Question

If $\varepsilon_{0}$ is permittivity of free space, $e$ is charge of proton, $G$ is universal gravitational constant and $m_{p}$ is mass of a proton then the dimensional formula for $\frac{e^{2}}{4 \pi \varepsilon_{0} G m_{ p }^{2}}$ is

• A. $\left[ M ^{1} L ^{1} T ^{-3} A ^{-1}\right]$
• B. $\left[ M ^{0} L ^{0} T ^{0} A ^{0}\right]$
• C. $\left[ M ^{1} L ^{3} T ^{-3} A ^{-1}\right]$
• D. $\left[ M ^{-1} L ^{-3} T ^{4} A ^{2}\right]$

Answer 1

Answer: (B)

Gravitational force $F_{1}=\frac{G M_{p}^{2}}{r^{2}}$
Electrostatic force $F_{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$
$\frac{F_{2}}{F_{1}}=\frac{e^{2}}{4 \pi \varepsilon_{0} G M_{P}^{2}}$
$\therefore $ Dimension less $\left[ M ^{\circ} L ^{\circ} T ^{\circ} A ^{\circ}\right]$