Discussion Forum

 A planet of radius R=   $\frac{1}{10}\times$  (radius of earth)  has the same mass density as earth. Scientists dig a well of depth $\frac{R}{5}$  on it and lower a wire of the same length and of linear mass density 10^-3 kgm^-1 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 =  $6\times10^{6}$  m and the acceleration due to gravity of the earth is 10 ms^-2)

Answer:

Explanation:

 Given,   $R_{planet}=\frac{R_{earth}}{10}$

and density ,   $\rho= \frac{M_{earth}}{\frac{4}{3}\pi R_{earth}^{3}}$

       $\Rightarrow$            $ M_{planet}=\frac{M_{planet}}{10^{3}}$

g_{surface of planet}   $=\frac{GM_{planet}}{R^{2}_{planet}}$

$=\frac{GM_{e}.10^{2}}{10^{3}.R^{2}_{e}}$

     $=\frac{GM_{e}.}{10^{}.R^{2}_{e}}$

   =g_{surface of earth/10}

g _{depth of planet}= g _{surafce of planet     $\left(\frac{X}{R}\right)$}

 where x= distance from centre of planet.

  $\therefore$  Total force on wire

                      $F=\int_{4R/5}^{R} \lambda dx g\left(\frac{X}{R}\right)$

  = $\frac{\lambda g}{R}\left[\frac{X^{2}}{2}\right]_{4R/5}^{R}$

  Here  g= g _{surface of planet}

                    R= R_planet

 Substituting  the given values , we get   F=108N