The Potential Of The Two Homogeneous Spherical Shells Of Same Surface Density At Their Respective Centres Are In The Ratio 3:4. If The Two Shells Coalesce Into A Single One The Surface Density Remains Unchanged. What Is The Potential At An Internal Point Of This Shell?

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Let us consider two spherical shells of radii \( r_1 \) and \( r_2 \). These two shells coalesce into one single spherical shell of radius \( r \). Let the surface density of the shells is \( \sigma \). The respective potentials of the shells at the centres be \( V_1 \), \( V_2 \) and \( V \).

Now, \( 4\pi{r_1}^2\sigma+4\pi{r_2}^2\sigma=4\pi{r}^2\sigma \)

or, \( {r_1}^2+{r_2}^2=r^2 \)

\( V_1=-G\frac{4\pi{r_1}^2\sigma}{r_1}=-G\ 4\pi{r_1}\sigma \) \( V_2=- G\frac{4\pi{r_2}^2\sigma}{r_2}=-G\ 4\pi{r_2}\sigma \) \( V=-G G\frac{4\pi{r}^2\sigma}{r}= -G\ 4\pi{r}\sigma \)

where, \( G \) is the gravitational constant.

Now we can write, \( \frac{V_1}{V_2}=\frac{r_1}{r_2}=\frac{3}{4} \) [Gieven, \( \frac{r_1}{r_2}=\frac{3}{4} \) ]

or, \( \frac{{r_1}^2}{{r_2}^2}=\frac{9}{6} \\or,\ \frac{{r_1}^2+{r_2}^2}{{r_2}^2}=\frac{9+16}{16}=\frac{25}{16} \)

or, \( \frac{{r_2}^2}{{r_1}^2+{r_2}^2}=\frac{16}{25} \)

Therefore, \( {r_1}^2:{r_2}^2:( {r_1}^2+{r_2}^2 )=9:16:25 \)

Or, \( {r_1}^2:{r_2}^2:{r}^2=9:16:25 \), [since \( {r_1}^2+{r_2}^2=r^2 \) ]

or, \( r_1 : r_2 : r=3:4:5 \)

or, \( V_1 : V_2 : V = r_1 : r_2 : r=3:4:5 \)

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