Kelvi

Kelvi

0 pts

0days until NEET 2026Every question counts. ✊

🌌

Did you know?

If you removed all empty space from atoms in your body, you'd compress to the size of a speck of dust.

Kelvi

0 pts

A physical quantity of the dimensions of length that can be formed out of $c, G$ and $\frac{e^2}{4\pi\varepsilon_0}$ is ( $c$ is the velocity of light, G is universal gravitational constant, and $e$ is charge)

hard

Units and Measurements

2017

physics

$c^2 \left[ G \frac{e^2}{4\pi\varepsilon_0} \right]^{1/2}$

Explanation

To solve this problem, we need to find a physical quantity with the dimensions of length using $c, G,$ and $\frac{e^2}{4\pi\varepsilon_0}. \\$

Our AI powered practice platform can help you achieve your doctor dream.

Practice 2000+ previous year NEET questions
More practice, more score
AI generated hints
Use hints to get start solving
AI companion chat to clear doubts 24*7
Ask any question, get instant answers
AI generated solutions
Get detailed step by step solutions
Check related NCERT content
Read while solving
Track your progress
Improve every day

Sign up / Login

$\frac{1}{c^2} \left[ \frac{e^2}{G 4\pi\varepsilon_0} \right]^{1/2}$

$\frac{1}{c} G \frac{e^2}{4\pi\varepsilon_0}$

$\frac{1}{c^2} \left[ G \frac{e^2}{4\pi\varepsilon_0} \right]^{1/2}$

First, let's determine the dimensions of each quantity:

\\

•

c

(velocity of light) has dimensions

[LT^{-1}]. \\

•

G

(universal gravitational constant) has dimensions

[M^{-1}L^3T^{-2}]. \\

•

\frac{e^2}{4\pi\varepsilon_0}

(Coulomb's constant times charge squared) has dimensions

[ML^3T^{-2}]. \\

We need to form a quantity with dimensions of length

[L]. \\

Assume the quantity is of the form

c^a G^b \left(\frac{e^2}{4\pi\varepsilon_0}\right)^c. \\

The dimensions of this expression are:

\\ [L^a T^{-a}] [M^{-b} L^{3b} T^{-2b}] [M^c L^{3c} T^{-2c}]. \\

Combine the dimensions:

\\ [M^{-b+c} L^{a+3b+3c} T^{-a-2b-2c}]. \\

For this to have dimensions of length

[L],

we need:

\\ 1. -b + c = 0 \quad (

no mass dimension

) \\ 2. a + 3b + 3c = 1 \quad (

length dimension

) \\ 3. -a - 2b - 2c = 0 \quad (

time dimension

) \\

From equation 1:

c = b. \\

Substitute

c = b

into equations 2 and 3:

\\ a + 3b + 3b = 1 \\ a + 6b = 1 \quad

(equation 2)

\\ -a - 2b - 2b = 0 \\ -a - 4b = 0 \quad

(equation 3)

\\

From equation 3:

a = -4b. \\

Substitute

a = -4b

into equation 2:

\\ -4b + 6b = 1 \\ 2b = 1 \\ b = \frac{1}{2}. \\

Then

c = b = \frac{1}{2}. \\

Substitute

b = \frac{1}{2}

into

a = -4b: \\ a = -4 \times \frac{1}{2} = -2. \\

Thus, the expression is:

c^{-2} G^{1/2} \left(\frac{e^2}{4\pi\varepsilon_0}\right)^{1/2}. \\

This corresponds to Option 4:

\frac{1}{c^2} \left[ G \frac{e^2}{4\pi\varepsilon_0} \right]^{1/2}. \\

Therefore, the correct option is Option 4.