Question 173: Units , Dimensions and Error Analysis

Question

If speed V, area A and force F are chosen as
fundamental units, then the dimension of
Young’s modulus will be

Accepted Answer

Y = k [F]
x
[A]
y
[V]
z
[M
1
L
1
T
Ã¢ÂÂ2
] = [MLT
Ã¢ÂÂ2
]
x
[L
2
]
y
[LT
Ã¢ÂÂ1
]
z
[M
1
L
1
T
Ã¢ÂÂ2
] = [M]
x
[L]
x+2y+z
[T]
Ã¢ÂÂ2xÃ¢ÂÂz
Comparing power of M, L and T
x = 1 Ã¢ÂÂ¦Ã¢ÂÂ¦(1)
x + 2y + z = Ã¢ÂÂ1 Ã¢ÂÂ¦Ã¢ÂÂ¦(2)
Ã¢ÂÂ2x Ã¢ÂÂ z = Ã¢ÂÂ2 Ã¢ÂÂ¦Ã¢ÂÂ¦(3)
After solving
x = 1
y = Ã¢ÂÂ1
z = 0
$$ 	herefore $$ Y = FA
Ã¢ÂÂ1
V
0

Answer

FA–1V0

Answer

FA2V–1

Answer

FA2V–2

Answer

FA2V–3

Units , Dimensions and Error Analysis

If speed V, area A and force F are chosen as fundamental units, then the dimension of Young’s modulus will be

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