Question 3: Units , Dimensions and Error Analysis

Question

In the expression $\mathrm{P}=\mathrm{El}^2 \mathrm{~m}^{-5} \mathrm{G}^{-2}$ where $\mathrm{E}, 1, \mathrm{~m}$ and G represent Energy, Angular Momentum, Mass and Gravitational Constant, the dimensions of P are

Accepted Answer

LetÃ¢ÂÂs write down the dimensions of each symbol and then combine them.
Energy, $E$:
$      [E] = \mathrm{M}^1 \,\mathrm{L}^2 \,\mathrm{T}^{-2}    $
Angular momentum, $\ell$:
$      [\ell] = \mathrm{M}^1 \,\mathrm{L}^2 \,\mathrm{T}^{-1}    $
Mass, $m$:
$      [m] = \mathrm{M}^1    $
Gravitational constant, $G$ (from $F = G\,m_1m_2/r^2$):
$      [G] = \frac{[\!F\!]\,[r]^2}{[m]^2}           = \frac{\mathrm{M}^1\mathrm{L}^1\mathrm{T}^{-2}\;\mathrm{L}^2}{\mathrm{M}^2}           = \mathrm{M}^{-1}\,\mathrm{L}^3\,\mathrm{T}^{-2}    $
Now for
$   P = E\,\ell^2\,m^{-5}\,G^{-2}, $
we add exponents of $\mathrm{M}$, $\mathrm{L}$, $\mathrm{T}$:
Ã¢ÂÂ¢ M-exponent:
$1$ (from $E$)
$+\;2	imes1$ (from $\ell^2$)
$-5	imes1$ (from $m^{-5}$)
$-2	imes(-1)$ (from $G^{-2}$)
= $1 + 2 - 5 + 2 = 0$
Ã¢ÂÂ¢ L-exponent:
$2$ (from $E$)
$+\;2	imes2$ (from $\ell^2$)
$-2	imes3$ (from $G^{-2}$)
= $2 + 4 - 6 = 0$
Ã¢ÂÂ¢ T-exponent:
$-2$ (from $E$)
$+\;2	imes(-1)$ (from $\ell^2$)
$-2	imes(-2)$ (from $G^{-2}$)
= $-2 - 2 + 4 = 0$
So
$   [P] = \mathrm{M}^0\,\mathrm{L}^0\,\mathrm{T}^0, $
which is dimensionless.
Answer: Option C.

Answer

$\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$

Answer

$\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$

Answer

$\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]$

Answer

$\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-2}\right]$

Units , Dimensions and Error Analysis

In the expression \mathrm{P}=\mathrm{El}^2 \mathrm{~m}^{-5} \mathrm{G}^{-2} where \mathrm{E}, 1, \mathrm{~m} and G represent Energy, Angular Momentum, Mass and Gravitational Constant, the dimensions of P are

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