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NEET 2010
Exam
NEET
Year
2010
Subject
Physics
Chapter
Units and Dimensions

NEET 2010 · Physics · Units and Dimensions

The dimension of (1)/(2) _0 E^2 , where _0 is permittivity of free...

The dimension of 12ε0E2\frac{1}{2} \varepsilon_0 \mathrm{E}^2, where ε0\varepsilon_0 is permittivity of free space and EE is electric field, is

Options

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

Correct answer

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

Step-by-step solution

\begin{aligned} & \text{ Dimensions of } \varepsilon_0=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right] \\ & \text{ Dimensions of } \mathrm{E}=\left[\mathrm{MLT}^{-3} \mathrm{~A}^{-1}\right] \\ & \therefore \text{ Dimensions of } \frac{1}{2} \varepsilon_0 \mathrm{E}^2=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right] \\ & \quad=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-3}\right] \quad \times\left[\mathrm{M}^2 \mathrm{~L}^2 \mathrm{~T}^{-6} \mathrm{~A}^{-2}\right] \end{aligned}
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