If in a system, pressure'p', density 'ρ' and acceleration 'a' are taken as fundamental quantities then dimensions of force are given as
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If in a system, pressure'p', density 'ρ' and acceleration 'a' are taken as fundamental quantities then dimensions of force are given as
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Answer:
Hii
Explanation:
F=P
α
v
β
T
γ
or, [M
1
L
1
T
−2
]=[M
α
L
−α+β
T
−2α−β+γ
]
i.e. [M
1
L
1
T
−2
]=[M
α
L
−α+β
T
−2α−β+γ
]
Hence we have,
α=1, −α+β=1 and −2α−β+γ=−2
Solving these we get,
α=1, β=2, γ=2
⇒F=Pv
2
T
2
Given:
Pressure p, density ρ and acceleration a are fundamental quantities.
To find:
Dimensions of force.
Solution:
Fundamental quantities mean that they are independent of one another.
Force = m*a , where m is the mass of the body and a is its acceleration.
F = [M L T-2]
Pressure = Force/area [M L-1 T-2]
Density = [M L-3 T0]
Acceleration = [M0 L T-2]
[M L T-2] = [M L-1 T-2]^x + [M0 L T-2]^z+ [M L-3 T0]^y
x+ y = 1
-1x -3y = 1
-2x -2z = -2
On solving these equation, we get:
x = 2
y =-1
z = -1
Dimensions of force = p²ρ^-1 a^-1.
Therefore, the dimension of force in the given fundamental quantities will be p²ρ^-1 a^-1.