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 $10^{-3}$ $kgm^{-1}$ 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 = $6 \times 10^6$ m and the acceleration due to gravity of earth is $10 \,ms^{-2}$)

• 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
$=\int\limits_{4 R / 5}^{2}(\lambda d r)\left(\frac{4}{3} G \pi r \rho\right)=\left(\frac{4}{3} G \pi \rho\right)\left(\frac{9 \lambda}{50}\right) R^{2}$
Here, $\rho=$ density of earth $=\frac{M_{c}}{\frac{4}{3} \pi R_{e}^{2}}$
Also, $R=\frac{R_{c}}{10} ;$ putting all values, $F=108 \,N$