There is an infinite thin flat sheet with mass density $\mathrm{\sigma }$ per unit area. The gravitational force, due to sheet, on a point mass m located at a distance x from the sheet is F.
Consider a large flat horizontal sheet of material density r and thickness t, placed on the surface of the earth. The density of the earth is ${\mathrm{\rho }}_0$. If it is found that gravitational field intensity just below the sheet is larger than field just above it,  Assume t << R. Then select the correct option$\left(s\right).$

(a) Consider a ring element on the sheet as shown in figure. Mass of the ring element is –

$\mathrm{dm}=\mathrm{\sigma }2\mathrm{\pi rdr}$
Force on mass m placed at P due to ring will be perpendicular to the sheet equal to
$\begin{array}{l}\mathrm{dF}=\frac{\mathrm{Gm}\cdot \mathrm{dm}}{{\mathrm{R}}^2}\cdot \cos \mathrm{\theta }=\frac{\mathrm{G}\cdot \mathrm{m}\cdot \mathrm{\sigma }\cdot 2\mathrm{\pi }}{{\mathrm{R}}^3}\mathrm{rdr}\cdot \mathrm{x}\\ =\mathrm{G}\cdot 2\mathrm{\pi \sigma m}\cdot \mathrm{x}\frac{\mathrm{rdr}}{{\left({\mathrm{x}}^2+{\mathrm{r}}^2\right)}^{3/2}}\end{array}$
Force will be equal to sum of forces due to all concentric rings which make up the sheet.
$\begin{array}{l}\therefore \mathrm{F}=\mathrm{G}2\mathrm{\pi \sigma mx}{\int }_{\mathrm{r}=0}^{\mathrm{r}=\mathrm{\infty }} \frac{\mathrm{rdr}}{{\left({\mathrm{x}}^2+{\mathrm{r}}^2\right)}^{3/2}}\\ =-\mathrm{G}2\mathrm{\pi \sigma mx}{\left[\frac{1}{\sqrt{{\mathrm{x}}^2+{\mathrm{r}}^2}}\right]}_{\mathrm{r}=0}^{\mathrm{r}=\mathrm{\infty }}\\ =\mathrm{G}2\mathrm{\pi \sigma m} \left[\text{ Independent of distance }\mathrm{x}\right]\end{array}$
Note: Students may also use gauss’ theorem for gravitation to arrive at the result. (b) From the result obtained above, we can say that field due to a thin large sheet, of mass density $\mathrm{\sigma }$ per unit area, is $\mathrm{G}.2\mathrm{\pi \sigma }$
If the sheet is of thickness t, then also the result will remain same because the sheet can be sliced into infinite number of thin sheets each producing field $\mathrm{G}.2\mathrm{\pi \sigma }$ [Remember this field is independent of distance from the sheet]
$$\begin{gathered} \text{ And }\mathrm{\sigma }=\mathrm{\rho t} \\ \therefore \text{ Field due to sheet }{\mathrm{E}}_{\text{sheet }}=\mathrm{G}2\mathrm{\pi \rho t} \end{gathered}$$

Resultant field at A is
$$\begin{gathered} {\mathrm{E}}_{\mathrm{A}}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}-2\mathrm{\pi G\rho t} \\ \text{ Field at }\mathrm{B}\text{ is; }{\mathrm{E}}_{\mathrm{B}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{t}{)}^2}+2\mathrm{\pi G\rho t} \\ \text{ Where mass of earth }\mathrm{M}=\frac{4}{3}{\mathrm{\pi R}}^3{\mathrm{\rho }}_0 \end{gathered}$$
As per question E_A > E_B
$$\begin{gathered} \frac{\mathrm{GM}}{{\mathrm{R}}^2}-2\mathrm{\pi G\rho t}>\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{t}{)}^2}+2\mathrm{\pi G\rho t} \\ \begin{array}{l}\Rightarrow \frac{4}{3}{\mathrm{\pi R}}^3{\mathrm{\rho }}_0\left[\frac{1}{{\mathrm{R}}^2}-\frac{1}{{\mathrm{R}}^2{\left(1+\frac{\mathrm{t}}{\mathrm{R}}\right)}^2}\right]>4\mathrm{\pi \rho t}\\ \Rightarrow {\mathrm{\rho }}_0\mathrm{R}\left[1-{\left(1+\frac{\mathrm{t}}{\mathrm{R}}\right)}^{-2}\right]>3\mathrm{\rho t}\end{array} \end{gathered}$$
$$\begin{gathered} \mathrm{We} \mathrm{can} \mathrm{write} {\left(1+\frac{\mathrm{t}}{\mathrm{R}}\right)}^{-2}\simeq 1-\frac{2\mathrm{t}}{\mathrm{R}} \\ \therefore {\mathrm{\rho }}_0\mathrm{R}\left[1-1+\frac{2\mathrm{t}}{\mathrm{R}}\right]>3\mathrm{\rho t}\Rightarrow {\mathrm{\rho }}_0>\frac{3}{2}\mathrm{\rho } \end{gathered}$$