{MLT}^{-2} \mathrm{ L}^2}\\ & =\left[\mathrm{M}^{-1} \mathrm{ L}^{-3} \mathrm{ T}^4 \mathrm{ A}^2\right]. \end{aligned} (where 4π4 \pi is dimensionless) = \left[\mathrm{M}^{-1} \mathrm{ L}^{-3} \mathrm{ T}^4 \mathrm{ A}^2\right]
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