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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. |
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Answer» Solution :We know that, dimensions of`(h)=[M^(1)L^(2)T^(-1)]` (From E=hf) dimensions of `(c) =[L^(1)T^(-1)](c)` (c is velocity) Dimensions of gravitational constant (G) `=[M^(-1)L^(3)T^(-2)] ""("From" F=(Gm_(1)m_(2))/(r^(2)))` (i) Let `m prop c^(a)h^(b)G^(c)` `rArr m=kc^(a)h^(b)G^(c)` where, k is a dimensionless constant of proportionality. Substituting dimensions of each term in Eq. (i), we get `[ML^(0)T^(0)]=[LT^(-1)]^(x)xx[M^(2)L^(-1)]^(y)[M^(-1)K^(-3)T^(-2)]^(z)` `=[M^(b-c)L^(a+2b+3c)T^(-a-b-2c)]` COMPARING powers of same TERMS on both sides, we get `b-c=1 ""...(ii)` `a+2b+3c=0 ""...(iii)` `-a-b-2c=0""...(IV)` Adding EQS. (i), (iii) and (iv), we get `2b=1 rArr b=(1)/(2)` Substituting value of b in eq. (i), we get `c=-(1)/(2)` From eq. (iv) `a=-b-2c` Substituting values of b and we get `a=-(1)/(2)-2(-(1)/(2))=(1)/(2)` Putting values of a, b and c in equ. (i), we get `m=kc^((1)/(2))h^((1)/(2))G^(-(1)/(2))` `:.m=ksqrt((ch)/(G))` (ii) Let `L prop c^(a)h^(b)G^(c)` `:.L=kc^(a)h^(b)G^(c) ""..(iv)` Where, k is a dimensionless constant. Substituting dimensions of each term in equ. (v), we get `[M^(0)LT^(0)]=[LT^(-1)]^(a)xx[ML^(2)T^(-1)]^(b)xx[M^(-1)L^(3)T^(-2)]^(c)` `=[M^(b-c)L^(a+2b+3c)T^(-a-b-2c)]` On comparing powers of same terms, we get `b-c=0 ""...(vi)` `a+2b+3c= ""...(vii)` `-a-b-2c=0""...(viii)` Adding equ. (vi), (vii) and (viii), we get `2b=1` `:.b=(1)/(2)` Substituting value of b in equ. (vi), we get `c=(1)/(2)` From equ. (viii), `a=-b-2c` Substituting values of b and c, we get `a=-(1)/(2)-2((1)/(2))=-(3)/(2)` Putting values of a, b and c in equ. (v), we get `L=kc^((3)/(2)h^((1)/(2))G^((1)/(2))` `L=ksqrt((hG)/(c^(3)))` (iii) Let `T prop c^(a)h^(b)G^(c)` `:.T=kc^(a)h^(b)G^(c)` where, k is a dimensionless constant, Substituting dimensions of each term in equ. (ix), we get `[M^(0)K^(0)T]=[LT^(-1)]^(a)xx[ML^(2)T^(-1)]^(b)xx[M^(-1)L^(3)T^(-2)]^(c)` `=[M^(b-c)L^(a+2b+3c)T^(-a-b-2c)]` On comparing powers of same terms, we get `b-c=0 ""...(x)` `a+2b+3c=0 ""...(xi)` `-a-b-2c=1 ""...(xii)` Adding equ. (x). (xi) and (xii), we get `2y=1` `:.y=(1)/(2)` Substituting value of b in equ. (x), we get `c=b=(1)(2)` From eq. (xii) `a=-b-2c-1` Substituting values of b and c we get `x=-(1)/(2)-2((1)/(2))-1=-(5)/(2)` Putting values of a,b and c in equ. (ix) we get `T=kc^((5)/(2))h^((1)/(2))G^((1)/(2))` `T=ksqrt((hG)/(c^(5)))` |
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