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1.Units, Dimensions and Measurement
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ધારો કે $[ {\varepsilon _0} ]$ એ શૂન્યાવકાશની પરમિટિવિટિ (પરાવૈદ્યુતિક) દર્શાવે છે.જો $M=$ દળ, $L=$ લંબાઇ, $T=$ સમય અને $A=$ વિદ્યુતપ્રવાહ દર્શાવે, તો .........
A$[{\varepsilon _0}]=[M^{-1}L^{-3}T^2A]$
B$[{\varepsilon _0}]=[M^{-1}L^{-3}T^4A^2]$
C$[{\varepsilon _0}]=[M^{-1}L^2T^{-1}A^{-2}]$
D$[{\varepsilon _0}]=[M^{-1}L^2T^{-1}A]$
(JEE MAIN-2013)
Solution
As we know $F = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{R^2}}} \Rightarrow {\varepsilon _0} = \frac{{{q_1}{q_2}}}{{4\pi F{R^2}}}$
Hence ${\varepsilon _0} = \frac{{{{\left[ {AT} \right]}^2}}}{{ML{T^{ – 2}}{L^2}}} = \left[ {{M^{ -1}}{L^{ – 3}}{T^{ 4}}{A^2}} \right]$
Hence ${\varepsilon _0} = \frac{{{{\left[ {AT} \right]}^2}}}{{ML{T^{ – 2}}{L^2}}} = \left[ {{M^{ -1}}{L^{ – 3}}{T^{ 4}}{A^2}} \right]$
Standard 11
Physics
