Exemplar MCQs

Hint

The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity cannot be zero at any_point. Explanation:

Because as we know that \(\mathrm{g}=\frac{4}{3} \pi \mathrm{p} G\), here if the density is non-uniform then \(\mathrm{g}\) will be different on the surface at different points. So, in this equation, if we substitute any value of density, we will get \(g \neq 0\) at any point.

Hint

As observed from earth, the sun appears to move in an approximate circular orbit. For the motion of another planet like mercury as observed from earth, this would not be true because the major gravitational force on mercury is due to sun.
Explanation:
As observed from the earth, the sun appears to move in an approximately circular orbit. The gravitational force of attraction between the earth and the sun always follows the inverse square law. All planets move around the sun due to the huge gravitational force of the sun acting on them. The gravitational force on the mercury due to the earth is much smaller as compared to that acting on it due to the sun and hence it revolves around the sun and not around the earth.

Hint

As the earth is revolving around the sun in a circular motion due to gravitational attraction. The force of attraction will be of radial nature i.e., angle between position vector \(r\) and force \(F\) is zero.
So, torque \(=|\tau|=|r \times F|\)
\(
=r F \sin 0^{\circ}=0
\)

Hint

Satellites orbiting the earth have finite life and sometimes debris of satellites fall to the earth. This is because of viscous forces causing the speed of satellite and hence height to gradually decrease.

Explanation:
As the total energy of the earth satellite bounded system is negative \(\left(\frac{-G M}{2 a}\right)\). where a is the radius of the satellite and \(M\) is mass of the earth.
Due to the viscous force acting on the satellite, energy decreases continuously and the radius of the orbit or height decreases gradually.

Hint

Both earth and moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon will not be strictly elliptical because the total gravitational force on it is not central.
Explanation:
The moon revolves around the earth in a nearly circular orbit. When it is observed from the sun, two types of forces are acting on the moon one is due to gravitational attraction between the sun and the moon and the other is due to gravitational attraction between the earth and the moon. So the moon is moving under the combined gravitational pull acting on it due to the earth and the sun. Hence, the total force on the moon is not central.

Hint

In our solar system, the inter-planetary region has chunks of matter (much smaller in size compared to planets) called asteroids. They will move in orbits like planets and obey Kepler’s laws.
Explanation:
Because as we know asteroids move in orbits like planet which follows Kepler’s law as they act upon central gravitational forces.

Hint

The answer is the option (d) The gravitational mass of a proton is equivalent to its inertial mass & is independent of the presence of neighbouring heavy objects.

Hint

Particles of masses \(2 \mathrm{M}, \mathrm{m}\) and \(\mathrm{M}\) are respectively at points \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) with \(\mathrm{AB}=1 / 2(\mathrm{BC}) . \mathrm{m}\) is much-much smaller than \(\mathrm{M}\) and at time \(t=0\), they are all at rest (Figure). At subsequent times before any collision takes place \(\mathbf{m}\) will move towards \(2 \mathrm{M}\).

Explanation:
The particle \(m\) at \(B\) will move towards \(A\) with the greater force, due to particle \(2 \mathrm{M}\) at \(\mathrm{A}\).
Force on \(\mathrm{m}\) at \(\mathrm{B}\) due to \(2 \mathrm{M}\) at \(\mathrm{A}\) is
\(F_{B A}=\frac{G(m \times 2 M)}{(A B)^2}\) towards \(B A\)
Force on mass \(\mathrm{m}\) at \(\mathrm{B}\) due to mass \(\mathrm{M}\) at \(\mathrm{C}\) is
\(F_{B C}=\frac{G(M \times M)}{(B C)^2}\) towards \(B C\)
Therefore, the resultant force on mass \(m\) at \(B\) due to masses \(M\) and \(2 M\) is \(F_{\text {net }}=F_{B A}-F_{B C} \ldots .\). (Because \(\mathrm{F}_{\mathrm{BA}}\) and \(\mathrm{F}_{\mathrm{BC}}\) are acting in opposite directions)
\(
F_{\text {net }}=\frac{2 G M m}{(A B)^2}-\frac{G M m}{(B C)^2}
\)
\(
\begin{aligned}
&\text { As }(B C)=2 A B \\
&\Rightarrow F_{\text {net }}=\frac{2 G M m}{(A B)^2}-\frac{G M m}{(2 A B)^2} \\
&=\frac{2 G M m}{(A B)^2}-\frac{G M m}{4(A B)^2}
\end{aligned}
\)
\(=\frac{7 G M m}{4(A B)^2}\) (along BA)
Hence, m will move towards BA (i.e., 2M).

Hint

(a) The acceleration due to gravity at an altitude (height) \(g h=g\left(1-\frac{2 h}{R}\right.\) ) Increasing height (h) decreases the value of gh option (a) is correct.
(b) Assuming the earth to be a sphere of uniform density. the acceleration due to gravity at a particular depth (d).gd \(=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)\). increasing depth (d) decreases the value of gd option (b) is incorrect.
(c) if \(\lambda\) is latitude on earth then \(g \lambda=g-\omega^2 R \cos ^2 \lambda\)
As value of \(\cos \lambda\) decrease from \(0^0\) to \(90^{\circ}\) (from 1 to 0 ) the acceleration due to gravity increases from equator \(\left(\lambda=0^0\right)\) to pole \(\left(\lambda=90^{\circ}\right)\) option (c) is correct.
(d) The acceleration due to gravity on surface of earth is \(\mathrm{g}=\frac{\mathrm{GM}_e}{\mathrm{R}_e^2}\)
So \(\mathrm{g}\) on earth depends on mass of earth . option (d) is incorrect.

Hint

If the law of gravitation becomes an inverse cube law. then we can write, for a planet of mass m revolving around the sun of mass \(M\),
\(\mathrm{F}=\frac{G M m}{a^3}=\frac{m v^2}{a} (\) Where \(\mathrm{a}\) is the radius of orbiting planet)
\(
\begin{aligned}
&\Rightarrow \mathrm{v}=\text { orbital speed }=\frac{\sqrt{G M}}{a} \\
&\Rightarrow v \propto \frac{1}{a}
\end{aligned}
\)
Time period of revolution of a planet \(\mathrm{T}=\frac{2 \pi a}{v}=\frac{2 \pi a}{\frac{\sqrt{G M}}{a}}=\frac{2 \pi a^2}{\sqrt{G M}}\) \(\Rightarrow T^2 \propto a^4\)

Hence, the orbit will not be elliptical. …..[For elliptical orbit \(\mathrm{T}^2 \propto \mathrm{a}^3\) ]
As force \(\mathrm{F}=\left(\frac{G M}{a^3}\right) m=g m\)
Where, \(g^{\prime}=\frac{G M}{a^3}\)
As \(\mathrm{g}^{\prime}\), acceleration due to gravity is constant, hence path followed by a projectile will be approximately parabolic. (as \(\mathrm{T} \propto \mathrm{a}^2\) )

Hint

a) Gravitation field due to the earth
\(
g=\frac{\mathrm{GM}_e}{\mathrm{R}^2}
\)
\(g \propto G\)
As G increases 10 times
\(
g^{\prime}=10 \mathrm{~g}
\)
Force on man due to earth (Weight of person) \(=m g=m \times 10 g=10 \mathrm{mg}\)
Force on the man due to the sun
\(
F=\frac{G M_s^{\prime} \mathrm{m}}{r^2}
\)
Given, \(M_s^{\prime}=\frac{M_s}{10}\)
\(
F=\frac{G\left(M_s\right) m}{10 r^2}
\)
As \(r>>>R\) (radius of the earth)
F will be very small. So, the effect of the sun will be neglected. Due to this reason gravity pull on the person will increase. Due to it, walking on ground would become more difficult.
Hence this option is correct.
b) Gravitational field due to the earth
\(
\begin{aligned}
&g=\frac{G M_e}{R_e^2} \\
&g \propto G
\end{aligned}
\)
As \(G\) increases 10 times
\(
g^{\prime}=10 g
\)
Acceleration due to gravity changes and its value increases.
Hence this option is incorrect.
c) As effective acceleration due to gravity increases raindrops will fall much faster.
Hence this option is correct.
d) To overcome the increased gravitational force of the Earth, the aeroplanes will have to travel much faster.
Hence this option is correct.

Hint

Coulomb’s electric force or Electrostatic force of attraction will produce opposite charges.
If the sun and the planets carry a huge amount of opposite charges, then the electrostatic force of attraction will be large. Gravitational force is also attractive in nature have both forces will be added and both are radial in nature.

Hint

We know the gravitational force between the earth and the sun.
\(F_G=\frac{G M m}{r^2}\), where \(\mathrm{M}\) is mass of the sun and \(\mathrm{m}\) is the mass of the earth.
When \(G\) decreases with time. the gravitational force \(F_G\) will become weaker with time. As is changing with time. Due to it, the earth will be going around the sun not strictly in a closed orbit and the radius also increases, since the attraction force is getting weaker.
Hence, after a long time, the earth will leave the solar system.

Hint

Given, \(F_1=-F_2=\frac{-r_{12}}{r_{12}^3} G M_0^2\left(\frac{m_1 m_2}{m_0^2}\right)^n\)
Acceleration due to gravity, \(g=\frac{|F|}{\text { mass }}\)
\(
=\frac{G M_0^2\left(m_1 m_2\right)^n}{r_{12}^2\left(M_0\right)^{2 n}} \times \frac{1}{(\text { mass })}
\)
Since. \(g\) depends upon the position vector, hence it will be different for different objects. As \(g\) is not constant, hence constant of proportionality will not be constant in Kepler’s third law. Hence, Kepler’s thir law will not be valid.
As the force is of central nature. \(\quad \ldots . .\left[\because\right.\) Force \(\left.\propto \frac{1}{r^2}\right]\)
Hence, the first two of Kepler’s laws will be valid.
For negative \(\mathrm{n}, \mathrm{g}=\frac{G M_0^2\left(m_1 m_2\right)^{-n}}{r_{12}^2\left(M_0\right)^{-2 n}} \times \frac{1}{(\text { mass })}\)
\(
\begin{aligned}
&=\frac{G M_0^{2(1+n)}}{r_{12}^2} \frac{\left(m_1 m_2\right)^{-n}}{(\mathrm{mass})} \\
&\mathrm{g}=\frac{G M_0^2}{r_{12}^2}\left(\frac{M_0^2}{m_1 m_2}\right) \times \frac{1}{\mathrm{mass}}
\end{aligned}
\)
As \(\mathrm{M}_0>\mathrm{m}_1\) or \(\mathrm{m}_2\)
\(g>0\), hence in this case situation will reverse i.e., an object lighter than water will sink in water.

Hint

The satellite which appears stationary relative to earth is called the geostationary satellite. it revolves around the earth in the west-east direction with the same angular velocity as done by the earth about its own axis in the west-east direction. A polar satellite revolves around the earth’s pole in northsouth direction. it is independent of earth’s rotation. option (a), (c) are correct.

Hint

The center of gravity is based on weight, whereas the center of mass is based on mass. So, when the gravitational field across an object is uniform, the two are identical. However, when the object enters a spatially-varying gravitational field, the COG will move closer to regions of the object in a stronger field, whereas the COM is unmoved.
More practically, the COG is the point over which the object can be perfectly balanced; the net torque due to gravity about that point is zero. In contrast, the COM is the average location of the mass distribution or it is the point where the whole mass of the body is supposed to be concentrated. If the object were given some angular momentum, it would spin about the COM.

For small objects, say of sizes less than \(100 \mathrm{~m}\) placed in the uniform gravitational field the centre of mass is very close to the centre of gravity of the body. But when the size of object increases, its weight changes and its CM and CG become far from each other. Like in the case of spherical balls, the CM and the CG are the same, but in the case of Mount Everest, its CM lies a bit above its CG. option d is correct.

Hint

We know that the distance of the Moon from the Earth is about 60 times the radius of the earth. So, acceleration due to gravity at that distance is \(0.0027 \mathrm{~m} / \mathrm{s}^2\). When the Moon is stopped for an instant and then released, it will fall towards the Earth with an initial acceleration of \(0.0027 \mathrm{~m} / \mathrm{s}^2\).

Hint

According to the previous question, we have:
Radius of the moon, \(R_m=\frac{R_e}{4}=\frac{6400000}{4}=1600000 \mathrm{~m}\)
So, when the Moon is just about to hit the surface of the Earth, its centre of mass is at a distance of \(\left(R_e{ }^{+}\right.\) \(R_m\) ) from the centre of the Earth.
Acceleration of the Moon just before hitting the surface of the earth is given by
\(
\begin{aligned}
&g^{\prime}=\frac{G M}{\left(R_e+R_m\right)^2}=\frac{G M}{R_e{ }^2\left(1+\frac{R_m}{R_e}\right)^2} \\
&\Rightarrow g^{\prime}=\frac{g}{\left(1+\frac{R_m}{R_e}\right)^2}=\frac{10}{\left(1+\frac{1}{4}\right)^2}=\frac{10 \times 16}{25} \\
&\Rightarrow g^{\prime}=6.4 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)

Hint

g is inversely proportional to the square of R and also, at some point force due to earth and mass cancels each other, and gravitational force can’t be negative.

At one point between the Earth and Mars, the gravitational field intensity is zero. So, at that point, the weight of the passenger is zero. The curve C indicates that the weight of the passenger is zero at a point between the Earth and Mars.

Hint

(The weight of the object on the Earth is \(W=m \frac{G M_e}{R_e{ }^2}\)
Here, \(m\) is the actual mass of the object; \(M_e\) is the mass of the earth and \(R_e\) is the radius of the earth.
Let \(R_p\) be the radius of the planet.
Mass of the planet, \(M_p=2 M_e\)
If \(\rho\) is the average density of the planet then
\(
\begin{aligned}
&\frac{4}{3} \pi R_p{ }^3 \times \rho=2 \times\left(\frac{4}{3} \pi R_e{ }^3 \times \rho\right) \\
&\Rightarrow R_p=(2)^{\frac{1}{3}} R_e
\end{aligned}
\)
Now, weight of the body on the planet is given by
\(
\begin{aligned}
&W_p=m\left(\frac{G M_p}{R_p^2}\right)=m\left(\frac{2 G M_e}{2^{\frac{2}{3}} R_e^2}\right) \\
&\Rightarrow W_p=2^{\frac{1}{3}} \times m\left(\frac{G M_e}{R_e^2}\right) \\
&\Rightarrow W_p=2^{\frac{1}{3}} \times W
\end{aligned}
\)

Hint

Work done \(=-(\) final potential energy \(–\) initial potential energy \()\)
\(
\begin{aligned}
&\Rightarrow W=-\left(\frac{G M m}{2 R}-\frac{G M m}{R}\right) \\
&\Rightarrow W=\frac{1}{2} \frac{G M m}{R}=\frac{1}{2} m R \times\left(\frac{G M}{R^2}\right) \\
&\Rightarrow W=\frac{1}{2} m R g\left[\because g=\frac{G M}{R^2}\right]
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { Work done, } W=\Delta K E+\Delta P E\\
&-3=\frac{1}{2} \times 1 \times\left(2^2-0^2\right)+m\left(V_A-V_{\infty}\right)\\
&-3=2+0+V_A-0\\
&\mathrm{V}_{\mathrm{A}}=-2-3\\
&=-5 \mathrm{~J} / \mathrm{kg}
\end{aligned}
\)

Hint

(A) is not correct because the gravitational potential inside the shell is constant =\(\mathrm{GM} / \mathrm{a}\), where a is the radius of the shell. As the distance increases from a, the potential gradually starts increasing till infinity to a value zero. So the curve is not discontinuous. (B) is correct because the gravitational field inside the shell is zero and just as it goes outside the value jumps to GM/a \({ }^2\) and then gradually decreases to zero at infinity. So the curve at \(r=a\) is discontinuous.

Hint

(A) is wrong because the value of \(\mathrm{V}\) at the center is \(-3 \mathrm{GM} / 2 \mathrm{a}\) and it increases continuously to \(-GM/a\) at the surface and from \(-GM/a\) at the surface to zero at the infinity. So it is continuous.
(B) is wrong because the value of \(E\) increases linearly from zero at the center to \(\mathrm{GM} / \mathrm{a}^2\) at the surface, then decreases from \(\mathrm{GM} / \mathrm{a}^2\) at the surface to zero at the infinity. So it is also continuous.

Hint

The value of \(g\) is maximum at the poles than any other point anywhere. As the value of \(g\) decreases due to height there might be some point about the pole where value of \(g\) might be equal to that at equator. Thus statement \(A\) is correct but \(\mathrm{B}\) is wrong because At points outside the earth value of acceleration due to gravity is less than \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)

Hint

Time period of earth satellite
\(
T=2 \pi \sqrt{\frac{(R+h)^3}{G M_e}}
\)

\(
\text { where } \mathrm{R}+\mathrm{h}=\text { orbital radius of satellite, }
\)
\(\mathrm{M_e} \rightarrow\) mass of earth
\(\mathrm{G} \rightarrow\) Gravitational constant
So, It is independent of mass of the satellite

Hint

Let the mass of the moon \(=m\), velocity \(v\) and its distance from the earth \(=a\).
\(
\begin{aligned}
&v=\sqrt{ }(G M / a), K . E .=K=1 / 2 m v^2=1 \frac{1}{2} \mathrm{~m} {\times} \mathrm{GM} / \mathrm{a} \\
&\rightarrow \mathrm{GMm} / \mathrm{a}=2 \mathrm{~K}
\end{aligned}
\)
The magnitude of gravitational potential energy of the moon
\(
\mathrm{U}=\mathrm{GMm} / \mathrm{a}=2 \mathrm{~K}
\)
Obviously U>K.

Hint

According to Kepler’s law, the line joining a planet and the Sun sweep equal areas in equal intervals of time.
Given: \(\operatorname{Area}(\mathrm{ASB})=\operatorname{Area}(\mathrm{CSD})\)
\(
\therefore \mathrm{t}_1=\mathrm{t}_2
\)

Hint

Only (c) is correct. The weight and the centrifugal force balance each other, hence the normal force is zero.

Hint

The reading on the spring balance is independent of the radius and thus both weight will be the same. Also, there is no gravitational force acting on a satellite.
Hence, \(\mathrm{W}_1=\mathrm{W}_2\).

Hint

(c) Escape velocity \(v_e=\sqrt{2 g R}\)
\(\therefore\) Kinetic energy
\(
=\frac{1}{2} m v_e^2=\frac{1}{2} m \times 2 g R=m g R .
\)

Hint

Potential energy of the particle at a distance \(R\) from the surface of the Earth is
\(
(P . E .)=\frac{G M m}{(R+R)}=\frac{1}{2} \frac{G M m}{R}
\)
Here, \(M\) is the mass of the earth; \(R\) is the radius of the earth and \(m\) is the mass of the body.
Let the particle be projected with speed \(v\) so that it just escapes the gravitational pull of the earth.
So, kinetic energy of the body \(=-\) [change in the potential energy of the body]
Now, kinetic energy of the body \(=-[\) final potential energy-initial potential energy]
\(
\Rightarrow \frac{1}{2} m v^2=-\left[\frac{G M m}{\infty}-\frac{G M m}{2 R}\right]
\)
\(
\Rightarrow v=\sqrt{\frac{G M }{R}}
\)

Hint

Its minimum escape speed will be \(v_e\) when projected vertically upward, but if projected otherwise the speed will be greater. So the escape speed will depend on the direction of the projection.

Hint

Consider a thin spherical shell and choose \(r=\) infinity where the potential is zero. At infinity \(V=0\), \(E=0\). So (a) is true. At any point outside the shell \(V \neq 0\) and \(E \neq 0\). So (d) is true. At any point inside the shell \(E=0\) but \(V \neq 0\). So (c) is true.

But if we choose reference point just outside the shell then here \(V=0\) but \(E \neq 0\). So (b) is true.

Keeping infinity as a reference point (b) cannot be true.

Hint

Inside a spherical shell, the gravitational field is zero and hence potential remains the same everywhere.

For uniform spherical shell,
\(
\begin{aligned}
&E_{\text {inside }}=0(r<a) \\
&I_{\text {outside }}=\frac{G M}{r^2}(r>a) \\
&I_{\text {surface }}=\frac{G M}{a^2}(r = a)
\end{aligned}
\)

Hint

The gravitational potential of a spherical shell of mass \(m\) and radius \(r\) is given below.
\(
V_G=-\frac{m g}{r}
\)
Here, \(g\) is acceleration due to gravity.
Now, according to the question, the spherical shell gradually shrinks maintaining its shape. It means the radius of the spherical shell decreases.

From the formula, we see that, if the radius decreases then the quantity on the right-hand side increases.
But due to the presence of a negative sign on the right-hand side, the potential decreases.
\(
r(\downarrow) \Rightarrow \frac{m g}{r}(\uparrow) \Rightarrow V_g(\downarrow)
\)
Therefore, we can say that when the radius decreases, the potential at the center also decreases.

Hint

When a planet is moving in an elliptical orbit, at some point, the line joining the centre of the Sun and the planet is perpendicular to the velocity of the planet. For that instant, work done by the gravitational force on the planet becomes zero. As there is no net increase in the speed of the planet after one complete revolution about the Sun, the work done by the gravitational force on the planet in one complete revolution is zero.
Note: For elliptical orbits angle between force and velocity is always 90 so there the work done is zero in any small part of the orbit.

Work done \(w=\vec{F} . \vec{ds}\), for a displacement over any samll part of the orbit \(\vec{F}\) and \(\vec{d}s\) are perpendicular to each other, hence work done is zero. Also for one complete revolution, displacement is zero, hence work done is zero.

Hint

The velocity of a satellite in an orbit, \(v=\sqrt{ }(G M / a)\). So the velocity of a satellite in an orbit is in independent of its mass. So (a) is true. The potential energy of the earth satellite system \(=-GMm/a\), so it is dependent on the mass of the satellite. Hence (b) is not true. Since the K.E. \(\left(1 / 2 m v^2\right)\) depends on the mass, so (c) is not true. Again the total energy of the system \(=-GMm/2a\). It is also dependent on mass, so (d) is not true.

Hint

The speed in an elliptical orbit is not constant and hence not the kinetic energy. So (a) and (c) are not true. The radius vector also does not sweep equal angle in equal time, angular speed is not constant. So (b) is not true. Since there is no external torque on the satellite the angular momentum at any point is constant. (d) is true.