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JEE Advanced 2014Paper 2
Exam
JEE Advanced
Year
2014
Paper
Paper 2
Subject
Physics
Chapter
Gravitation

JEE Advanced 2014 · Paper 2 · Physics · Gravitation

A planet of radius R = (1)/(10) × (radius of Earth) has the same...

A planet of radius R=110×R = \dfrac{1}{10} \times (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth R5\dfrac{R}{5} on it and lower a wire of the same length and of linear mass density 10−3kgm−1{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 =6×106m= 6 \times {10}^{6}m and the acceleration due to gravity of Earth is 10ms−210m{s}^{- 2} )

Options

  1. A96 N
  2. B108 N
  3. C120 N
  4. D150 N

Correct answer

B. 108 N

Step-by-step solution

Inside planet
gi=gsrR=43G\pirρ{g}_{i} = {g}_{s}\dfrac{r}{R} = \dfrac{4}{3}G\pirρ
Force to keep the wire at rest (F)
= weight of wire
=∫4R5R(\lambdadr)(43G\pirρ)=(43Gπρ)(9λ50)R2= \mathop{∫}_{\dfrac{4R}{5}}^{R}\left(\lambdadr\right)\left(\dfrac{4}{3}G\pirρ\right) = \left(\dfrac{4}{3}G\piρ\right)\left(\dfrac{9\lambda}{50}\right){R}^{2}
Here, ρ=ρ = density of earth =Me43πRe2= \dfrac{{M}_{e}}{\dfrac{4}{3}\pi{R}_{e}^{2}}
Also, R=Re10R = \dfrac{{R}_{e}}{10} ; putting all values, F=108NF = 108N
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