Question

A planet of radius $R=\frac{1}{10}$ x (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 $10^{-3}$ $kg{m}^{-1}$ 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 = $6\times 10^6$ m and the acceleration due to gravity of earth is $10m{s}^{-2}$)

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

Answer 1

Answer: (B)

Inside planet
$g_i=g_s\frac{r}{R}=\frac{4}{3}G\pi r\rho$
Force to keep the wire at rest (F)
$=$ weight of wire
$=\underset{4R/5}{\overset{2}{\int }}(\lambda dr)\left(\frac{4}{3}G\pi r\rho \right)=\left(\frac{4}{3}G\pi \rho \right)\left(\frac{9\lambda }{50}\right)R^2$
Here, $\rho =$ density of earth $=\frac{M_c}{\frac{4}{3}\pi R_e^2}$
Also, $R=\frac{R_c}{10};$ putting all values, $F=108N$