Past NEET Papers

Hint

(a) Weight of a body on the surface of the earth,
\(
W_S=m g =72 \mathrm{~N}
\)
Acceleration due to gravity, \(g\) varies with height,
\(
\begin{aligned}
&h=\frac{R}{2} \text { (given) } \\
&W_h=\frac{W_s}{\left(1+\frac{h}{R}\right)^2}=\frac{72}{\left(1+\frac{R / 2}{R}\right)^2}=\frac{72}{(3 / 2)^2} \\
&=\frac{4}{9} \times 72=32 \mathrm{~N}
\end{aligned}
\)

Hint

(b)
Radius of earth \(=R\)
Let at depth d, gravitational acceleration becomes \(\frac{g}{n}\).
i.e., \(g_d=\frac{g}{n} \Rightarrow g\left(1-\frac{d}{R}\right)=\frac{g}{n}\)
\(
\begin{aligned}
&\Rightarrow 1-\frac{d}{R}=\frac{1}{n} \Rightarrow 1-\frac{1}{n}=\frac{d}{R} \\
&\Rightarrow \frac{n-1}{n}=\frac{d}{R} \Rightarrow d=\left(\frac{n-1}{n}\right) R
\end{aligned}
\)

Hint

From Kepler’s third law, the time period of revolution of satellite around the earth is
\(
T^2 \propto r^3 \text { or } T \propto r^{3 / 2} \dots(i)
\)
where, \(r\) is the radius of satellite’s orbit.
Here, \(r_1=6 R_E+R_E, T_1=24 \mathrm{~h}\)
\(
r_2=2.5 R_E+R_E, T_2=\text { ? }
\)
where \(R_E=\) radius of earth
So, from Eq. (i), we get
\(\frac{T_1}{T_2}=\left(\frac{r_1}{r_2}\right)^{3 / 2}\)
\(\frac{24}{T_2}=\left(\frac{6 R_E+R_E}{2.5 R_E+R_E}\right)^{3 / 2}=\left(\frac{7}{3.5}\right)^{3 / 2}\)
\(\Rightarrow \quad T_2=\frac{24}{(2)^{3 / 2}}=\frac{24}{2 \sqrt{2}}=\frac{12}{\sqrt{2}}=6 \sqrt{2} \mathrm{~h}\)

Hint

Initial potential energy at earths surface is
\(
U_i=\frac{-G M m}{R}
\)
Final potential energy at height \(h=R\)
\(
U_f=\frac{-G M m}{2 R}
\)
As work done \(=\) Change in \(\mathrm{PE}\)
\(
\begin{aligned}
&\therefore W=U_f-U_i \\
&=\frac{G M m}{2 R}=\frac{g R^2 m}{2 R}=\frac{m g R}{2}\left(\because G M=g R^2\right)
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
\text { (c) } & \text { As } U=-\frac{G M m}{r} \\
\therefore & (\mathrm{P} . \mathrm{E})_{\mathrm{A}}=-\frac{G M m}{R} ;(\mathrm{P} . \mathrm{E})_{\mathrm{B}}=-\frac{G M m}{R+h} \\
\therefore & \Delta U=(\mathrm{P} . \mathrm{E})_{\mathrm{B}}-(\mathrm{P} . \mathrm{E})_{\mathrm{A}} \\
& =-\frac{G M m}{R+h}+\frac{G M m}{R}=\frac{G M m h}{(R)(R+h)}
\end{array}
\)

Hint

The figure above shows an ellipse traced by a planet around the Sun, S. The closed point \(A\) is known as perihelion (perigee) and the far thest point \(C\) is known as aphelion (apogee).
Since, as per the result the Kepler’s second law of area, that the planet will move slowly \(\left(v_{\min }\right)\) only when it is farthest from the Sun and more rapidly \(\left(v_{\max }\right)\) when it is nearest to the Sun.
Thus, \(\quad v_A=v_{\max }, v_C=v_{\text {min }}\)
Therefore, we can write
\(
v_A>v_B>v_C \quad \text {…(i) }
\)
Kinetic energy of the planet at any point is given as, \(K=\frac{1}{2} m v^2\)
Thus, at \(A_1 K_A=\frac{1}{2} m v_A^2\)
At B, \(\quad K_B=\frac{1}{2} m v_B^2\)
AtC, \(\quad K_c=\frac{1}{2} m v_c^2\)
From Eq. (i), we can write
\(
K_A>K_B>K_C
\)

Hint

Let the original mass of the Sun was \(M_s\) and gravitational constant \(G^{\prime}\).
According to the question,
New mass of Sun, \(M_s^{\prime}=\frac{M_s}{10}\)
New gravitational constant, \(G^{\prime}=10 G\)
As, the acceleration due to gravity is given as
\(
g=\frac{G M_E}{R^2} \dots(i)
\)
where, \(M_E\) is the mass of Earth and \(R\) is the radius of the Earth.

Now, new acceleration due to gravity,
\(
\begin{aligned}
&\qquad g^{\prime}=\frac{G M_e}{R^2} \\
&\qquad=\frac{10 M_e G}{R^2} \dots(ii)
\end{aligned}
\)
\(\therefore \quad g^{\prime}=10 g \quad\) [from Eqs. (i)and (ii)] This means the acceleration due to gravity has been increased. Hence, force of gravity acting on a body placed on or surface of the Earth increases.
Due to this, rain drops will fall faster, walking on ground would become more difficult. As, time period of the simple pendulum is
\(
T=2 \pi \sqrt{\frac{1}{g}} \text { Or } T \propto \frac{1}{\sqrt{g}}
\)

Thus, the time period of the pendulum also decreases with the increase in \(g\).

Hint

Thinking Process \(g_n=\) Acceleration due to gravity at height h above earth’s surface \(=g\left(\frac{R}{R+h}\right)^2=g\left(1-\frac{2 h}{R}\right)\),  \(g_d=\) Acceleration at depth \(d\) below earth’s surface
\(
=g\left(1-\frac{d}{R}\right)
\)
Given, when \(h=1 \mathrm{~km}, g_d=g_h\) or \(g\left(1-\frac{d}{R}\right)=g\left(1-\frac{2 h}{R}\right)\) \(\Rightarrow \quad d=2 h\) or \(d=2 \mathrm{~km}\)

Hint

(a) Both the astronauts are in the condition of weightlessness. The gravitational force between them pulls towards each other. Hence Astronauts move towards each other under mutual gravitational force.

Hint

(a) As we know, gravitational potential (V) and acceleration due to gravity \((g)\) with height
\(\mathrm{V}=\frac{-G M}{(R+h)}=-5.4 \times 10^7\)
and \(g=\frac{G M}{(R+h)^2}=6\)
Dividing (1) by (2)
\(
\begin{aligned}
\Rightarrow & \frac{\frac{-G M}{(R+h)}}{\frac{G M}{(R+h)^2}}=\frac{-5.4 \times 10^7}{6} \\
\Rightarrow & \frac{5.4 \times 10^7}{(R+h)}=6 \\
\Rightarrow &(R+h)=9000 \mathrm{~km}, \mathrm{so}, h=2600 \mathrm{~km}
\end{aligned}
\)

Hint

Acceleration due to gravity at a depthd below the surface of the earth is given by
\(
g_{\text {depph }}=g_{\text {surface }}\left(1-\frac{d}{R}\right)
\)
Also, for a point at height habove surface,
\(
g_{\text {height }}=g_{\text {surlace }}\left[\frac{R^2}{(R+h)^2}\right]
\)
Therefore, we can say that value of \(g\) increases from centre of maximum at the surface and then decreases as depicted in graph (c).

Hint

(a) As we know, escape velocity,
\(
\begin{aligned}
&v_e=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}} \cdot\left(\frac{4}{3} \pi \mathrm{R}^3 \rho\right)} \\
&v_e \propto \mathrm{R} \sqrt{\rho} \\
&\therefore \quad \frac{\mathrm{V}_{\mathrm{e}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{p}}} \sqrt{\frac{\rho_{\mathrm{e}}}{\rho_{\mathrm{p}}}} \\
&\Rightarrow \quad \frac{\mathrm{V}_{\mathrm{e}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{R}_{\mathrm{e}}}{2 \mathrm{R}_{\mathrm{e}}} \sqrt{\frac{\rho_{\mathrm{e}}}{2 \rho_{\mathrm{e}}}} \\
&\therefore \quad \text { Ratio } \frac{\mathrm{V}_{\mathrm{e}}}{\mathrm{V}_{\mathrm{p}}}=1: 2 \sqrt{2}
\end{aligned}
\)

Hint

(b)
\(\because\) Total energy of a satellite at height his
\(=K E+P E=\frac{G M m}{2(R+h)}-\frac{G M m}{(R+h)}\)
\(=\frac{-G M m}{2(R+h)}=\frac{-G M m R^2}{2 R^2(R+h)}=\frac{-m g_0 R^2}{2(R+h)}\)
\(\left(\because g_0=\frac{G M}{R^2}\right)\)

Hint

The gravitational force of attraction between the planet and sun provide the centripetal force
i.e. \(\quad \frac{G M m}{r^2}=\frac{m v^2}{r}\)
\(\Rightarrow \quad v=\sqrt{\frac{G M}{r}}\)
The time period of planet will be
\(
\begin{aligned}
T &=\frac{2 \pi r}{v} \\
\Rightarrow T^2 &=\frac{4 \pi^2 r^2}{\frac{G M}{r}}=\frac{4 \pi^2 r^3}{G M} \dots(i)
\end{aligned}
\)
Also from Kepler’s third law
\(
T^2=K r^3 \dots(ii)
\)
From Eqs. (i) and (ii), we get
\(
\frac{4 \pi^2 r^3}{G M}=K r^3 \Rightarrow G M K=4 \pi^2
\)

Hint

At time of collision the distance between their centers \(=3 \mathrm{R}\)
So total distance travelled by both \(=12 R-3 R=9 R\)
Since the bodies move under mutual forces, center of mass will remain stationary so (assume, the smaller body cover a distance \(x\) before collision)
\(
\begin{aligned}
&m_1 x_1=m_2 x_2 \\
&m x=5 m(9 R-x) \\
&x=45 R-5 x \\
&6 x=45 R \\
&x=\frac{45}{6} R = 7.5 R
\end{aligned}
\)

Hint

Given, the height of a satellite
\(
h=0.25 \times 10^6 \mathrm{~m}
\)
Earth’s radius, \(R_e=6.38 \times 10^8 \mathrm{~m}\)
For the satellite revolving around the earth, orbital velocity of the satellite
\(
\begin{aligned}
v_0 &=\sqrt{\frac{G M_e}{R_e}}=\sqrt{\frac{G M_e}{R_e\left[1+\frac{h}{R_e}\right]}} \\
\Rightarrow v_0 &=\sqrt{\frac{g R_e}{1+\frac{h}{R_e}}}
\end{aligned}
\)

\(
\left[\because \frac{\mathrm{GM_e}}{\mathrm{R_e}^2}=\mathrm{g}\right]
\)
Substitutes the values of \(g, R_e\) and \(h\), we get
\(
\begin{aligned}
&v_0=\sqrt{60 \times 10^8} \mathrm{~m} / \mathrm{s} \\
&v_0=7.76 \times 10^3=7.76 \mathrm{~km} / \mathrm{s}
\end{aligned}
\)

Hint

(d)
As we know that, force on satellite is only gravitational force which will always be towards the centre of earth. Thus, the acceleration of \(S\) is always directed towards the centre of the earth.

Hint

For the black hole, the escape speed is more than c (speed of light). We should compare the escape speed with the \(c\) (Note that the escape speed should be at least just greater than c).
Escape velocity
\(
\begin{aligned}
&=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=c=\text { speed of light } \\
&\Rightarrow \mathrm{R}=\frac{2 \mathrm{GM}}{\mathrm{c}^2} \\
&=\frac{2 \times 6.6 \times 10^{-11} \times 5.98 \times 10^{24}}{\left(3 \times 10^8\right)^2} \mathrm{~m} \\
&=10^{-2} \mathrm{~m}
\end{aligned}
\)

Hint

(b) First when \((r<R) \mathrm{E} \propto r\) and then when \(r \geq\) R
\(E \propto \frac{1}{r^2}\). Hence graph (b) correctly depicts.
The field strength of uniform solid sphere within it decreases linearly with ‘ \(r\) ‘ and becomes zero as we reach at the centre of the sphere.

Hint

The resulting gravitational potential,
\(
\begin{aligned}
V &=-2 G\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots\right] \\
\Rightarrow V &=-2 G\left[1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3} \cdots\right] \\
\Rightarrow V &=-2 G\left(1+\frac{1}{2}\right)^{-1} \\
\Rightarrow V &=-\frac{2 G}{\left(1-\frac{1}{2}\right)}=-\frac{2 G}{\left(\frac{1}{2}\right)}=-4 G
\end{aligned}
\)

Hint

(b) Initial P. E., \(U_i=\frac{-\mathrm{GMm}}{\mathrm{R}}\),
Final P.E., \(U_f=\frac{-\mathrm{GMm}}{3 \mathrm{R}}\left[\because \mathrm{R}^{\prime}=\mathrm{R}+2 \mathrm{R}=3 \mathrm{R}\right]\)
\(\therefore\) Change in potential energy,
\(
\begin{aligned}
&\Delta \mathrm{U}= U_f-U_i =\frac{-\mathrm{GMm}}{3 \mathrm{R}}+\frac{\mathrm{GMm}}{\mathrm{R}} \\
&=\frac{\mathrm{GMm}}{\mathrm{R}}\left(1-\frac{1}{3}\right)=\frac{2}{3} \frac{\mathrm{GMm}}{\mathrm{R}}=\frac{2}{3} \mathrm{mgR} \\
& \\
&\qquad\left(\because \frac{\mathrm{GMm}}{\mathrm{R}}=\mathrm{mgR}\right)
\end{aligned}
\)

Hint

(c)
From Kepler’s third law
\(
\left[\begin{array}{c}
T^2 \propto r^3 \\
\text { where } T=\text { time period of satellite } \\
r=\text { radius of elliptical orbit(semi } \\
\text { major axis) }
\end{array}\right]
\)
Hence, \(\quad T_1^2 \propto r_1^3\) and \(T_2^2 \propto r_2^3\)
So, \(\frac{T_2^2}{T_1^2}=\frac{r_2^3}{r_1^3}=\frac{(3 R)^3}{(6 R)^3}[latex] or [latex]\frac{T_2^2}{T_1^2}=\frac{1}{8}\)
\(
T_2^2=\frac{1}{8} T_1^2 \Rightarrow T_2=\frac{24}{2 \sqrt{2}}=6 \sqrt{2} \mathrm{~h}
\)

Hint

(c)
According to the question,
\(
\frac{\mathrm{GMm}}{(R+h)^2}=\frac{1}{16} \frac{\mathrm{GMm}}{R^2}
\)
where, \(m=\) mass of the body and \(\frac{G M}{R^2}=\) gravitational acceleration
\(\frac{1}{(R+h)^2}=\frac{1}{16 R^2}\)
or \(\frac{R}{R+h}=\frac{1}{4}\)
or \(\frac{R+h}{R}=4\)
\(h=3 R\)

Hint

(b) A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It will stay in any position as the horizontal component of the earth’s magnetic field becomes zero at the geomagnetic pole.

Hint

(a)
According to Newton’s law of gravitation force,
\(
F=\frac{G M m}{R^2}
\)
Force on planet of mass \(M_p\) and body of mass \(m\) is given by
\(
F=\frac{G M_p m}{\left(D_p / 2\right)^2}
\)
\(\left[\begin{array}{l}\text { where, } D_p=\text { diameter of planet } \\ \text { and } R_p=\text { radius of planet }=\frac{D_p}{2}\end{array}\right]\)
\(
F=\frac{4 G M_p m}{D_p^2}
\)
As we know that, \(F=m a\)
So, acceleration due to gravity
\(
a=\frac{F}{m}=\frac{4 G M_\rho}{D_\rho^2}
\)

Hint

(d) \(v_e=\sqrt{\frac{2 G M}{R}} \Rightarrow v_0=\sqrt{\frac{G M}{R}}\)
\(
v_e=\sqrt{2} v_0
\)
The orbital velocity of a satellite at a height \(h\) above the surface of earth,
\(
v_0=\sqrt{\frac{G M}{(R+h)}}=\sqrt{\frac{g R^2}{(R+h)}}\left(\because G M=g R^2\right)
\)
Here, \(M=\) mass of earth,
\(\mathrm{R}=\) radius of earth,
\(\mathrm{g}=\) acceleration due to If the satellite is very close to the surrface of earth gravity earth on surface of then \(h=0\)
\(
\therefore \quad v_0=\sqrt{\frac{g R^2}{R}}=\sqrt{g R}=\sqrt{\frac{G M}{R}}
\)

Hint

(b) The Gravitational field due to a thin spherical shell of radius \(R\) at distance \(r\).
\(
\left.E=\frac{G M}{r^2} \text { (If } r>R\right)
\)
For \(r=R\) i.e., on the surface of the shell \(E=\frac{G M}{R^2}\)
For \(r<R\) i.e., inside the shell \(E=0\)

Hint

Apply conservation of angular momentum.
From the law of conservation of angular momentum, \(L_1=L_2\)
So, \(m r_1 v_1=m r_2 v_2\)
\(
\begin{aligned}
&{\left[\begin{array}{rl}
\text { where, } m & =\text { mass the of planet } \\
r & =\text { radius of orbit } \\
v & =\text { velocity of the planet }
\end{array}\right]} \\
&r_1 v_1=r_2 v_2 \Rightarrow \frac{v_1}{v_2}=\frac{r_2}{r_1}
\end{aligned}
\)

Hint

We know that Power, \(
\text { Power, } P=\vec{F} \cdot \vec{v}= Fv \cos \theta\)

So, just before hitting, \(\theta\) is zero, power will be maximum.
Hence, the power exerted by the gravitational force is greatest at the instant just before the body hits the earth.

Hint

(b) Here, Mass of a particle = M, Mass of a spherical shell =M, Radius of a spherical shell = a

\(\mathrm{V}_{\mathrm{P}}=\mathrm{V}_{\text {spherical shell }}+\mathrm{V}_{\text {particle }}\)
\(
=\frac{G M}{a}+\frac{G M}{a / 2}=\frac{3 G M}{a}
\)

Hint

(a) The velocity u should be equal to the escape velocity. That is,
\(
\mathrm{u}=\sqrt{2 \mathrm{gR}}
\)
But \(g=\frac{G M}{R^2}\)
\(
\therefore \mathrm{u}=\sqrt{2 \cdot \frac{\mathrm{GM}}{\mathrm{R}^2} \cdot \mathrm{R}} =\sqrt{\frac{2 G M}{R}}
\)

Hint

(a) Potential at the given point \(=\) Potential at the point due to the shell + Potential due to the particle
\(
=-\frac{G M}{a}-\frac{2 G M}{a}=-\frac{3 G M}{a}
\)

Alternate:

gravitational potential at point \(a / 2\) distance from centre is given by,
\(
V=-\frac{G M}{a}-\frac{G M}{a / 2}=-\frac{3 G M}{a}
\)

Hint

(b) Orbital velocity of a satellite in a circular orbit of radius a is given by
\(
\begin{aligned}
&v=\sqrt{\frac{G M}{a}} \\
&\Rightarrow v \propto \sqrt{\frac{1}{a}} \\
&\Rightarrow \quad \frac{v_2}{v_1}=\sqrt{\frac{a_1}{a_2}} \\
&\therefore \quad v_2=v_1 \sqrt{\frac{4 R}{R}}=2 v_1=6 \mathrm{~V}
\end{aligned}
\)

Hint

Concept Apply Kepler’s second law. The line joining the sun to the planet sweeps out equal areas in equal time interval i.e. areal velocity is constant. \(\frac{d A}{d t}=\) constant or \(\frac{A_1}{t_1}=\frac{A_2}{t_2}\)
where, \(A_1=\) area under \(S C D\)
\(A_2=\) area under \(A B S\)
\(
\Rightarrow \quad t_1=\frac{A_1}{A_2} t_2
\)
Given, \(A_1=2 A_2\)
\(
\therefore \quad t_1=2 t_2
\)

Hint

The appearance of weightlessness occurs in space when the gravitational attraction of the earth on a body in space is equal to the centripetal force.

\(
\therefore \quad \frac{m v^2}{r}=m g
\)
or \(\quad v=\sqrt{r g}=\sqrt{20 \times 10}=14.14 \mathrm{~m} / \mathrm{s}\)

Hence, the speed of the car should be between \(14 \mathrm{~ms}^{-1}\) and \(15 \mathrm{~ms}^{-1}\).

Hint

(b) Since the orbital velocity of the satellite is
\(v=\sqrt{\frac{G M}{r}}\), it does not depend upon the mass of the satellite.
Therefore, both satellites will move with same speed.

Hint

(a)
At a platform at a height \(h\), Escape energy \(=\) binding energy of sphere
or \(\frac{1}{2} m\left(f v_e\right)^2=\frac{G M m}{R+h}\)
or \(f V_e=\sqrt{\frac{2 G M}{R+h}}=\sqrt{\frac{2 G M}{2 R}} \ldots\) (i) \(\quad(\because h=R)\)
But at surface of the earth,
\(
v_e=\sqrt{\frac{2 G M}{R}} \dots(ii)
\)
Dividing Eq. (ii) by Eq. (i).
Hence, \(\frac{f v_e}{v_e}=\frac{\sqrt{\frac{G M}{R}}}{\sqrt{\frac{2 G M}{R}}} \Rightarrow f=\frac{1}{\sqrt{2}}\)

Hint

The acceleration due to gravity on the new planet can be found using the relation
\(
g=\frac{G M}{R^2} \dots(i)
\)
but \(M=\frac{4}{3} \pi R^3 \rho, \rho\) being density.
Thus, Eq. (i) becomes
\(\frac{G \times \frac{4}{3} \pi R^3 \rho}{R^2}=G \times \frac{4}{3} \pi R \rho\)
\(\Rightarrow \quad g \propto R\)
\(\therefore \quad \frac{g^{\prime}}{g}=\frac{R^{\prime}}{R}\)
\(\Rightarrow \quad \frac{g^{\prime}}{g}=\frac{3 R}{R}=3 \Rightarrow g^{\prime}=3 g\)

Hint

(b)

Potential energy, \(U=-\frac{G M_e m}{R_e}\) where,
\(M_e=\) mass of the earth
\(m=\) mass of satellite
\(R_e=\) radius of the earth
\(G=\) gravitational constant
or \(|U|=\frac{G M_e m}{R_e}\)
Kineticenergy, \(K==\frac{1}{2} m v^2=\frac{1}{2} m\left(\sqrt{\frac{G M_e}{R_e}}\right)^2=\frac{1}{2} \frac{G M_e m}{R_e}\)
Thus, \(\quad \frac{K}{|U|}=\frac{1}{2} \frac{G M_e m}{R_e} \times \frac{R_e}{G M_e m}=\frac{1}{2}\)

Hint

The acceleration due to gravity on an object of mass \(m\)
\(
g=\frac{F}{m}
\)
but from Newton’s law of gravitation
\(
F=\frac{G M m}{R^2}
\)
where, Mis the mass of the earth and \(R\) is the radius of the earth.
\(
\therefore \quad g=\frac{G M m / R^2}{m}=\frac{G M}{R^2}
\)
\(
\left[\because M_p=\frac{4}{3} \pi R_p^3 \rho_p \text { and } M_e=\frac{4}{3} \pi R_e^3 \rho_e\right]
\)
Given, \(\quad \rho_{\text {planet }}=2 \rho_{\text {earh }}\)
Also, \(\quad g_{\text {planet }}=g_{\text {earth }}\)
\(
\frac{G M_p}{R_p^2}=\frac{G M_e}{R_e^2}
\)
As, \(\operatorname{Density}(\rho)=\frac{\operatorname{Mass}(M)}{\operatorname{Volume}(V)}\)
So, \(\frac{G \times \frac{4}{3} \pi R_p^3 \rho_p}{R_p^2}=\frac{G \times \frac{4}{3} \pi R_e^3 \rho_e}{R_e^2}\)
or \(\quad R_p \rho_p=R_e \rho_e\)
or \(\quad R_p \times 2 \rho_e=R_e \rho_e\)
or \(\quad R_p=\frac{R_e}{2}=\frac{R}{2}\)

Hint

(d)
According to Newton’s law of gravitation, the force between two spheres is given by,
\(
F=\frac{G M m}{r^2}
\)
Well, the force of gravitation will be the same as f because gravitational force is dependent on the masses of the body and the distance between them and does not depend on the medium between the masses.

Hint

(d) Applying the conservation of total mechanical energy principle
\(
\begin{aligned}
&\frac{1}{2} m v^2=m g_A h_A=m g_B h_B \\
&\Rightarrow g_A h_A=g_B h_B
\end{aligned}
\)
\(
\Rightarrow h_B=\left(\frac{g_A}{g_B}\right) h_A=9 \times 2=18 \mathrm{~m}
\)

Alternate:
The velocity of the mass while reaching the surface of both the planets will be same.
\(
\begin{aligned}
&\Rightarrow \sqrt{2 g^{\prime} h^{\prime}}=\sqrt{2 g h} \\
&\Rightarrow \sqrt{2 g h^{\prime}}=\sqrt{2 \times 9 g \times 2} \\
&\Rightarrow 2 h^{\prime}=36 \Rightarrow h^{\prime}=18 \mathrm{~m}
\end{aligned}
\)

Hint

(b) Gravitational potential energy (GPE) on the surface of earth,
\(
E_1=-\frac{G M m}{R}
\)
GPE at \(3 R, E_2=-\frac{G M m}{(R+3 R)}=-\frac{G M m}{4 R}\)
\(\therefore\) Change in GPE
\(
\begin{aligned}
&=E_2-E_1=-\frac{G M m}{4 R}+\frac{G M m}{R}=\frac{3 G M m}{4 R} \\
&=\frac{3 g R^2 m}{4 R} \quad\left(\because g=\frac{G M}{R^2}\right) \\
&=\frac{3}{4} m g R
\end{aligned}
\)

Hint

The expression for the speed with which a body should be projected so as to reach a height \(h\) is
\(
v=\sqrt{\frac{2 g h}{1+(h / R)}}
\)
Here, \(h=R\) (given)
\(
v=\sqrt{\frac{2 g R}{1+(R / R)}}=\sqrt{\frac{2 \times \frac{G M}{R^2} \times R}{2}}=\sqrt{\frac{G M}{R}}
\)

Hint

(c) Mass of the satellite \(=\mathrm{m}\) and height of satellite from earth \((h)=6.4 \times 10^6 \mathrm{~m}\).
We know that gravitational potential energy of the satellite at height \(\mathrm{h}\),
\(
\begin{aligned}
\mathrm{GPE} &=-\frac{G M_e m}{R_e+h}=-\frac{g R_e^2 m}{2 R_e} \\
&=-\frac{g R_e m}{2}=-0.5 m g R_e
\end{aligned}
\)
(where, \(G M_e=g R_e^2\) and \(h=R_e\) )

Hint

(a) We know that effective gravity \(g^{\prime}\) at depth below earth surface is given by
\(
g^{\prime}=g\left(1-\frac{d}{R}\right)
\)
Here, \(d=100 \mathrm{~km}, R=6400 \mathrm{~km}\),
\(
\therefore \quad g^{\prime}=9.8\left(1-\frac{100}{6400}\right)=9.65 \mathrm{~m} / \mathrm{s}^2
\)

Hint

(a) \(m g=72 \mathrm{~N} \quad\) (body weight on the surface)
\(
g=\frac{G M}{R^2}
\)
At a height \(h=\frac{R}{2}\),
\(
\begin{aligned}
g^{\prime} &=\frac{G M}{\left(R+\frac{R}{2}\right)^2} \\
&=\frac{4 G M}{9 R^2}
\end{aligned}
\)
Body weight at height \(h=\frac{R}{2}\),
\(
\begin{aligned}
&m g^{\prime}=m \times \frac{4}{9}\left(\frac{G M}{R^2}\right) \\
&=m \times \frac{4}{9} \times g=\frac{4}{9} m g \\
&=\frac{4}{9} \times 72=32 \mathrm{~N}
\end{aligned}
\)

Hint

(a) \(v_{\text {earth }}=\sqrt{\frac{2 G M_e}{R_e}}\)
\(
v_{\text {planet }}=\sqrt{\frac{2 G M_p}{R_p}}=\sqrt{\frac{2 G M_e}{R_e / 4}}=\sqrt{\frac{8 G M_e}{R_e}}
\)
\(
\begin{aligned}
&\frac{v_{\text {planet }}}{v_{\text {earth }}}=\sqrt{\frac{8 G M_e}{R_e}} \times \sqrt{\frac{R_e}{2 G M_e}}=2 \\
&\therefore v_{\text {planet }}=2 \times v_{\text {earth }}=2 \times 11.2=22.4 \mathrm{~km} / \mathrm{s}
\end{aligned}
\)

Hint

(b)
The binding energy of sphere of mass \(m\) (say) on the surface of the earth kept at rest is \(\frac{G M_e m}{R_e}\). To escape it from the earth’s surface, this much energy in the form of kinetic energy is supplied to it.
So, \(\frac{1}{2} m v_e^2=\frac{G M_e m}{R_e}\)
or \(\quad v_e=\) escape velocity \(=\sqrt{\frac{2 G M_e}{R_e}}\)
where, \(R_e=\) radius of earth,
\(M_e=\) mass of the earth.

Hint

(b)
According to Kepler’s third law
\(T^2 \propto r^3\)
where, \(T=\) Time period of revolution
\(r=\) Semi major axis
\(\therefore \quad \frac{T_A^2}{T_B^2}=\frac{r_A^3}{r_B^3}\)
\(\therefore \quad \frac{r_A}{r_B}=\left(\frac{T_A}{T_B}\right)^{2 / 3}=(8)^{2 / 3}=2^{3 \times \frac{2}{3}}=4\)
or \(\quad r_A=4 r_B\)

Hint

(b) Escape velcocity
\(
\begin{aligned}
&v_e=\sqrt{\frac{2 G M_e}{R_e}}, v_e^{\prime}=\sqrt{\frac{2 G M_e^{\prime}}{R_e^{\prime}}} \\
&\therefore \frac{v_e^{\prime}}{v_e}=\sqrt{\frac{M_e^{\prime}}{M_e} \times \frac{R_e}{R_e^{\prime}}} \\
&\text { Given } M_e^{\prime}=2 M_e \text { and } R_e^{\prime}=\frac{R_e}{2} \\
&\therefore \frac{v_e^{\prime}}{v_e}=\sqrt{\frac{2 M_e}{M_e} \times \frac{R_e}{R_e / 2}}=\sqrt{4}=2 \\
&v_e^{\prime}=2 v_e=2 \times 11.2=22.4 \mathrm{~km} / \mathrm{s}
\end{aligned}
\)

Hint

(b) The orbital speed of the satellite is independent of the mass of the satellite, so the ball will behave as a satellite and will continue to move with the same speed in the original orbit.

Hint

(d)
Let \(m\) be the mass of the body, it is placed on the spherical body of mass \(M\), radius \(R\), and centre 0. If acceleration due to gravity is \(g\) and the density of the spherical body is uniform such that its mass can be supposed to be concentrated at its centre 0.

Let \(F\) be the force of attraction between the body of mass m and the spherical body of mass \(M\).
According to Newton’s law of gravitation
\(
F=\frac{G M m}{R^2}
\)
From gravity pull F \(=m g\)
\(\therefore \quad m g=\frac{G M m}{R^2}\) or \(g=\frac{G M}{R^2}\)
\(\therefore \quad M=\frac{g R^2}{G}\)

Alternate:

We know that \(m g=\frac{G M m}{R^2}\)
\(
\therefore g=\frac{G M}{R^2} \Rightarrow M=\frac{g R^2}{G}
\)

Hint

(a)
Escape velocity on surface of the earth is given by
\(
v_e=\sqrt{2 g R_e}=\sqrt{\frac{2 G M_e}{R_e}} \quad\left(\because g=\frac{G M_e}{R_e^2}\right)
\)
where, \(M_{\mathrm{e}}=\) mass of earth
\(R_e=\) radius of the earth
\(G=\) gravitational constant
\(
\therefore \quad v_{\mathrm{e}} \propto \sqrt{\frac{M_e}{R_e}}
\)
If \(v_p\) is escape velocity from the surface of the planet, then \(\frac{v_e}{v_p}=\sqrt{\frac{M_e}{R_e}} \times \sqrt{\frac{R_p}{M_p}}\) where, \(M_p\) is mass of the planet and \(R_p\) is radius of the planet.
but \(R_p=3 R_e\)
(given)
and \(M_p=3 M_e\)
\(
\therefore \frac{v_e}{v_p}=\sqrt{\frac{M_e}{R_e}} \times \sqrt{\frac{3 R_e}{3 M_e}}=\frac{1}{1}=1 \text { or } v_p=v_e
\)

Hint

(c) \(F=\left(\frac{d M}{d t}\right) v=\alpha v^2 \quad\left(\because \frac{d M}{d t}=\alpha v\right)\)
\(\therefore\) Retardation \(=\frac{-F}{M}=-\frac{\alpha v^2}{M}\)

Hint

(c) \(T^2 \propto R^3\) (According to Kepler’s law) \(T_1^2 \propto\left(10^{13}\right)^3\) and \(T_2^2 \propto\left(10^{12}\right)^3\)
\(
\therefore \quad \frac{T_1^2}{T_2^2}=(10)^3 \text { or } \frac{T_1}{T_2}=10 \sqrt{10}
\)

Hint

(d)
According to Kepler’s third law, the square of the time period of revolution of a planet around the sun is directly proportional to the cube of semi major-axis of its elliptical orbit i.e.
\(
T^2 \propto r^3
\)
where, \(T=\) time taken by the planet to go once around the sun.
\(r=\) semi-major axis of the elliptical orbit
\(
\therefore \quad \frac{T_1^2}{T_2^2}=\frac{(r)^3}{(2 r)^3}=\frac{1}{8} \Rightarrow \frac{T_1}{T_2}=\frac{1}{2 \sqrt{2}}
\)

Hint

(b)
As, \(v_e=\sqrt{2 g R}=\sqrt{\frac{2 G M}{R}}\)
Hence, escape velocity does not depend on the angle of projection. Escape velocity will remain the same.

Hint

Let \(v_0\) be orbital speed and be is the radius of orbit of a geostationary satellite. So, time period of satellite
\(
\begin{aligned}
T &=\frac{2 \pi r}{V_0} \\
A s, v_0=\sqrt{\frac{G M}{r}} &=\sqrt{\frac{g R^2}{r}} \quad\left(\because g=\frac{G M}{R^2}\right)
\end{aligned}
\)
\(
\therefore \quad T=\frac{2 \pi r}{\left(g R^2 / r\right)^{1 / 2}} =\frac{2 \pi r^{3 / 2}}{\sqrt{g R^2}}
\)
but \(T=\frac{2 \pi}{\omega} \Rightarrow T=\frac{2 \pi r^{3 / 2}}{\sqrt{g R^2}}=\frac{2 \pi}{\omega}\)
Hence, \(r^{3 / 2}=\frac{\sqrt{g R^2}}{\omega}\) or \(r^3=\frac{g R^2}{\omega^2}\)
or
\(
r=\left(\frac{g R^2}{\omega^2}\right)^{1 / 3}
\)

Hint

(d)
When the rocket accelerates upward with acceleration a, then the effective acceleration of the rocket is \((g+a)\)
As, \(T=2 \pi \sqrt{\frac{1}{g}}=2 \pi \sqrt{\left(\frac{1}{g+a}\right)}\)
Hence, the period of oscillation of the seconds pendulum decreases when the rocket moves up with uniform acceleration.

Hint

(d) Let satellite of mass \(m\) be revolving closely around the earth of mass M and radius \(R\).
Total energy of satellite
\(
\begin{aligned}
&=P E+K E=-\frac{G M m}{R}+\frac{1}{2} m v^2 \\
&=-\frac{G M m}{R}+\frac{m}{2} \frac{G M}{R}\left[\text { as } v=\sqrt{\frac{G M}{R}}\right]
\end{aligned}
\)
\(
\begin{aligned}
&\quad=-\frac{G M m}{2 R} \\
&\therefore \text { Total energy of satellite }=-\frac{1}{2} m v^2
\end{aligned}
\)

Alternate:

(d) Total energy \(=-\mathrm{KE}=\frac{\mathrm{PE}}{2}\)
\(
\mathrm{K} . \mathrm{E}=\frac{-1}{2} m v^2
\)

Hint

When the planet moves in circular or elliptical orbit, then torque is always acting parallel to displacement or velocity. So, angular momentum is conserved. When the field is attractive, then potential energy is given by
\(
v=-\frac{G M m}{R}
\)
Negative sign shows that the potential energy is due to attractive gravitational force. Kinetic energy changes as velocity increases when distance is less. Hence, option (c) is correct.

Hint

b)
According to question, gravitational force between two objects
\(
F=\frac{k}{R}
\)
In equilibrium, the gravitational force provides the required centripetal force to the particle
\(
\therefore \quad \frac{m v^2}{R}=\frac{k}{R}
\)
Hence, \(\quad v \propto R^0\)

Hint

(a) Since, escape velocity \(\left(v_e=\sqrt{2 g R_e}\right)\) is independent of angle of projection, so it will not change.

Hint

(c) \(T^2 \propto R^3\) (Kepler’s law)
\(
\frac{T_1^2}{T_2^2}=\left(\frac{10^{13}}{10^{12}}\right)^3 \Rightarrow \frac{T_1}{T_2}=10 \sqrt{10}
\)

Hint

(c) Applying the properties of ellipse, we have
\(
\frac{2}{R}=\frac{1}{r_1}+\frac{1}{r_2}=\frac{r_1+r_2}{r_1 r_2}
\)

\(
R=\frac{2 r_1 r_2}{r_1+r_2}
\)

Hint

\(
\begin{aligned}
& F=m E_G \\
& 3=\frac{60}{1000} E_G \\
& E_G=50 \mathrm{~N} / \mathrm{kg}
\end{aligned}
\)

Hint

Let the electric field at point \(Q\) be zero
So,
\(
\begin{aligned}
& \frac{G m}{x^2}=\frac{G(9 m)}{(R-x)^2} \\
& \frac{(R-x)^2}{x^2}=9 \\
& x=\frac{R}{4}
\end{aligned}
\)
\(
\begin{aligned}
& V_p=\frac{-G m}{x}-\frac{G(9 m)}{R-x} \\
& V_p=\frac{-G m}{\frac{R}{4}}-\frac{G(9 m)}{\frac{3 R}{4}}=\frac{-4 G m}{R}-\frac{12 G m}{R}=\frac{-16 G m}{R}
\end{aligned}
\)