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 $10 \,ms^{-2}$)

Updated On: Jun 14, 2022
  • 96 N
  • 108 N
  • 120 N
  • 150 N
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

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, $F=108 \,N$
Was this answer helpful?
0
0

Questions Asked in JEE Advanced exam

View More Questions

Concepts Used:

Gravitation

In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.

Newton’s Law of Gravitation

According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,

  • F ∝ (M1M2) . . . . (1)
  • (F ∝ 1/r2) . . . . (2)

On combining equations (1) and (2) we get,

F ∝ M1M2/r2

F = G × [M1M2]/r2 . . . . (7)

Or, f(r) = GM1M2/r2

The dimension formula of G is [M-1L3T-2].