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

  • Option 1)

    c^{2}\left [ G\frac{e^{2}}{4\pi \varepsilon _{o}} \right ]^{\frac{1}{2}}

  • Option 2)

    \frac{1}{c^{2}}\left [ \frac{e^{2}}{G4\pi \varepsilon _{o}} \right ]^{\frac{1}{2}}

  • Option 3)

    \frac{1}{c}G\frac{e^{2}}{4\pi \varepsilon _{o}}

  • Option 4)

    \frac{1}{c^{2}}\left [ G\frac{e^{2}}{4\pi \varepsilon _{o}} \right ]^{\frac{1}{2}}

 

Answers (1)

As we learnt in

Time period of a simple pendulam -

T=Km^{a}l^{b}g^{0}

Equating exponents of similar quantities

a=0     b=1/2    c=-1/2

herefore T= 2pi sqrt{l/g}

- wherein

T= time : period

l= length

g=: acceleration : due: to: gravity

 

 \frac{e^{2}}{4\pi\epsilon _{^{\circ}} }= [F \times d^{2}]= Mc^{3}T^{-2}

G= ML^{3}T^{-2}

C= L^{-1}

l\alpha(\frac{e^{2}}{4\pi \epsilon _{^{\circ}}})^{P}G^{2}C^{r}

[L']= [MC^{3}T^{-2}]^{P} [M^{-1}L^{3}T^{-2}]^{q} [LT^{-1}]^{r}

on comparing both sides we get

P=\frac{1}{2}, q=\frac{1}{2}, r =-2

\therefore l= \frac{1}{C^{2}}[ \frac{Ge^{2}}{4\pi \epsilon _{^{\circ}}} ]^{1/2}

 


Option 1)

c^{2}\left [ G\frac{e^{2}}{4\pi \varepsilon _{o}} \right ]^{\frac{1}{2}}

Incorrect

Option 2)

\frac{1}{c^{2}}\left [ \frac{e^{2}}{G4\pi \varepsilon _{o}} \right ]^{\frac{1}{2}}

Incorrect

Option 3)

\frac{1}{c}G\frac{e^{2}}{4\pi \varepsilon _{o}}

Incorrect

Option 4)

\frac{1}{c^{2}}\left [ G\frac{e^{2}}{4\pi \varepsilon _{o}} \right ]^{\frac{1}{2}}

Correct

Posted by

Vakul

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