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  • If velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

If velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

Answers (1)

Let us first write down the dimensions of-

h=\frac{E}{v}

    = \frac{[ML^{2}T^{-2}]}{[T^{-1}]}

     =[ML^{2}T^{-1}]

C=[LT^{-1}]

G = [M^{-1}L^{3}T^{-2}]

(i) Let us consider that Mass m is directly proportional to [h]^{a}[C]^{b}[G]^{c}

[M^{1}L^{0}T^{0}] = k [ML^{2}T^{-1}]^{a}[LT^{-1}]^{b}[M^{-1}L^{3}T^{-1}]^{c}

 [M^{1}L^{0}T^{0}] = k [M^{a-c}L^{2a+b+3c}T^{-a-b-2c}]

Comparing the powers on R.H.S. & L.H.S.

a-c=1 

Thus, a=c+1

2a+b+3c=0\; \; \; \; \; \; \; .....(i)

-a-b-2c=0 \; \; \; \; \; \; \; ..... (ii)

Substituting the value of a in eq (i) & (ii)

2(c+1) + b + 3c = 0

2c + 2 + b + 3c =0

b+5c = -2 \; \; \; \; \; \; \; \; ..... (iii)

-(c+1)-b-2c = 0

-b-3c=1 \; \; \; \; \; \; \; ...... (iv)

By adding (iii) & (iv), we get,

c=\frac{1}{2}

now, a = c+1

Thus, a = \frac{-1}{2}+1 = \frac{1}{2}

Substituting these values in eq. (iii)

b + 5(-1/2) = -2

b = -2 + \frac{2}{5}

b = \frac{1}{2}

Thus, m = kh^{\frac{1}{2}}C^{\frac{1}{2}}G^{\frac{-1}{2}}

M=k\sqrt{\frac{hc}{G}}

Now, let L\; \alpha \; C^{a}h^{b}G^{c}

Thus, [L^{1}] = k[LT^{-1}]^{a}[ML^{2}T^{-1}]^b[M^{-1}L^{3}T^{2}]^{c}

                     =k[M^{b-c}L^{a+2b+3c}T^{-a-b-2c}]

Equating the powers on both sides

b-c=0

Thus,b=c

a+2b+3c = 1 \; \; \; \; \; \; \; ...... (i)

-a-b-2c = 0 \; \; \; \; \; \; .... (ii)

Now, since b=c

a+2b+3b=1 becomes

a+5b=1 \; \; \; \; \; \; .....(iii)

& eq. (ii)

a+b+2c = 0 becomes

a + 3b = 0\; \; \; \; \; \; \; \; ...... (iv)

Subtracting eq (iv) from eq (iii) we get,

b=\frac{1}{2}

Thus, c=\frac{1}{2}

And eq (iv) will be

a +3(\frac{1}{2}) = 0

a=\frac{-3}{2}

Therefore, I = k\; C^{\frac{3}{2}}h^{\frac{1}{2}}G^{\frac{1}{2}}

L=k\sqrt{\frac{hG}{c^{3}}}

Let us consider T\; \alpha \; G^{a}h^{b}C^{c}

Thus, [M^{0}L^{0}T^{-1}] = k[M^{-1}L^{3} T^{-2}]^{a}[ML^{2}T^{-1}][LT^{-1}]^{c}

                                    = k[M^{-a+b} L^{3a+2b+c}T^{-2a-b-c}]
-a+b=0 \; \; \; \; \; \; \; ... (i)

3a + 2b + c = 0 \; \; \; \; \; \; ....... (ii)

-2a-b-c = 1 \; \; \; \; \; \; \; \; .... (iii)

On adding eq (ii) & (iii) we get,

a+b=1 \; \; \; \; \; \: \: \: \: \: ..... (iv)

From eq (i) &(iv)

b=\frac{1}{2}

a=\frac{1}{2} 

Thus, c=\frac{-5}{2}

T = k\; G^{\frac{1}{2}}h^{\frac{1}{2}}C^{\frac{-5}{2}}

Thus, T = k\sqrt{\frac{hG}{c^{5}}}

Posted by

infoexpert22

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