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1.Units, Dimensions and Measurement
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જો $E$ અને $G$ એ અનુક્રમે ઊર્જા અને ગુરૂત્વાકર્ષી અચળાંક દર્શાવે તો $\frac{\mathrm{E}}{\mathrm{G}}$નું પરિમાણ $.....$ થશે.
A$[\mathrm{M}]\left[\mathrm{L}^{-1}\right]\left[\mathrm{T}^{-1}\right]$
B$\left[\mathrm{M}^{2}\right]\left[\mathrm{L}^{-1}\right]\left[\mathrm{T}^{0}\right]$
C$[\mathrm{M}]\left[\mathrm{L}^{0}\right]\left[\mathrm{T}^{0}\right]$
D$\left[\mathrm{M}^{2}\right]\left[\mathrm{L}^{-2}\right]\left[\mathrm{T}^{-1}\right]$
(NEET-2021)
Solution
$E=$ energy $=\left[M L^{2} T^{-2}\right]$
$G=$ Gravitational constant $=\left[\mathrm{M}^{-1} L^{3} \mathrm{~T}^{-2}\right]$
So, $\frac{E}{G}=\frac{[E]}{[G]}=\frac{M L^{2} T^{-2}}{M^{-1} L^{3} T^{-2}}=\left[M^{2} L^{-1} T^{0}\right]$
$G=$ Gravitational constant $=\left[\mathrm{M}^{-1} L^{3} \mathrm{~T}^{-2}\right]$
So, $\frac{E}{G}=\frac{[E]}{[G]}=\frac{M L^{2} T^{-2}}{M^{-1} L^{3} T^{-2}}=\left[M^{2} L^{-1} T^{0}\right]$
Standard 11
Physics
