Question 172: Units , Dimensions and Error Analysis

Question

If momentum (P), area (A) and time (T) are
taken to be the fundamental quantities then the
dimensional formula for energy is

Accepted Answer

Let

[E] =
K[P]
x
[A]
y
[T]
z
[ML
2
T
Ã¢ÂÂ2
] = [MLT
Ã¢ÂÂ1
]
x
[L
2
]
y
[T]
z
[ML
2
T
Ã¢ÂÂ2
] = [M
x
][L
x+2y
][T
Ã¢ÂÂx+z
]
Comparing both side we get,
x = 1
x + 2y = 2
$$ \Rightarrow $$ 1 + 2y= 2 or y = $$\frac{1}{2} $$
z Ã¢ÂÂ x = Ã¢ÂÂ2 $$ \Rightarrow $$ zÃ¢ÂÂ1 = Ã¢ÂÂ2 or z = Ã¢ÂÂ1
$$ 	herefore $$ [E] = $$\left[ {P{A^{{1 \over 2}}}{T^{ - 1}}} ight]$$

Answer

[P2AT–2]

Answer

$$\left[ {{P^{{1 \over 2}}}A{T^{ - 1}}} \right]$$

Answer

$$\left[ {P{A^{{1 \over 2}}}{T^{ - 1}}} \right]$$

Answer

[PA–1T–2]

Units , Dimensions and Error Analysis

If momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is

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