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Gravitation

Section: Physics

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Question : 13 of 22

Marks: +1, -0

A planet of radius R =

\frac{1}{10}

× (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}\,\mathrm{kg}\,\mathrm{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 10^{6}

m and the acceleration due to gravity of Earth is 10

\mathrm{m}\,\mathrm{s}^{-2}

[JEE Adv 2014 P2]

96 N
108 N
120 N
150 N

Solution:

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_{\frac{4R}{R}}^{R} \; (\lambda dr)\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_e}{\frac{4}{3}\pi R_e^2}

Also, R =

\frac{R_e}{10}

; putting all values , F = 108 N

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