Question

Planck's constant(h) , speed of light in vacuum(c) and Newton's gravitational constant(G) are three fundamental constants. which of the following combinations of these has the dimension of length ?
$\bold{a) \: \: \frac{ \sqrt{hG} }{ {c}^{ \frac{ 5 }{2} } } }$
$\bold{ b) \: \: \frac{ \sqrt{hG } }{ {c}^{ \frac{3}{2} } } }$
$\bold{ c) \: \: \sqrt{ \frac{hc}{G} } }$
$\bold{d) \: \: \sqrt{ \frac{Gc}{ {h}^{ \frac{3}{2} } } } }$

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Answer 1

$\maltese\: \underline{\underline{\sf AnsWer :}}\:\maltese$

DIMENSIONAL ANALYSIS :

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Let us assume that the Length (L) depends on the planks constant (h) , Speed of light in vaccum (c) and Newton's Gravitational constant (G).

$\longrightarrow \: \sf L\propto [h]^x [c]^y [G]^z$

Adding k as dimensionaless constant we have :

$\longrightarrow \: \sf L = k [h]^x [c]^y [G]^z$

$\longrightarrow \: \sf [L] = [h]^x [c]^y [G]^z$

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DIMENSIONS :

$\dag\:\underline{\tt Dimensions \: of \: planks \: constant \: (h) = [ML^2T^{-1}]}$

$\dag\:\underline{\tt Dimensions \: of \: speed \: of \: light \: in \: vaccum \: (c) = [M^0L^{1} T^{-1}]}$

$\dag\:\underline{\tt Dimensions \: of \: Gravitational\: constant\: (G) = [M^{-1}L^{3} T^{-2}]}$

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$\longrightarrow \: \sf [M^0L^1T^0] = [ML^2T^{-1}]^x [M^0L^{1} T^{-1}]^y [M^{-1}L^{3} T^{-2}]^z$

$\longrightarrow \: \sf [M^0L^1T^0] = [M]^{(x - z)} [L] ^{(2x + y + 3z)} [T]^{ (- x - y - 2z)}$

On comparing the powers of LHS and RHS we have :

$\longrightarrow \: \sf x - z = 0$

$\longrightarrow \: \sf x = z \: \: \: ...(i)$

Now,

$\longrightarrow \: \sf 2x + y + 3z = 1$

From equation (i) replace the value of x = z in above equation :

$\longrightarrow \: \sf 2z + y + 3z = 1$

$\longrightarrow \: \sf 5z + y = 1 \: \: \: \: ...(ii)$

Also,

$\longrightarrow \: \sf - x - y - 2z = 0$

$\longrightarrow \: \sf - z - y - 2z = 0$

$\longrightarrow \: \sf - 3z - y = 0$

$\longrightarrow \: \sf y = - 3z \: \: \: \: \: ...(iii)$

By Substituting the value of y from equation (iii) to equation (ii) :

$\longrightarrow \: \sf 5z + y = 1$

$\longrightarrow \: \sf 5z - 3z = 1$

$\longrightarrow \: \sf 2z = 1$

$\longrightarrow \: \bf z = \dfrac{1}{2}$

From equation (i) we observe that x = z, therefore:

$\longrightarrow \: \sf x = z \:$

$\longrightarrow \: \bf x = \dfrac{1}{2} \:$

By taking equation (iii) we have :

$\longrightarrow \: \sf y = - 3z$

$\longrightarrow \: \sf y = - 3 \bigg( \frac{1}{2} \bigg)$

$\longrightarrow \: \bf y = \frac{ - 3}{2}$

Now,

$\longrightarrow \: \sf [L] = [h]^{ \frac{1}{2} } [c]^{ \frac{ - 3}{2} } [G]^{ \frac{1}{2} }$

$\longrightarrow \: \sf [L] = \dfrac{[h]^{ \frac{1}{2} } [G]^{ \frac{1}{2} }}{[c]^{ \frac{3}{2} }}$

$\longrightarrow \: \sf L= \dfrac{ \sqrt{h} \sqrt{G }}{c^{ \frac{3}{2} }}$

$\longrightarrow \: \underline{ \boxed{ \bold{ L= \dfrac{ \sqrt{hG }}{c^{ \frac{3}{2} }}}}}$

Hence, option (B) is the correct answer.

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