Solution:

Given: $\mathrm{m} _{1}=200 \mathrm{Kg}$

$$ \begin{aligned} & \mathrm{m} _{2}=500 \mathrm{Kg} \\ & \mathrm{r}=10 \mathrm{~m} \\ & \mathrm{G}=6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^{2} \mathrm{Kg}^{-2} \\ & \therefore \quad \mathrm{F}=\frac{\mathrm{Gm} _{1} \mathrm{~m} _{2}}{\mathrm{r}^{2}}=\frac{6.67 \times 10^{-11} \times 200 \times 500}{(10)^{2}}=6.67 \times 10^{-8} \mathrm{~N} \end{aligned} $$

(Please note that this force is negligible in magnitude, even though the masses are heavy. Our feeling that they are heavy is because of their weight which is none other than the gravitational attraction on then by earth).

Solution:

The figure represents the situation given in the question. The distance between the centres of earth and the body $=\mathrm{R}$.

$\therefore \mathrm{r}=\mathrm{R}=6400 \times 10^{3} \mathrm{~m}$

$\mathrm{m} _{1}=\mathrm{m}=50 \mathrm{Kg}$

$\mathrm{m} _{2}=\mathrm{M}=6 \times 10^{24} \mathrm{Kg}$

Now, $F=\frac{G m _{1} m _{2}}{r^{2}}$

$$ \begin{aligned} & =\frac{\mathrm{G} \mathrm{mm}}{\mathrm{R}^{2}} \\ & =\frac{6.67 \times 10^{-11} \times 50 \times 6 \times 10^{24}}{\left(6400 \times 10^{3}\right)^{2}} \\ & =\frac{6.67 \times 3}{64 \times 64} \times 10^{5} \\ & =488.5 \mathrm{~N} \end{aligned} $$

(Note that this attractive force on $50 \mathrm{Kg}$ is practically large and it is directed towards the centre of earth. This is called the weight of the body. We all feel this gravitational attraction or ‘weight’ which enables us to walk and move on the surface of the earth).

Solution:

$\mathrm{F}=\frac{\mathrm{Gm} _{1} \mathrm{~m} _{2}}{\mathrm{r}^{2}}$

$$ \begin{aligned} & =\frac{6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times 9.1 \times 10^{-31}}{\left(10^{-15}\right)^{2}} \\ & =101.36399 \times 10^{-39} \mathrm{~N} \\ & =1.01 \times 10^{-37} \mathrm{~N} \end{aligned} $$

(Note that this force is extremely negligible. In comparison, the electrostatic force between the same proton and electron is:

$\mathrm{F} _{\mathrm{e}}=\frac{1}{4 \pi \varepsilon _{0}} \cdot \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r}^{2}}$ $\left[\frac{1}{4 \pi \varepsilon _{0}}=9 \times 10^{9}\right.$ SI units

$$ \begin{aligned} & =\frac{9 \times 10^{9} \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\left(10^{-15}\right)^{2}} \\ & =23.04 \times 10 \\ & =230.4 \mathrm{~N} \quad \text { (Attractive) } \end{aligned} $$

(This is very large compared to the gravitational force between them).

Solution:

(i) Figure represents the position of the fourth particle at D. It is clear that two forces on $\mathrm{m}$ at $\mathrm{D}$; due to mass at $\mathrm{B}$ and $\mathrm{C}$, are equal and opposite. So they cancel out.

$\therefore$ The third force, directed from $\mathrm{D}$ to $\mathrm{A}$ is the resultant force on $\mathrm{m}$ at D.

$\therefore \quad F _{D}=\frac{G m \times m}{(A D)^{2}}=\frac{G m^{2}}{\left(\frac{3}{4} a^{2}\right)}=\frac{4}{3} \frac{G m^{2}}{a^{2}}$

(ii) When the fourth particle is placed at the centroid $\mathrm{O}$, the three forces on the mass $\mathrm{m}$ at $\mathrm{O}$ due to the other three have equal magnitude of $\mathrm{F}$ each. The angle between any two of these forces is $120^{\circ}$. Therefore the third force (i.e. along OA) will become the equilibrant of the other two. Hence the net force would be zero.

Solution:

We have, $F=\frac{G m _{1} m _{2}}{r^{2}}$

or

$\mathrm{F} \propto \mathrm{m} _{1} \mathrm{~m} _{2}$

and

$\mathrm{F} \propto \frac{1}{\mathrm{r}^{2}}$

(i)

(ii)

Solution:

Here, $\mathrm{r} _{12}=4 \mathrm{~m}$ and $\mathrm{r} _{31}=\mathrm{r} _{13}=3 \mathrm{~m}$

$$ \mathrm{r} _{23}=\mathrm{r} _{32}=\sqrt{4^{2}+3^{2}}=5 \mathrm{~m} $$

$\mathrm{m} _{1}=1 \mathrm{Kg}, \mathrm{m} _{2}=2 \mathrm{Kg}$ and $\mathrm{m} _{3}=2 \mathrm{Kg}$

The gravitational potential energy of the system is:

$$ \begin{aligned} \mathrm{U}=- & \frac{\mathrm{Gm} _{1} \mathrm{~m} _{2}}{\mathrm{r} _{12}}+-\frac{\mathrm{Gm} _{2} \mathrm{~m} _{3}}{\mathrm{r} _{23}}+-\frac{\mathrm{Gm} _{3} \mathrm{~m} _{1}}{\mathrm{r} _{31}} \\ & =-6.67 \times 10^{-11}\left(\frac{1 \times 2}{4}+\frac{2 \times 2}{5}+\frac{2 \times 1}{3}\right) \\ & =-6.67 \times 10^{-11} \frac{59}{30} \end{aligned} $$

$$ =-1.31 \times 10^{-10} \mathrm{~J} $$

The binding energy of the system is:

B.E $=-(\mathrm{U})$

$$ \begin{aligned} & =-\left(-1.31 \times 10^{-10} \mathrm{~J}\right) \\ & =1.31 \times 10^{-10} \mathrm{~J} \end{aligned} $$

Solution:

(i) For the surface,

$$ \begin{aligned} \mathrm{V}= & -\frac{\mathrm{GM}}{\mathrm{R}}=-\frac{\mathrm{G}}{\mathrm{R}} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho \\ & =-\frac{4}{3} \pi \mathrm{GR}^{2} \rho \\ & =-\frac{4}{3} \times 3.14 \times 6.67 \times 10^{-11} \times\left(6.37 \times 10^{6}\right)^{2} \times 5.5 \times 10^{3} \\ & =-6.235 \times 10^{7} \mathrm{~J} \mathrm{Kg}^{-1} \end{aligned} $$

(ii) At a depth of $\frac{\mathrm{R}}{2}$,

$$ \begin{aligned} V= & -\frac{G M}{2 R^{3}}\left(3 R^{2}-r^{2}\right) \\ & =-\frac{G M}{2 \times R^{3}}\left(3 R^{2}-\frac{R^{2}}{4}\right) \end{aligned} $$

$$ \begin{aligned} & =-\frac{\mathrm{GM}}{2 \times \mathrm{R}^{3}} \times \frac{11}{4} \times \mathrm{R}^{2} \\ & =-\frac{11}{8} \frac{\mathrm{GM}}{\mathrm{R}} \\ & =-\frac{11}{8} \times 6.235 \times 10^{7} \\ & =-8.57 \times 10^{7} \mathrm{~J} \mathrm{Kg}^{-1} \end{aligned} $$

(iii) At the centre,

$$ \begin{aligned} \mathrm{V}= & -\frac{3}{2} \frac{\mathrm{GM}}{\mathrm{R}} \\ & =-\frac{3}{2} \times 6.235 \times 10^{7} \\ & =-9.35 \times 10^{7} \mathrm{~J} \mathrm{Kg}^{-1} \end{aligned} $$

Solution:

We have, $\mathrm{g}=\frac{4}{3} \pi \mathrm{GR} \rho$

$\therefore \frac{\mathrm{g} _{\mathrm{P}}}{\mathrm{g} _{\mathrm{E}}}=\frac{\mathrm{R} _{\mathrm{P}} \rho _{\mathrm{P}}}{\mathrm{R} _{\mathrm{E}} \rho _{\mathrm{E}}}=\left(\frac{\mathrm{R} _{\mathrm{P}}}{\mathrm{R} _{\mathrm{E}}}\right)\left(\frac{\rho _{\mathrm{P}}}{\rho _{\mathrm{E}}}\right)$ $=\frac{3}{4} \times \frac{2}{3}=\frac{1}{2}$

$\therefore \frac{\mathrm{W} _{\mathrm{P}}}{\mathrm{W} _{\mathrm{E}}}=\frac{\mathrm{mg} _{\mathrm{P}}}{\mathrm{mg} _{\mathrm{E}}}=\frac{1}{2}$

or $\quad \mathrm{W} _{\mathrm{P}}=\frac{\mathrm{W} _{\mathrm{E}}}{2}=\frac{600}{2} \mathrm{~N}=300 \mathrm{~N}$

Solution:

Here, $\mathrm{g}=\frac{4}{3} \pi \mathrm{GR} \rho$

$\therefore \frac{\mathrm{g} _{1}}{\mathrm{~g} _{2}}=\frac{\mathrm{R} _{1}}{\mathrm{R} _{2}} \cdot \frac{\rho _{1}}{\rho _{2}}$ $=2: 9$

Now, $\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \quad(\ell=$ constant $)$

$\therefore \frac{\mathrm{T} _{1}}{\mathrm{~T} _{2}}=\frac{\sqrt{\mathrm{g} _{2}}}{\sqrt{\mathrm{g} _{1}}}$

$=\sqrt{9}: \sqrt{2}$

Solution:

We have $\frac{\Delta \mathrm{g}}{\mathrm{g}}=-\frac{2 \Delta \mathrm{g}}{\mathrm{R}}+\frac{\Delta \mathrm{M}}{\mathrm{M}}$

Here, $\Delta \mathrm{M}=0$

$$ \begin{aligned} & \text { and } \frac{\Delta \mathrm{R}}{\mathrm{R}}=-5 \% \\ & \begin{aligned} \therefore \quad \frac{\Delta \mathrm{g}}{\mathrm{g}} & =-2 \times-5 \% \\ & =10 \% \end{aligned} \end{aligned} $$

$\therefore$ g increases by $10 \%$

Solution:

Note that the decrease in $\mathrm{g}$ is large. Therefore the height $\mathrm{h}$ is large.

So, we don’t use the formula:

$\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$, which is applicable only for $\mathrm{h}«\mathrm{R}$.

We know, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$

and $\quad g^{\prime}=\frac{G M}{(R+h)^{2}}$

$\therefore \frac{g^{\prime}}{g}=\frac{R^{2}}{(R+h)^{2}}$

or $\quad \frac{1}{9}=\frac{\mathrm{R}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}$

$\therefore \frac{1}{3}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}$

or $\mathrm{h}=2 \mathrm{R}$

$=2 \times 6400 \mathrm{~km}$

$=12800 \mathrm{~km}$

Solution:

We have $g^{\prime}=g\left(1-\frac{d}{R}\right)$

$\therefore \frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\mathrm{d}}{\mathrm{R}}$

Now, $\frac{\Delta \mathrm{g}}{\mathrm{g}}$ is same $\frac{\Delta \mathrm{W}}{\mathrm{W}}$, the $\%$ ge decrease in weight.

$\therefore \%$ ge decrease in weight $=\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\mathrm{d}}{\mathrm{R}} \times 100 \%$

$$ \begin{aligned} & =\frac{32}{6400} \times 100 \% \\ & =0.5 \% \end{aligned} $$

Solution:

At a depth $d, g^{\prime}=g\left(1-\frac{d}{R}\right)$

At a height $\mathrm{h}$, since $\mathrm{h}«\mathrm{R} ; \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$

Equating, $\mathrm{d}=2 \mathrm{~h}=128 \mathrm{~km}$

Solution:

Total initial energy of the system is:

$$ \begin{aligned} & \mathrm{E} _{\mathrm{i}}=\mathrm{K} \cdot \mathrm{E} _{\mathrm{i}}+\mathrm{P} \cdot \mathrm{E} _{\mathrm{i}} \\ & =0+\frac{-\mathrm{Gm} _{1} \mathrm{~m} _{2}}{\mathrm{~d}} \end{aligned} $$

When they start moving, their P.E. will decrease and the K.E will increase (from zero)

When $\mathrm{d}$ becomes $\frac{\mathrm{d}}{2}$, the find energy becomes:

$$ \begin{aligned} & \mathrm{E} _{\mathrm{f}}=\mathrm{K} \cdot \mathrm{E} _{\mathrm{f}}+\Delta \cdot \mathrm{E} _{\mathrm{f}} \\ &=\left(\frac{1}{2} \mathrm{~m} _{1} \mathrm{v} _{1}^{2}+\frac{1}{2} \mathrm{~m} _{2} \mathrm{v} _{2}^{2}\right)+\frac{-\mathrm{Gm} _{1} \mathrm{~m} _{2}}{\mathrm{~d} / 2} \\ &=\frac{1}{2} \mathrm{~m} _{1} \mathrm{v} _{1}^{2}+\frac{1}{2} \mathrm{~m} _{2} \mathrm{v} _{2}^{2}-\frac{2 \mathrm{Gm} _{1} \mathrm{~m} _{2}}{\mathrm{~d}} \end{aligned} $$

Now, $\mathrm{E} _{\mathrm{f}}=\mathrm{E} _{\mathrm{i}}$ (Conservation of energy)

$\therefore \quad \frac{1}{2} \mathrm{~m} _{1} \mathrm{v} _{1}^{2}+\frac{1}{2} \mathrm{~m} _{2} \mathrm{v} _{2}^{2}=\frac{\mathrm{Gm} _{1} \mathrm{~m} _{2}}{\mathrm{~d}}$

or

$$ \mathrm{m} _{1} \mathrm{v} _{1}^{2}+\mathrm{m} _{2} \mathrm{v} _{2}^{2}=\frac{2 \mathrm{Gm} _{1} \mathrm{~m} _{2}}{\mathrm{~d}} $$

Now, (from the conservation of linear momentum)

$$ \mathrm{m} _{1} \mathbf{v} _{1}+\mathrm{m} _{2} \mathbf{v} _{2}=0 $$

$\therefore \quad \mathbf{v} _{1}=-\left(\frac{\mathrm{m} _{2}}{\mathrm{~m} _{1}}\right) \mathbf{v} _{2}$

or $\quad \mathrm{v} _{1}=-\frac{\mathrm{m} _{2}}{\mathrm{~m} _{1}} \mathrm{v} _{2}$ (as they move along the same line)

Therefore, $\mathrm{m} _{1}\left(-\frac{\mathrm{m} _{2}}{\mathrm{~m} _{1}} \mathrm{v} _{2}\right)^{2}+\mathrm{m} _{2} \mathrm{v} _{2}^{2}=\frac{2 \mathrm{Gm} _{1} \mathrm{~m} _{2}}{\mathrm{~d}}$ or $\quad\left(\frac{\mathrm{m} _{2}^{2}}{\mathrm{~m} _{1}}+\mathrm{m} _{2}\right) \mathrm{v} _{2}^{2}=\frac{2 \mathrm{Gm} _{1} \mathrm{~m} _{2}}{\mathrm{~d}}$

$\therefore \quad \mathrm{v} _{2}^{2}=\frac{2 \mathrm{G} \mathrm{m} _{1}^{2}}{\left(\mathrm{~m} _{1}+\mathrm{m} _{2}\right) \mathrm{d}}=\frac{2 \times 6.67 \times 10^{-11} \times 10^{2}}{30 \times 1}$

or $\quad \mathrm{v} _{2}^{2}=4.45 \times 10^{-10}$

$$ \mathrm{v} _{2}=2.1 \times 10^{-5} \mathrm{~ms}^{-1} $$

$\mathrm{v} _{1}=-\frac{\mathrm{m} _{2}}{\mathrm{~m} _{1}} \times \mathrm{v} _{2}$

$$ \begin{aligned} & =-2 \times 2.1 \times 10^{-5} \\ & =-4.2 \times 10^{-5} \mathrm{~ms}^{-1} \end{aligned} $$

Solution:

By Keplers law: $\mathrm{T}^{2} \propto \mathrm{r}^{3}$

$$ \begin{aligned} & \frac{\mathrm{T} _{\mathrm{S}}^{2}}{\mathrm{~T} _{\mathrm{E}}^{2}}=\frac{\mathrm{r} _{\mathrm{S}}^{3}}{\mathrm{r} _{\mathrm{E}}^{3}} \\ \therefore \quad \mathrm{r} _{\mathrm{s}}= & \left(\frac{\mathrm{T} _{\mathrm{S}}}{\mathrm{T} _{\mathrm{E}}}\right)^{2 / 3} \times \mathrm{r} _{\mathrm{E}} \\ & =\left(\frac{29.4 \times \mathrm{T} _{\mathrm{E}}}{\mathrm{T} _{\mathrm{E}}}\right)^{2 / 3} \times 1.5 \times 10^{8} \mathrm{~km} \\ & =1.43 \times 10^{9} \mathrm{~km} \end{aligned} $$

Solution:

Here , $E_i= E_f \quad or \quad K.E_i + P.E_i = K.E_f + P.E_f$

$\therefore \frac{1}{2} \mathrm{mv} _{\mathrm{i}}^{2}+\frac{-\mathrm{GMm}}{\mathrm{R}}=\frac{1}{2} \mathrm{mv} _{\mathrm{f}}^{2}+\frac{-\mathrm{GMm}}{\propto}$

or $\quad\left(3 v _{e}\right)^{2}-v _{e}^{2}=v _{f}^{2}$

$\therefore \quad \mathrm{v} _{\mathrm{f}}=\sqrt{8} \mathrm{v} _{\mathrm{e}}$

$=2.828 \times 11.2 \mathrm{kms}^{-1}$

$=31.67 \mathrm{kms}^{-1}$

Solution:

At point $\mathrm{O}$ the intensities due to masses $\mathrm{m} _{1}$ and $\mathrm{m} _{2}$ are:

$$ \mathrm{I} _{1}=\frac{\mathrm{Gm} _{1}}{\mathrm{r} _{1}^{2}}=\frac{\mathrm{G} \times 100}{(1 / 2)^{2}}=400 \mathrm{G}, \text { towards } \mathrm{m} _{1} $$

and, $\mathrm{I} _{2}=400 \mathrm{G}$, towards $\mathrm{m} _{2}$

So, $\mathrm{I} _{1}=-\mathrm{I} _{2}$

$\therefore \mathrm{I} _{0}=0$

$$ \begin{aligned} \mathrm{V} _{0}=\mathrm{V} _{1} & +\mathrm{V} _{2}=\frac{-\mathrm{Gm} _{1}}{\left(\frac{1}{2}\right)}+\frac{-\mathrm{Gm} _{2}}{\left(\frac{1}{2}\right)} \\ & =-6.67 \times 10^{-11} \times 400 \\ & =-2.66 \times 10^{-8} \mathrm{~J} \mathrm{~kg}^{-1} \end{aligned} $$

As $\mathrm{I} _{0}=0$, there is no force on the $10 \mathrm{~g}$ body placed at 0 . So it will remain in equilibrium at 0 . However, if we move it aside a little, the attractive force due to $\mathrm{m} _{1}$ or $\mathrm{m} _{2}$ will become greater and therefore, it move towards that mass from which attractive force is greater, and will never come back. Therefore, the equilibrium is unstable.

Solution:

We have $\mathrm{I}=-\frac{\mathrm{dV}}{\mathrm{d} x}$

$\therefore \mathrm{V}=-\int \mathrm{I} \mathrm{d} x$

$$ =-\mathrm{A} \int _{x=\infty}^{x} x^{-4} \mathrm{~d} x $$

$$ \begin{aligned} & =-\mathrm{A}\left[\frac{x^{-4+1}}{-4+1}\right] _{\infty}^{x} \\ & =\frac{\mathrm{A}}{3} x^{-3}=\frac{\mathrm{A}}{3 x^{3}} \end{aligned} $$

Solution:

Let $\mathrm{P}$ be this point, which will be in between the two shells.

There is no field due to the outer shell at $\mathrm{P}$.

$\therefore \mathrm{I} _{\mathrm{p}}=$ field of $\mathrm{m} _{1}$ only

$$ =\frac{G m _{1}}{\left(\frac{r _{1}+r _{2}}{2}\right)^{2}}=\frac{4 G m _{1}}{\left(r _{1}+r _{2}\right)^{2}} $$

Now, as $\mathrm{P}$ is inside the outer shell, the potential due to $\mathrm{m} _{2}$ is $\mathrm{V} _{2}=\frac{-\mathrm{Gm} _{2}}{\mathrm{r} _{2}}$

(Inside a shell potential due to that shell is constant)

Therefore, the net potential at $\mathrm{P}$ is:

$\mathrm{v} _{\mathrm{p}}=\mathrm{v} _{1}+\mathrm{v} _{2}$

$$ \begin{aligned} & =\frac{-\mathrm{Gm} _{1}}{\left[\left(\mathrm{r} _{1}+\mathrm{r} _{2}\right) / 2\right]}+\frac{-\mathrm{Gm} _{2}}{\mathrm{r}^{2}} \\ & =-\mathrm{G}\left[\frac{2 \mathrm{~m} _{1}}{\mathrm{r} _{1}+\mathrm{r} _{2}}+\frac{\mathrm{m} _{2}}{\mathrm{r} _{2}}\right] \end{aligned} $$

Solution:

$\mathrm{v} _{0}=\frac{1}{2} \mathrm{v} _{\mathrm{e}}$

$\mathrm{v} _{0}^{2}=\frac{1}{4} \mathrm{v} _{\mathrm{e}}^{2}$

or $\quad \frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}=\frac{1}{4} \times \frac{\mathrm{GM}}{\mathrm{R}}$

$\therefore 4 \mathrm{R}=\mathrm{R}+\mathrm{h}$

or $\mathrm{h}=3 \mathrm{R}=3 \times 6400 \mathrm{~km}=19200 \mathrm{~km}$

Solution:

Suppose $\mathrm{F} \propto \frac{1}{\mathrm{r}^{\mathrm{n}}}$ i.e. $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{r}^{\mathrm{n}}}$

Then $\frac{\mathrm{GMm}}{\mathrm{r}^{\mathrm{n}}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}} \quad(\mathrm{v}=$ orbital velocity $)$ $\therefore \mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}^{\mathrm{n}-1}}}$

Here, $\mathrm{n}=5$

$\therefore \quad v=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}^{5-1}}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}^{4}}}=\frac{1}{\mathrm{r}^{2}} \sqrt{\mathrm{GM}}$

$\therefore \quad \mathrm{v} \propto \frac{1}{\mathrm{r}^{2}}$

Now, $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}}=\frac{2 \pi \mathrm{r}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{r}^{\mathrm{n}-1}}}}=\frac{2 \pi}{\sqrt{\mathrm{GM}}} \cdot \sqrt{\mathrm{r}^{\mathrm{n}-1} \times \mathrm{r}^{2}}$

or $\quad \mathrm{T}=\frac{2 \pi}{\sqrt{\mathrm{GM}}} \cdot \sqrt{\mathrm{r}^{\mathrm{n}+1}}$

or $\quad \mathrm{T}=\frac{2 \pi}{\sqrt{\mathrm{GM}}} \cdot \mathrm{r}^{\left(\frac{\mathrm{n}+1}{2}\right)}$

Hence, $\mathrm{n}=5$

$\therefore \mathrm{T}=\frac{2 \pi}{\sqrt{\mathrm{GM}}} \cdot \mathrm{r}^{\frac{5+1}{2}}$

$=\frac{2 \pi}{\mathrm{GM}} \cdot \mathrm{r}^{3}$

or

$$ \mathrm{T} \propto \mathrm{r}^{3} $$

Correct answer: (3)

Solution:

By Kepler’s law, $\mathrm{T}^{2} \propto \mathrm{r}^{3}$

In the first case, $4^{2} \propto(2 \mathrm{R})^{3}$

$$ (\mathrm{r}=\mathrm{R}+\mathrm{h}, \therefore \mathrm{r}=\mathrm{R}+\mathrm{R}) $$

In the second case, $\mathrm{T}^{2} \propto(8 \mathrm{R})^{3} \quad$ ( $\mathrm{h}$ is increased by $6 \mathrm{R}, \therefore \mathrm{h}=\mathrm{R}+6 \mathrm{R}=7 \mathrm{R} \quad \therefore \mathrm{r}=\mathrm{R}+7 \mathrm{R}$ )

$$ \begin{aligned} & \therefore \quad \frac{4^{2}}{\mathrm{~T}^{2}}=\frac{8 \mathrm{R}^{3}}{8^{3} \mathrm{R}^{3}} \\ & \therefore \quad \mathrm{T}^{2}=8^{2} \times 4^{2} \\ & \mathrm{~T}=8 \times 4 \\ & =32 \text { Hours } \end{aligned} $$

Correct answer: (4)

Solution:

$$ \mathrm{h}=\frac{\mathrm{R}}{4} $$

At see level, $\mathrm{u} _{\mathrm{i}}=-\frac{\mathrm{GMm}}{\mathrm{R}}=-\left(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right) \mathrm{mR}=-(\mathrm{mgR})$

At $\mathrm{h}=\frac{\mathrm{R}}{4}, \mathrm{u} _{\mathrm{f}}=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}=-\frac{\mathrm{GMm}}{\frac{5}{4} \mathrm{R}}=-\left(\frac{4}{5} \mathrm{mgR}\right)$

$\therefore \Delta \mathrm{u}=\mathrm{u} _{\mathrm{f}}-\mathrm{u} _{\mathrm{i}}=\frac{1}{5} \mathrm{MgR}$

Hence, %age increase is $\frac{\Delta \mathrm{u}}{\left|\mathrm{u} _{\mathrm{i}}\right|} \times 100=20 \%$

Correct answer: (3)

Solution:

A satellite is ejected into the orbit with a velocity $v _{0}=\sqrt{\frac{G M}{r}}$. This satellite will have two motions-(i) free fall with an acceleration equal to $g$ at distance $r$ and (ii) forward motion due to $v _{0}$. The forward motion is such that the distance from the earth’s surface remains a constant, or in other words, the satellite orbits the earth.

As the satellite and astronaut are in free fall, the apparent weight will be zero, or the astronaut feels weightlessness.

Correct answer: (2)

Solution:

Let A and B be the satellites. The relative velocity of $A$ with respect to $\mathrm{B}$, as observed from ground, is:

$$ \mathrm{v}=\mathrm{v} _{1}+\mathrm{v} _{2} $$

The time interval of approach of A to $B$ is:

$\Delta t=\frac{2 \pi r _{1}}{v _{1}+v _{2}}$

$$ =\frac{2 \pi r _{1}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{r} _{1}}}+\sqrt{\frac{\mathrm{GM}}{\mathrm{r} _{2}}}} $$

$$ \left(\because \mathrm{v} _{0}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}\right) $$

or $\quad \Delta t=\frac{2 \pi \mathrm{r} _{1}}{\sqrt{\mathrm{GM}}} \cdot \frac{1}{\left(\frac{1}{\sqrt{\mathrm{r} _{1}}}+\frac{1}{\sqrt{\mathrm{r} _{2}}}\right)}$

or $\quad \Delta t=\frac{\mathrm{T} _{1}}{\sqrt{\mathrm{r} _{1}}} \cdot \frac{\sqrt{\mathrm{r} _{1}} \cdot \sqrt{\mathrm{r} _{2}}}{\sqrt{\mathrm{r} _{1}}+\sqrt{\mathrm{r} _{2}}}$

$$ \left(\because \mathrm{T}=\frac{2 \pi \mathrm{r}}{\sqrt{\mathrm{GM}}} \cdot \sqrt{\mathrm{r}}\right) $$

or $\quad \Delta \mathrm{t}=\mathrm{T} _{1}\left(\frac{\sqrt{\mathrm{r} _{2}}}{\sqrt{\mathrm{r} _{1}}+\sqrt{\mathrm{r} _{2}}}\right)$

Correct answer: (3)

Solution:

The centripetal acceleration of the moon is equal to the acceleration due to gravity of earth at the moon.

$\therefore \quad \mathrm{a} _{\mathrm{c}}=\mathrm{g} _{1}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}$

$$ \begin{array}{ll} =\frac{\mathrm{GM}}{(59+1)^{2} \mathrm{R}^{2}} & {[\because \mathrm{r}=\mathrm{h}+\mathrm{R}=(59+1) \mathrm{R}=60 \mathrm{R}]} \\ =\frac{\mathrm{g}}{(60)^{2}} \\ =\frac{9.81}{60 \times 60} & \\ =2.72 \times 10^{-3} \mathrm{~ms}^{-2} \end{array} $$

Correct answer: (2)

Solution:

At $\mathrm{O}$, there are two gravitational potentials:

(i) Due to the particle, $V _{1}=\frac{-G M}{r+R}$

$\therefore$ Net gravitational potential at $\mathrm{O}$ is:

$$ \begin{aligned} & \mathrm{V}= \mathrm{V} _{1}+\mathrm{V} _{2}=-\mathrm{GM}\left(\frac{1}{\mathrm{r}+\mathrm{R}}+\frac{1}{\mathrm{R}}\right) \\ &=-\mathrm{GM}\left(\frac{2 \mathrm{R}+\mathrm{r}}{(\mathrm{r}+\mathrm{R}) \mathrm{R}}\right) \end{aligned} $$

At $\mathrm{O}$, the field intensity due to the shell is zero. So, gravitational field at ’ $\mathrm{O}$ ’ is due to the particle only.

$\therefore \quad \mathrm{I}=\frac{\mathrm{GM}}{(\mathrm{r}+\mathrm{R})^{2}}$, towards the particle.

Correct answer: (1)

Solution:

Let ’ $m$ ’ be the mass of the particle.

At $\mathrm{C},\left(\mathrm{r}=\frac{3}{2} \mathrm{a}\right)$, the force due to shell $\mathrm{B}$ is zero because $\mathrm{C}$ is inside shell $\mathrm{B}$.

$\therefore \quad \mathrm{F} _{\mathrm{C}}=\frac{\mathrm{GMm}}{\left(\frac{3}{2} \mathrm{a}\right)^{2}}+0=\frac{4}{9} \frac{\mathrm{GMm}}{\mathrm{a}^{2}} \hspace{40mm} . . . . . . .(1)$

At point $\mathrm{D},(\mathrm{r}=3 \mathrm{a})$, the shells $\mathrm{A}$ and $\mathrm{B}$ behave like point masses at $O$ because $D$ lies outside both $A$ and $B$.

$\therefore$ Total mass at $\mathrm{O}$ is $4 \mathrm{M}$

Hence, force of the particle of mass $m$, at $D$ is:

$$ \begin{equation*} \mathrm{F} _{\mathrm{D}}=\frac{\mathrm{G} 4 \mathrm{Mm}}{(3 \mathrm{a})^{2}}=\frac{4}{9} \frac{\mathrm{GMm}}{\mathrm{a}^{2}} \tag{2} \end{equation*} $$

From (1) and (2), we get

$\mathrm{F} _{\mathrm{C}}=\mathrm{F} _{\mathrm{D}}$

$\therefore$ Change in force is zero.

Correct answer: (4)

Solution:

The gravitational field intensity at $\mathrm{O}$, due to the ring is:

$$ \begin{gathered} \mathrm{I}=\frac{\mathrm{Gm} x}{\left(x^{2}+\mathrm{r}^{2}\right)^{3 / 2}}=\frac{\mathrm{Gm} \sqrt{8} \mathrm{r}}{\left(8 \mathrm{r}^{2}+\mathrm{r}^{2}\right)^{3 / 2}} \\ =\frac{\mathrm{Gm} \sqrt{8} \mathrm{r}}{\left(3^{2} \mathrm{r}^{2}\right)^{3 / 2}} \\ =\frac{\sqrt{8}}{27} \frac{\mathrm{Gm}}{\mathrm{r}^{2}} \end{gathered} $$

$\therefore$ Force on the sphere $=$ Force on mass $\mathrm{M}$ at $\mathrm{O}$

$$ =\mathrm{MI} $$

$$ =\frac{\sqrt{8}}{27}\left(\frac{\mathrm{GMm}}{\mathrm{r}^{2}}\right) $$

Correct answer: (3)

Solution:

$$ \mathrm{I} _{\mathrm{A}}=\frac{\mathrm{GM}}{\mathrm{r}^{2}}=\frac{\mathrm{GM}}{\left(\frac{3}{2} \mathrm{R}\right)^{2}} $$

or $\quad \mathrm{I} _{\mathrm{A}}=\frac{4}{9} \frac{\mathrm{GM}}{\mathrm{R}^{2}}$

$\mathrm{I} _{\mathrm{B}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(\frac{\mathrm{r}}{\mathrm{R}}\right)$, when $\mathrm{r}<\mathrm{R}$ for solid sphere.

or $\quad \mathrm{I} _{\mathrm{B}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(\frac{\mathrm{R} / 2}{\mathrm{R}}\right)$

or $\quad \mathrm{I} _{\mathrm{B}}=\frac{1}{2} \frac{\mathrm{GM}}{\mathrm{R}^{2}}$

$\therefore \quad \frac{\mathrm{I} _{\mathrm{A}}}{\mathrm{I} _{\mathrm{B}}}=\frac{4 / 9}{1 / 2}=\frac{8}{9}$

or $\mathrm{I} _{\mathrm{A}}: \mathrm{I} _{\mathrm{B}}=8: 9$

Correct answer: (1)

Solution:

As the spheres A and B are symmetrically located, their presence or absence will not contribute to the net field at $\mathrm{O}$, as their individual fields cancel out.

Now, as sphere $\mathrm{C}$ is not paired, the removal of sphere $\mathrm{C}$ will create a net field due to a similar sphere located geometrically opposite to $\mathrm{C}$, i.e. at $\mathrm{R}\left(-\frac{\hat{\mathrm{j}}}{2}\right)$. This will produce a net field at $\mathrm{O}$, directed along $(-\mathbf{j})$. All other regions of the sphere together will produce zero net field at $\mathrm{O}$.

So, the field at $\mathrm{O}$ will be the field due to a existing small sphere, of radius $\frac{\mathrm{R}}{4}$, located at $\mathrm{R}\left(-\frac{\mathbf{j}}{2}\right)$.

i.e. $\mathrm{I} _{\mathrm{net}}=\frac{\mathrm{GM}^{\prime}}{(\mathrm{R} / 2)^{2}}$

But $\mathrm{M}^{\prime}=\frac{4}{3} \pi\left(\frac{\mathrm{R}}{4}\right)^{3} \times \rho$

$=\frac{4}{3} \pi \frac{\mathrm{R}^{3}}{64} \cdot \frac{\mathrm{M}}{\frac{4}{3} \pi \mathrm{R}^{3}}$

$=\frac{M}{64}$

$\therefore \quad \mathrm{I} _{\text {net }}=\frac{\mathrm{G}}{\frac{\mathrm{R}^{2}}{4}} \times \frac{\mathrm{M}}{64}$

or $\quad I _{\text {net }}=\frac{1}{16} \frac{\mathrm{GM}}{\mathrm{R}^{2}}$, along $(-\mathbf{j})$

or $\quad I _{n e t}=\frac{G M}{16 R^{2}}(-\mathbf{j})$

Correct answer: (3)

Solution:

Let $r$ be the maximum distance to which the body moves, from the center of the earth. The increase in potential energy of the body will be:

$$ \Delta \mathrm{U}=\mathrm{U} _{\mathrm{f}}-\mathrm{U} _{\mathrm{f}}=-\frac{\mathrm{GMm}}{\mathrm{r}}+\frac{\mathrm{GMm}}{\mathrm{R}} $$

or $\quad \Delta \mathrm{U}=\mathrm{GMm}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right)$

Now, this increase in potential energy occurs due to the K.E.of projection.

$\therefore \quad \mathrm{K} . \mathrm{E}-\Delta \mathrm{U}$

or $\quad \frac{1}{2} \mathrm{mv}^{2}=\operatorname{GMm}\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{r}}\right)$

or $\quad \frac{1}{2} \mathrm{~m} \alpha^{2} \mathrm{v} _{\mathrm{e}}^{2}=\operatorname{GMm}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right)$

or $\quad \frac{1}{2} \mathrm{~m} \alpha^2 \cdot \frac{2 \mathrm{GM}}{\mathrm{R}}=\mathrm{GMm}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right)$ $$ \because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} $$ $\therefore \quad \alpha^2=\mathrm{R}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right)$ or $\quad \alpha^2=1-\frac{\mathrm{R}}{\mathrm{r}}$ or $\quad \frac{R}{r}=1-\alpha^2$ $$ \therefore \quad r=\frac{R}{1-\alpha^2} $$

Correct answer: (4)

Solution:

Let the mass of body be $\mathrm{m}$. The potential energy of the particle at the middle point $\mathrm{O}$ is:

$$ \begin{aligned} \mathrm{U}= & \frac{-\mathrm{GM} _{1} \mathrm{~m}}{(\mathrm{~d} / 2)}+\frac{-\mathrm{GM} _{2} \mathrm{~m}}{(\mathrm{~d} / 2)} \\ & =\frac{-2 \mathrm{GM}}{\mathrm{d}}\left(\mathrm{M} _{1}+\mathrm{M} _{2}\right) \end{aligned} $$

The K.E $=\frac{1}{2} \mathrm{mv}^{2}$, provided to the body should increase the potential energy to zero, when it reaches infinte distance from both earth and moon.

$\therefore \quad \frac{1}{2} \mathrm{mv}^{2}=\frac{2 \mathrm{GM}}{\mathrm{d}}\left(\mathrm{M} _{1}+\mathrm{M} _{2}\right)$

or $\quad v=2 \sqrt{\frac{\left(M _{1}+M _{2}\right) G}{d}}$

Correct answer: (3)

Solution:

Since charge occurs on its own, there is no external force and therefore no external torque.

The angular momentum of earth remains conserved, $\mathrm{I} _{1} \omega _{1}=\mathrm{I} _{2} \omega _{2}$

Moment of inertia of earth, being a solid sphere is $\mathrm{I}=\frac{2}{5} \mathrm{MR}^{2}$

$\therefore \quad \mathrm{I} _{1}=\frac{2}{5} \mathrm{MR} _{1}^{2}$ and $\mathrm{I} _{2}=\frac{2}{5} \mathrm{MR} _{2}^{2}$

and $\quad \omega _{1}=\frac{2 \pi}{\mathrm{T} _{1}} \quad$ (where $\mathrm{T} _{1}=24$ hours)

$$ \omega _{2}=\frac{2 \pi}{\mathrm{T} _{2}} \quad \text { (where } \mathrm{T} _{2} \text { is the new time period of changed earth around its own axis) } $$

Here, $\mathrm{I} _{1} \omega _{1}=\mathrm{I} _{2} \omega _{2}$

or $\quad \frac{2}{5} \mathrm{MR} _{1}^{2}\left(\frac{2 \pi}{\mathrm{T} _{1}}\right)=\frac{2}{5} \mathrm{MR} _{2}^{2}\left(\frac{2 \pi}{\mathrm{T} _{2}}\right)$

or $\frac{\mathrm{T} _{1}}{\mathrm{~T} _{2}}=\frac{\mathrm{R} _{1}^{2}}{\mathrm{R} _{2}^{2}}$

or $\frac{\mathrm{T} _{2}}{\mathrm{~T} _{1}}=\frac{\mathrm{R} _{2}^{2}}{\mathrm{R} _{1}^{2}}$

But $\quad \mathrm{R} _{2}=\mathrm{R} _{1}-\frac{\mathrm{R} _{1}}{\mathrm{n}}=\mathrm{R} _{1}\left(1-\frac{1}{\mathrm{n}}\right)$

$\therefore \frac{\mathrm{T} _{2}}{\mathrm{~T} _{1}}=\frac{\mathrm{R} _{1}^{2}\left(1-\frac{1}{\mathrm{n}}\right)^{2}}{\mathrm{R} _{1}^{2}}$

or $\frac{\mathrm{T} _{2}}{\mathrm{~T} _{1}}=\left(1-\frac{1}{\mathrm{n}}\right)^{2}$ $\therefore \frac{\mathrm{T} _{2}-\mathrm{T} _{1}}{\mathrm{~T} _{1}}=\left[\left(1-\frac{1}{\mathrm{n}}\right)^{2}-1\right]=\frac{1-2 \mathrm{n}}{\mathrm{n}^{2}}$

Hence, $\%$ ge change in duration of the day $=\frac{1-2 \mathrm{n}}{\mathrm{n}^{2}} \times 100 \%$

Correct answer: (2)

Solution:

The gravitational potential, $\mathrm{V}=-\frac{\mathrm{GM}}{\mathrm{r}}$. Therefore the net gravitational potential at the origin will be:

$$ \mathrm{V} _{\mathrm{net}}=\mathrm{V} _{1}+\mathrm{V} _{2}+\mathrm{V} _{3}+\ldots \ldots \ldots \ldots \ldots $$

or

$$ \mathrm{V} _{\mathrm{net}}=-\mathrm{GM}\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots\right] $$

or

$$ \mathrm{V} _{\mathrm{net}}=-\mathrm{GM}\left(\frac{1}{1-\frac{1}{2}}\right) $$

or $\mathrm{V} _{\text {net }}=-2 \mathrm{GM}$

Now, $\mathrm{I}=\frac{\mathrm{GM}}{\mathrm{r}^{2}}(\mathrm{i})$ (as the field $\mathrm{I}$ is directed towards the mass)

$\therefore \quad I_{\text {net }} $=$G M[\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{8^2}+\ldots \ldots .]\hat{(i)} $

$ =G M[\frac{1}{1}+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots \ldots .]\hat{(i)} $

$ =G M[\frac{1}{1-\frac{1}{4}}]\hat{(i)} $

$ =\frac{4}{3} G M \hat{(i)} $

$ =\frac{4}{3} G M(i)$

Correct answer: (1)

Solution:

At the height $h$, the value of $g$ is:

$$ g=g _{0}\left(1-\frac{2 h}{R}\right), \text { when } \mathrm{h}«\mathrm{R} $$

$\therefore \%$ age decrease in $\mathrm{g}=\frac{2 \mathrm{~h}}{\mathrm{R}} \times 100 \%$

$$ \begin{aligned} & =\frac{2 \times 400}{6400 \times 1000} \times 10 \% \\ & =\frac{1}{80} \% \end{aligned} $$

Now, $\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$

or $\mathrm{T}^{2}=4 \pi^{2} \frac{\ell}{\mathrm{g}} \quad(\ell=$ constant $)$

$\therefore \frac{\Delta \mathrm{T}}{\mathrm{T}}=-\frac{1}{2} \frac{\Delta \mathrm{g}}{\mathrm{g}}=-\frac{1}{2} \times \frac{-1}{80}=\frac{1}{160} \%$

So, the time period of the pendulum increases by $\frac{1}{160} \%$

$\therefore$ Loss in 1 second $=\frac{1}{16000}$

or $\quad$ Loss in 1 day $=\frac{1}{16000} \times 86400$

$=5.4$ second

Correct answer: (2)

Solution:

The gravitational potential at any point on or inside the spherical shell is

$$ \mathrm{V} _{1}=-\frac{\mathrm{GM}}{\mathrm{a}} $$

The gravitational potential at $\mathrm{P}$ due to mass $\mathrm{M}$ at $\mathrm{O}$ is:

$$ \mathrm{V} _{2}=-\frac{\mathrm{GM}}{\mathrm{a} / 2} $$

or $\quad \mathrm{V} _{2}=-\frac{2 \mathrm{GM}}{\mathrm{a}}$

$\therefore$ Net gravitational potential at $\mathrm{P}$ is:

$$ \begin{aligned} & V _{P}=V _{1}+V _{2} \\ & =-\frac{3 G M}{a} \end{aligned} $$

The gravitational field (I) at any point inside the shell due to itself is zero.

$\therefore \mathrm{I}$ at point $\mathrm{P}$ is due to the mass $\mathrm{M}$ at $\mathrm{O}$ only.

or $\quad I _{P}=\frac{-G M}{(a / 2)^{2}}$

or $\mathrm{I} _{\mathrm{P}}=\frac{-4 \mathrm{GM}}{\mathrm{a}^{2}}=\frac{4 \mathrm{GM}}{\mathrm{a}^{2}}$ towards $\mathrm{O}$

Correct answer: (3)

Solution:

On the surface of the earth, $g=\frac{G M}{R^{2}}$

or $\quad \mathrm{g}=\mathrm{G} \cdot \frac{4}{3} \frac{\pi \mathrm{R}^{3} \rho}{\mathrm{R}^{2}}$

or $\quad \mathrm{g}=\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R} \rho \hspace{40mm}. . . . . . . (1) $

On the surface of the planet, of radius $r$,

$$ \begin{array}{r} \mathrm{g}^{\prime}=\frac{\mathrm{g}}{4}=\frac{\mathrm{GM}^{\prime}}{\mathrm{r}^{2}} \\ \therefore \frac{\mathrm{g}}{4}=\mathrm{G} \cdot \frac{\frac{4}{3} \pi \mathrm{r}^{3}(2 \rho)}{\mathrm{r}^{2}} \end{array} $$

or $\quad \mathrm{g}=\mathrm{G} \cdot \frac{4}{3} \pi \rho \mathrm{r} \times 8 \hspace{40mm}. . . . . . . (2) $

Equating (1) and (2), we get

$$ r=\frac{R}{8} $$

Correct answer: (4)

Solution:

Given : Kinetic energy of the satellite $=\mathrm{K}$

We know that the potential energy of a satellite is: $U=-2 K$

Now, if kinetic energy of this satellite is doubled, the total energy of the satellite would become:

$\mathrm{E}=2 \mathrm{~K}+-2 \mathrm{~K}=0$

when $\mathrm{E}=0$, the satellite cannot remain bound to earth (for which $-\mathrm{E} \neq 0$ is must).

Hence the orbital radius of the satellite would become infinite.

Correct answer: (3)

Solution:

For satellite around the earth, $v _{0}=\sqrt{\frac{\mathrm{GM} _{\mathrm{e}}}{\mathrm{e}}}$

$\therefore$ Kinetic energy of the satellite around earth is: $\mathrm{K} _{\mathrm{es}}=\frac{1}{2} \mathrm{mv} _{0}^{2}=\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM} _{\mathrm{e}}}{\mathrm{r}}$

or $\quad \mathrm{K} _{\mathrm{es}}=\frac{1}{2} \frac{\mathrm{GM} _{\mathrm{e}} \mathrm{m}}{\mathrm{r}}$

Now, total energy, $\mathrm{E} _{\mathrm{es}}=-\mathrm{K} _{\mathrm{es}}=-\frac{1}{2} \frac{\mathrm{GM} _{\mathrm{e}} \mathrm{m}}{\mathrm{r}}$

For the planet, $M _{P}=6 M _{e}$

$\therefore \quad v _{0}=\sqrt{\frac{\mathrm{G} \times 6 \mathrm{M} _{\mathrm{e}}}{\mathrm{r}}}$, for the satellite of the planet.

$\therefore$ Kinetic energy of the satellite of the planet is:

$$ \mathrm{K} _{\mathrm{PS}}=\frac{1}{2} \mathrm{mv} _{0}^{2}=\frac{1}{2} \times 6 \frac{\mathrm{GM} _{\mathrm{e}} \mathrm{m}}{\mathrm{r}} $$

Now, the total energy of this satellite is : $\mathrm{E} _{\mathrm{PS}}=-\mathrm{K} _{\mathrm{PS}}$

$$ \mathrm{E} _{\mathrm{PS}}=-\frac{1}{2} \times \frac{6 \mathrm{GM} _{\mathrm{e}} \mathrm{m}}{\mathrm{r}} $$

Hence, $\frac{\mathrm{E} _{\mathrm{es}}}{\mathrm{E} _{\mathrm{PS}}}=1: 6$

Correct answer: (1)

Solution:

The mass of a small element of the rod, of length $d x$ is:

$$ \mathrm{dM}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{d} x $$

The gravitational field intensity at point $\mathrm{O}$ due to one such element at a distance of $x$ from $\mathrm{O}$ is:

$$ \mathrm{dI}=\frac{\mathrm{GdM}}{x^{2}}=\frac{\mathrm{GM}}{\mathrm{L}} x^{-2} \mathrm{~d} x $$

$\therefore$ The total intensity at $\mathrm{O}$ due to the entire rod is:

$$ \begin{aligned} \mathrm{I}=\int \mathrm{dI} & =\frac{\mathrm{GM}}{\mathrm{L}} \int _{x=\mathrm{r}}^{x=\mathrm{r}+\mathrm{L}} x^{-2} \mathrm{~d} x \\ & =\frac{\mathrm{GM}}{\mathrm{L}}\left(-\frac{1}{x}\right) _{\mathrm{r}}^{\mathrm{r}+\mathrm{L}}=\frac{\mathrm{GM}}{\mathrm{L}}\left(\frac{1}{x}\right) _{\mathrm{r}+\mathrm{L}}^{\mathrm{r}} \end{aligned} $$

$$ =\frac{\mathrm{GM}}{\mathrm{L}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{r}+\mathrm{L}}\right]=\frac{\mathrm{GM}}{\mathrm{L}}\left[\frac{\mathrm{L}}{\mathrm{r}(\mathrm{r}+\mathrm{L})}\right] $$

or $\mathrm{I}=\frac{\mathrm{GM}}{\mathrm{r}(\mathrm{r}+\mathrm{L})}$

$\therefore$ Force, $\mathrm{F}=\mathrm{mI}=\frac{\mathrm{GMm}}{\mathrm{r}(\mathrm{r}+\mathrm{L})}$

Correct answer: (3)

Solution:

Let $R _{E}=R$ and $\rho _{E}=\rho$

Then $\mathrm{R} _{\mathrm{\rho}}=\frac{3}{2} \mathrm{R}$ and $\rho _{\mathrm{P}}=\frac{5}{4} \rho$

Now, $\mathrm{g} _{\mathrm{E}}=\mathrm{g}=\mathrm{G} \frac{4}{3} \pi \mathrm{R} \rho$

and $\quad \mathrm{g} _{\rho}=\mathrm{G} \frac{4}{3} \pi\left(\frac{3}{2} \mathrm{R}\right)\left(\frac{5}{4} \rho\right)$

or $\quad g _{\rho}=\frac{15}{8} \mathrm{~g}$

Suppose, $\mathrm{W}$ be the effort put in by the astronaut. Then:

$$ \mathrm{W}=\mathrm{mgh} _{\mathrm{E}} $$

and $\mathrm{W}=\mathrm{mg} _{\rho} \mathrm{h} _{\rho}$

$\therefore \mathrm{gh} _{\mathrm{E}}=\mathrm{g} _{\mathrm{\rho}} \mathrm{h} _{\mathrm{\rho}}$

or $\mathrm{g} \times 1.5=\frac{15}{8} \mathrm{~g} \times \mathrm{h} _{\rho}$

or $\quad \mathrm{h} _{\rho}=0.8 \mathrm{~m}$

Correct answer: (1)

Solution:

Let $v _{m}$ be the escape velocity on the surface of moon. Then:

$$ \mathrm{v} _{\mathrm{m}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} $$

where $\mathrm{M}=$ mass of moon and $\mathrm{R}=$ radius of moon.

Now, for earth, $M _{E}=81 \mathrm{M}$ and $R _{E}=4 R$

$\therefore \mathrm{v} _{\mathrm{e}}=\sqrt{\frac{2 \mathrm{G} \times 81 \mathrm{M}}{4 \mathrm{R}}}$

or $\quad \mathrm{v} _{\mathrm{e}}=\frac{9}{2} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ or $\quad \mathrm{v} _{\mathrm{e}}=\frac{9}{2} \mathrm{v} _{\mathrm{m}}$

$\therefore \quad \mathrm{v} _{\mathrm{m}}=\frac{2 \mathrm{v} _{\mathrm{e}}}{9}=\frac{2 \times 11}{9}=2.44 \mathrm{~ms}^{-1}$

Correct answer: (2)

Solution:

Given: $\mathrm{R}=6400 \mathrm{~km}$

$$ \begin{aligned} & \mathrm{h}=9800 \mathrm{~km} \\ & =\frac{3}{2} \mathrm{R} \end{aligned} $$

Now, $\mathrm{g} _{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}=\frac{4}{25}\left(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right)=\frac{4}{25} \mathrm{~g}$

Also, $g _{d}=g\left(1-\frac{d}{R}\right)$

Given $\mathrm{g} _{\mathrm{d}}=\mathrm{g} _{\mathrm{n}}$; therefore

$$ g\left(1-\frac{d}{R}\right)=\frac{4}{25} g $$

or $\frac{\mathrm{d}}{\mathrm{R}}=1-\frac{4}{25}$

or $\quad \mathrm{d}=\frac{21}{25} \mathrm{R}$

$=5376 \mathrm{~km}$

Correct answer: (2)

Solution:

Given $\mathrm{f} \propto \frac{1}{\mathrm{r}^{3}}$

$$ \therefore \quad \mathrm{f}=\frac{\mathrm{GMm}}{\mathrm{r}^{3}} $$

Now, for orbital velocity $\mathrm{v} _{0}$ of the satellite, $\mathrm{f}=$ centripetal force

$$ \frac{\mathrm{GMm}}{\mathrm{r}^{3}}=\frac{\mathrm{mv} _{0}^{2}}{\mathrm{r}} $$

$$ \text { or } \mathrm{v} _{0}^{2}=\frac{\mathrm{GM}}{\mathrm{r}^{2}} $$

or $\quad \mathrm{v} _{0}=\frac{\sqrt{\mathrm{GM}}}{\mathrm{r}}$

or $\quad \mathrm{v} _{0} \propto \frac{1}{\mathrm{r}}$

Hence the graph is:

Now, $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v} _{0}}$ or $\quad \mathrm{T}=\frac{2 \pi \mathrm{r} \times \mathrm{r}}{\sqrt{\mathrm{GM}}}$

or $\quad \mathrm{T}=\left(\frac{2 \pi}{\sqrt{\mathrm{GM}}}\right) \mathrm{r}^{2}$

$\therefore \mathrm{T} \alpha \mathrm{r}^{2}$

Hence the graph is porabolic as shown below

Correct answer: (3)

Solution:

We know that: $v _{0}=\sqrt{\frac{G M}{R+h}}=\left(\sqrt{\frac{g}{R+h}}\right) R$

and $\mathrm{v} _{\mathrm{e}}=\sqrt{2 \mathrm{gR}}$

Given: $\mathrm{v} _{0}=\frac{1}{3} \mathrm{v} _{\mathrm{e}}$

$\therefore \mathrm{R} \sqrt{\frac{\mathrm{g}}{\mathrm{R}+\mathrm{h}}}=\frac{1}{3} \sqrt{2 \mathrm{gR}}$

or $\quad \mathrm{R}^{2}\left(\frac{\mathrm{g}}{\mathrm{R}+\mathrm{h}}\right)=\frac{1}{9}(2 \mathrm{gR})$

or $\quad 9 R^{2}=2 R^{2}+2 R h$

or $\quad 7 \mathrm{R}=2 \mathrm{~h}$

$\therefore \mathrm{h}=3.5 \mathrm{R}$

Now, when the satellite is stopped in the orbit, its $\mathrm{K} . \mathrm{E}=0$ and gravitational potential energy $=\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}$

This potential energy decreases to $\frac{-\mathrm{GMm}}{\mathrm{R}}$, when the satellite hits the ground. Hence, its kinetic energy will increase from 0 to $\frac{1}{2} \mathrm{mv}^{2}$, such that:

$$ \begin{array}{ll} & \frac{-G M m}{R+h}=\frac{1}{2} m v^{2}+\left(\frac{-G M m}{R}\right) \\ \text { or } \quad & \frac{-G M}{R+h}+\frac{G M}{R}=\frac{1}{2} v^{2} \\ \text { or } \quad & \frac{-g R^{2}}{R+h}+g R=\frac{1}{2} v^{2} \\ \text { or } \quad & \frac{g R h}{R+h}+\frac{1}{2} v^{2} \\ \therefore \quad & v^{2}=\frac{2 g h}{1+\frac{h}{R}} \end{array} $$

Here, $\mathrm{h}=3.5 \mathrm{R}$

$\therefore \quad \mathrm{v}^{2}=\frac{2 \mathrm{~g} \times 3.5 \mathrm{R}}{4.5}=\frac{2 \times .9 .8 \times 3.5 \times 6.4 \times 10^{6}}{45}$

or $ \quad \mathrm{v}^{2}=100 \times 10^{6} \mathrm{~m}^{2} \mathrm{~s}^{-2}$

$\therefore \mathrm{v}^{2}=100 \mathrm{~km}^{2} \mathrm{~s}^{-2}$

or

$$ \mathrm{v}=10 \mathrm{kms}^{-1} $$

Correct answer: (2)

Solution:

The weight of the passenger on earth is $588 \mathrm{~N}$ and on mars is $240 \mathrm{~N}$. At some point, on the way, the net force on the passenger, due to earth’s and mar’s gravity together, should be zero. There the weight of the passenger becomes zero. But weight cannot be less than zero in magnitude. This point, where net weight of the person becomes zero, is obviously closer to mars, as its gravity is lesser. So after this point, where the weight is zero, the time of travel will be lesser (with constant speed). Hence the correct plot is (b).

Correct answer: (3)

Solution:

The diagram represents the extreme position of the pendulum, for small amplitude of oscillation. The restoring force on the bob of the pendulum is:

$$ \mathrm{F}=-\mathrm{mg} \sin (\theta+\phi) $$

or $\quad \mathrm{F}=-\operatorname{mg}(\theta+\phi)$

($\theta \quad and \quad \phi$ are small)

or $\quad \mathrm{F}=-\mathrm{mg}\left(\frac{x}{\ell}+\frac{x}{\mathrm{R}}\right)$

or $\quad \mathrm{F}=\frac{-\mathrm{mg}}{\mathrm{R}}\left(\frac{\mathrm{R}}{\ell}+1\right) x$

$\therefore$ Acceleration, $\mathrm{a}=\frac{-\mathrm{g}}{\mathrm{R}}\left(\frac{\mathrm{R}}{\ell}+1\right) x$

This is of the form, $\mathrm{a}=-\omega^{2} x$ for SHM.

$\therefore \omega=\sqrt{\frac{\mathrm{g}}{\mathrm{R}}\left(\frac{\mathrm{R}}{\ell}+1\right)}$

or $\mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}\left(\frac{\mathrm{R}}{\ell}+1\right)}}$

Here, $\ell=\frac{\mathrm{R}}{2}$

$\therefore \quad T=2 \pi \sqrt{\frac{R}{g(2+1)}}=2 \pi \sqrt{\frac{R}{3 g}}$

Correct answer: (2)

Solution:

Refer to diagram

Here, $\mathrm{r} _{\mathrm{A}}=2 \mathrm{R}+\mathrm{R}=3 \mathrm{R}$

$$ \mathrm{r} _{\mathrm{B}}=5 \mathrm{R}+\mathrm{R}=6 \mathrm{R} $$

We have $\mathrm{T}^{2} \propto \mathrm{r}^{3}$, by Kepler’s law.

$$ \therefore \frac{\mathrm{T} _{\mathrm{A}}^{2}}{\mathrm{~T} _{\mathrm{B}}^{2}}=\frac{\mathrm{r} _{\mathrm{A}}^{3}}{\mathrm{r} _{\mathrm{B}}^{2}} $$

or $\frac{\mathrm{T} _{\mathrm{A}}^{2}}{\mathrm{~T} _{\mathrm{B}}^{2}}=\frac{3^{3}}{6^{3}}$

$$ \begin{aligned} & =\frac{1}{2^{3}} \\ & =\frac{1}{8} \end{aligned} $$

$\therefore \frac{\mathrm{T} _{\mathrm{A}}}{\mathrm{T} _{\mathrm{B}}}=\frac{1}{\sqrt{8}}=\frac{1}{2.828}$

Correct answer: (3)

Solution:

The acceleration due to gravity (a) at the depth $d$ is:

$$ a=-g\left(\frac{R-d}{R}\right) $$

(-ve sign indicating that a is towards $\mathrm{O}$ ).

Let $\mathrm{y}=\mathrm{R}-\mathrm{d}$, be the displacement of ball, from $\mathrm{O}$.

Then $a=-\left[\frac{g}{R} y\right]$

This is of the form:

$$ \begin{aligned} & a=-\omega^{2} y \text { for SHM } \\ \therefore \quad \omega & =\sqrt{\frac{g}{R}} \\ \text { or } \quad T & =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{R}{g}} \end{aligned} $$

( $\mathrm{T}$ does not depend on the position of releasing of the ball).

Correct answer: (1)

Solution:

$$ \begin{aligned} & \mathrm{v} _{\mathrm{e}}=\sqrt{2 \mathrm{gR}} \\ & \therefore \frac{\mathrm{v} _{\mathrm{A}}}{\mathrm{v} _{\mathrm{B}}}=\frac{\sqrt{2 \mathrm{~g} _{\mathrm{A}} \mathrm{R} _{\mathrm{A}}}}{\sqrt{2 \mathrm{~g} _{\mathrm{B}} \mathrm{R} _{\mathrm{B}}}}=\sqrt{\frac{\mathrm{g} _{\mathrm{A}}}{\mathrm{g} _{\mathrm{B}}} \cdot \frac{\mathrm{R} _{\mathrm{A}}}{\mathrm{R} _{\mathrm{B}}}}=\sqrt{\mathrm{K} _{1} \mathrm{~K} _{2}} \end{aligned} $$

But this is not given as option

Therefore, we use the formula;

$$ \mathrm{v} _{\mathrm{e}}=\sqrt{2 \mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^{2} \rho}=\mathrm{R} \sqrt{\rho}\left(\sqrt{\frac{8}{2} \pi \mathrm{G}}\right) $$

Hence, $\frac{\mathrm{v} _{\mathrm{A}}}{\mathrm{v} _{\mathrm{B}}}=\frac{\mathrm{R} _{\mathrm{A}}}{\mathrm{R} _{\mathrm{B}}} \cdot \sqrt{\frac{\rho _{\mathrm{A}}}{\rho _{\mathrm{B}}}}$

$$ =\mathrm{K} _{2} \sqrt{\mathrm{K} _{3}} $$

Correct answer: (4)

Solution:

Total initial energy $=$ Total find energy

$\therefore \frac{1}{2} \mathrm{mv} _{0}^{2}+\frac{-\mathrm{GMm}}{\mathrm{R}}=\frac{1}{2} \mathrm{mv}^{2}+\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}$

or $\quad \mathrm{v} _{0}^{2}-\mathrm{v}^{2}=2 \mathrm{GM}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}\right]$

$$ \begin{aligned} & =2 \mathrm{GM}\left[\frac{\mathrm{h}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\right] \\ & =2 \mathrm{gR}^{2}\left(\frac{\mathrm{h}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\right) \quad\left(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right) \\ & =2 \mathrm{gh}\left(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}\right) \\ & =2 \mathrm{gh}\left(\frac{1}{1+\frac{\mathrm{h}}{\mathrm{R}}}\right) \\ & =2 \mathrm{gh}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{-1} \\ & =2 \mathrm{gh}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right) \quad(\because \mathrm{h}«\mathrm{R}) \end{aligned} $$

Correct answer: (2)

Solution:

Let the side of the square be ’ $a$ ‘.

Binding Energy $=-($ Gravitational Potential Energy $)$

$$ =-U $$

Now,

$$ \begin{aligned} & -U=\frac{G}{a}\left[(2 \times 5)+\left(\frac{2 \times 10}{\sqrt{2}}\right)+(5 \times 10)+(2 \times 20)+\left(\frac{5 \times 20}{\sqrt{2}}\right)+(10 \times 20)\right] \\ & =\frac{G}{a}[10+10 \sqrt{2}+50+40+50 \sqrt{2}+200] \\ & =\frac{G}{a}[300+60 \sqrt{2}] \\ & =\frac{G}{a} \times 60(5+\sqrt{2}) \\ & =\frac{G \times 60}{6.414} \times 6.414=\mathrm{G} \times 60=6.67 \times 10^{-11} \times 60 \\ & \quad=4.0 \times 10^{-9} \mathrm{~J} \end{aligned} $$

Correct answer: (3)

Solution:

We have the formula for $h _{\text {max }}$ as:

$$ \mathrm{h} _{\max }=\frac{\mathrm{v} _{0}^{2}}{2 \mathrm{~g}-\left(\frac{\mathrm{v} _{0}^{2}}{\mathrm{R}}\right)} $$

(where $\mathrm{v} _{0}=$ velocity of projection from ground)

Here, $\mathrm{v} _{0}=40 \% \mathrm{v} _{\mathrm{e}}$

$$ \begin{aligned} = & 0.4 \times \sqrt{2 \mathrm{gR}} \\ \therefore \quad \mathrm{h} _{\max } & =\frac{0.16 \times 2 \mathrm{gR}}{2 \mathrm{~g}-0.16 \times 2 \mathrm{~g}} \\ & =\left(\frac{0.32}{2-0.32}\right)^{\mathrm{R}} \\ & =\left(\frac{0.32}{1.68}\right)^{\mathrm{R}} \\ & =1.9 \times 6400 \mathrm{Km} \\ & =12160 \mathrm{Km} \end{aligned} $$

Correct answer: (2)

Solution:

Let $\mathrm{I}=0$ at a distance of $x$ from $\mathrm{m} _{1}$

Here, $\frac{\mathrm{Gm} _{1}}{x^{2}}=\frac{\mathrm{GM} _{2}}{(\mathrm{r}-x)^{2}}$

or $\quad \frac{(\mathrm{r}-x)^{2}}{x^{2}}=\frac{\mathrm{m} _{2}}{\mathrm{~m} _{1}}$

or $\quad \frac{\mathrm{r}-x}{x}=\frac{\sqrt{\mathrm{m} _{2}}}{\sqrt{\mathrm{m} _{1}}}$

or $\quad \frac{\mathrm{r}}{\mathrm{x}}=1+\frac{\sqrt{\mathrm{m} _{2}}}{\sqrt{\mathrm{m} _{1}}} \quad$ or $\quad \frac{\mathrm{r}}{x}=\frac{\sqrt{\mathrm{m} _{1}}+\sqrt{\mathrm{m} _{2}}}{\sqrt{\mathrm{m} _{1}}}$

Hence, $x=\left[\frac{\sqrt{\mathrm{m} _{1}}}{\sqrt{\mathrm{m} _{1}}+\sqrt{\mathrm{m} _{2}}}\right] \mathrm{r} \hspace{40mm}. . . . . . . (1)$

$\therefore \mathrm{r}-x=\frac{\sqrt{\mathrm{m} _{2}} \mathrm{r}}{\sqrt{\mathrm{m} _{1}}+\sqrt{\mathrm{m} _{2}}} \hspace{40mm}. . . . . . . (2)$

The gravitational potential at the point where $\mathrm{I}=0$ is:

$\mathrm{V}=\mathrm{V} _{1}+\mathrm{V} _{2}$

or $\mathrm{V}=\frac{-\mathrm{Gm} _{1}}{x}+\frac{-\mathrm{Gm} _{2}}{\mathrm{r}-x}$

or $\mathrm{V}=-\mathrm{G}\left(\frac{\mathrm{m} _{1}}{x}+\frac{\mathrm{m} _{2}}{\mathrm{r}-x}\right)$

or $V=-G\left[\frac{m _{1}\left(\sqrt{m _{1}}+\sqrt{m _{2}}\right)}{\sqrt{m _{1} r}}+\frac{m _{2}\left(\sqrt{m _{1}}+\sqrt{m _{2}}\right)}{\sqrt{m _{2}} \mathrm{r}}\right]$ ( using equations (1) and (2) )

$$ \begin{aligned} & =\frac{-\mathrm{G}}{\mathrm{r}}\left[\sqrt{\mathrm{m} _{1}}\left(\sqrt{\mathrm{m} _{1}}+\sqrt{\mathrm{m} _{2}}\right)+\sqrt{\mathrm{m} _{2}}\left(\sqrt{\mathrm{m} _{1}}+\sqrt{\mathrm{m} _{2}}\right)\right] \\ & =\frac{-\mathrm{G}}{\mathrm{r}}\left[\mathrm{m} _{1}+\sqrt{\mathrm{m} _{1} \mathrm{~m} _{2}}+\mathrm{m} _{2}+\sqrt{\mathrm{m} _{1} \mathrm{~m} _{2}}\right] \\ & =-\frac{\mathrm{G}}{\mathrm{r}}\left[\mathrm{m} _{1}+\mathrm{m} _{2}+2 \sqrt{\mathrm{m} _{1} \mathrm{~m} _{2}}\right] \end{aligned} $$

Correct answer: (2)

Solution:

By Kepler’s law, $\mathrm{T}^{2} \propto \mathrm{r}^{3}$

$\therefore \frac{\mathrm{T} _{1}^{2}}{\mathrm{~T} _{2}^{2}}=\frac{\mathrm{r} _{1}^{3}}{\mathrm{r} _{2}^{3}}$

$$ \left(\mathrm{T} _{1}=24 \text { Hrs, } \mathrm{T} _{2}=?\right) $$

$\frac{(24)^{2}}{\mathrm{~T} _{2}^{2}}=\left(\frac{7 \mathrm{R}}{3.5 \mathrm{R}}\right)^{3}$

$$ \left(\mathrm{r} _{1}=7 \mathrm{R}, \mathrm{r} _{2}=2.5 \mathrm{R}+\mathrm{R}=3.5 \mathrm{R}\right) $$

$\therefore \quad \mathrm{T} _{2}^{2}=\frac{24 \times 24}{8}=3^{2} \times 8$

$\therefore \mathrm{T} _{2}=3 \times \sqrt{8}$

$$ =3 \times 2.808 $$

$$ =8.484 \mathrm{Hrs} $$

Correct answer: (3)

Solution:

Though $G$ has become time dependent the gravitational force is still radial. Hence torque due to gravitational force is zero. The angular momentum would, therefore be still conserved.

Correct answer: (2)

Solution:

$\mathrm{F} _{\mathrm{S}}=$ gravitational force due to sun on moon

$$ =G \frac{M _{S} M _{m}}{\left(r _{S m}\right)^{2}} $$

$\mathrm{F} _{\mathrm{E}}=$ gravitational force due to earth on moon

$=\mathrm{G} \frac{\mathrm{M} _{\mathrm{E}} \mathrm{M} _{\mathrm{m}}}{\left(\mathrm{r} _{\mathrm{Em}}\right)^{2}}$

$=\frac{F_{S}}{F_{E}}=(\frac{M _{S}}{M _{E}})(\frac{r _{E _{m}}}{r _{S m}})^{2}$

Given $\frac{M _{S}}{M _{E}}=3.2 \times 10^{5} ; \frac{r _{S m}}{r _{E m}}=400$

$\therefore \frac{\mathrm{F} _{\mathrm{S}}}{\mathrm{F} _{\mathrm{E}}}=\frac{3.2 \times 10^{5}}{(400)^{2}}=2$

Correct answer: (2)

Solution:

Consider a small element PQ of the rod, of length $\mathrm{d} x$, as shown in Fig. $\mathrm{dm}=$ mass of element $\mathrm{PQ}=\lambda \mathrm{d} x=\lambda x \mathrm{~d} x$

$\mathrm{dF}=$ gravitational force on mass $\mathrm{m}$ at $\mathrm{O}$ due to elements $\mathrm{PQ}$

$$ =\frac{\mathrm{Gmdm}}{x^{2}}=\operatorname{Gm} \lambda _{0} \frac{\mathrm{d} x}{x} $$

$\mathrm{F}=\int _{x=\mathrm{L} _{0}}^{\mathrm{L} _{0}+\mathrm{L}} \mathrm{dF}=\mathrm{Gm} \lambda _{0} \int _{\mathrm{L} _{0}}^{\mathrm{L} _{0}+\mathrm{L}} \frac{\mathrm{d} x}{x}=\mathrm{Gm} \lambda _{0} \ln \left(\frac{\mathrm{L} _{0}+\mathrm{L}}{\mathrm{L} _{0}}\right)$

$=2.303 \mathrm{Gm} \lambda _{0} \log \left(1+\frac{\mathrm{L}}{\mathrm{L} _{0}}\right)$

Correct answer: (3)

Solution:

Fig. shows three particles each of mass $\mathrm{M}$ located at $\mathrm{O}, \mathrm{A}$ and $\mathrm{B}$.

We want net gravitational force on mass at $\mathrm{B}$.

$\mathbf{F} _{0}=$ gravitational force on $\mathrm{B}$ due to mass at $\mathrm{O}$.

$$ =\frac{\mathrm{GM}^{2}}{\mathrm{a}^{2}}(-\mathbf{j}) $$

$F _{A}=$ magnitude of gravitational force on $B$ due to mass $A$

$$ =\frac{\mathrm{GM}^{2}}{(\sqrt{2} \mathrm{a})^{2}}=\frac{\mathrm{GM}^{2}}{2 \mathrm{a}^{2}} $$

Resolving $\mathbf{F} _{\mathrm{A}}$ into rectangular components along $\mathrm{x}$ and $\mathrm{y}$-axis; we can write

$$ \begin{aligned} \mathbf{F} _{\mathrm{A}}= & \left(\frac{\mathrm{GM}^{2}}{2 \mathrm{a}^{2}} \cos 45\right) \mathbf{i}+\left(\frac{\mathrm{GM}^{2}}{2 \mathrm{a}^{2}} \sin 45^{0}\right)(-\mathbf{j}) \\ = & \frac{\mathrm{GM}^{2}}{\mathrm{a}^{2}}\left[\frac{\mathbf{i}}{2 \sqrt{2}}-\frac{\mathbf{i}}{2 \mathrm{r} 2}\right] \end{aligned} $$

$\mathbf{F}=$ net gravitational force on mass at point $\mathrm{B}$

$$ =\mathbf{F} _{0}+\mathbf{F} _{\mathrm{A}}=\frac{\mathrm{GM}^{2}}{\mathrm{a}^{2}}\left[\frac{\mathbf{i}}{2 \sqrt{2}}-\left(1+\frac{1}{2 \sqrt{2}}\right) \mathbf{j}\right] $$

Correct answer: (4)

Solution:

$\mathrm{OA}=\left[(\mathrm{a})^{2}+(\sqrt{3 \mathrm{a}})^{2}\right]^{1 / 2}=2 \mathrm{a}$

$\left|\mathbf{F} _{\mathrm{A}}\right|=\left|\mathbf{F} _{\mathrm{B}}\right|=\frac{2 \mathrm{Gm}^{2}}{(2 \mathrm{a})^{2}}=\frac{\mathrm{Gm}^{2}}{2 \mathrm{a}^{2}}$

Resolve $\mathbf{F} _{\mathrm{A}}$ and $\mathbf{F} _{\mathrm{B}}$ into rectangular components.

The new unbalanced force

$\mathbf{F}=\left(2 \mathrm{~F} _{\mathrm{A}} \cos \theta\right) \mathbf{j}=\frac{\sqrt{3}}{2}\left(\frac{\mathrm{Gm}^{2}}{\mathrm{a}^{2}}\right) \mathbf{j} \quad\left[\because \theta=30^{\circ}\right]$

The initial acceleration $\mathbf{a} _{0}$ is

$$ \mathbf{a} _{0}=\frac{\mathbf{F}}{\mathrm{m}}=\frac{\sqrt{3} \mathrm{Gm}}{2 \mathrm{a}^{2}} \mathbf{j} $$

Correct answer: (4)

Solution:

$$ \begin{gathered} \mathbf{I} _{\mathrm{A}}=\frac{\mathrm{Gm}}{\mathrm{a}^{2}}(\mathbf{i}) ; \mathbf{I} _{\mathrm{B}}=\frac{\mathrm{Gm}}{2 \mathrm{a}^{2}}\left[\frac{\mathbf{i}+\mathbf{j}}{\sqrt{2}}\right] \\ \mathbf{I} _{\mathrm{C}}=\frac{\mathrm{Gm}}{\mathrm{a}^{2}} \mathbf{j} \\ \mathbf{I}=\mathbf{I} _{\mathrm{A}}+\mathbf{I} _{\mathrm{B}}+\mathbf{I} _{\mathrm{C}} \\ =\frac{\mathrm{Gm}}{\mathrm{a}^{2}}\left[1+\frac{1}{2 \sqrt{2}}\right][\mathbf{i}+\mathbf{j}] \end{gathered} $$

Correct answer: (1)

Solution:

Let $\mathrm{PQR}$ be the equilateral triangle side a. Let $\mathrm{O}$ be centeroid of the triangle. The circular orbit circumscribing the triangle is a circle of center $\mathrm{O}$ and radius $\mathrm{r}=\mathrm{OP}=\frac{\mathrm{a}}{\sqrt{3}}$.

Let $f=\frac{\mathrm{Gm}^{2}}{\mathrm{a}^{2}}$ be gravitational force on $\mathrm{P}$ due to mass at $\mathrm{Q}$ or $\mathrm{R}$. The net force on mass at

$$ \mathrm{P}=\mathrm{F}=\sqrt{3} \mathrm{f}=\sqrt{3} \frac{\mathrm{Gm}^{2}}{\mathrm{a}^{2}} $$

For circular orbit;

centripetal force $=$ net gravitational force

or $\frac{\mathrm{mv}^{2}}{(\mathrm{a} / \sqrt{3})}=\sqrt{3} \frac{\mathrm{Gm}^{2}}{\mathrm{a}^{2}}$

or $\quad v=\sqrt{\frac{G m}{a}}$

Correct answer: (2)

Solution:

From Fig. $\mathrm{O} _{1} \mathrm{~A}=\mathrm{O} _{2} \mathrm{~A}=\sqrt{(\mathrm{d})^{2}+(\sqrt{3} \mathrm{~d})^{2}}=2 \mathrm{~d}$

$\mathrm{f}=$ gravitational force on mass $\mathrm{m}$; at a due to $\mathrm{O} _{1}$ or $\mathrm{O} _{2}$

$$ =\frac{\mathrm{GMm}}{(2 \mathrm{~d})^{2}}=\frac{\mathrm{GMm}}{4 \mathrm{~d}^{2}} $$

$F=$ net gravitational force on mass $m$ at $A=2 f \cos \theta=2\left(\frac{G M m}{4 d^{2}}\right) \times \frac{\sqrt{3}}{2}$

$$ =\frac{\sqrt{3} \mathrm{GMm}}{4 \mathrm{~d}^{2}} $$

$\mathrm{a} _{0}=$ initial acceleration of mass $\mathrm{m}$ at $\mathrm{A}=\frac{\mathrm{F}}{\mathrm{m}}$

$$ =\frac{\sqrt{3} \mathrm{GM}}{4 \mathrm{~d}^{2}} $$

$\mathrm{E} _{\mathrm{e}}=$ initial total energy of mass $\mathrm{m}$ at $\mathrm{A}$

$$ =-2\left[\frac{\mathrm{GMm}}{2 \mathrm{~d}}\right] $$

$\mathrm{E} _{\mathrm{f}}=$ Total final energy of mass $\mathrm{m}$ at $\mathrm{O}$

$$ =\frac{1}{2} \mathrm{mv}^{2}+\left(-\frac{2 \mathrm{GMm}}{\mathrm{d}}\right) $$

From law of conservation of energy, $\mathrm{E} _{\mathrm{e}}=\mathrm{E} _{\mathrm{f}}$

$\therefore \quad-\frac{2 \mathrm{GMm}}{2 \mathrm{~d}}=\frac{1}{2} \mathrm{mv}^{2}+\left(-\frac{2 \mathrm{GMm}}{\mathrm{d}}\right)$

or $\quad \frac{2 \mathrm{GM}}{2 \mathrm{~d}}=\frac{1}{2} \mathrm{v}^{2}$

or $\quad v=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{d}}}$

Correct answer: (2)

Solution:

Consider the semi-circle as made up of large number of point masses (identical). The magnitude of $\mathbf{d I}$ is same but direction is different. The total intensity $\mathbf{I}=\int \mathrm{d} \mathbf{I}$

$$ |\mathrm{d} \mathbf{I}|=\frac{\mathrm{G}(\lambda \mathrm{Rd} \theta)}{\mathrm{R}^{2}}=\left|\mathrm{d} \mathbf{I}^{\prime}\right| $$

Resolve $\mathrm{dI}$ and $\mathrm{d} \mathbf{I}^{\prime}$ into rectangular components along $\mathrm{x}$ and $\mathrm{y}-$ axis. The $\mathrm{x}-$ components cancel and $\mathrm{y}-$ component and up.

There resultant intensity I is along y-axis. Also

$$ \begin{aligned} |\mathbf{I}|= & 2 \int _{0}^{\pi / 2} \mathrm{dI} \cos \theta=\frac{2 \lambda \mathrm{G}}{\mathrm{R}} \int _{0}^{\pi / 2} \cos \theta \mathrm{d} \theta \\ & =\frac{2 \lambda \mathrm{G}}{\mathrm{R}}|\sin \theta| _{0}^{\pi / 2}=\frac{2 \lambda \mathrm{G}}{\mathrm{R}} \\ & =\frac{2(\pi \mathrm{R} \lambda) \mathrm{G}}{\pi \mathrm{R}^{2}}=\frac{2 \mathrm{GM}}{\pi \mathrm{R}^{2}} \\ \therefore \quad \mathbf{I} & =\frac{2 \mathrm{GM}}{\pi \mathrm{R}^{2}} \mathbf{j} \end{aligned} $$

Correct answer: (2)

Solution:

Let ’ $\mathrm{g}$ ’ acceleration due to gravity on earth.

Given $\mathrm{u}=0 ; \mathrm{S}=19.6 \mathrm{~m} ; \mathrm{a}=\mathrm{g} ; \mathrm{t}=2 \mathrm{~s}$. Therefore

$$ \begin{aligned} & 19.6=\frac{1}{2} g \times(2)^{2} \\ & \text { or } \quad g=9.8 \mathrm{~ms}^{-2} \\ & M=\text { mass of earth }=\frac{4}{3} \pi R^{3} \rho \\ & \text { Now, } g=\frac{G M}{R^{2}} \\ & \therefore \quad \mathrm{G}=\frac{g R^{2}}{M}=\frac{g R^{2}}{\frac{4 \pi}{3} R^{3} \rho}=\frac{3 g}{4 \pi R \rho} \end{aligned} $$

Given $\rho=5.5 \mathrm{~g} / \mathrm{cc}=\frac{5.5 \times 10^{-3} \mathrm{Kg}}{\left(10^{-2} \mathrm{~m}\right)^{3}}=5.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$

$$ \mathrm{R}=6400 \mathrm{Km}=6.4 \times 10^{6} \mathrm{~m} $$

$$ \begin{aligned} \therefore \quad \mathrm{G} & =\frac{3 \times 9.8}{4 \times 3.14 \times\left(6.4 \times 10^{6}\right) \times 5.5 \times 10^{3}} \text { SI unit } \\ & =6.6 \times 10^{-11} \frac{\mathrm{Nm}^{2}}{\mathrm{~kg}^{2}} \end{aligned} $$

Correct answer: (1)

Solution:

$1 \mathrm{Kgf}=\mathrm{g}$ Newton $=9.8 \mathrm{~N}$

$\therefore \mathrm{g}=9.8 \frac{\mathrm{m}^{2}}{\mathrm{~s}}$

$\mathrm{G}=6.67 \times 10^{-7} \times 10^{-4} \frac{\mathrm{Nm}^{2}}{\mathrm{~kg}^{2}}=6.67 \times 10^{-11} \frac{\mathrm{Nm}^{2}}{\mathrm{~kg}}$

Let $\mathrm{M}$ be mass of earth and $\mathrm{R}=6400 \mathrm{Km}$ its radius.

$$ \begin{aligned} & \mathrm{G}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \quad \text { or } \quad \mathrm{M}=\frac{8 \mathrm{R}^{2}}{\mathrm{G}} \\ & \therefore \quad \mathrm{M}=\frac{9.8 \times\left(6.4 \times 10^{6}\right)}{6.67 \times 10^{-11}} \mathrm{~kg} \\ & \quad \simeq 6.02 \times 10^{24} \mathrm{~kg} \end{aligned} $$

Correct answer: (4)

Solution:

We know for present earth

$$ \begin{equation*} \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \tag{1} \end{equation*} $$

Let $M _{1}$ and $M _{2}$ be mass of part of earth of radius $\frac{R}{2}$ and the part of earth radius varying from $\frac{R}{2}$ to $R$ respectively. Then

$$ \begin{aligned} M _{1}= & \frac{4 \pi}{3}\left(\frac{R}{2}\right)^{3} \rho _{0}=\frac{M}{8} \\ M _{2}= & \int _{R / 2}^{R} 4 \pi r^{2}\left(\rho _{0} r\right) d r=4 \pi \rho _{0} \int _{R / 2}^{R} r^{3} d r \\ & =4 \pi \rho _{0}\left[\frac{15 R^{4}}{64}\right]=\left(\frac{4 \pi}{3} R^{3} \rho _{0}\right)\left(\frac{45}{64} R\right) \\ & =\frac{45}{64} M R \end{aligned} $$

$M^{\prime}=$ The total mass of new earth

$$ =\mathrm{M} _{1}+\mathrm{M} _{2}=\frac{\mathrm{M}}{8}\left[1+\frac{45}{8} \mathrm{R}\right] $$

$$ =\frac{\mathrm{M}}{8}[1+5.625 \mathrm{R}] $$

$\mathrm{g}^{\prime}=$ Acceleration due to gravity on surface of new earth $=\frac{\mathrm{GM}^{\prime}}{\mathrm{R}^{2}}$

$$ \begin{aligned} & =\frac{\mathrm{GM}}{8 \mathrm{R}^{2}}[1+5.625 \mathrm{R}] \\ & =\frac{\mathrm{g}}{8}[1+5.625 \mathrm{R}] \end{aligned} $$

Correct answer: (1)

Solution:

Gravitational field is zero at all points inside hollow sphere. For points on surface and outside, sphere behaves as if entire mass is concentrated at center O. i.e. I $\propto \frac{1}{x^{2}}$

The gravitational potential $(\mathrm{V})$ is constant and equals potential on surface; for all points inside hollow sphere. For points outside $\mathrm{V} \propto \frac{1}{x}$.

Correct answer: (1)

Solution:

The gravitational field I at points inside solid sphere is

$$ \mathrm{I}=-\frac{\mathrm{GM}}{\mathrm{R}^{3}} x \quad \text { or } \quad \mathrm{I} \propto x $$

For points outside sphere

$$ \mathrm{I}=-\frac{\mathrm{GM}}{x^{2}} \quad \text { or } \quad \mathrm{I} \propto \frac{1}{x^{2}} $$

The gravitational potential $\mathrm{V}$ at points inside solid sphere is

$$ V=-G M \frac{\left(3 R^{2}-x^{2}\right)}{2 R^{3}} $$

For points outside solid sphere

$$ \mathrm{V}=\frac{-\mathrm{GM}}{x} $$

Correct answer: (3)

Solution:

Mass of body is independent of height (h) above surface of earth. Therefore mass of body remains $50 \mathrm{~kg}$. Let $g$ and $g _{h}$ be acceleration due to gravity on surface of earth and a height $h$ above surface of earth. Then

$$ \begin{aligned} & \frac{g _{h}}{g}=\left(\frac{R}{R+h}\right)^{2} \\ \therefore \quad & g _{h}=9.8\left(\frac{64}{66}\right)^{2} \mathrm{~ms}^{-2} \\ & =9.215 \mathrm{~ms}^{-2} \end{aligned} $$

The weight of body at height ’ $\mathrm{h}$ ’ $=\mathrm{mg} _{\mathrm{h}}$

$$ \begin{aligned} & =50 \times 9.215 \mathrm{~N} \\ & =460.7 \mathrm{~N} \end{aligned} $$

Correct answer: (3)

Solution:

Fig. (a) show the given point masses. The gravitational field is zero at point $\mathrm{O} \cdot \mathrm{AO}=x=$ distance of point from mass $\mathrm{M}$. Then

$$ \begin{gathered} \frac{\mathrm{GM}}{x^{2}}=\frac{\mathrm{G} 4 \mathrm{M}}{(3 \mathrm{~d}-x)^{2}} \\ \text { or } \quad \frac{3 \mathrm{~d}-x}{x}=2 \text { or } x=\mathrm{d} \end{gathered} $$

Fig. (a)

The net gravitational potential at $\mathrm{O}$ is

$\mathrm{v}=-\frac{\mathrm{GM}}{\mathrm{d}}-\frac{\mathrm{G}(4 \mathrm{~m})}{2 \mathrm{~d}}$

$=-\frac{3 \mathrm{GM}}{\mathrm{d}}$

Correct answer: (3)

Solution:

$\mathrm{U} _{1}=\mathrm{W} _{1}=$ The gravitational P.E of system in Fig. $(\mathrm{a})=\frac{1}{2} \sum _{\mathrm{i}=1}^{4} \sum _{\mathrm{j}=1}^{4}-\frac{\mathrm{Gm}^{2}}{\mathrm{r} _{\mathrm{ij}}}$

$$ \mathrm{i} \neq \mathrm{j} $$

$$ =\frac{1}{2}\left[\left(\mathrm{U} _{12}+\mathrm{U} _{13}+\mathrm{U} _{14}\right)+\left(\mathrm{U} _{21}+\mathrm{U} _{23}+\mathrm{U} _{24}\right)+\left(\mathrm{U} _{31}+\mathrm{U} _{32}+\mathrm{U} _{34}\right)+\left(\mathrm{U} _{41}+\mathrm{U} _{42}+\mathrm{U} _{43}\right)\right] $$

$\mathrm{r} _{12}=\mathrm{r} _{13}=\mathrm{r} _{14}=\frac{\ell}{\sqrt{3}}$ and $\mathrm{r} _{23}=\mathrm{r} _{34}=\mathrm{r} _{24}=\ell$

$\therefore \quad \mathrm{W} _{1}=\left[\left(\mathrm{U} _{12}+\mathrm{U} _{13}+\mathrm{U} _{14}\right)+\left(\mathrm{U} _{23}+\mathrm{U} _{24}+\mathrm{U} _{34}\right)\right]$

$$ =-\mathrm{Gm}^{2}\left[\left(\frac{1}{\mathrm{r} _{12}}+\frac{1}{\mathrm{r} _{13}}+\frac{1}{\mathrm{r} _{14}}\right)+\left(\frac{1}{\mathrm{r} _{23}}+\frac{1}{\mathrm{r} _{24}}+\frac{1}{\mathrm{r} _{34}}\right)\right] $$

$$ =-\mathrm{Gm}^{2}\left[\frac{3 \sqrt{3}}{\ell}+\frac{3}{\ell}\right] \simeq-8.2 \frac{\mathrm{Gm}^{2}}{\ell} $$

$\mathrm{U} _{1}^{\prime}=\mathrm{W} _{1}^{\prime}=$ work done in forming assembly shown in Fig. (b)

Now, $\mathrm{r} _{12}=\mathrm{r} _{14}=\frac{\ell}{2} ; \mathrm{r} _{13}=\frac{\sqrt{3} \ell}{2} ; \mathrm{r} _{23}=\mathrm{r} _{34}=\mathrm{r} _{24}=\ell$

$\therefore \quad \mathrm{W} _{1}^{\prime}=-\mathrm{Gm}^{2}\left[\left(\frac{2}{\ell}+\frac{2}{\sqrt{3} \ell}+\frac{2}{\ell}\right)+\frac{3}{\ell}\right]=-8.15 \frac{\mathrm{Gm}^{2}}{\ell}$

$\mathrm{W} _{2}=$ additional work done in charging arrangement from Fig. (a) to Fig. (b) $=\mathrm{U} _{1}^{\prime}-\mathrm{U} _{1}=0.05 \frac{\mathrm{Gm}^{2}}{\ell}$

$\therefore \frac{\mathrm{W} _{1}}{\mathrm{~W} _{2}}=\frac{8.2}{0.05} \simeq 164$

Correct answer: (4)

Solution:

Let $F _{1}$ be magnitude of gravitational force on $m$ due to complete sphere of radius $R$. Then

$$ \mathrm{F} _{1}=\frac{\mathrm{GMm}}{x^{2}} $$

Let $\mathrm{m}^{\prime}$ be mass of the sphere cut-out $\mathrm{m}^{\prime}=\frac{4 \pi}{3}\left(\frac{\mathrm{R}}{2}\right)^{3} \rho=\frac{\mathrm{M}}{8}$.

The distance of mass $\mathrm{m}$ from center $\mathrm{O} _{1}$ of the cut-out part is $\left(x-\frac{\mathrm{R}}{2}\right)$.

$\mathrm{F} _{2}=$ magnitude of gravitational force on $m$ due to cut-out part $=\frac{\mathrm{G} \frac{\mathrm{M}}{8} \mathrm{~m}}{\left(x-\frac{\mathrm{R}}{2}\right)^{2}}$

$\mathrm{F}=$ magnitude of net gravitational force on $\mathrm{m}$ due to the remaining part of sphere $=\mathrm{F} _{1}-\mathrm{F} _{2}$

$$ \begin{aligned} & =\frac{\mathrm{GMm}}{x^{2}}-\frac{\mathrm{GMm}}{8 x^{2}\left(1-\frac{\mathrm{R}}{2 x}\right)^{2}} \\ & =\frac{\mathrm{GMm}}{x^{2}}\left[1-\frac{1}{8}\left(1-\frac{\mathrm{R}}{2 x}\right)^{-2}\right] \end{aligned} $$

Correct answer: (4)

Solution:

(i) For points $0 \leq x \leq \mathrm{R}$; (internal points for both shells)

$$ \mathrm{V}=\mathrm{V} _{1}+\mathrm{V} _{2} $$

$\mathrm{V} _{1}=$ potential at internal point due to inner shell $=-\frac{\mathrm{GM}}{\mathrm{R}}$

$V _{2}=$ potential at internal point due to outer shell $=-\frac{G(2 M)}{2 R}=-\frac{G M}{R}$

$$ \therefore \mathrm{V}=-\frac{2 \mathrm{GM}}{\mathrm{R}} $$

Therefore, $\mathrm{V}$ is constant $\left(=\frac{-2 \mathrm{GM}}{\mathrm{R}}\right)$ for $0 \leq x \leq \mathrm{R}$

(ii) For points $\mathrm{R} \leq x \leq 2 \mathrm{R}$. (points outside inner shell but inside outer shell), the new potential $\mathrm{V}$ is

$$ \mathrm{V}=-\frac{\mathrm{GM}}{x}-\frac{\mathrm{G}(2 \mathrm{M})}{2 \mathrm{R}} $$

$$ =-\frac{\mathrm{GM}}{x}-\frac{\mathrm{GM}}{\mathrm{R}} $$

For point on surface of outer shell $x=2 \mathrm{R}$

$\therefore \quad \mathrm{V}=-\frac{\mathrm{GM}}{2 \mathrm{R}}-\frac{\mathrm{GM}}{\mathrm{R}}=-\frac{3 \mathrm{GM}}{2 \mathrm{R}}$

Therefore, $\mathrm{V}$ increases non-linearly from

$\mathrm{V}=\frac{-2 \mathrm{GM}}{\mathrm{R}}$, at $x=\mathrm{R}$, to $\mathrm{V}=\frac{-3 \mathrm{GM}}{2 \mathrm{R}}$, at $x=2 \mathrm{R}$

(iii) For $x>2 \mathrm{R}$; the point is external point for both shells; therefore

$$ \mathrm{V}=-\frac{\mathrm{G}(3 \mathrm{M})}{x} $$

Correct answer: (1)

Solution:

(i) For points $0<x<\mathrm{R}$; the points are inside both hollow spheres, the intensity of gravitational field due to each sphere is zero.

(ii) For points $\mathrm{R}<x<2 \mathrm{R}$, the points are outside inner sphere but inside outer sphere. The intensity $\mathrm{I} _{1}$ due to inner sphere various as $\frac{1}{x^{2}}$. The intensity $I _{2}$ due to outer sphere is zero. At point on the inner surface of outer sphere; the resultant intensity I is

$$ \mathrm{I}=\frac{\mathrm{GM}}{(2 \mathrm{R})^{2}}=\frac{\mathrm{GM}}{4 \mathrm{R}^{2}} $$

However for point on outer surface of outer sphere the intensity I is

$$ \begin{aligned} I= & \frac{G M}{(2 R)^{2}}+\frac{G(2 M)}{(2 R)^{2}} \\ & =\frac{G M}{4 R^{2}}+\frac{G M}{2 R^{2}}=\frac{3}{4} \frac{G M}{R^{2}} \end{aligned} $$

For points $x>2 \mathrm{R}$;

$$ \mathrm{I}=\frac{\mathrm{G}(3 \mathrm{M})}{x^{2}} $$

Correct answer: (3)

Solution:

$\mathrm{E} _{\mathrm{i}}=$ Total initial energy of rocket

$$ =\frac{1}{2} \mathrm{mv}^{2}+\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right) $$

Let $h$ be maximum vertical height attained by rocket. The instantaneous speed of rocket is zero here and its distance from center of earth $=\mathrm{R}+\mathrm{h}$.

Therefore $\mathrm{E} _{\mathrm{f}}=$ Total final energy of rocket.

$$ =0+\left(-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\right) $$

From law of conservation of energy, $\mathrm{E} _{\mathrm{i}}=\mathrm{E} _{\mathrm{f}}$

$\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{GMm}}{\mathrm{R}}=\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}$

or $\quad \frac{1}{2} \mathrm{mv}^{2}=\mathrm{GMm}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}\right]$

or $\quad \mathrm{v}^{2}=2 \mathrm{GM}\left[\frac{\mathrm{h}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\right]$

$=\frac{2 g R^{2} h}{2(R+h)}=\frac{2 g h R}{(R+h)}$

$\therefore\left(6 \times 10^{3}\right)^{2}=\frac{2 \times 9.8 \mathrm{~h} \times 6.4 \times 10^{6}}{\left(6.4 \times 10^{6}+\mathrm{h}\right)}$

$\therefore \mathrm{h} \approx 2.85 \times 10^{6} \mathrm{~m}$

$=2.58 \times 10^{3} \mathrm{~km}$

$\sim 2580 \mathrm{~km}$

Correct answer: (3)

Solution:

Let $\mathrm{m}$ be mass of meteorite and $\mathrm{M}$ be mass of earth. $\mathrm{E} _{\mathrm{i}}=$ The total initial energy of meteorite $=-\frac{\mathrm{GMm}}{10 \mathrm{R}}$. Let $\mathrm{v}$ be the speed with which meteorite hits earth. $\mathrm{E} _{\mathrm{f}}=$ The total final energy of meteorite $=$ $\frac{1}{2} \mathrm{mv}^{2}+\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)$

From law of conservation of energy; $\mathrm{E} _{\mathrm{i}}=\mathrm{E} _{\mathrm{f}}$

$$ -\frac{\mathrm{GMm}}{10 \mathrm{R}}=\frac{1}{2} \mathrm{mv}^{2}-\frac{\mathrm{GMm}}{\mathrm{R}} $$

or $\quad \frac{9}{10} \frac{\mathrm{GMm}}{\mathrm{R}}=\frac{1}{2} \mathrm{mv}^{2}$

or $\quad \mathrm{v}=\sqrt{\frac{9}{5} \frac{\mathrm{GM}}{\mathrm{R}}}=3 \sqrt{\frac{\mathrm{gR}^{2}}{5 \mathrm{R}}}$ $\simeq 3 \sqrt{\frac{8 \mathrm{R}}{5}}$

Correct answer: (1)

Solution:

Let $\mathrm{E}$ be the total energy of satellite when $\mathrm{r}$ is the instantaneous radius of its circular path around earth. We know

$$ \mathrm{E}=-\frac{\mathrm{GMm}}{2 \mathrm{r}} \quad\left[\text { off T.E }=\frac{1}{2} \mathrm{P} . \mathrm{E}\right] $$

$\therefore \frac{\mathrm{dE}}{\mathrm{dt}}=\frac{\mathrm{GMm}}{2 \mathrm{r}^{2}}\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)$

Given $\frac{\mathrm{dE}}{\mathrm{dt}}=-\mathrm{k}$; therefore

$$ \mathrm{dt}=-\left[\frac{\mathrm{GMm}}{2 \mathrm{k}}\left(\frac{\mathrm{dr}}{\mathrm{r}^{2}}\right)\right] $$

Integrating,

$$ \mathrm{t}=\int _{0}^{\mathrm{t}} \mathrm{dt}=-\frac{\mathrm{GMm}}{2 \mathrm{k}} \int _{\mathrm{R}+\mathrm{h}}^{\mathrm{R}} \frac{\mathrm{dr}}{\mathrm{r}^{2}}=\frac{\mathrm{GMmh}}{2 \mathrm{k}(\mathrm{R}+\mathrm{h}) \mathrm{R}}=\frac{\mathrm{gRhm}}{2 \mathrm{k}(\mathrm{R}+\mathrm{h})} \quad\left(\because \mathrm{GM}=\mathrm{gR}^{2}\right) $$

Correct answer: (2)

Solution:

(a)

Fig. (a) shows initial position of stars. $\mathrm{E} _{\mathrm{i}}$, the total initial energy $=\frac{\mathrm{GM}^{2}}{\mathrm{r}} \hspace{40mm}. . . . . . . (1)$

Fig. (b) shows the position when stars just collide with one another. Let $\mathrm{v}$ be speed of each star at this moment. Then $\mathrm{E} _{\mathrm{f}}=$ The total find energy

$$ \begin{equation*} =2\left(\frac{1}{2} \mathrm{Mv}^{2}\right)+\left(-\frac{\mathrm{GM}^{2}}{2 \mathrm{R}}\right) \tag{2} \end{equation*} $$

From law of conservation of energy $\mathrm{E} _{\mathrm{i}}=\mathrm{E} _{\mathrm{f}}$

$\therefore-\frac{\mathrm{GM}^{2}}{\mathrm{r}}=\mathrm{Mv}^{2}-\frac{\mathrm{GM}^{2}}{2 \mathrm{R}}$

or $\quad \mathrm{v}^{2}=\mathrm{GM}\left[\frac{1}{2 \mathrm{R}}-\frac{1}{\mathrm{r}}\right]$

$$ =6.67 \times 10^{-11} \times 3 \times 10^{3}\left[\frac{1}{2 \times 10^{8}}-\frac{1}{10^{12}}\right] $$

or

$$ \begin{aligned} & \mathrm{v} \simeq 4.48 \times 10^{7} \mathrm{~ms}^{-1} \\ & =4.48 \times 10^{4} \mathrm{~km} / \mathrm{s} \end{aligned} $$

Correct answer: (2)

Solution:

Let $\mathrm{V}$ be the speed of space vehicle as it just misses the planet. From law of conservation of energy.

Total initial energy $=$ Total final energy

$$ \begin{equation*} \frac{1}{2} \mathrm{mv}^{2}+0=\frac{1}{2} \mathrm{mV}^{2}+\left(-\frac{\mathrm{GMm}}{\mathrm{r}}\right) \tag{1} \end{equation*} $$

The motion of space vehicle is under gravitational force of planet on vehicle. Regarding planet and space vehicle as one system, there is no external force and therefore no external torque. The angular momentum of system is conserved. Therefore

$$ \begin{equation*} \mathrm{mvR}=\mathrm{mVr} \quad \text { or } \quad \mathrm{V}=\mathrm{v}\left(\frac{\mathrm{R}}{\mathrm{r}}\right) \tag{2} \end{equation*} $$

From eqns. (1) and (2) we have

$$ \frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{mv}^{2}\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{2}-\frac{\mathrm{GMm}}{\mathrm{r}} $$

or $\quad R^{2}=\frac{r^{2}}{v^{2}}\left[v^{2}+\frac{2 G M}{r}\right]$

$\therefore \quad \mathrm{R}=\frac{\mathrm{r}}{\mathrm{v}}\left[\mathrm{v}^{2}+\frac{2 \mathrm{GM}}{\mathrm{r}}\right]^{1 / 2}$

Correct answer: (2)

Solution:

$E=$ initial K.E energy of rocket $=\frac{1}{2} \mathrm{mv} _{0}^{2}$

$\mathrm{E}^{\prime}=\mathrm{K} . \mathrm{E}$ of rocket as it emerges out of the atmosphere around earth $=\mathrm{E}-\frac{\mathrm{E}}{4}=\frac{3}{4} \mathrm{E}$

$\frac{3}{4}\left(\frac{1}{2} \mathrm{mv} _{0}^{2}\right)=$ Work done against gravity of earth, in rising to height $\mathrm{h}$

$$ =\int _{\mathrm{R}}^{\mathrm{R}+\mathrm{h}}-\frac{\mathrm{GMm}}{x^{2}} \mathrm{~d} x=\mathrm{GMm}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}\right] $$

or $\quad \frac{3}{8} \mathrm{mv} _{0}^{2}=\frac{\mathrm{GMmh}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}=\frac{\mathrm{gR}^{2} \mathrm{mh}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}$

or $\quad \frac{3 \mathrm{v} _{0}^{2}}{8}=\frac{\mathrm{gRh}}{(\mathrm{R}+\mathrm{h})}$

or $\frac{\mathrm{R}+\mathrm{h}}{\mathrm{h}}=\frac{8 \mathrm{gR}}{3 \mathrm{v} _{0}^{2}}$

$$ \frac{\mathrm{R}}{\mathrm{h}}=\left(\frac{8 \mathrm{gR}}{3 \mathrm{v} _{0}^{2}}-1\right) $$

$\therefore \mathrm{h}=\frac{\mathrm{R}}{\left(\frac{8 \mathrm{gR}}{3 \mathrm{v} _{0}^{2}}-1\right)}$

Correct answer: (3)

Solution:

For a circular orbit of radius $r \simeq R$;

$$ \frac{\mathrm{GMm}}{\mathrm{R}^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{R}} $$

$\mathrm{v}=$ velocity of satellite in orbit. Let $\mathrm{T}$ be time period of satellite in orbit. Then $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}}$

$\therefore \frac{\mathrm{GMm}}{\mathrm{R}^{2}}=\frac{\mathrm{m}}{\mathrm{R}}\left(\frac{2 \pi \mathrm{R}}{\mathrm{T}}\right)^{2}$

or $\quad \mathrm{T}=\frac{2 \pi \mathrm{R}^{3 / 2}}{\sqrt{\mathrm{GM}}}$

Also $M=\frac{4 \pi}{3} R^{3} \rho$. Therefore,

$$ \begin{gathered} \mathrm{T}=\frac{2 \pi \mathrm{R}^{3 / 2}}{\sqrt{\mathrm{G} \frac{4 \pi}{3} \mathrm{R}^{3} \rho}} \\ =\sqrt{\frac{3 \pi}{\mathrm{G} \rho}} \end{gathered} $$

$\mathrm{T}$ is independent of $\mathrm{R}$.

Correct answer: (2)

Solution:

Let $v$ be the speed of mass $m$ when it is moving in a circular orbit of radius $r$. Then

$$ \begin{array}{r} \frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{\mathrm{C}}{\mathrm{r}^{\mathrm{n}}} \\ \text { or } \quad \mathrm{v}^{2}=\frac{\mathrm{C}}{\mathrm{mr}^{\mathrm{n}-1}} \tag{1} \end{array} $$

The time period $\mathrm{T}$ of mass $\mathrm{m}$ is

$$ \mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}} $$

$$ \begin{align*} & =\frac{2 \pi r}{\sqrt{\frac{C}{m}} r^{-(n-1) / 2}} \\ & =2 \pi \sqrt{\frac{m}{C}} r^{(n+1) / 2} \tag{2} \end{align*} $$

$\therefore \mathrm{T} \propto \mathrm{r}^{(n+1) / 2}$

Correct answer: (2)

Solution:

$\omega _{1}=$ angular speed of earth around its own axis $=\frac{2 \pi}{\mathrm{T} _{1}}$

$\omega _{2}=$ angular speed of satellite in orbit $=\frac{2 \pi}{\mathrm{T} _{2}}$

$\omega _{\mathrm{r}}=$ The relative angular speed of satellite with respect to earth $=\omega _{2}-\omega _{1}$

$$ =2 \pi\left(\frac{1}{\mathrm{~T}^{2}}-\frac{1}{\mathrm{~T} _{1}}\right)=2 \pi \frac{\left(\mathrm{T} _{1}-\mathrm{T} _{2}\right)}{\mathrm{T} _{1} \mathrm{~T} _{2}} $$

The time interval between successive positions when satellite is vertically above observer fixed to equator on earth $=\frac{2 \pi}{\omega _{\mathrm{r}}}=\frac{\mathrm{T} _{1} \mathrm{~T} _{2}}{\mathrm{~T} _{1}-\mathrm{T} _{2}}$

Correct answer: (2)

Solution:

For body to move in circular orbit at height $h$;

$$ \frac{\mathrm{mv}^{2}}{\mathrm{R}+\mathrm{h}}=\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^{2}} $$

or $\quad v=\sqrt{\frac{G M}{R+h}}=\sqrt{\frac{g R^{2}}{R+h}} \hspace{40mm}. . . . . . . (1)$

The time-period, T; of body in orbit is:

$$ \begin{equation*} \mathrm{T}=\frac{2 \pi(\mathrm{R}+\mathrm{h})}{\mathrm{V}}=\frac{2 \pi(\mathrm{R}+\mathrm{h})^{3 / 2}}{\sqrt{\mathrm{gR}^{2}}} \tag{2} \end{equation*} $$

From equations (1) and (2) we have,

$$ \frac{\mathrm{v}}{\mathrm{T}}=\frac{\mathrm{gR}^{2}}{2 \pi(\mathrm{R}+\mathrm{h})^{2}} $$

Correct answer: (1)

Solution:

Let $T _{0}$ and $T _{i}$ be the time period of the particle in the outer most and inner most ring respectively. Let $R _{0}$ and $\mathrm{R} _{\mathrm{i}}$ be the corresponding radii of the orbit. Using Kepler’s third law.

$$ \begin{aligned} & \left(\frac{\mathrm{T} _{\mathrm{i}}}{\mathrm{T} _{0}}\right)^{2}=\left(\frac{\mathrm{R} _{\mathrm{i}}}{\mathrm{R} _{0}}\right)^{3} \\ & \text { or } \quad \frac{\mathrm{T} _{\mathrm{i}}}{\mathrm{T} _{0}}=\left(\frac{\mathrm{R} _{\mathrm{i}}}{\mathrm{R} _{0}}\right)^{3 / 2}=\left(\frac{7 \times 10^{4}}{1.35 \times 10^{5}}\right)^{3 / 2}=\left(\frac{7}{13.5}\right)^{3 / 2} \\ & \simeq 0.373 \end{aligned} $$

Correct answer: (4)

Solution:

$\mathrm{v} _{\text {sat }}=$ The satellite velocity required at height of $600 \mathrm{~km}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}=\sqrt{\frac{\mathrm{gR^2}}{\mathrm{r}}}$

where $\mathrm{r}=6400+600=7000 \mathrm{Km}$

$$ \begin{aligned} & \therefore \quad \mathrm{v} _{\mathrm{sat}}=\sqrt{\frac{10 \times\left(6.4 \times 10^{6}\right)^{2}}{7 \times 10^{6}} \mathrm{~ms}^{-1}} \\ & \quad \simeq 7.648 \times 10^{3} \mathrm{~ms}^{-1} \end{aligned} $$

$\mathrm{p} _{1}=$ Linear momentum of satellite at $\mathrm{h}=600 \mathrm{Km}$

$=\mathrm{m} \times \mathrm{v}=10^{2} \times 4 \times 10^{3}=4 \times 10^{5} \mathrm{Kgms}^{-1}$, vertically upwards

$\mathrm{p} _{2}=$ Linear momentum required by satellite to put into circular orbit at $\mathrm{h}=600 \mathrm{Km}$ $\mathrm{mv} _{\text {sat }}=7.648 \times 10^{5} \mathrm{~ms}^{-1} ;$ horizontal The impulse required equals change in linear momentum $=\mathbf{p} _{2}-\mathbf{p} _{1}$

$\therefore$ Magnitude of impulse $=\sqrt{\mathrm{p} _{1}^{2}+\mathrm{p} _{2}^{2}} \simeq 8.63 \times 10^{5} \mathrm{kgms}^{-1}$

$$ \tan \alpha=\frac{\mathrm{p} _{1}}{\mathrm{p} _{2}}=\frac{4}{7.648}=0.57 ; \alpha \simeq 29.7^{0} $$

Correct answer: (4)

Solution:

Let $\omega _{1}$ and $\omega _{2}$ be angular speed of the two satellites in their respective orbits. Then

$$ \mathrm{mr} _{1} \omega _{1}^{2}=\frac{\mathrm{GMm}}{\mathrm{r} _{1}^{2}} ; \mathrm{mr} _{2} \omega _{2}^{2}=\frac{\mathrm{GMm}}{\mathrm{r} _{2}^{2}} $$

or $\mathrm{r} _{1}^{3} \omega _{1}^{2}=\mathrm{r} _{2}^{3} \omega _{2}^{2}$

$\omega _{1}=\frac{2 \pi}{\mathrm{T} _{1}}=\frac{2 \pi}{1} \mathrm{rad} \mathrm{hr}^{-1} ; \quad \omega _{2}=\frac{2 \pi}{\mathrm{T} _{2}}=\frac{2 \pi}{2 \sqrt{2}} \mathrm{radhr}^{-1}$

Also $\mathrm{r} _{1}=10^{4} \mathrm{~km}$; therefore

$$ \begin{aligned} & r _{2}^{3}=\left(10^{4}\right)^{3} \times\left(\frac{\omega _{1}}{\omega^{2}}\right)^{2} \\ & =\left(10^{4}\right)^{3} \times(2 \sqrt{2})^{2} \end{aligned} $$

or $\mathrm{r} _{2}=2 \times 10^{4} \mathrm{Km}$

$\mathrm{v} _{1}=$ linear speed of $\mathrm{S} _{1}=\mathrm{r} _{1} \omega _{1}=2 \pi \times 10^{4} \mathrm{~km} \mathrm{hr}^{-1}$

$\mathrm{v} _{2}=$ linear speed of $\mathrm{S} _{2}=\mathrm{r} _{2} \omega _{2}=2 \times 10^{4} \times \frac{2 \pi}{2 \sqrt{2}}$

$$ =\sqrt{2} \pi \times 10^{4} \mathrm{~km} \mathrm{hr}^{-1} $$

relative speed of $S _{2}$ w.r.t $S _{1}=v _{2}-v _{1}$

$$ =(\sqrt{2}-2) \pi \times 10^{4} \mathrm{~km} \mathrm{hr}^{-1} $$

Correct answer: (3)

Solution:

Let $\mathrm{v}$ be instantaneous velocity of body when it is at a distance $x$ from centre of earth. Then instantaneous acceleration $=\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-\frac{\mathrm{GM}}{x^{2}}$

or $\quad \frac{\mathrm{dv}}{\mathrm{d} x} \cdot \frac{\mathrm{d} x}{\mathrm{dt}}=-\frac{\mathrm{GM}}{x^{2}} \quad$ or $\quad \mathrm{vdv}=-\frac{\mathrm{GM}}{x^{2}} \mathrm{~d} x$

Integrating,

$$ \begin{aligned} & \quad \frac{\mathrm{v}^{2}}{2}=\frac{\mathrm{GM}}{x} \quad \text { or } \quad \mathrm{v}=\sqrt{\frac{2 \mathrm{GM}}{x}}=\frac{\mathrm{d} x}{\mathrm{dt}} \\ & \therefore \quad \mathrm{dt}=\frac{1}{\sqrt{2 \mathrm{GM}}} x^{1 / 2} \mathrm{~d} x \end{aligned} $$

Integrate,

$$ \begin{aligned} \int _{0}^{\mathrm{t}} \mathrm{dt} & =\frac{1}{\sqrt{2 \mathrm{GM}}} \int _{\mathrm{R}}^{2 \mathrm{R}} x^{1 / 2} \mathrm{~d} x \\ & =\frac{1}{\sqrt{2 \mathrm{gR}^{2}}}\left[\frac{(2 \mathrm{R})^{3 / 2}-\mathrm{R}^{3 / 2}}{3 / 2}\right] \\ & =\frac{2}{3} \sqrt{\frac{\mathrm{R}}{2 \mathrm{~g}}}\left[2^{3 / 2}-1\right] \\ & =\frac{1}{3} \sqrt{\frac{2 \mathrm{R}}{\mathrm{g}}}[2 \sqrt{2}-1] \end{aligned} $$

Correct answer: (1)

Solution:

$\mathrm{W} _{1}=$ Work done by 1st stage of rocket

$$ =\int _{\mathrm{R}}^{\frac{11 \mathrm{R}}{10}} \frac{\mathrm{GMm}}{\mathrm{x}^{2}} \mathrm{dx}=\mathrm{GMm}\left[\frac{1}{\mathrm{R}}-\frac{10}{11 \mathrm{R}}\right] $$

$$ =\mathrm{gR}^{2} \mathrm{~m} \times \frac{1}{11 \mathrm{R}}=\frac{\mathrm{mgR}}{11} $$

The second stage does work $\mathrm{W} _{2}$ to give necessary speed to the rocket so that it becomes satellite at height $\frac{\mathrm{R}}{10}$. Therefore $\mathrm{W} _{2}=\frac{1}{2} \mathrm{mv} _{\text {sat }}^{2}$,

where $\mathrm{v} _{\text {sat }}$ is satellite velocity at the height and its value is

$$ \mathrm{v} _{\mathrm{sod}}=\sqrt{\frac{\mathrm{GM}}{\frac{11 \mathrm{R}}{10}}} $$

$\therefore \quad \mathrm{W} _{2}=\frac{1}{2} \mathrm{GMm} \times \frac{10}{11 \mathrm{R}}=\frac{5 \mathrm{mgR}^{2}}{11 \mathrm{R}}=\frac{5}{11} \mathrm{mgR}$

Hence $\frac{\mathrm{W} _{1}}{\mathrm{~W} _{2}}=\frac{1}{5}$

Correct answer: (2)

Solution:

Let $P$ and $Q$ be the two diagonally opposite points on earth. $S _{1}$ and $S _{2}$ are the two geostationary satellites above $\mathrm{P} \quad and \quad \mathrm{Q}$.

$r=$ radius of orbit of geostationary satellite.

$$ \mathrm{mr} \omega^{2}=\frac{\mathrm{GMm}}{\mathrm{r}^{2}} $$

$\mathrm{m}=$ mass of satellite; $\mathrm{M}=$ mass of earth.

$$ \begin{aligned} & \omega=\frac{2 \pi}{\mathrm{T}}=\frac{2 \pi}{24 \times 3600} \mathrm{rad} / \mathrm{s} \\ \therefore & \mathrm{r}^{3}=\frac{\mathrm{GM}}{\omega^{2}}=\frac{\mathrm{gR}^{2}}{\omega^{2}}=9.8 \times\left(6.4 \times 10^{6}\right)^{2} \times(24 \times 3600)^{2} \\ \text { or } \quad \mathrm{r} & \simeq 4.2 \times 10^{7} \mathrm{~m} \end{aligned} $$

From Fig.; $\mathrm{PS} _{\mathrm{i}}=\sqrt{\mathrm{OS} _{\mathrm{i}}^{2}-\mathrm{PO}^{2}}=\sqrt{\mathrm{r}^{2}-\mathrm{R}^{2}}$

The length of path of signal in going from $\mathrm{P}$ to $\mathrm{Q}$ via satellite $\mathrm{S} 1 \quad and \quad \mathrm{~S} _{2}=2 \mathrm{PS} _{1}+\mathrm{S} _{1} \mathrm{~S} _{2}=2 \mathrm{PS} _{1}+2 \mathrm{R}$

$$ \simeq 9.574 \times 10^{7} \mathrm{~m} $$

$\therefore$ Time taken $\mathrm{t}=\frac{9.574 \times 10^{7}}{3 \times 10^{8}} \mathrm{~s}$

$\simeq 0.32 \mathrm{~s}$

Correct answer: (2)

Solution:

The escape velocity from the planet is

$$ \mathrm{v}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}+\mathrm{h}}} $$

or $\mathrm{R}+\mathrm{h}=\frac{2 \mathrm{GM}}{\mathrm{v}^{2}}$

$\therefore \frac{1}{\mathrm{v}^{2}}=\left(\frac{\mathrm{R}}{2 \mathrm{GM}}\right)+\left(\frac{1}{2 \mathrm{GM}}\right) \mathrm{h}$

$\frac{1}{\mathrm{v}^{2}}$ graph is a straight line

Slope of st. line $=\frac{\mathrm{n}-\ell}{\mathrm{m}}=\frac{1}{2 \mathrm{GM}} \hspace{40mm}. . . . . . . (1)$

Intercept on $\frac{1}{\mathrm{v}^{2}}$ axis $=\ell=\frac{\mathrm{R}}{2 \mathrm{GM}} \hspace{40mm}. . . . . . . (2)$

$\therefore \quad M=\frac{m}{2 G(n-\ell)} ; \quad$ and $R=\frac{m \ell}{(n-\ell)}$

Correct answer: (2)

Solution:

Fig. shows earth with $\mathrm{A} _{1} \mathrm{~A}$ as smooth tunnel. Initially particle of mass $\mathrm{m}$ is at $\mathrm{P}, \mathrm{r}=\frac{\mathrm{R}}{2}=\mathrm{OP}$. Let $\mathrm{V}$ be net gravitational potential at $\mathrm{P}$ due to earth. Then

$$ \begin{equation*} \mathrm{V}=\mathrm{V} _{1}+\mathrm{V} _{2} \tag{1} \end{equation*} $$

$\mathrm{V} _{1}=$ Gravitational potential $\mathrm{P}$ due to a solid sphere of radius $\frac{\mathrm{R}}{2}$

$$ =-\frac{\mathrm{GM} _{1}}{\left(\frac{\mathrm{R}}{2}\right)} $$

Also $\mathrm{M} _{1}=\frac{\mathrm{M}}{4}$. There are

$$ \begin{equation*} \mathrm{V} _{1}=-\frac{\mathrm{GM}}{2 \mathrm{R}} \tag{2} \end{equation*} $$

Let $V _{2}$ be gravitational potential at $\mathrm{P}$ due to remaining part of earth. We consider this part as made of

concentric spherical shells having radius varying from $\frac{R}{2}$ to $R$. Consider an elementary shell of radius $r$, thickness dr as shown in Fig.

$\mathrm{dm}=$ mass of shell considered $=4 \pi \mathrm{r}^{2} \mathrm{dr}$

$\rho=$ density of earth

Since point $\mathrm{P}$ is INSIDE the shell considered, the potential $\mathrm{dV} _{2}$ at $\mathrm{P}$ is same as on surface of shell considered. Therefore

$$ \begin{align*} & \mathrm{dV} _{2}=-\frac{\mathrm{Gdm}}{\mathrm{r}}=-4 \pi \mathrm{G} \rho \mathrm{rdr} \\ \therefore \quad \mathrm{V} _{2} & =\int _{\mathrm{R} / 2}^{\mathrm{R}} \mathrm{dV} _{2}=-4 \pi \mathrm{G} \rho \int _{\mathrm{R} / 2}^{\mathrm{R}} \mathrm{rdr}=-4 \pi \mathrm{G} \rho\left[\frac{3 \mathrm{R}^{2}}{8}\right] \\ & =-\left(\frac{4 \pi}{3} \mathrm{R}^{3} \rho\right) \mathrm{G}\left(\frac{9 \mathrm{R}^{2}}{8 \mathrm{R}^{3}}\right)=-\frac{9}{8}\left(\frac{\mathrm{GM}}{\mathrm{R}}\right) \tag{3} \end{align*} $$

From eqns. (1), (2) and (3)

$$ \begin{equation*} \mathrm{V}=-\frac{\mathrm{GM}}{2 \mathrm{R}}-\frac{9 \mathrm{GM}}{8 \mathrm{R}}=-\frac{13 \mathrm{GM}}{8 \mathrm{R}} \tag{4} \end{equation*} $$

The total energy $E _{1}$ of particle at $P$, when it is given a speed $v$ is:

$$ \begin{equation*} \mathrm{E} _{1}=\frac{1}{2} \mathrm{mv}^{2}+\mathrm{mV}=\frac{1}{2} \mathrm{mv}^{2}-\frac{13 \mathrm{GMm}}{8 \mathrm{R}} \tag{5} \end{equation*} $$

The total energy $\mathrm{E} _{2}$ of particle as it reaches surface of earth is

$$ \mathrm{E} _{2}=0+\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right) \quad[\mathrm{K} . \mathrm{E} \text { of particle at } \mathrm{A} \text { is zero }] $$

From law of conservation of energy $\mathrm{E} _{1}=\mathrm{E} _{2}$. Therefore

$$ \frac{1}{2} \mathrm{mv}^{2}-\frac{13 \mathrm{GMm}}{8 \mathrm{R}}=-\frac{\mathrm{GMm}}{\mathrm{R}} $$

or $\quad \mathrm{v}^{2}=\frac{5 \mathrm{GM}}{4 \mathrm{R}}=\frac{1.25 \mathrm{gR}^{2}}{\mathrm{R}}$

$\therefore \quad \mathrm{v}=\sqrt{1.25 \mathrm{gR}}$

Correct answer: (3)

Solution:

Let $\mathrm{P}$ be point, $\mathrm{OP}=\frac{\mathrm{R}}{\sqrt{2}}$ inside tunnel where particle is released from rest. The gravitational potential energy of particle at $\mathrm{P}$

$$ \begin{aligned} \mathrm{U} _{1}= & -\frac{\mathrm{G}(\mathrm{M} / 2) \mathrm{m}}{\mathrm{R} / \sqrt{2}}-4 \pi \rho \mathrm{G} \int _{\mathrm{R} / \sqrt{2}}^{\mathrm{R}} \mathrm{rdr} \\ & =-\frac{\mathrm{GMm}}{\sqrt{2} \mathrm{R}}-\left(\frac{4 \pi}{3} \mathrm{R}^{3} \rho\right) \mathrm{G}\left(\frac{3 \mathrm{R}^{2}}{4 \mathrm{R}^{3}}\right) \\ & =-\frac{\mathrm{GMm}}{\sqrt{2} \mathrm{R}}-\frac{3 \mathrm{Gm}}{4 \mathrm{R}}=-\left(\frac{1}{\sqrt{2}}+\frac{3}{4}\right) \frac{\mathrm{GMm}}{\mathrm{R}} \end{aligned} $$

The gravitational potential energy $U _{2}$ of mass $m$ at center of earth is

$$ \mathrm{U} _{2}=-\frac{3 \mathrm{GM}}{2 \mathrm{R}} $$

Let $v$ be speed of mass at centre of earth. From law of conservation of energy

$$ \begin{aligned} & 0+\left[-\left(\frac{1}{\sqrt{2}}+\frac{3}{4}\right) \frac{\mathrm{GMm}}{\mathrm{R}}\right]=\frac{1}{2} \mathrm{mv}^{2}+\left[-\frac{3}{2} \frac{\mathrm{GMm}}{\mathrm{R}}\right] \\ & \therefore \quad \mathrm{v}^{2}=\left(\frac{3}{2}-\sqrt{2}\right) \frac{\mathrm{GM}}{\mathrm{R}}=\left(\frac{3}{2}-\sqrt{2}\right) \mathrm{gR} \\ & \text { or } \quad \mathrm{v}=\left[\left(\frac{3}{2}-\sqrt{2}\right) \mathrm{gR}\right]^{1 / 2} \end{aligned} $$

Correct answer: (4)

Solution:

$\mathrm{v} _{\text {esp }}=$ Escape velocity from surface of earth $=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$

$\mathrm{M}=$ mass of earth; $\mathrm{R}=$ radius of earth.

$\mathrm{v} _{\mathrm{e}}=$ initial speed of projection of body $=3 \mathrm{v} _{\text {esp }}$

Let $\mathrm{v}$ be its speed when it has just escaped gravitational pull of earth. From law of conservation of energy.

$$ \begin{aligned} & \frac{1}{2} \mathrm{~m}\left(2 \mathrm{v} _{\text {esp }}\right)^{2}-\frac{\mathrm{GMm}}{\mathrm{R}}=\frac{1}{2} \mathrm{mv}^{2}+0 \\ & 2 \mathrm{mv} _{\mathrm{esp}}^{2}-\frac{1}{2} \mathrm{mv} _{\mathrm{esp}}^{2}=\frac{1}{2} \mathrm{mv}^{2} \\ & \therefore \mathrm{v}=\sqrt{3} \mathrm{v} _{\mathrm{esp}} \\ & \therefore \frac{\mathrm{v}}{\mathrm{v} _{\mathrm{i}}} \times 100=\frac{\sqrt{3}}{2} \times 100=86.6 \% \end{aligned} $$

Correct answer: (2)

Solution:

Since the ball thrown with a speed of $38 \mathrm{~ms}^{-1}$ just fails to return to the surface of the planet, the escape velocity $\left(v _{\text {esp }}\right)$ from planet is $38 \mathrm{~ms}^{-1}$. Let $\mathrm{M}$ be mass and $\mathrm{R}$ radius of planet. Then

$\mathrm{v} _{\text {esp }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{\frac{2 \mathrm{G}\left(\frac{4 \pi}{3} \mathrm{R}^{3} \rho\right)}{\mathrm{R}}}$

$=\sqrt{\frac{8 \pi \mathrm{G} \rho}{3}} \cdot \mathrm{R}$

$$ \begin{aligned} \therefore \quad \mathrm{R} & =\mathrm{v} _{\text {esp }} \sqrt{\frac{3}{8 \pi \mathrm{G} \rho}} \\ & =38 \sqrt{\frac{3}{8 \times 3.14 \times 6.67 \times 10^{-11} \times 2.9 \times 10^{3}}} \\ & \simeq 2987 \mathrm{~m} \\ & =2.987 \mathrm{~km} \end{aligned} $$

Correct answer: (2)

Solution:

Let $\mathrm{r} _{1}$ and $\mathrm{r} _{2}$ be distance of mass at $\mathrm{A}$ from $\mathrm{O} _{1}$ and $\mathrm{O} _{2}$. Then $\mathrm{r} _{1}=\frac{\mathrm{d}}{3}$ and $\mathrm{r} _{2}=\frac{2 \mathrm{~d}}{3}$. Let $\mathrm{v}$ be minimum speed of mass (m) at A ,so that it just escapes gravitational pull of $M _{1}$ and $M _{2}$. From law of conservation of energy.

$$ \begin{aligned} & \frac{1}{2} \mathrm{mv}^{2}+\left(-\frac{\mathrm{GMm}}{\mathrm{r} _{1}}\right)+\left(-\frac{\mathrm{GM} _{2} \mathrm{~m}}{\mathrm{r} _{2}}\right)=0 \\ & \text { or } \frac{1}{2} \mathrm{v}^{2}=\frac{\mathrm{GM} _{1}}{(\mathrm{~d} / 3)}+\frac{\mathrm{GM}^{2}}{(2 \mathrm{~d} / 3)}=\frac{3 \mathrm{G}}{2 \mathrm{~d}}\left(2 \mathrm{M} _{1}+\mathrm{M} _{2}\right) \\ & \text { or } \mathrm{v}=\sqrt{\frac{3 \mathrm{G}\left(2 \mathrm{M} _{1}+\mathrm{M} _{2}\right)}{\mathrm{d}}} \end{aligned} $$

Correct answer: (3)

Solution:

Since $v _{0}$ is velocity of circular orbit of radius $r _{1}$;

$$ \begin{equation*} \frac{\mathrm{mv} _{0}^{2}}{\mathrm{r} _{1}}=\frac{\mathrm{GMm}}{\mathrm{r} _{1}^{2}} \quad \text { or } \quad \mathrm{v} _{0}^{2}=\frac{\mathrm{GM}}{\mathrm{r} _{1}} \tag{1} \end{equation*} $$

$\mathrm{L} _{\mathrm{A}}=$ angular momentum of planet at $\mathrm{A}=\mathrm{mv} _{1} \mathrm{r} _{1}$

$\mathrm{E} _{\mathrm{A}}=$ Total energy of planet at $\mathrm{A}=\frac{1}{2} \mathrm{~m}\left(1.1 \mathrm{v} _{0}\right)^{2}-\frac{\mathrm{GMm}}{\mathrm{r} _{1}}$

$$ =\frac{121}{200} \frac{\mathrm{GMm}}{\mathrm{r} _{1}}-\frac{\mathrm{GMm}}{\mathrm{r} _{1}}=-\frac{79}{200} \frac{\mathrm{GMm}}{\mathrm{r} _{1}} \hspace{40mm}. . . . . . . (2) $$

Let $\mathrm{v} _{2}$ be speed of planet at $\mathrm{B}$. Then

$\mathrm{L} _{\mathrm{B}}=$ angular momentum of planet at $\mathrm{B}=\mathrm{mv} _{2} \mathrm{r} _{2}$

From law of conservation of angular momentum $\mathrm{L} _{\mathrm{A}}=\mathrm{L} _{\mathrm{B}}$

or $\mathrm{mv} _{1} \mathrm{r} _{1}=\mathrm{mv} _{2} \mathrm{r} _{2}$ or $\mathrm{v} _{2}=\frac{1 . \mathrm{lv} _{0} \mathrm{r} _{1}}{\mathrm{r} _{2}} \hspace{40mm}. . . . . . . (3)$

$E _{B}=$ Total energy of planet at $B$

$$ =\frac{1}{2} \mathrm{mv} _{2}^{2}-\frac{\mathrm{GMm}}{\mathrm{r}^{2}}=\frac{1}{2} \mathrm{~m}\left(\frac{121}{100}\right)\left(\frac{\mathrm{r} _{1}}{\mathrm{r} _{2}}\right)^{2} \frac{\mathrm{GM}}{\mathrm{r} _{1}}-\frac{\mathrm{GMm}}{\mathrm{r} _{2}} $$

Using law of conservation of energy

$$ \frac{121}{200} \mathrm{GMm}\left(\frac{\mathrm{r} _{1}}{\mathrm{r} _{2}^{2}}\right)-\frac{\mathrm{GMm}}{\mathrm{r} _{2}}=-\frac{79}{200} \frac{\mathrm{GMm}}{\mathrm{r} _{1}} $$

or $\quad 121\left(\frac{r _{1}}{r _{2}}\right)^{2}-200\left(\frac{r _{1}}{r _{2}}\right)=-79$

or $\quad \frac{\mathrm{r} _{1}}{\mathrm{r} _{2}}=200 \pm \frac{\sqrt{4000-38236}}{242}$

$\simeq 0.653$

$\therefore \frac{\mathrm{r} _{2}}{\mathrm{r} _{1}} \sim 1.5$

Correct answer: (3)

Solution:

$\mathrm{v} _{\text {sat }}$ is velocity of satellite at hight $h$. Then

$$ \mathrm{v} _{\mathrm{sat}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}}=\sqrt{\frac{\mathrm{gR}}{\mathrm{R}+\mathrm{h}}} $$

$$ =\sqrt{\frac{10 \times\left(6.4 \times 10^{6}\right)^{2}}{7.4 \times 10^{6}}} \mathrm{~ms}^{-1} \simeq 7.44 \times 10^{3} \mathrm{~ms}^{-1} $$

Let $\mathrm{v} _{\text {esp }}$ be escape velocity from a height ’ $h$ ’ above surface of earth.

$$ \begin{aligned} \mathrm{v} _{\text {esp }} & =\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}+\mathrm{h}}}=\sqrt{\frac{2 \mathrm{gR}^{2}}{\mathrm{R}+\mathrm{h}}}=\sqrt{2} \mathrm{v} _{\text {sat }} \\ & =1.414 \times 7.44 \times 10^{3} \mathrm{~ms}^{-1} \\ & \simeq 10.52 \times 10^{3} \mathrm{~ms}^{-1} \end{aligned} $$

Therefore additional velocity needed

$$ \begin{aligned} & =(10.52-7.44) \times 10^{3} \mathrm{~ms}^{-1} \\ & \simeq 3.08 \times 10^{3} \mathrm{~ms} \\ & \simeq 3.1 \mathrm{~km} / \mathrm{s} \end{aligned} $$

Correct answer: (3)

Solution:

Let $M _{S}$ and $M _{E}$ denote mass of sun and earth. Let $v$ be speed of earth around sun; in orbit of radius $R$. Then

$$ \frac{\mathrm{GM} _{\mathrm{S}} \mathrm{M} _{\mathrm{E}}}{\mathrm{R}^{2}}=\frac{\mathrm{M} _{\mathrm{E}} \mathrm{v}^{2}}{\mathrm{R}} $$

or $\quad \mathrm{v}^{2}=\frac{\mathrm{GMs}}{\mathrm{R}}$

$\mathrm{T}=$ Time-period of earth around sun $=\frac{2 \pi \mathrm{R}}{\mathrm{V}}$

$$ \begin{equation*} \therefore \quad \mathrm{T}^{2}=\frac{4 \pi^{2} \mathrm{R}^{3}}{\mathrm{GM} _{\mathrm{S}}} \tag{1} \end{equation*} $$

Given $M _{S}^{\prime}=1.5 M _{s} ; R^{\prime}=2 R$. Let $T^{\prime}$ be the time-period of earth, in orbit, now. Then from equation(1)

$$ \begin{aligned} & \quad\left(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}\right)^{2}=\left(\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\right)^{3}\left(\frac{\mathrm{M} _{\mathrm{S}}}{\mathrm{M} _{\mathrm{S}}^{\prime}}\right)=(2)^{3} \frac{2}{3} \\ & \therefore \quad \frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\frac{4}{\sqrt{3}}=\frac{4}{1.73} \simeq 2.34 \end{aligned} $$