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Gravitation question

2014 · Shift 2 · Q44
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  5. /2014 · Shift 2 · Q44

Gravitation question

2014 · Shift 2 · Q44

JEE AdvancedPhysicsGravitationMCQ+3 / −1
A planet of radius R = 110×{1 \over {10}} \times101​×(radius of Earth) has the same mass density as Earth. Scientists dig a well of depth R5{R \over 5}5R​ on it and lower a wire of the same length and of 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 in place is (take the radius of Earth = 6 ×\times× 106 m and the acceleration due to gravity of Earth is 10 ms -2)
  1. A
    96 N
  2. B
    108 N
  3. C
    120 N
  4. D
    150 N
View written solutionFree

Correct answer: B

  1. Find gravity on the planet

For a spherical planet of the same density as Earth, g∝Rg \propto Rg∝R because g=GMR2,M=43πR3ρ  ⟹  g=43πGρR.g = \frac{GM}{R^2}, \quad M = \frac{4}{3}\pi R^3\rho \implies g = \frac{4}{3}\pi G\rho R.g=R2GM​,M=34​πR3ρ⟹g=34​πGρR.

Given Rplanet=110RE,R_{planet} = \frac{1}{10}R_E,Rplanet​=101​RE​, and same density, gplanet=110gE=110×10=1 m s−2.g_{planet} = \frac{1}{10}g_E = \frac{1}{10}\times 10 = 1\ \text{m s}^{-2}.gplanet​=101​gE​=101​×10=1 m s−2.

  1. Depth of the well and length of wire inside

Depth of well: d=R5.d = \frac{R}{5}.d=5R​.

Radius of planet: R=110×6×106=6×105 m.R = \frac{1}{10}\times 6\times 10^6 = 6\times 10^5\ \text{m}.R=101​×6×106=6×105 m.

Hence, d=6×1055=1.2×105 m.d = \frac{6\times 10^5}{5} = 1.2\times 10^5\ \text{m}. d=56×105​=1.2×105 m.

The wire has the same length as the well, so L=d=1.2×105 m.L = d = 1.2\times 10^5\ \text{m}. L=d=1.2×105 m.

Linear mass density: λ=10−3 kg m−1.\lambda = 10^{-3}\ \text{kg m}^{-1}. λ=10−3 kg m−1.

Total mass of wire: m=λL=10−3×1.2×105=120 kg.m = \lambda L = 10^{-3}\times 1.2\times 10^5 = 120\ \text{kg}. m=λL=10−3×1.2×105=120 kg.

  1. Gravity variation inside the planet

Inside a uniform-density planet, g(r)=gsurfacerR.g(r) = g_{surface}\frac{r}{R}.g(r)=gsurface​Rr​.

If depth below surface is xxx, then r=R−x,r = R-x,r=R−x, so g(x)=gsurface(1−xR).g(x) = g_{surface}\left(1-\frac{x}{R}\right).g(x)=gsurface​(1−Rx​).

Here, gsurface=1 m s−2.g_{surface}=1\ \text{m s}^{-2}. gsurface​=1 m s−2.

  1. Weight of the wire using integration

Take a small element of wire of length dxdxdx at depth xxx. Its mass is dm=λ dx.dm = \lambda\,dx.dm=λdx.

Its weight is dF=g(x) dm=λgsurface(1−xR)dx.dF = g(x)\,dm = \lambda g_{surface}\left(1-\frac{x}{R}\right)dx.dF=g(x)dm=λgsurface​(1−Rx​)dx.

So total force needed to hold the wire is F=∫0Lλgsurface(1−xR)dx.F = \int_0^L \lambda g_{surface}\left(1-\frac{x}{R}\right)dx.F=∫0L​λgsurface​(1−Rx​)dx.

Substitute values: F=λgsurface[x−x22R]0L.F = \lambda g_{surface}\left[ x - \frac{x^2}{2R}\right]_0^L.F=λgsurface​[x−2Rx2​]0L​.

Thus, F=λgsurface(L−L22R).F = \lambda g_{surface}\left(L - \frac{L^2}{2R}\right).F=λgsurface​(L−2RL2​).

Now L=R/5L=R/5L=R/5, so

= \lambda g_{surface}R\left(\frac{1}{5} - \frac{1}{50}\right) = \lambda g_{surface}R\cdot \frac{9}{50}. $$ Substitute: $$\lambda = 10^{-3},\quad g_{surface}=1,\quad R=6\times 10^5.$$ Therefore, $$F = 10^{-3}\times 1\times 6\times 10^5\times \frac{9}{50}.$$ First, $$10^{-3}\times 6\times 10^5 = 600.$$ Then, $$F = 600\times \frac{9}{50} = 12\times 9 = 108\ \text{N}. $$ --- 5. **Check options** - A: $96\ \text{N}$ - B: $108\ \text{N}$ - C: $120\ \text{N}$ - D: $150\ \text{N}$ The correct option is: $$\boxed{\text{B: }108\ \text{N}}$$
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