JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Gravitation

A planet of radius

R = \frac{1}{10} \times

(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

1 0^{- 3} k g 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 1 0^{6} m

and the acceleration due to gravity on Earth is

10 m s^{- 2}

)

Answer: (B)

Solution

The planet's radius is

R = \frac{6 \times 1 0 ^{6}}{10} = 6 \times 1 0^{5}

m. Since density is the same,

g_{planet} = g_{Earth} \times \frac{R _{planet}}{R _{Earth}} = 10 \times \frac{1}{10} = 1

m/s².

Inside the planet at depth

d

from the surface (distance

r

from center),

g (r) = g_{0} \frac{r}{R}

.

The wire extends from

r = \frac{4 R}{5}

r = R

. The force is:

F = \int_{4 R /5}^{R} λ g (r) d r = \int_{4 R /5}^{R} λ \frac{g _{0} r}{R} d r = \frac{λ g _{0}}{R} [\frac{r ^{2}}{2}]_{4 R /5}^{R} = \frac{λ g _{0}}{2 R} (R^{2} - \frac{16 R ^{2}}{25}) = \frac{9 λ g _{0} R}{50}

.

Substituting:

F = \frac{9 \times 1 0 ^{- 3} \times 1 \times 6 \times 1 0 ^{5}}{50} = \frac{5400}{50} = 108

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option