If the energy E=G^(p)h^(q)c^(r ) where G is the universal gravitational constant, h is the planck's constant and c is the velocity of light, then the values of p, q and r are respectively:a) -1/2,1/2 and 5/2 b) 1/2, -1/2 and -5/2 c) -1/2,1/2 and 3/2 d) 1/2, -1/2 and -3/2

If the energy E=G^(p)h^(q)c^(r ) where G is the universal gravitationa

1.	If the energy E=G^(p)h^(q)c^(r ) where G is the universal gravitational constant, h is the planck's constant and c is the velocity of light, then the values of p, q and r are respectively:a) -1/2,1/2 and 5/2 b) 1/2, -1/2 and -5/2 c) -1/2,1/2 and 3/2 d) 1/2, -1/2 and -3/2
Answer» `-(1)/(2), (1)/(2) and (5)/(2)` `(1)/(2), -(1)/(2) and -(5)/(2)` `-(1)/(2), (1)/(2) and (3)/(2)` `(1)/(2), -(1)/(2) and -(3)/(2)` Solution :`E=G^(p)h^(Q)c^(r )` Using dimensions we can write, `[ML^(2)T^(-1)]=[M^(-1)L^(3)T^(-2)]^(p)[ML^(2)T^(-1)]^(q)[LT^(-1)]^(r )` `therefore [M^(1)L^(2)T^(-2)]=[M^(-p)L^(3p)T^(-2p)][M^(q)L^(2Q)T^(-q)][L'T^(-r )]` `=M^(-p+q)L^(3p_r)T^(-2p-q-r)` Comparing the powers of M, L and T we GET `therefore -p+q=1"...(2)"` `3p+2q+r=2"...(3)"` `-2p-q-r=-2` `therefore 2p+q+r=2"...(4)"` Solving (3) and (4) we get `{:(" 3"p+2q+r=2),("2"p+q+r=2),("(-)(-)(-)(-)"),(bar(""p+q=0" ")"...(5)"),(""-p+q=1"...(2)"),(bar(""2q=1" ")):}` `q=(1)/(2)` From (5) `therefore p=-q=-(1)/(2)` `therefore p=-(1)/(2)` From (4) `therefore r=2-2p-q` `=2-2(-(1)/(2))-(1)/(2)` `=2+1-(1)/(2)=2(1)/(2)=(5)/(2)` `r=(5)/(2)` `therefore p=-(1)/(2), q=(1)/(2) and r=(5)/(2)`