☺Please answer this guys ✌☝
●Class 11 UNIT & DIMENSIONS
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Heya.....^_^
Here's ur ans.⬇
c = Speed of light
Dimension formula (c) = LT^-1
h = Planck's constant
D.F (h) = ML^2T^-1
G= Newton's gravitational constant
D.F (G) = M^-1 L^3 T^-2
Dimension formula of mass = c^x h^y G^z
(In new system of unit
Assume x, y and z as there power)
M L^0 T^0 =( LT^-1)^x (ML^2T^-1) ^y (M^-1L^3T^-2)
M L^0 T^0 = M^(y-z) L^(x+2y+3z) T^(-x-y-2z)
On comparing
x + 2y + 3z = 0 ... 1st equation
y-z = 1 ... 2nd equation
-x-y-2z = 0. ... 3rd equation
From 2nd equation
y = 1+z
Substituting it in 1st equation
x + 2(1+z) + 3z = 0
x+2+5z = 0
x = -5z -2
Now. , Substituting the value of x and y in 3rd equation
-(-5z-2) - (1+z) -2z = 0
5z + 2 - 1- z -2z = 0
2z +1 = 0
↪ z = -1/2
Therefore
↪ x = -5 × (-1/2 ) -2 = 1/2
↪ y = 1+(-1/2 )= 1/2
Hence , Dimension formula of mass ⬇
Similarly , u can find it for length and time
Hope it helps
Here's ur ans.⬇
c = Speed of light
Dimension formula (c) = LT^-1
h = Planck's constant
D.F (h) = ML^2T^-1
G= Newton's gravitational constant
D.F (G) = M^-1 L^3 T^-2
Dimension formula of mass = c^x h^y G^z
(In new system of unit
Assume x, y and z as there power)
M L^0 T^0 =( LT^-1)^x (ML^2T^-1) ^y (M^-1L^3T^-2)
M L^0 T^0 = M^(y-z) L^(x+2y+3z) T^(-x-y-2z)
On comparing
x + 2y + 3z = 0 ... 1st equation
y-z = 1 ... 2nd equation
-x-y-2z = 0. ... 3rd equation
From 2nd equation
y = 1+z
Substituting it in 1st equation
x + 2(1+z) + 3z = 0
x+2+5z = 0
x = -5z -2
Now. , Substituting the value of x and y in 3rd equation
-(-5z-2) - (1+z) -2z = 0
5z + 2 - 1- z -2z = 0
2z +1 = 0
↪ z = -1/2
Therefore
↪ x = -5 × (-1/2 ) -2 = 1/2
↪ y = 1+(-1/2 )= 1/2
Hence , Dimension formula of mass ⬇
Similarly , u can find it for length and time
Hope it helps
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