JEE Challenger
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PhyMedAdv 2014 • Paper 2

Force Required to Hold Wire Suspended in Deep Well on Planet

A spherical planet having radius R=110×(radius of Earth)R = \frac{1}{10} \times (\text{radius of Earth}) possesses the same uniform mass density as Earth. Suppose a vertical well of depth R5\frac{R}{5} is excavated into the planet, and a wire of identical length and linear mass density 10−3 kg m−110^{-3}\,\text{kg}\,\text{m}^{-1} is lowered into it. Assuming the wire does not touch the walls or base, determine the force exerted at the top end of the wire by a person holding it in equilibrium. (Take the radius of Earth =6×106 m= 6 \times 10^6\,\text{m} and the acceleration due to gravity on the Earth's surface as 10 m s−210\,\text{m}\,\text{s}^{-2})

Options

A

96 N96\text{ N}

B

108 N108\text{ N}

Correct
C

120 N120\text{ N}

D

150 N150\text{ N}

Topics & Concepts

GravitationGravitation

Step-by-Step Solution

To find the force required to hold the wire in equilibrium, we need to calculate the total gravitational force exerted on the wire suspended in the well.

1. Acceleration due to Gravity Inside the Planet

The mass density ρ\rho of the planet is uniform and equal to that of the Earth. The gravitational acceleration gEg_E at the surface of Earth (radius RER_E) is given by: gE=43πGρREg_E = \frac{4}{3}\pi G \rho R_E

For a point inside the planet at a distance rr from its center, the gravitational field strength depends only on the mass enclosed within a sphere of radius rr: g(r)=43πGρrg(r) = \frac{4}{3}\pi G \rho r

Substituting 43πGρ=gERE\frac{4}{3}\pi G \rho = \frac{g_E}{R_E}, we obtain: g(r)=(gERE)rg(r) = \left(\frac{g_E}{R_E}\right) r

2. Setting Up the Integration

The planet's radius is R=RE10R = \frac{R_E}{10}. The well has a depth of R5\frac{R}{5}, so the wire extends vertically from the surface at r=Rr = R down to r=R−R5=45Rr = R - \frac{R}{5} = \frac{4}{5}R.

Consider a small element of the wire of length drdr at a distance rr from the center of the planet. The mass of this element is: dm=μ drdm = \mu \, dr where μ=10−3 kg m−1\mu = 10^{-3} \text{ kg m}^{-1} is the linear mass density of the wire.

The gravitational force on this small element is: dF=dm⋅g(r)=μ(gERE)r drdF = dm \cdot g(r) = \mu \left(\frac{g_E}{R_E}\right) r \, dr

3. Calculating the Total Gravitational Force

The total force FF required to hold the wire in equilibrium is equal to the integral of dFdF from r=45Rr = \frac{4}{5}R to r=Rr = R: F=∫4R/5Rμ(gERE)r dr=μ(gERE)[r22]4R/5RF = \int_{4R/5}^{R} \mu \left(\frac{g_E}{R_E}\right) r \, dr = \mu \left(\frac{g_E}{R_E}\right) \left[ \frac{r^2}{2} \right]_{4R/5}^{R}

F=μ gE2RE(R2−(45R)2)=μ gE2RE(R2−1625R2)=950μ(gERE)R2F = \frac{\mu \, g_E}{2 R_E} \left( R^2 - \left(\frac{4}{5}R\right)^2 \right) = \frac{\mu \, g_E}{2 R_E} \left( R^2 - \frac{16}{25}R^2 \right) = \frac{9}{50} \mu \left(\frac{g_E}{R_E}\right) R^2

Since R=RE10R = \frac{R_E}{10}, we have R2=RE2100R^2 = \frac{R_E^2}{100}. Substituting this gives: F=950μ(gERE)(RE2100)=95000μgEREF = \frac{9}{50} \mu \left(\frac{g_E}{R_E}\right) \left(\frac{R_E^2}{100}\right) = \frac{9}{5000} \mu g_E R_E

4. Calculation

Given values:

  • μ=10−3 kg m−1\mu = 10^{-3} \text{ kg m}^{-1}
  • gE=10 m s−2g_E = 10 \text{ m s}^{-2}
  • RE=6×106 mR_E = 6 \times 10^6 \text{ m}

Substituting the numerical values into the equation: F=95000×10−3×10×(6×106)F = \frac{9}{5000} \times 10^{-3} \times 10 \times (6 \times 10^6)

F=9×600005000=9×12=108 NF = \frac{9 \times 60000}{5000} = 9 \times 12 = 108 \text{ N}

Thus, the force exerted at the top end of the wire is 108 N108\text{ N}.

Correct Answer: B

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