Question

A planet of radius $R=1/10\times$ (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 ${10}^{-3}kg{m}^{-3}$ 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 on Earth is $10m{s}^{-2}$)

• A. $96N$
• B. $108N$
• C. $120N$
• D. $150N$

Answer 1

The correct option is B $108N$

$R_e$ is the radius of the earth
$g_e=\frac{G{M}_e}{R_e^2}=G\frac{4}{3}\pi R_e\rho$
Now g at the mentioned planet is given as
$g_p=\frac{GM}{r^2}$, here $m=\frac{4}{3}\pi r^3\rho$
Thus $g_p=\frac{4}{3}r\rho$
Thus we get $F={\int }_{4R/5}^R\lambda g_{p}dr=108N$