Gravitation Ques 8
- 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 $10^{-3} kg m^{-1}$ 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 $=6 \times 10^{6} m$ and the acceleration due to gravity of earth is $10 ms^{-2}$ )
(2014 Adv.)
(a) $96 N$
(b) $108 N$
(c) $120 N$
(d) $150 N$
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Answer:
Correct Answer: 8.(b)
Solution:
Formula:
- Given, $R _{\text {planet }}=\frac{R _{\text {earth }}}{10}$ and density, $\rho=\frac{M _{\text {earth }}}{\frac{4}{3} \pi R _{\text {earth }}^{3}}$
$$ \begin{aligned} = & \frac{M _{\text {planet }}}{\frac{4}{3} \pi R _{\text {planet }}^{3}} \Rightarrow M _{\text {planet }}=\frac{M _{\text {earth }}}{10^{3}} \\ g _{\text {surface of planet }}= & \frac{G M _{\text {planet }}}{R _{\text {planet }}^{2}}=\frac{G M _e \cdot 10^{2}}{10^{3} \cdot R _e^{2}}=\frac{G M _e}{10 R _e^{2}} \\ & =\frac{g _{\text {surface of earth }}}{10} \\ g _{\text {depth of planet }} & =g _{\text {surface of planet }} \frac{x}{R} \end{aligned} $$
where, $x=$ distance from centre of planet.
$\therefore$ Total force on wire
$$ F=\int _{4 R / 5}^{R} \lambda d x g \frac{x}{R}=\frac{\lambda g}{R}{\frac{x^{2}}{2}} _{4 R / 5}^{R} $$
Here, $\quad g=g _{\text {surface of planet }}, R=R _{\text {planet }}$
Substituting the given values, we get $F=108 N$
BETA
