Planck's constant h,speed of light c,and grav | vedclass

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(C) The dimensions of the constants are:$h = [M L^2 T^{-1}]$$c = [L T^{-1}]$$G = [M^{-1} L^3 T^{-2}]$To find the mass $M$,we assume $M = k h^a c^b G^d

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$(A, C, D)$

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(C) The dimensions of the constants are:$h = [M L^2 T^{-1}]$$c = [L T^{-1}]$$G = [M^{-1} L^3 T^{-2}]$To find the mass $M$,we assume $M = k h^a c^b G^d$. Substituting dimensions:$[M] = [M L^2 T^{-1}]^a [L T^{-1}]^b [M^{-1} L^3 T^{-2}]^d$$[M] = M^{a-d} L^{2a+b+3d} T^{-a-b-2d}$Comparing powers:$a - d = 1 \implies a = 1 + d$$2a + b + 3d = 0$$-a - b - 2d = 0 \implies b = -a - 2d = -(1+d) - 2d = -1 - 3d$Substituting into the second equation: $2(1+d) + (-1-3d) + 3d = 0 \implies 2 + 2d - 1 = 0 \implies d = -1/2$.Then $a = 1 - 1/2 = 1/2$ and $b = -1 - 3(-1/2) = 1/2$.Thus,$M \propto h^{1/2} c^{1/2} G^{-1/2} = \sqrt{\frac{hc}{G}}$.Similarly for length $L = k h^x c^y G^z$:$[L] = [M L^2 T^{-1}]^x [L T^{-1}]^y [M^{-1} L^3 T^{-2}]^z$$x - z = 0 \implies x = z$$2x + y + 3z = 1$$-x - y - 2z = 0 \implies y = -x - 2z = -3z$$2z - 3z + 3z = 1 \implies 2z = 1 \implies z = 1/2$.Thus $x = 1/2, y = -3/2, z = 1/2$.$L \propto h^{1/2} c^{-3/2} G^{1/2} = \sqrt{\frac{hG}{c^3}}$.From these relations:$M \propto \sqrt{h}, M \propto \sqrt{c}, M \propto 1/\sqrt{G}$$L \propto \sqrt{h}, L \propto \sqrt{G}, L \propto 1/\sqrt{c^3}$Therefore,options $(A), (C), (D)$ are correct.

Planck's constant $h$,speed of light $c$,and gravitational constant $G$ are used to form a unit of length $L$ and a unit of mass $M$. Then the correct option$(s)$ is(are):
$(A)$ $M \propto \sqrt{c}$
$(B)$ $M \propto \sqrt{G}$
$(C)$ $L \propto \sqrt{h}$
$(D)$ $L \propto \sqrt{G}$

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Planck's constant $h$,speed of light $c$,and gravitational constant $G$ are used to form a unit of length $L$ and a unit of mass $M$. Then the correct option$(s)$ is(are):$(A)$ $M \propto \sqrt{c}$$(B)$ $M \propto \sqrt{G}$$(C)$ $L \propto \sqrt{h}$$(D)$ $L \propto \sqrt{G}$