- A$25-50$
- ✓$75-100$
- C$125-150$
- D$175-200$
For equilibrium :
$mg ^{\prime}=\frac{1}{2} \rho v ^2 A$
$mg \left(1-\frac{2 h }{ R }\right)=\frac{1}{2} \times 1.2 e ^{-\frac{ h }{10}} v ^2 A$
$\Rightarrow 7500 \times 9.8\left(1-\frac{2 h }{ R }\right)=0.6 e ^{-\frac{ h }{10}}\left(8 \times 10^3\right)^2 \times 30$
$\downarrow$ Neglect this term for a while and we get $h \simeq 95 \,km$
So, with $\left(1-\frac{2 h}{R}\right)$, we will get little lesser value or $h$.
$\therefore(B)$ is the answer.
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$E=\frac{3}{5} \frac{Z(Z-1) e^2}{4 \pi \varepsilon_0 R}$
The measured masses of the neutron, ${ }_1^1 \mathrm{H},{ }_7^{15} \mathrm{~N}$ and ${ }_8^{15} \mathrm{O}$ are $1.008665 \mathrm{u}, 1.007825 \mathrm{u}$, $15.000109 \mathrm{u}$ and $15.003065 \mathrm{u}$, respectively. Given that the radii of both the ${ }_7^{15} \mathrm{~N}$ and ${ }_8^{15} \mathrm{O}$ nuclei are same, $1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^2$ ( $c$ is the speed of light) and $e^2 /\left(4 \pi \varepsilon_0\right)=1.44 \mathrm{MeV} \mathrm{fm}$. Assuming that the difference between the binding energies of ${ }_7^{15} \mathrm{~N}$ and ${ }_8^{15} \mathrm{O}$ is purely due to the electrostatic energy, the radius of either of the nuclei is
$\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right.$ )
