if force (f) ,velocity (v) and mass (M) are choosen as fundamental quatities then the dimension of length in terms of these quantities would be
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11th
Physics
Units and Measurement
Dimensions and Dimensional Analysis
Time (T), velocity (C) and ...
PHYSICS
Time (T), velocity (C) and angular momentum (h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be :
MEDIUM
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ANSWER
m=T
a
C
b
h
c
In dimensions,
[m]=[T
1
]
a
[l
1
T
−1
]
b
[M
1
L
2
T
−1
]
c
=[T
a
][L
b
T
−b
][M
c
L
2c
T
−c
]
[M
1
]=[T
a−b−c
][L
b+2c
][M
c
]
c=1
a−b−c=0
b+2c=0
b+2=0
b=−2
a+2−1=0
a+1=0
a=−1
[m]=[T
−1
C
−2
h
1
]
Given that Force (F) , velocity (V) and Mass (M) are choosen as fundamental quantities.
We have to find the dimension of length in terms of these quantities.
Let the dimensional formula of Length in terms of Force, Velocity and mass be raised to the power of x , y and n.
So,
⇒ [L] = [F]ˣ [V]ʸ [M]ⁿ
We know, Dimensional formula of :
- [F] = [MLT⁻²]
- [V] = [LT⁻¹]
- [M] = [M]
⇒ [L] = [MLT⁻²]ˣ [LT⁻¹]ʸ [M]ⁿ
⇒ [L] = [M]ˣ [L]ˣ [T]⁻²ˣ [L]ʸ [T]⁻ʸ [M]ⁿ
⇒ [L] = [M]⁽ˣ⁺ⁿ⁾ [L]⁽ˣ⁺ʸ⁾ [T]⁽⁻²ˣ⁻ʸ⁾
⇒ [M⁰L¹T⁰] = [M]⁽ˣ⁺ⁿ⁾ [L]⁽ˣ⁺ʸ⁾ [T]⁽⁻²ˣ⁻ʸ⁾
Comparing each fundamental quantities both sides, we have
[M]⁰ = [M]⁽ˣ⁺ⁿ⁾
⇒ 0 = x + n
⇒ x = -n ...(1)
[L]¹ = [L]⁽ˣ⁺ʸ⁾
⇒ 1 = x + y
⇒ x = 1 - y ...(2)
[T]⁰ = [T]⁽⁻²ˣ⁻ʸ⁾
⇒ 0 = - 2x - y
⇒ 0 = -2(1 - y) - y [ from (1) ]
⇒ 0 = -2 + 2y - y
⇒ y = 2
Substituting [y = 2] in (2),
⇒ x = 1 - y
⇒ x = 1 - 2
⇒ x = -1
Putting, [x = -1] in (1), we get n
⇒ x = -n
⇒ -1 = -n
⇒ n = 1
We assumed the dimensional formula as:
⇒ [L] = [F]ˣ [V]ʸ [M]ⁿ
Substituting values of x , y and n
⇒ [L] = [F]⁻¹ [V]² [M]¹
Hence, The dimensional formula of length in terms of force, velocity, and mass can be given as [F]⁻¹ [V]² [M]¹ .