1. Gravitation is best described as the interaction by which
ⓐ. only magnets attract other objects
ⓑ. any two masses attract each other
ⓒ. only charged bodies exert force on each other
ⓓ. only Earth attracts nearby bodies
Correct Answer: any two masses attract each other
Explanation: Gravitation is a universal attractive interaction between masses. It is not limited to Earth and a falling object; the same kind of interaction acts between Earth and the Moon, the Sun and planets, and any two material bodies. The force may be very small between ordinary objects because the gravitational constant \(G\) is small, but the interaction is still present. Magnetism and electrostatic forces need magnetic or electric properties, while gravitation is linked with mass. The word universal means the idea applies to all masses, not only to objects near Earth.
2. A stone released from rest near Earth falls downward mainly because
ⓐ. air pushes it downward with constant force
ⓑ. the stone loses all its mass after release
ⓒ. the stone moves only when it touches Earth
ⓓ. Earth exerts a gravitational pull on it
Correct Answer: Earth exerts a gravitational pull on it
Explanation: A freely falling stone accelerates because Earth attracts it gravitationally. The force acts even when the stone is not in contact with the ground, so contact is not required for the fall. Air may affect the motion through resistance, but it is not the basic cause of downward acceleration. The mass of the stone does not disappear during the fall. Near Earth, the direction of gravitational acceleration is toward Earth’s centre.
3. Two metal spheres are placed far apart in deep space, away from other bodies. Their mutual gravitational force is best described as
ⓐ. zero because there is no air between them
ⓑ. present only if the spheres are electrically charged
ⓒ. attractive along the line joining their centres
ⓓ. repulsive because both objects have mass
Correct Answer: attractive along the line joining their centres
Explanation: Gravitation does not require air, a string, or physical contact. Two masses attract each other even in empty space. For a pair of bodies, the force is directed along the straight line joining their centres when they are treated as spherical or point-like. The gravitational force is attractive, not repulsive. Electric charge is not needed for gravitational attraction because mass is the relevant property.
4. Use the arrangement described below: Object \(P\) is on the left and object \(Q\) is on the right. The gravitational force on \(P\) due to \(Q\) acts
ⓐ. toward the right, along \(PQ\)
ⓑ. perpendicular to the line \(PQ\)
ⓒ. toward the left, away from \(Q\)
ⓓ. vertically upward, independent of \(Q\)
Correct Answer: toward the right, along \(PQ\)
Explanation: Gravitational force is a central attractive force. Since \(Q\) is to the right of \(P\), the force on \(P\) due to \(Q\) is directed toward \(Q\). The line of action is the line joining the two masses, not a perpendicular direction. A vertical direction would be suitable only in a special arrangement where the attracting mass lies vertically above or below. Direction in gravitation is decided by the position of the attracting body.
5. In the notation of gravitation, the symbol \(G\) represents
ⓐ. the local acceleration due to gravity near a planet
ⓑ. the universal gravitational constant
ⓒ. the orbital speed of a satellite
ⓓ. the weight of a body in newtons
Correct Answer: the universal gravitational constant
Explanation: The symbol \(G\) denotes the universal gravitational constant used in Newton’s law of gravitation. It has the same value for any pair of masses in the same system of units. The symbol \(g\), not \(G\), usually represents the local acceleration due to gravity near Earth or another planet. Weight is commonly written as \(W\), and orbital speed is often written as \(v_o\). Confusing \(G\) and \(g\) changes both the meaning and the unit of the quantity.
6. Near the surface of Earth, \(g\) is used to denote
ⓐ. Earth’s centre-to-surface radius
ⓑ. the universal gravitational constant
ⓒ. free-fall acceleration near Earth
ⓓ. the gravitational mass of Earth
Correct Answer: free-fall acceleration near Earth
Explanation: The symbol \(g\) represents acceleration due to gravity at a given place. Near Earth’s surface, a freely falling body has acceleration approximately equal to \(g\) if air resistance is neglected. This quantity is local because its value can change slightly with height, depth, latitude, or the planet considered. It is not the same as \(G\), which is a universal constant. The direction associated with \(g\) near Earth is toward Earth’s centre.
7. The SI unit of \(g\) is
ⓐ. \( \text{N m}^2\,\text{kg}^{-2} \)
ⓑ. \( \text{m s}^{-2} \)
ⓒ. \( \text{J kg}^{-1} \)
ⓓ. \( \text{kg} \)
Correct Answer: \( \text{m s}^{-2} \)
Explanation: The quantity \(g\) is an acceleration. The SI unit of acceleration is \( \text{m s}^{-2} \), which means metre per second squared. The unit \( \text{kg} \) is used for mass, not acceleration. The unit \( \text{N m}^2\,\text{kg}^{-2} \) belongs to \(G\), while \( \text{J kg}^{-1} \) is used for gravitational potential. Remembering the unit helps separate the local quantity \(g\) from the universal constant \(G\).
8. The missing SI unit in \(G = 6.67\times10^{-11}\,\_\_\_\_\) is
ⓐ. \( \text{N kg}^{-1} \)
ⓑ. \( \text{N m}^2\,\text{kg}^{-2} \)
ⓒ. \( \text{m s}^{-2} \)
ⓓ. \( \text{kg m}^{-2}\,\text{s}^{-1} \)
Correct Answer: \( \text{N m}^2\,\text{kg}^{-2} \)
Explanation: Newton’s law of gravitation is \(F=G\frac{m_1m_2}{r^2}\). Rearranging gives \(G=\frac{Fr^2}{m_1m_2}\). Since \(F\) is measured in \( \text{N} \), \(r^2\) in \( \text{m}^2 \), and \(m_1m_2\) in \( \text{kg}^2 \), the unit of \(G\) becomes \( \text{N m}^2\,\text{kg}^{-2} \). The SI unit of \(g\) is not the unit of force or acceleration by itself. The unit must make \(G\frac{m_1m_2}{r^2}\) come out in \( \text{N} \).
9. When the separation between two point masses is doubled while both masses remain unchanged, the gravitational force becomes
ⓐ. \(2\) times the original force
ⓑ. \(\frac{1}{2}\) of the original force
ⓒ. \(\frac{1}{4}\) of the original force
ⓓ. \(4\) times the original force
Correct Answer: \(\frac{1}{4}\) of the original force
Explanation: Gravitational force follows an inverse-square dependence on separation. For unchanged masses, \(F\propto\frac{1}{r^2}\). If \(r\) becomes \(2r\), then the new force is proportional to \(\frac{1}{(2r)^2}=\frac{1}{4r^2}\). So the force becomes one-fourth of its original value. Doubling the distance does not merely halve the force because the distance is squared in the denominator.
10. For two spherical bodies separated by a gap, the distance \(r\) in the gravitational force formula is measured
ⓐ. from the top of one body to the bottom of the other
ⓑ. from centre to centre
ⓒ. from surface to surface
ⓓ. from the heavier body’s surface to the lighter body’s centre
Correct Answer: from centre to centre
Explanation: In \(F=G\frac{m_1m_2}{r^2}\), \(r\) is the separation between the centres of the two masses when spherical bodies are treated externally. This is why Earth’s radius and height above the surface must be handled carefully in later gravitation problems. Surface-to-surface gap alone does not represent the full separation of the mass centres. The same centre-based idea is used for planets, moons, and many satellite calculations. Using the wrong distance changes the force because \(r\) is squared.
11. In a classroom comparison, mass and weight are best paired as
ⓐ. mass in \( \text{N} \), weight in \( \text{kg} \)
ⓑ. mass in \( \text{kg} \), weight in \( \text{N} \)
ⓒ. both mass and weight in \( \text{m s}^{-2} \)
ⓓ. both mass and weight in \( \text{kg} \)
Correct Answer: mass in \( \text{kg} \), weight in \( \text{N} \)
Explanation: Mass is a measure of the amount of matter or inertia of a body, and its SI unit is \( \text{kg} \). Weight is the gravitational force on a body, so its SI unit is \( \text{N} \). Near Earth, weight is related to mass by \(W=mg\). A body’s mass does not change just because it is taken to another planet, but its weight can change because \(g\) changes. The unit difference is a simple way to avoid mixing the two quantities.
12. The following table lists some common symbols in gravitation. Match each symbol with its usual meaning.
| Symbol | Meaning |
|---|
| P. \(G\) | I. Local acceleration due to gravity |
| Q. \(g\) | II. Universal gravitational constant |
| R. \(M\) | III. Mass of a large attracting body |
| S. \(r\) | IV. Centre-to-centre separation |
ⓐ. P-IV, Q-I, R-II, S-III
ⓑ. P-I, Q-II, R-III, S-IV
ⓒ. P-II, Q-III, R-I, S-IV
ⓓ. P-II, Q-I, R-III, S-IV
Correct Answer: P-II, Q-I, R-III, S-IV
Explanation: The symbol \(G\) stands for the universal gravitational constant. The symbol \(g\) usually represents the local acceleration due to gravity. A capital \(M\) is often used for a large attracting mass such as Earth or the Sun, while \(m\) is often used for a smaller test body. The symbol \(r\) generally represents the distance from the centre of the attracting body or the centre-to-centre separation between two masses. These symbols become especially important when formulas such as \(F=G\frac{Mm}{r^2}\) are used.
13. Consider the statements about gravitational force.
I. It is always attractive.
II. It requires direct surface contact between two bodies.
III. For two bodies, it acts along the line joining their centres.
The suitable choice is
ⓐ. I and III only
ⓑ. I only
ⓒ. II and III only
ⓓ. I and II only
Correct Answer: I and III only
Explanation: Gravitational force between masses is attractive in ordinary Class 11 gravitation. It acts along the line joining the centres of the bodies when they are treated as point masses or spherical bodies externally. Direct contact is not required; Earth attracts the Moon even across space. Statement II confuses gravitation with contact forces such as normal reaction or friction. The central direction of the force is the starting point for later orbital-motion reasoning.
14. A small artificial satellite keeps moving around Earth instead of travelling in a straight line because
ⓐ. the satellite’s mass becomes zero in orbit
ⓑ. there is no force acting on the satellite
ⓒ. Earth’s gravity provides an inward pull
ⓓ. air pressure continuously pushes it toward Earth
Correct Answer: Earth’s gravity provides an inward pull
Explanation: A satellite in orbit needs an inward force to keep changing the direction of its velocity. Earth’s gravitational attraction provides this inward force. The satellite is not force-free; it is continuously falling around Earth while also moving forward. Its mass does not become zero in orbit. The important starting idea is that orbital motion is linked with gravity and circular-motion reasoning, not with absence of force.
15. Calling gravitation universal mainly means that
ⓐ. it becomes a contact force for small objects
ⓑ. it applies to every pair of masses
ⓒ. it acts only near Earth’s surface
ⓓ. it acts only on planets
Correct Answer: it applies to every pair of masses
Explanation: The universality of gravitation means every mass attracts every other mass. The same law explains a falling object, the Moon’s motion around Earth, and planets moving around the Sun. For small everyday masses, the force is usually too small to notice without sensitive instruments. The interaction does not become a contact force just because the objects are small. Universal does not mean equally strong in all situations; it means governed by the same basic rule.
16. A notebook line says, “\(G\) and \(g\) are the same because both appear in gravitation.” The best correction is
ⓐ. \(G\) is universal, while \(g\) is local acceleration
ⓑ. \(G\) is local acceleration, while \(g\) is universal
ⓒ. both \(G\) and \(g\) are forces measured in \( \text{N} \)
ⓓ. both \(G\) and \(g\) are masses measured in \( \text{kg} \)
Correct Answer: \(G\) is universal, while \(g\) is local acceleration
Explanation: The symbols \(G\) and \(g\) have different meanings. The constant \(G\) appears in Newton’s universal law of gravitation and has unit \( \text{N m}^2\,\text{kg}^{-2} \). The quantity \(g\) is acceleration due to gravity at a place and has unit \( \text{m s}^{-2} \). The value of \(g\) depends on the planet and location, while \(G\) is treated as universal. Similar-looking symbols should not be treated as the same physical quantity.
17. Two bodies attract each other with gravitational force \(F\). If one mass is doubled, the other mass is tripled, and the separation is doubled, the new force is
ⓐ. \(\frac{3}{2}F\)
ⓑ. \(3F\)
ⓒ. \(6F\)
ⓓ. \(\frac{2}{3}F\)
Correct Answer: \(\frac{3}{2}F\)
Explanation: \( \textbf{Original relation:} \)
\[
F=G\frac{m_1m_2}{r^2}
\]
\( \textbf{Changed quantities:} \) one mass becomes \(2m_1\), the other mass becomes \(3m_2\), and separation becomes \(2r\).
\( \textbf{New force:} \)
\[
F' = G\frac{(2m_1)(3m_2)}{(2r)^2}
\]
\( \textbf{Simplify the numerator:} \)
\[
(2m_1)(3m_2)=6m_1m_2
\]
\( \textbf{Simplify the denominator:} \)
\[
(2r)^2=4r^2
\]
So,
\[
F'=G\frac{6m_1m_2}{4r^2}=\frac{3}{2}G\frac{m_1m_2}{r^2}
\]
Since \(G\frac{m_1m_2}{r^2}=F\), the new force is \(F'=\frac{3}{2}F\).
\( \textbf{Final answer:} \) the force becomes \(\frac{3}{2}F\), because the mass product increases by \(6\) while the squared distance increases by \(4\).
18. An object is at a height \(h\) above Earth’s surface. If Earth’s radius is \(R_E\), the centre-to-centre distance from Earth to the object is
ⓐ. \(h\)
ⓑ. \(R_E-h\)
ⓒ. \(R_E+h\)
ⓓ. \(R_E\)
Correct Answer: \(R_E+h\)
Explanation: In gravitational formulas involving Earth and an external object, distance is measured from Earth’s centre to the object. The object is already one Earth radius \(R_E\) from Earth’s centre when it is on the surface. Raising it by height \(h\) adds to that centre distance. Hence the relevant separation is \(R_E+h\), not just the height above the surface. This distinction becomes very important in altitude, orbital, and escape-speed calculations.
19. A book resting on a table experiences Earth’s gravitational pull downward and the table’s normal reaction upward. The gravitational force differs from the normal reaction because gravitational force
ⓐ. has unit \( \text{kg} \) instead of \( \text{N} \)
ⓑ. acts between the book and Earth without contact
ⓒ. acts upward when a book is kept on a table
ⓓ. requires contact between the book and Earth’s surface
Correct Answer: acts between the book and Earth without contact
Explanation: Earth pulls the book gravitationally even though the book is not in contact with Earth’s centre. Gravity is a field force, while the normal reaction is a contact force between the book and the table. The gravitational pull on the book is directed downward toward Earth’s centre. Both gravitational force and normal reaction are forces, so both are measured in \( \text{N} \). The table’s support balances the book only in this resting situation; it does not remove Earth’s gravitational pull.
20. For a body falling freely close to Earth, ignoring air resistance, the acceleration is nearly independent of
ⓐ. the direction toward Earth’s centre
ⓑ. the mass of the falling body
ⓒ. the existence of a gravitational field
ⓓ. Earth’s gravitational pull
Correct Answer: the mass of the falling body
Explanation: Near Earth, a freely falling body has acceleration approximately \(g\) when air resistance is neglected. Although the gravitational force on a larger mass is larger, the larger mass also has greater inertia. These two effects combine so that the acceleration due to gravity does not depend on the falling body’s own mass. This is why a light body and a heavy body can have the same free-fall acceleration in ideal conditions. Air resistance can change the observed motion, but it is not part of the ideal gravitational result.
