Answer

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**Hint:**We can find the gravitational potential due to cylindrical mass distribution by comparing it with the electric field due to cylindrical charge distribution.

The gravitational field provides the necessary centripetal force. So, on comparing the equations of gravitational field and centripetal force, we can find the desired relation.

Formula used:

Gravitational force $ =\dfrac{2G\lambda }{r} $ where

$ G=\dfrac{1}{4\pi E} $

Centripetal force $ =m{{w}^{2}}r $ .

**Complete step by step solution**

According to Gauss law for cylindrical charge distribution,

$ \begin{align}

& E\times 2\pi rl=\dfrac{\lambda L}{E} \\

& E=\dfrac{\lambda }{2\pi rE} \\

\end{align} $

The gravitational potential due to cylindrical mass distribution can be found by comparing it with electric field due to cylindrical charge distribution.

So gravitational field $ =\dfrac{\lambda }{2\pi rE} $

$ =\dfrac{2\lambda }{4\pi Er} $

$ E=\dfrac{2G\lambda }{r} $

Here $ G=\dfrac{1}{4\pi E} $

Now centripetal force is given by:

$ F=m{{w}^{2}}r.........(1) $

Also,

As gravitational field is inversely proportional to distance from center, the centripetal force is

$ \begin{align}

& F=mE \\

& F=m\dfrac{2G\lambda }{r}.........(2) \\

\end{align} $

On comparing (1) and (2), we get

$ m{{w}^{2}}r=m\dfrac{2G\lambda }{r} $

or $ w=\dfrac{1}{2}\sqrt{2G\lambda } $

Now as we know that Time period T is given by

$ \begin{align}

& T=\dfrac{2\pi }{w} \\

& T=\dfrac{r}{\sqrt{2G\lambda }} \\

\end{align} $

or $ T\text{ }\propto \text{ }r $ .

**Note**

Gauss law for cylindrical charge distribution is equivalent to electric field due to line charge.

To find electric charge at a point, a cylindrical surface is drawn about the line charge as its axis. This cylindrical surface may be treated as the Gaussian surface for the line charge.

Electric flux through the Gaussian surface=magnitude of electric field $ \times $ area of curved surface of cylindrical of radius r and length l.

Flux= $ E\times 2\pi rl $ ……… (A)

According to gauss theorem, we have

Electric flux

$ =\dfrac{q}{E} $

Now, charge enclosed by Gaussian surface $ q=\lambda l $

So, Flux $ =\dfrac{\lambda l}{E} $ ……. (B)

From equations (A) and (B) we get

$ \begin{align}

& E\times 2\pi rl=\dfrac{\lambda l}{E} \\

& E=\left( \dfrac{1}{2\pi E} \right)\left( \dfrac{\lambda }{r} \right)=\left( \dfrac{1}{4\pi E} \right)\left( \dfrac{2\lambda }{r} \right) \\

\end{align} $ .

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