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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 $10^{-3} \ kgm ^{-1}$ 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 on Earth is $10 \ ms ^{-2}$ )
$96 \ N$
$108 \ N$
$120 \ N$
$150 \ N$
Solution
Given, $R_{\text {planet }}=\frac{R_{\text {earth }}}{10}$ and
density, $\rho=\frac{M_{\text {earth }}}{\frac{4}{3} \pi R_{\text {earth }}^3}=\frac{M_{\text {Planet }}}{\frac{4}{3} R_{\text {planet }}^3} \quad \Rightarrow \quad M_{\text {planet }}=\frac{M_{\text {earth }}}{10^3}$
$g_{\text {surtace of planet }}=\frac{G M_{\text {planet }}^2}{R_{\text {planet }}^2}=\frac{G M_e \cdot 10^2}{10^3 \cdot R_e^2}=\frac{G M_e}{10 R_e^2}=\frac{g_{\text {surface of earth }}}{10}$
$g_{\text {depphotplanet }}=g_{\text {surfasec c planet }}\left(\frac{x}{R}\right) \quad \text { where } x=\text { distance from centre of planet }$
$T =\int_{4 R / 5}^R \lambda dx q\left(\frac{x}{R}\right)=\frac{\lambda g}{R}\left[\frac{x^2}{2}\right]_{4 R / 5}^R=108 \ N$
