કોઇ પદ્ધતિમાં પ્રકાશનો વેગ (c), ગુરુત્વાકર્ષણ | vedclass

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Question

Let time, $T \propto c^{x} G^{y} h^{z}$

$\Rightarrow T=k c^{x} G^{y} h^{z}$

Taking dimensions on both sides $\left[M^{0} L^{0} T^{1}\right]=\lef

Vedclass · Accepted Answer

${G^{1/2}}{h^{1/2}}{c^{ - 5/2}}$

Vedclass · Answer

Let time, $T \propto c^{x} G^{y} h^{z}$

$\Rightarrow T=k c^{x} G^{y} h^{z}$

Taking dimensions on both sides $\left[M^{0} L^{0} T^{1}ight]=\left[L T^{-1}ight]^{x}\left[M^{-1} L^{3} T^{-2}ight]^{y}\left[M L^{2} T^{-1}ight]^{z}$

$i . e$

$\left[M^{0} L^{0} T^{1}ight]=\left[M^{-y+z} L^{x+3 y+2 z} T^{-x-2 y-z}ight]$

Equating power of $M, L,$ Ton both sides, we get

$-y+z=0 \quad \ldots(1)$

$x+3 y+2 z=0 \quad \ldots(2)$

$-x-2 y-z=1 \quad \ldots .(3)$

From $(1) \Rightarrow z=y$

Adding (2) and $(3) \Rightarrow y+z=1$

or $2 y=1 \quad[	ext { From }]$

i.e, $y=\frac{1}{2}$

$	herefore z=y=\frac{1}{2}$

Putting these values in (2) we get $x+\frac{3}{2}+1=0$ or $x=\frac{-5}{2}$

Hence$,[T]=\left[G^{1 / 2} h^{1 / 2} c^{-5 / 2}ight]$

લિસ્ટ $-I$	લિસ્ટ $-II$
$(a)$ કેપેસીટન્સ, $C$	$(i)$ ${M}^{1} {L}^{1} {T}^{-3} {A}^{-1}$
$(b)$ શૂન્યાવકાશની પરમિટિવિટી, $\varepsilon_{0}$	$(ii)$ ${M}^{-1} {L}^{-3} {T}^{4} {A}^{2}$
$(c)$ શૂન્યાવકાશની પરમીએબીલીટી, $\mu_{0}$	$(iii)$ ${M}^{-1} L^{-2} T^{4} A^{2}$
$(d)$ વિદ્યુતક્ષેત્ર, $E$	$(iv)$ ${M}^{1} {L}^{1} {T}^{-2} {A}^{-2}$

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