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Previous Years AIEEE/JEE Mains Questions
(1) $\frac{\mathrm{R}} {2}$ (2) $\sqrt{2} \mathrm{R}$ (3) 2R (4) $\frac{\mathrm{R}}{\sqrt{2}}$ [AIEEE – 2009]
$\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}} \Rightarrow\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}=\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=9$ $\Rightarrow 1+\frac{\mathrm{h}}{\mathrm{R}}=3 \Rightarrow \mathrm{h}=2 \mathrm{R}$
$(1)-\frac{6 \mathrm{Gm}}{\mathrm{r}}$ $(2)-\frac{9 \mathrm{Gm}}{\mathrm{r}}$ (3) zero $(4)-\frac{4 \mathrm{Gm}}{\mathrm{r}}$ [AIEEE – 2011]
(1) $\sqrt{\frac{\mathrm{Gm}}{\mathrm{R}}}$ (2) $\sqrt{\frac{\mathrm{Gm}}{4 \mathrm{R}}}$ (3) $\sqrt{\frac{\mathrm{Gm}}{3 \mathrm{R}}}$ (4) $\sqrt{\frac{\mathrm{Gm}}{2 \mathrm{R}}}$ [AIEEE-2011]
(1) $6.4 \times 10^{10}$ Joules (2) $6.4 \times 10^{11}$ Joules (3) $6.4 \times 10^{8}$ Joules (4) $6.4 \times 10^{9}$ Joules [AIEEE-2012]
$\mathrm{PE}_{\mathrm{i}}+\mathrm{KE}_{\mathrm{i}}=\mathrm{PE}_{\mathrm{f}}+\mathrm{KE}_{\mathrm{f}}$ $-\mathrm{mgR}+\mathrm{KE}_{\mathrm{i}}=0+0$ $\mathrm{KE}_{\mathrm{i}}=+\mathrm{mgR}=1000 \times 10 \times 6.4 \times 10^{6}$ work done $=6.4 \times 10^{10} \mathrm{J}$
(1) $\frac{5 \mathrm{GmM}}{6 \mathrm{R}}$ (2) $\frac{2 \mathrm{GmM}}{3 \mathrm{R}}$ (3) $\frac{\mathrm{GmM}}{2 \mathrm{R}}$ (4) $\frac{\mathrm{GmM}}{3 \mathrm{R}}$ [JEE-Mains 2013]
From energy conservation $\frac{\mathrm{GMm}}{\mathrm{R}}+\mathrm{KE}=\frac{-\mathrm{GMm}}{3 \mathrm{R}}+\frac{1}{2} \mathrm{mV}^{2} \ldots(\mathrm{i})$ …(i) From force balance at A, $\frac{\mathrm{GMm}}{(3 \mathrm{R})^{2}}=\frac{\mathrm{mv}^{2}}{3 \mathrm{R}} \Rightarrow \mathrm{V}^{2}=\frac{\mathrm{GM}}{3 \mathrm{R}}$ ………..(ii) from (i) & (ii) $\mathrm{KE}_{\text {suface }}=\frac{5}{6} \frac{\mathrm{GMm}}{\mathrm{R}}$
(1) $\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$ (2) $\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$ (3) $\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ (4) $\sqrt{2 \sqrt{2} \frac{\mathrm{GM}}{\mathrm{R}}}$ [JEE-Mains 2014]
Net force on one particle $\mathrm{F}_{\mathrm{net}}=\mathrm{F}_{1}+2 \mathrm{F}_{2} \cos 45^{\circ}=$ Centripetal force $\Rightarrow \frac{\mathrm{GM}^{2}}{(2 \mathrm{R})^{2}}+\left[\frac{2 \mathrm{GM}^{2}}{(\sqrt{2} \mathrm{R})^{2}} \cos 45^{\circ}\right]=\frac{\mathrm{MV}^{2}}{\mathrm{R}}$ $\mathrm{V}=\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$ $\mathrm{V}=-\frac{\mathrm{GM}}{2 \mathrm{R}^{3}}\left[3 \mathrm{R}^{2}-\frac{\mathrm{R}^{2}}{4}\right]+\frac{3 \mathrm{G}}{2} \frac{\mathrm{M}}{8 \frac{\mathrm{R}}{2}}$ $=\frac{-11 \mathrm{GM}}{8 \mathrm{R}}+\frac{3 \mathrm{GM}}{8 \mathrm{R}}=-\frac{\mathrm{GM}}{\mathrm{R}}$
(1) $\frac{-2 \mathrm{GM}}{3 \mathrm{R}}$ (2) $\frac{-2 \mathrm{GM}}{\mathrm{R}}$ (3) $\frac{-\mathrm{GM}}{2 \mathrm{R}}$ (4) $\frac{-\mathrm{GM}}{\mathrm{R}}$ [JEE-Mains 2015]
By principle of superosition $\mathrm{V}=-\frac{\mathrm{GM}}{2 \mathrm{R}^{3}}\left[3 \mathrm{R}^{2}-\frac{\mathrm{R}^{2}}{4}\right]+\frac{3 \mathrm{G}}{2} \frac{\mathrm{M}}{8 \frac{\mathrm{R}}{2}}$ $=\frac{-11 \mathrm{GM}}{8 \mathrm{R}}+\frac{3 \mathrm{GM}}{8 \mathrm{R}}=-\frac{\mathrm{GM}}{\mathrm{R}}$
(1) $\sqrt{\mathrm{gR}}(\sqrt{2}-1)$ (2) $\sqrt{2 \mathrm{gR}}$ (3) $\sqrt{\mathrm{gR}}$ (4) $\sqrt{\mathrm{gR} / 2}$ [JEE-Mains 2016]
$\mathrm{V}_{0}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ or $\sqrt{\mathrm{gR}}$ $\mathrm{V}_{\mathrm{e}} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ or $\sqrt{2 \mathrm{gR}}$ $\therefore$ Increase in velocity $=\sqrt{\mathrm{gR}}[\sqrt{2}-1]$
[JEE-Mains 2017]
$\mathrm{g}=\frac{\mathrm{GMx}}{\mathrm{R}^{3}}$ inside the Earth (straight line) $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{r}^{2}}$ outside the Earth where M is Mass of Earth option (2)
Kamino kaha hai previous year questiond