If _0 is permittivity of free space, e is charge of proton, G is … - Vidyadip

MCQ

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

✓

Answer

Correct option: B.

$\left[ M ^0 L ^0 T ^0 A ^0\right]$

b
(b)

Gravitational force $F_1=\frac{G M_P^2}{r^2}$

Electrostatic force $F_2=\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2}$

$\frac{F_2}{F_1}=\frac{e^2}{4 \pi \varepsilon_0 G M_\rho^2}$

$\therefore$ Dimension less $\left[ M ^{\circ} L ^{\circ} T ^{\circ} A ^{\circ}\right]$

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