A planet of radius `R=(1)/(10)xx(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 a 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 inplace is (take the radius of Earth`=6xx10^6m` and the acceleration due to gravity on Earth is `10ms^(-2)`

Options

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

Answer

B

Explanation

Let `R_(p)` be the radius of the planet. Then
`R_(p) = (R_(e))/(10) = (6 xx 10^(6))/(10) = 6 xx 10^(5) m`
As `g = (GM)/(R^(2)) = (G)/(R^(2)) xx (4)/(3) pi R^(3) rho = (4)/(3) pi G rho R`
So `g prop R`
`:. (g_(P))/(g_(e)) = (R_(P))/(R_(e)) = (1)/(10)`
or `g_(P) = (g_(e))/(10) = (10)/(10) = 1 ms^(-2)`
Acceleration due to gravity at depth `x` on a planet is
`g_(x) = (1-(x)/(R )) g_(P)`
Let `lambda` be the mass per unit lenght of wire.
Given, `lambda = 10^(-3) kg m^(-1)`.
Force on a small segment of wire of lenght `dx` at a depth `x`from surface is
`dF = (lambda dx)g_(x) = lambda dx xx (1- (x)/(R )) g_(P)`
`= lambda (1- (x)/(R )) g_(P) dx`
Total force on wire is
`:. F = underset(0) overset(R//5) int lambda (1- (x)/(R )) g_(P) dx = lambda [x - (x^(2))/(2R)]_(0)^(R//5) g_(P)`
`= lambda [(R)/(5) - (R )/(50)] g_(P) = 10^(-3) [(9R)/(50)] g_(P)`
`= 10^(-3) xx (9)/(50) xx 6 xx 10^(5) xx 1= 108 N`