A planet of radius R = (1/10) (radius of Earth) has the same mass density as earth. Scientists dig a well of depth R/5 on it and lower a wire of the same length and of linear mass density ${10}^{-3}kg/m$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is:
(Take the radius of Earth $=6\times {10}^{6}m$ and the acceleration due to gravity of Earth is  $10m/s^2$ )
 

As $R=\frac{R_E}{10}---\left(1\right)$
  $$\begin{gathered} \rho =\frac{M}{\left(4\pi /3\right)R^3}=\frac{M_E}{\left(4\pi /3\right)R_E^3}=density \\ \Rightarrow M=\frac{M_E}{{10}^3} \end{gathered}$$
  $$\begin{gathered} g=\frac{\mathrm{GM}}{{\mathrm{R}}^2} \\ g=\frac{\mathrm{G}({\mathrm{M}}_{\mathrm{E}}/{10}^3)}{{({\mathrm{R}}_{\mathrm{E}}/10)}^2} \\ g=\frac{{\mathrm{GM}}_{\mathrm{E}}}{10{\mathrm{R}}_{\mathrm{E}}^2} \\ \Rightarrow g=\frac{{\mathrm{g}}_{\mathrm{E}}}{10}---\left(2\right) \end{gathered}$$
  $$\begin{gathered} g_{depth}=g\left(1-\frac{d}{R}\right) \\ {g}_{depth}=g\frac{\left(R-d\right)}{R} \\ {g}_{depth}=\frac{gx}{R}---\left(3\right) \\ mass per unit length=\lambda =\frac{dm}{dx} \\ \Rightarrow dm=\lambda dx---\left(4\right) \end{gathered}$$
(where x is the distance from the centre of the planet)
$$\begin{gathered} force F=mg \\ force of attraction element of wire, dF=\left(dm\right)g_{depth} \\ substitute, \\ F=\underset{4R/5}{\overset{R}{\int }}\left(\lambda dx\right)\left(\frac{gx}{R}\right) \\ F=\frac{\lambda g}{R}\underset{4R/5}{\overset{R}{\int }}xdx \\ F=\frac{9\lambda gR}{50} \\ substitute equation \left(1\right),\left(2\right) \\ F=\frac{9\lambda }{50}\left(\frac{g_E}{10}\right)\left(\frac{R_E}{10}\right)=\frac{9\lambda g_E{R}_E}{5000} \\ F=\frac{9\left({10}^{-3}kg/m^3\right)\left(10m/s^2\right)\left(6\times {10}^{6}m\right)}{5000} \\ F=108N \end{gathered}$$