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Question

Physics Question on Gravitation

A planet of radius R=110R=\frac{1}{10} x (radius of earth) has the same mass density as earth. Scientists dig a well of depth R5\frac{R}{5} on it and lower a wire of the same length and of linear mass density 10310^{-3} kgm1kgm^{-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 = 6×1066 \times 10^6 m and the acceleration due to gravity of earth is 10ms210 \,ms^{-2})

A

96 N

B

108 N

C

120 N

D

150 N

Answer

108 N

Explanation

Solution

Inside planet
gi=gsrR=43Gπrρg _{ i }= g _{ s } \frac{ r }{ R }=\frac{4}{3} G \pi r \rho
Force to keep the wire at rest (F)
== weight of wire
=4R/52(λdr)(43Gπrρ)=(43Gπρ)(9λ50)R2=\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, ρ=\rho= density of earth =Mc43πRe2=\frac{M_{c}}{\frac{4}{3} \pi R_{e}^{2}}
Also, R=Rc10;R=\frac{R_{c}}{10} ; putting all values, F=108NF=108 \,N