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101. What is the mathematical expression of the universal law of gravitation?
ⓐ. \(F = G \frac{m_1 + m_2}{r^2}\)
ⓑ. \(F = G \frac{m_1 m_2}{r}\)
ⓒ. \(F = G \frac{m_1 m_2}{r^2}\)
ⓓ. \(F = \frac{G}{m_1 m_2 r^2}\)
Correct Answer: \(F = G \frac{m_1 m_2}{r^2}\)
Explanation: The correct formula states that the gravitational force \(F\) is directly proportional to the product of the two masses and inversely proportional to the square of the distance between them. This inverse-square relationship is crucial for explaining stable planetary orbits and why gravity weakens with distance. The incorrect options either use addition of masses or wrong dependence on \(r\).
102. In the formula \(F = G \frac{m_1 m_2}{r^2}\), what does \(r\) represent?
ⓐ. Radius of one of the masses
ⓑ. Distance between the centers of the two masses
ⓒ. Diameter of the larger body
ⓓ. Average orbital speed of the bodies
Correct Answer: Distance between the centers of the two masses
Explanation: The separation \(r\) is always measured from center to center, treating each body as a point mass. This assumption is valid when objects are spherical and masses are uniformly distributed. Using radius or diameter instead would give incorrect force values.
103. What is the physical meaning of the constant \(G\) in the gravitational formula?
ⓐ. It sets the strength of the gravitational interaction
ⓑ. It equals acceleration due to gravity on Earth
ⓒ. It changes with location in space
ⓓ. It is dependent on the mass of objects involved
Correct Answer: It sets the strength of the gravitational interaction
Explanation: The universal gravitational constant \(G\) is the proportionality factor that determines the magnitude of the gravitational force. Its value is \(6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\), and it is universal — the same everywhere, regardless of the objects involved. It should not be confused with \(g\), which varies from place to place on Earth.
104. If the mass of one object is doubled while keeping distance constant, how does the force change?
ⓐ. It remains unchanged
ⓑ. It doubles
ⓒ. It quadruples
ⓓ. It halves
Correct Answer: It doubles
Explanation: Since gravitational force is directly proportional to the product of the two masses (\(F \propto m_1 m_2\)), doubling one mass directly doubles the force. If both masses are doubled, the force would quadruple. This proportionality shows the sensitivity of gravity to the amount of matter present.
105. If the distance between two objects becomes three times larger, the gravitational force becomes:
ⓐ. 3 times larger
ⓑ. 1/3 as large
ⓒ. 1/9 as large
ⓓ. 9 times larger
Correct Answer: 1/9 as large
Explanation: Force varies inversely with the square of the distance. If \(r\) becomes \(3r\), then \(F \propto 1/(3r)^2 = 1/9r^2\). This rapid decrease explains why gravity is strong near Earth’s surface but extremely weak between stars unless they are very massive.
106. Why do we use the term “point masses” in the derivation of the gravitational law?
ⓐ. Because only small particles follow the law
ⓑ. Because the law does not apply to real objects
ⓒ. To simplify calculations by assuming spherical symmetry
ⓓ. Because mass has no physical dimension
Correct Answer: To simplify calculations by assuming spherical symmetry
Explanation: Newton’s shell theorem shows that a spherical body attracts external objects as if all its mass were concentrated at its center. Hence, for planets and stars, we treat them as point masses when applying \(F = G \frac{m_1 m_2}{r^2}\). This greatly simplifies calculations.
107. What is the dimensional formula of the gravitational constant \(G\)?
ⓐ. \([M L T^{-2}]\)
ⓑ. \([M^{-1} L^3 T^{-2}]\)
ⓒ. \([M L^2 T^{-1}]\)
ⓓ. \([M^0 L^0 T^0]\)
Correct Answer: \([M^{-1} L^3 T^{-2}]\)
Explanation: From \(F = G \frac{m_1 m_2}{r^2}\), we rearrange to \(G = \frac{Fr^2}{m_1 m_2}\). Substituting dimensions: \(F = [M L T^{-2}], r^2 = [L^2], m_1 m_2 = [M^2]\). Hence \(G = [M^{-1} L^3 T^{-2}]\). This formula shows how \(G\) connects mass, distance, and force.
108. What is the value of gravitational force between two 1 kg masses separated by 1 meter?
ⓐ. \(6.67 \times 10^{-11} \, N\)
ⓑ. \(9.8 \, N\)
ⓒ. \(1 \, N\)
ⓓ. \(0 \, N\)
Correct Answer: \(6.67 \times 10^{-11} \, N\)
Explanation: Using \(F = G \frac{1 \times 1}{1^2}\), we get \(F = 6.67 \times 10^{-11} \, N\). This tiny value shows why gravity is negligible between small everyday objects, but significant when at least one mass is extremely large, like Earth.
109. Why is the gravitational force between two ordinary objects (like a book and a pen) negligible in daily life?
ⓐ. Because gravitational constant \(G\) is very small
ⓑ. Because they do not have mass
ⓒ. Because gravitation acts only on planets
ⓓ. Because electromagnetic force cancels it
Correct Answer: Because gravitational constant \(G\) is very small
Explanation: The tiny value of \(G\) means that unless at least one object has a huge mass (like Earth), the force is too small to notice. Between small objects, other forces like friction and electromagnetism dominate.
110. Why is the inverse-square law form \(1/r^2\) essential in the gravitational formula?
ⓐ. It makes gravity independent of distance
ⓑ. It ensures that gravity weakens with distance but never vanishes
ⓒ. It causes gravity to remain constant everywhere
ⓓ. It proves gravity is stronger than all other forces
Correct Answer: It ensures that gravity weakens with distance but never vanishes
Explanation: The \(1/r^2\) dependence means gravitational force decreases rapidly as objects separate, but it never truly becomes zero. This property allows for stable planetary orbits and explains why celestial bodies still influence each other even at great distances.
111. In the equation \(F = G \frac{m_1 m_2}{r^2}\), what does \(F\) represent?
ⓐ. Frictional force between objects
ⓑ. Gravitational force between two masses
ⓒ. Electrostatic attraction
ⓓ. Centrifugal force of rotation
Correct Answer: Gravitational force between two masses
Explanation: \(F\) denotes the mutual gravitational force of attraction between two masses \(m_1\) and \(m_2\). It is always attractive and acts along the line joining the centers of the two bodies. This force governs both falling objects on Earth and orbital motion of celestial bodies.
112. In the gravitational formula, what do \(m_1\) and \(m_2\) represent?
ⓐ. Volumes of the two objects
ⓑ. Radii of the objects
ⓒ. Masses of the two interacting bodies
ⓓ. Densities of the objects
Correct Answer: Masses of the two interacting bodies
Explanation: \(m_1\) and \(m_2\) are the masses of the two objects between which gravitational force acts. The product \(m_1 m_2\) shows that heavier objects exert stronger gravitational forces, while lighter objects experience weaker ones.
113. Why is the product \(m_1 m_2\) important in the formula?
ⓐ. Because force depends only on one mass
ⓑ. Because both masses equally contribute to gravitational attraction
ⓒ. Because it cancels out the effect of distance
ⓓ. Because it gives acceleration directly
Correct Answer: Because both masses equally contribute to gravitational attraction
Explanation: Gravitational force is mutual. The product \(m_1 m_2\) ensures that the force depends on the combined effect of both masses. Larger masses create stronger attraction, which is why Earth’s gravity is so dominant compared to small objects.
114. In the formula, what does \(r\) represent?
ⓐ. The radius of Earth
ⓑ. The distance between the centers of the two masses
ⓒ. The radius of the larger body only
ⓓ. The surface-to-surface separation between objects
Correct Answer: The distance between the centers of the two masses
Explanation: The gravitational force is calculated using center-to-center separation, not surface-to-surface distance. For spherical bodies like planets, this assumption is valid due to Newton’s shell theorem, which treats the mass as if concentrated at the center.
115. Why is \(r^2\) (square of the distance) used instead of just \(r\)?
ⓐ. To match experimental observations of force reduction with distance
ⓑ. To make calculations easier
ⓒ. Because gravity cancels out otherwise
ⓓ. To balance dimensions of the equation
Correct Answer: To match experimental observations of force reduction with distance
Explanation: Experimental data showed force decreases with the square of distance, not linearly. The inverse-square law ensures that doubling the distance reduces the force to one-fourth, which is consistent with astronomical and laboratory evidence.
116. What does the symbol \(G\) stand for in the equation?
ⓐ. Local gravity on Earth
ⓑ. Universal gravitational constant
ⓒ. Acceleration of free fall
ⓓ. Geometrical mean of masses
Correct Answer: Universal gravitational constant
Explanation: \(G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\) is a fundamental constant of nature that quantifies the strength of gravity. It is universal, the same everywhere in the universe, unlike \(g\), which changes with location on Earth.
117. Why is \(G\) called “universal”?
ⓐ. Because it only applies to celestial bodies
ⓑ. Because its value changes with altitude
ⓒ. Because it applies to all masses and distances in the universe
ⓓ. Because it applies only in vacuum
Correct Answer: Because it applies to all masses and distances in the universe
Explanation: The gravitational constant \(G\) does not depend on place, time, or the type of matter. It is valid for apples falling on Earth, satellites in orbit, and stars orbiting in galaxies, making it universal in scope.
118. What is the SI unit of \(G\)?
ⓐ. \(\text{Nm/kg}\)
ⓑ. \(\text{Nm}^2/\text{kg}^2\)
ⓒ. \(\text{kg m/s}^2\)
ⓓ. \(\text{m}^2/\text{kg}\)
Correct Answer: \(\text{Nm}^2/\text{kg}^2\)
Explanation: From the formula \(G = \frac{Fr^2}{m_1 m_2}\), substituting SI units: force \(F = \text{N}\), distance squared \(r^2 = \text{m}^2\), and mass squared \(= \text{kg}^2\). Thus the unit of \(G\) is \(\text{Nm}^2/\text{kg}^2\).
119. How is \(g\) (acceleration due to gravity) related to \(G\)?
ⓐ. \(g = \frac{G M}{R}\)
ⓑ. \(g = \frac{GM}{R^2}\)
ⓒ. \(g = \frac{G}{M R^2}\)
ⓓ. \(g = G^2 \frac{M}{R^2}\)
Correct Answer: \(g = \frac{GM}{R^2}\)
Explanation: By applying Newton’s law with Earth’s mass \(M\) and radius \(R\), the gravitational acceleration at Earth’s surface is \(g = \frac{GM}{R^2}\). This explains why \(g\) depends on Earth’s size and mass, unlike \(G\) which is universal.
120. Why is it important to understand each term in the gravitational equation?
ⓐ. To memorize formulas only
ⓑ. To apply the law correctly to real-life and astronomical problems
ⓒ. To make calculations harder
ⓓ. To cancel Newton’s laws
Correct Answer: To apply the law correctly to real-life and astronomical problems
Explanation: Each term—force, masses, distance, and constant—has a precise meaning. Misinterpreting them leads to wrong results. A clear understanding allows applications ranging from calculating satellite orbits to estimating masses of planets and stars.
121. Which of the following is a direct example of universal gravitation in everyday life?
ⓐ. A ball falling to the ground when dropped
ⓑ. A magnet attracting iron
ⓒ. An electric current flowing through a wire
ⓓ. Water boiling at 100°C
Correct Answer: A ball falling to the ground when dropped
Explanation: The ball is attracted towards Earth due to gravitational force. This everyday experience is a direct application of Newton’s law of gravitation, where the Earth’s huge mass dominates the interaction.
122. Which natural phenomenon is primarily explained by the gravitational pull of the Moon?
ⓐ. Seasons on Earth
ⓑ. Tides in oceans
ⓒ. Rotation of Earth
ⓓ. Earthquakes
Correct Answer: Tides in oceans
Explanation: The gravitational pull of the Moon and, to a lesser extent, the Sun causes the periodic rise and fall of ocean levels. This is a large-scale application of universal gravitation observable on Earth’s surface.
123. Why does the Earth revolve around the Sun instead of flying away into space?
ⓐ. Because of atmospheric pressure
ⓑ. Because of Sun’s gravitational pull
ⓒ. Because of magnetic fields of planets
ⓓ. Because of Earth’s rotation
Correct Answer: Because of Sun’s gravitational pull
Explanation: The Sun exerts a gravitational force on Earth that acts as a centripetal force, keeping Earth bound in an elliptical orbit. Without this force, Earth would move in a straight line into space.
124. Which force provides the centripetal force for satellites orbiting Earth?
ⓐ. Magnetic force
ⓑ. Electrostatic force
ⓒ. Gravitational force
ⓓ. Normal force
Correct Answer: Gravitational force
Explanation: Satellites remain in orbit because Earth’s gravitational force acts as the necessary centripetal force. This application of universal gravitation enables modern technologies such as communication and GPS.
125. The weight of an object on the surface of Earth is an application of which law?
ⓐ. Newton’s First Law of Motion
ⓑ. Universal Law of Gravitation
ⓒ. Hooke’s Law
ⓓ. Archimedes’ Principle
Correct Answer: Universal Law of Gravitation
Explanation: Weight is defined as \(W = mg\), which comes from \(F = G \frac{Mm}{R^2}\). Thus, weight is a practical example of universal gravitation, depending on Earth’s mass \(M\) and radius \(R\).
126. Gravitational force is responsible for which of the following cosmic structures?
ⓐ. Formation of galaxies
ⓑ. Stability of star systems
ⓒ. Orbital motion of planets
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Gravitation governs large-scale structures of the universe. It holds galaxies together, stabilizes solar systems, and keeps planets in orbit around stars. Without it, cosmic order would not exist.
127. How did Newton use universal gravitation to explain Kepler’s laws?
ⓐ. By deriving orbital shapes and periods from inverse-square law
ⓑ. By rejecting Kepler’s observations
ⓒ. By assuming circular orbits only
ⓓ. By applying Hooke’s law
Correct Answer: By deriving orbital shapes and periods from inverse-square law
Explanation: Newton mathematically proved that an inverse-square gravitational force leads to elliptical orbits, conservation of areal velocity, and the relation \(T^2 \propto a^3\), thereby explaining Kepler’s laws.
128. Why do objects inside the International Space Station appear weightless?
ⓐ. Because there is no gravity in space
ⓑ. Because the ISS is too far from Earth’s gravitational pull
ⓒ. Because ISS and astronauts are in free fall under Earth’s gravity
ⓓ. Because of centrifugal force only
Correct Answer: Because ISS and astronauts are in free fall under Earth’s gravity
Explanation: Although gravity still acts strongly at ISS height, both the station and astronauts fall together with the same acceleration. This free-fall condition makes them appear weightless.
129. Which application of universal gravitation allows us to calculate the mass of distant planets?
ⓐ. By measuring their brightness
ⓑ. By observing the orbits of their moons or satellites
ⓒ. By measuring surface temperature
ⓓ. By calculating their density directly
Correct Answer: By observing the orbits of their moons or satellites
Explanation: Using \(T^2 = \frac{4\pi^2 r^3}{GM}\), the mass \(M\) of a planet can be calculated from the orbital radius \(r\) and period \(T\) of its moons. This method is widely used in astronomy.
130. Which of the following space technologies directly depends on calculations from universal gravitation?
ⓐ. Communication satellites
ⓑ. GPS navigation systems
ⓒ. Weather forecasting satellites
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Every artificial satellite relies on precise gravitational calculations for orbital speed and stability. Applications like GPS, communication, and weather monitoring would not function without applying Newton’s law of gravitation.
131. What is the gravitational constant \(G\) commonly known as?
ⓐ. Acceleration due to gravity
ⓑ. Newton’s constant of gravitation
ⓒ. Kepler’s constant
ⓓ. Earth’s gravitational field strength
Correct Answer: Newton’s constant of gravitation
Explanation: The gravitational constant \(G\) is also called Newton’s constant of gravitation. It is a universal proportionality constant that relates gravitational force to the product of two masses and the inverse square of the distance between them.
132. What is the approximate numerical value of \(G\) in SI units?
ⓐ. \(9.8 \, \text{m/s}^2\)
ⓑ. \(6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\)
ⓒ. \(3 \times 10^8 \, \text{m/s}\)
ⓓ. \(1.6 \times 10^{-19} \, \text{C}\)
Correct Answer: \(6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\)
Explanation: The small value of \(G\) explains why gravitational forces between everyday objects are negligible compared to other fundamental forces. It becomes significant when at least one mass is very large, like a planet or a star.
133. Who first measured the value of the gravitational constant \(G\)?
ⓐ. Isaac Newton
ⓑ. Galileo Galilei
ⓒ. Henry Cavendish
ⓓ. Johannes Kepler
Correct Answer: Henry Cavendish
Explanation: In 1798, Henry Cavendish used a torsion balance experiment to measure \(G\). This experiment is historically described as “weighing the Earth,” since knowing \(G\) allowed the calculation of Earth’s mass.
134. Why is \(G\) called a universal constant?
ⓐ. Because it changes with location on Earth
ⓑ. Because it applies equally everywhere in the universe
ⓒ. Because it varies with mass of the bodies
ⓓ. Because it only applies to planetary motion
Correct Answer: Because it applies equally everywhere in the universe
Explanation: Unlike \(g\), which varies with altitude and location on Earth, \(G\) has the same value everywhere. It is independent of mass, distance, medium, or time, making it truly universal.
135. Which of the following best differentiates \(G\) from \(g\)?
ⓐ. \(G\) is constant, while \(g\) changes with location
ⓑ. \(G\) applies only to small bodies, \(g\) to large ones
ⓒ. \(G\) is larger in magnitude than \(g\)
ⓓ. \(G\) is variable, while \(g\) is universal
Correct Answer: \(G\) is constant, while \(g\) changes with location
Explanation: \(G\) is universal and constant. \(g\), the acceleration due to gravity, depends on the mass and radius of a planet. For example, \(g\) is \(9.8 \, \text{m/s}^2\) on Earth but much smaller on the Moon.
136. What is the dimensional formula of \(G\)?
ⓐ. \([M L T^{-2}]\)
ⓑ. \([M^{-1} L^3 T^{-2}]\)
ⓒ. \([M L^2 T^{-1}]\)
ⓓ. \([M^0 L^0 T^0]\)
Correct Answer: \([M^{-1} L^3 T^{-2}]\)
Explanation: From \(F = G \frac{m_1 m_2}{r^2}\), rearranging gives \(G = \frac{Fr^2}{m_1 m_2}\). Substituting the dimensional formulas yields \([M^{-1} L^3 T^{-2}]\). This shows how \(G\) relates mass, length, and time units.
137. Why is the value of \(G\) so difficult to measure accurately?
ⓐ. Because gravitational force is extremely weak compared to other forces
ⓑ. Because mass cannot be measured
ⓒ. Because \(G\) changes over time
ⓓ. Because distance cannot be measured
Correct Answer: Because gravitational force is extremely weak compared to other forces
Explanation: Gravity between laboratory-sized masses is incredibly small, making precise experiments difficult. Even minor disturbances like vibrations or air currents can affect the results, which is why Cavendish’s experiment was so remarkable.
138. Which of the following scientific advancements became possible after knowing the value of \(G\)?
ⓐ. Determination of Earth’s mass and density
ⓑ. Calculation of orbital velocity of satellites
ⓒ. Estimation of masses of planets and stars
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Once \(G\) was known, scientists could apply Newton’s law of gravitation to calculate the mass of Earth, planets, stars, and even galaxies. It also made satellite mechanics and space science possible.
139. What happens to the gravitational force between two bodies if the value of \(G\) were larger than its actual value?
ⓐ. Force would become weaker
ⓑ. Force would remain the same
ⓒ. Force would become stronger
ⓓ. Force would disappear
Correct Answer: Force would become stronger
Explanation: Since \(F \propto G\), an increase in \(G\) would make gravitational force stronger. If \(G\) were much larger, planets and stars might collapse more easily, altering the structure of the universe.
140. What is the fundamental significance of the gravitational constant in physics?
ⓐ. It explains atomic bonding
ⓑ. It sets the strength of the gravitational interaction in the universe
ⓒ. It determines the speed of light
ⓓ. It defines the unit of charge
Correct Answer: It sets the strength of the gravitational interaction in the universe
Explanation: \(G\) determines how strongly masses attract each other. It is fundamental to our understanding of planetary motion, satellite technology, astrophysics, and cosmology. Without \(G\), gravitational calculations would not be possible.
141. Who first experimentally determined the value of the gravitational constant \(G\)?
ⓐ. Isaac Newton
ⓑ. Galileo Galilei
ⓒ. Henry Cavendish
ⓓ. Johannes Kepler
Correct Answer: Henry Cavendish
Explanation: In 1798, Henry Cavendish used a torsion balance apparatus to measure \(G\). Though Newton proposed the law of gravitation, he did not know the actual value of \(G\). Cavendish’s experiment is often called “weighing the Earth” because it allowed the calculation of Earth’s mass and density.
142. Which apparatus did Cavendish use to measure \(G\)?
ⓐ. Pendulum
ⓑ. Spring balance
ⓒ. Torsion balance
ⓓ. Hydrometer
Correct Answer: Torsion balance
Explanation: Cavendish used a torsion balance consisting of a horizontal rod suspended by a thin wire with small lead spheres attached at each end. Larger lead spheres placed nearby exerted gravitational attraction, twisting the wire. The angle of twist provided a measure of the force and thus \(G\).
143. What physical principle allowed Cavendish to measure such a tiny force in his experiment?
ⓐ. Conservation of momentum
ⓑ. Torque on a suspended wire due to gravitational pull
ⓒ. Electromagnetic induction
ⓓ. Law of inertia
Correct Answer: Torque on a suspended wire due to gravitational pull
Explanation: The gravitational attraction between large and small lead spheres produced a small torque on the torsion wire. Measuring the equilibrium twist of the wire allowed Cavendish to determine the tiny gravitational force and hence \(G\).
144. Why was Cavendish’s experiment considered groundbreaking?
ⓐ. It proved Kepler’s laws
ⓑ. It directly measured the Earth’s radius
ⓒ. It allowed the calculation of Earth’s mass and density
ⓓ. It disproved Newton’s laws
Correct Answer: It allowed the calculation of Earth’s mass and density
Explanation: Once \(G\) was known, scientists could apply \(g = \frac{GM}{R^2}\) to determine Earth’s mass \(M\). Cavendish’s measurement made it possible to “weigh the Earth” for the first time in history.
145. Which of the following was the biggest challenge in Cavendish’s experiment?
ⓐ. Measuring the masses of the spheres
ⓑ. Controlling temperature fluctuations
ⓒ. Detecting the extremely small gravitational force between laboratory masses
ⓓ. Measuring the length of the wire
Correct Answer: Detecting the extremely small gravitational force between laboratory masses
Explanation: The gravitational force between small objects is on the order of \(10^{-7} \, N\) or less, which is extremely tiny. Cavendish had to carefully eliminate disturbances like air currents and vibrations to detect such small forces.
146. What is the order of magnitude of the gravitational force Cavendish measured between the lead spheres?
ⓐ. \(10^3 \, N\)
ⓑ. \(10^{-1} \, N\)
ⓒ. \(10^{-7} \, N\)
ⓓ. \(10^{-15} \, N\)
Correct Answer: \(10^{-7} \, N\)
Explanation: The gravitational attraction between the lead spheres in Cavendish’s torsion balance was extremely weak, of the order \(10^{-7} \, N\), much smaller than typical everyday forces.
147. How did Cavendish calculate the value of \(G\) from his experiment?
ⓐ. By comparing gravitational force with electric force
ⓑ. By measuring the torque and angular displacement of the torsion wire
ⓒ. By measuring acceleration due to gravity directly
ⓓ. By using planetary orbits
Correct Answer: By measuring the torque and angular displacement of the torsion wire
Explanation: Cavendish observed the equilibrium twist of the wire caused by the attraction between spheres. Knowing the restoring torque of the wire, the masses, and their separation, he calculated \(G\) from the balance condition.
148. Which modern techniques are used today to improve the measurement of \(G\)?
ⓐ. Laser interferometry and vacuum chambers
ⓑ. Radioactive decay experiments
ⓒ. Nuclear magnetic resonance
ⓓ. Satellite imaging
Correct Answer: Laser interferometry and vacuum chambers
Explanation: Modern experiments use highly sensitive torsion balances, laser interferometers to detect minuscule displacements, and vacuum chambers to eliminate air resistance, providing more accurate values of \(G\).
149. Why is \(G\) considered the least precisely known of the fundamental physical constants?
ⓐ. Because it changes with altitude
ⓑ. Because gravitational force is extremely weak and hard to measure
ⓒ. Because it depends on electromagnetic effects
ⓓ. Because it varies with time of day
Correct Answer: Because gravitational force is extremely weak and hard to measure
Explanation: Gravitational forces between laboratory masses are tiny compared to other forces, making \(G\) very hard to measure accurately. Despite centuries of effort, its uncertainty is still larger than other constants like \(c\) or \(h\).
150. What was the historical significance of Cavendish’s experiment beyond finding \(G\)?
ⓐ. It confirmed Einstein’s relativity
ⓑ. It measured the charge of the electron
ⓒ. It gave the first accurate value of Earth’s mean density
ⓓ. It discovered radioactivity
Correct Answer: It gave the first accurate value of Earth’s mean density
Explanation: Using the measured \(G\) and known value of \(g\), Cavendish calculated Earth’s mass and density. This was the first time humanity obtained a reliable estimate of the Earth’s total mass, marking a milestone in geophysics.
151. What is the SI unit of the gravitational constant \(G\)?
ⓐ. \(\text{Nm/kg}\)
ⓑ. \(\text{Nm}^2/\text{kg}^2\)
ⓒ. \(\text{m/s}^2\)
ⓓ. \(\text{kg m/s}^2\)
Correct Answer: \(\text{Nm}^2/\text{kg}^2\)
Explanation: From \(F = G \frac{m_1 m_2}{r^2}\), rearranging gives \(G = \frac{Fr^2}{m_1 m_2}\). Substituting SI units: force (N), distance squared (\(m^2\)), and mass squared (\(kg^2\)), the unit becomes \(\text{Nm}^2/\text{kg}^2\).
152. What is the dimensional formula of \(G\)?
ⓐ. \([M L T^{-2}]\)
ⓑ. \([M^{-1} L^3 T^{-2}]\)
ⓒ. \([M L^2 T^{-1}]\)
ⓓ. \([M^0 L^0 T^0]\)
Correct Answer: \([M^{-1} L^3 T^{-2}]\)
Explanation: Starting from \(F = G \frac{m_1 m_2}{r^2}\), we get \(G = \frac{Fr^2}{m_1 m_2}\). With \(F = [M L T^{-2}], r^2 = [L^2], m^2 = [M^2]\), the result is \([M^{-1} L^3 T^{-2}]\).
153. In CGS units, what is the unit of \(G\)?
ⓐ. \(\text{dyne cm}^2/\text{g}^2\)
ⓑ. \(\text{dyne cm}/\text{g}\)
ⓒ. \(\text{cm/s}^2\)
ⓓ. \(\text{erg/cm}\)
Correct Answer: \(\text{dyne cm}^2/\text{g}^2\)
Explanation: In CGS, force is measured in dyne, distance in cm, and mass in g. Substituting into the formula gives \(G\) the unit \(\text{dyne cm}^2/\text{g}^2\).
154. Which of the following best explains the negative power of \(M\) in the dimensional formula of \(G\)?
ⓐ. Because gravitational force does not depend on mass
ⓑ. Because \(G\) decreases when masses increase
ⓒ. Because mass appears in the denominator when solving for \(G\)
ⓓ. Because mass has no physical dimension
Correct Answer: Because mass appears in the denominator when solving for \(G\)
Explanation: Since \(G = \frac{Fr^2}{m_1 m_2}\), the two masses are in the denominator. This gives \(M^{-1}\) in the dimensional formula.
155. Which of the following dimensional checks confirms the correctness of Newton’s gravitational formula?
ⓐ. \([M^1 L^1 T^{-2}]\)
ⓑ. \([M^0 L^0 T^0]\)
ⓒ. \([M^1 L^2 T^{-2}]\)
ⓓ. Both sides reduce to dimensions of force \([M L T^{-2}]\)
Correct Answer: Both sides reduce to dimensions of force \([M L T^{-2}]\)
Explanation: Substituting dimensions of \(G, m_1, m_2, r\) into the formula gives \([M L T^{-2}]\), which matches the dimensional formula of force. This verifies the equation is dimensionally consistent.
156. Why are units and dimensions of \(G\) important in physics?
ⓐ. They help to prove Einstein’s relativity
ⓑ. They allow consistency checks and conversion between systems of units
ⓒ. They determine the value of \(g\) at Earth’s surface
ⓓ. They measure density of planets
Correct Answer: They allow consistency checks and conversion between systems of units
Explanation: Units and dimensions provide a way to confirm the correctness of physical formulas and ensure equations are balanced. They also help in converting values of \(G\) between SI, CGS, and other unit systems.
157. If force is expressed in dyne, mass in grams, and distance in centimeters, the numerical value of \(G\) will be:
ⓐ. Larger than in SI units
ⓑ. Smaller than in SI units
ⓒ. Exactly the same as in SI units
ⓓ. Zero
Correct Answer: Larger than in SI units
Explanation: In CGS units, \(G = 6.67 \times 10^{-8} \, \text{dyne cm}^2/\text{g}^2\), which appears numerically larger than the SI value because of the smaller unit system. The physical constant itself is the same, only the numerical value changes with units.
158. What is the dimensionless nature of the gravitational constant \(G\)?
ⓐ. It is dimensionless because it has no physical meaning
ⓑ. It is not dimensionless; it carries dimensions linking mass, length, and time
ⓒ. It is dimensionless in vacuum only
ⓓ. It is dimensionless in CGS system
Correct Answer: It is not dimensionless; it carries dimensions linking mass, length, and time
Explanation: \(G\) has dimensions \([M^{-1} L^3 T^{-2}]\). This means it connects quantities of different dimensions (mass, length, time), showing it is not dimensionless.
159. Which of the following constants has units most similar to \(G\)?
ⓐ. Planck’s constant \(h\)
ⓑ. Permittivity of free space \(\epsilon_0\)
ⓒ. Speed of light \(c\)
ⓓ. Boltzmann’s constant \(k\)
Correct Answer: Permittivity of free space \(\epsilon_0\)
Explanation: Both \(G\) and \(\epsilon_0\) act as proportionality constants in inverse-square force laws (gravitation and electrostatics, respectively). Their roles are analogous, though their units differ.
160. How does dimensional analysis of \(G\) help in deriving related physical equations?
ⓐ. By showing which variables are unnecessary
ⓑ. By ensuring derived formulas are dimensionally consistent
ⓒ. By changing values of constants
ⓓ. By proving Newton’s laws wrong
Correct Answer: By ensuring derived formulas are dimensionally consistent
Explanation: Dimensional analysis confirms that derived relationships, such as \(g = \frac{GM}{R^2}\) or \(T^2 = \frac{4\pi^2}{GM} a^3\), are dimensionally correct. This prevents mistakes in physics derivations and maintains logical consistency.
161. What role does \(G\) play in the equation \(F = G \frac{m_1 m_2}{r^2}\)?
ⓐ. It changes the direction of force
ⓑ. It acts as the proportionality constant that scales the gravitational force
ⓒ. It determines the orbital eccentricity of planets
ⓓ. It reduces the effect of distance
Correct Answer: It acts as the proportionality constant that scales the gravitational force
Explanation: Without \(G\), the equation would only show proportionality. \(G\) provides the correct scaling factor, linking the masses and distance to produce the actual magnitude of gravitational force measurable in experiments.
162. How is the value of \(G\) used in calculating Earth’s mass?
ⓐ. By measuring Earth’s circumference
ⓑ. By combining \(G\) with known values of \(g\) and \(R\)
ⓒ. By directly weighing Earth on a balance
ⓓ. By comparing Earth’s orbit with other planets
Correct Answer: By combining \(G\) with known values of \(g\) and \(R\)
Explanation: Using \(g = \frac{GM}{R^2}\), Earth’s mass \(M = \frac{gR^2}{G}\). Once \(G\) was known, this formula gave the first accurate value of Earth’s mass, showing the essential role of \(G\) in planetary calculations.
163. In orbital mechanics, which formula shows the direct use of \(G\)?
ⓐ. \(T^2 = \frac{4\pi^2}{GM}a^3\)
ⓑ. \(v = u + at\)
ⓒ. \(F = ma\)
ⓓ. \(E = mc^2\)
Correct Answer: \(T^2 = \frac{4\pi^2}{GM}a^3\)
Explanation: This is Newton’s version of Kepler’s Third Law. \(G\) appears explicitly in calculating orbital periods of planets and satellites. Without \(G\), such astronomical and spaceflight calculations would not be possible.
164. Why is \(G\) essential in determining the motion of artificial satellites?
ⓐ. It helps measure air resistance
ⓑ. It determines the centripetal force required for stable orbits
ⓒ. It increases the satellite’s weight
ⓓ. It prevents satellites from falling back
Correct Answer: It determines the centripetal force required for stable orbits
Explanation: Earth’s gravitational pull provides the centripetal force. By applying \(F = G \frac{Mm}{r^2}\), the orbital velocity and period can be calculated, ensuring correct satellite placement and functioning.
165. Which formula uses \(G\) to calculate the acceleration due to gravity on the surface of a planet?
ⓐ. \(g = \frac{GM}{R^2}\)
ⓑ. \(g = GMR\)
ⓒ. \(g = \frac{R^2}{GM}\)
ⓓ. \(g = \frac{G}{MR^2}\)
Correct Answer: \(g = \frac{GM}{R^2}\)
Explanation: Here \(M\) is the planet’s mass and \(R\) its radius. This relation shows how \(G\) links large-scale planetary properties with the local gravitational field strength experienced at the surface.
166. How is \(G\) applied in astrophysics to estimate the mass of stars?
ⓐ. By measuring their color and brightness
ⓑ. By analyzing orbits of planets or binary stars around them
ⓒ. By measuring their temperature
ⓓ. By using Doppler effect only
Correct Answer: By analyzing orbits of planets or binary stars around them
Explanation: Using orbital data (distance and period) of planets or companion stars, the mass of stars is calculated from \(T^2 = \frac{4\pi^2}{GM}a^3\). Thus \(G\) plays a direct role in astrophysical mass determination.
167. Why is \(G\) considered a link between terrestrial gravity and cosmic phenomena?
ⓐ. Because it changes with location on Earth
ⓑ. Because it explains both falling objects and orbital motion
ⓒ. Because it only applies to planets
ⓓ. Because it only applies in vacuum
Correct Answer: Because it explains both falling objects and orbital motion
Explanation: The same constant \(G\) applies when calculating the weight of an apple and the orbital path of Earth around the Sun. It unifies terrestrial experiences with astronomical scales, showing the universality of gravitation.
168. Which of the following calculations would be impossible without knowing \(G\)?
ⓐ. Determining the density of liquids
ⓑ. Calculating the mass of Earth
ⓒ. Finding the velocity of sound
ⓓ. Measuring the speed of light
Correct Answer: Calculating the mass of Earth
Explanation: Earth’s mass is derived from \(M = \frac{gR^2}{G}\). Without \(G\), we could not relate the surface gravity \(g\) and Earth’s radius \(R\) to its total mass.
169. In Newton’s law, why is \(G\) called the “scaling factor”?
ⓐ. Because it cancels distance effects
ⓑ. Because it sets the actual magnitude of gravitational force
ⓒ. Because it reduces equations to unitless form
ⓓ. Because it makes gravity negligible
Correct Answer: Because it sets the actual magnitude of gravitational force
Explanation: The inverse-square law describes proportionality, but \(G\) determines the numerical strength. Its small value explains why gravity is weak between small objects but dominant at planetary scales.
170. How does \(G\) connect with Einstein’s general theory of relativity?
ⓐ. \(G\) disappears in relativity
ⓑ. \(G\) acts as the coupling constant between matter-energy and spacetime curvature
ⓒ. \(G\) determines only atomic behavior
ⓓ. \(G\) is unrelated to relativity
Correct Answer: \(G\) acts as the coupling constant between matter-energy and spacetime curvature
Explanation: In Einstein’s field equations, \(G\) links the energy-momentum tensor (matter and energy content) to spacetime curvature. Thus, \(G\) plays a crucial role not only in Newtonian gravitation but also in modern gravitational theory.
171. How is \(G\) used to calculate the orbital velocity of a satellite around Earth?
ⓐ. \(v = \sqrt{\frac{GM}{R}}\)
ⓑ. \(v = \sqrt{\frac{GM}{R+h}}\)
ⓒ. \(v = \sqrt{\frac{G}{M(R+h)}}\)
ⓓ. \(v = \frac{GM}{R^2}\)
Correct Answer: \(v = \sqrt{\frac{GM}{R+h}}\)
Explanation: The orbital velocity is derived by equating centripetal force \(\frac{mv^2}{R+h}\) to gravitational force \(\frac{GMm}{(R+h)^2}\). Solving gives \(v = \sqrt{\frac{GM}{R+h}}\). Here \(G\) links Earth’s mass to the satellite’s orbital speed.
172. What is the expression for escape velocity of a body from Earth using \(G\)?
ⓐ. \(v_e = \sqrt{\frac{GM}{R}}\)
ⓑ. \(v_e = \sqrt{\frac{2GM}{R}}\)
ⓒ. \(v_e = \frac{GM}{R^2}\)
ⓓ. \(v_e = \sqrt{GR^2M}\)
Correct Answer: \(v_e = \sqrt{\frac{2GM}{R}}\)
Explanation: Escape velocity is obtained by equating kinetic energy \(\frac{1}{2}mv^2\) with gravitational potential energy \(\frac{GMm}{R}\). Solving gives \(v_e = \sqrt{\frac{2GM}{R}}\). \(G\) determines how strong the binding energy is for Earth.
173. How is the gravitational potential energy between two masses expressed using \(G\)?
ⓐ. \(U = -\frac{GM}{r}\)
ⓑ. \(U = -\frac{Gm_1 m_2}{r}\)
ⓒ. \(U = \frac{GMm}{r^2}\)
ⓓ. \(U = -G m_1 m_2 r^2\)
Correct Answer: \(U = -\frac{Gm_1 m_2}{r}\)
Explanation: The potential energy associated with two masses separated by distance \(r\) is \(U = -\frac{Gm_1 m_2}{r}\). The negative sign indicates that the gravitational system is bound, and energy is required to separate the masses to infinity.
174. Which formula shows how \(G\) determines the gravitational field strength at the surface of a planet?
ⓐ. \(g = \frac{GM}{R^2}\)
ⓑ. \(g = \frac{G}{MR^2}\)
ⓒ. \(g = \sqrt{GM R^2}\)
ⓓ. \(g = \frac{GR}{M^2}\)
Correct Answer: \(g = \frac{GM}{R^2}\)
Explanation: Gravitational field strength (acceleration due to gravity) depends on planet’s mass \(M\), its radius \(R\), and the universal constant \(G\). This connects surface gravity directly with cosmic parameters.
175. How is \(G\) applied in calculating the time period of a satellite’s orbit?
ⓐ. \(T = 2\pi \sqrt{\frac{R}{g}}\)
ⓑ. \(T = 2\pi \sqrt{\frac{(R+h)^3}{GM}}\)
ⓒ. \(T = 2\pi \sqrt{\frac{GM}{(R+h)^3}}\)
ⓓ. \(T = \frac{GM}{R+h}\)
Correct Answer: \(T = 2\pi \sqrt{\frac{(R+h)^3}{GM}}\)
Explanation: From Newton’s version of Kepler’s Third Law, \(T^2 = \frac{4\pi^2 (R+h)^3}{GM}\). Taking the square root gives the orbital period. \(G\) again provides the essential scaling factor.
176. Which equation involving \(G\) is used to calculate the mass of Earth?
ⓐ. \(M = \frac{g}{GR}\)
ⓑ. \(M = \frac{gR^2}{G}\)
ⓒ. \(M = \frac{GR}{g}\)
ⓓ. \(M = \frac{GR^2}{g}\)
Correct Answer: \(M = \frac{gR^2}{G}\)
Explanation: Using \(g = \frac{GM}{R^2}\), we rearrange to get \(M = \frac{gR^2}{G}\). This equation shows how \(G\) allows the determination of Earth’s total mass from measurable values of \(g\) and \(R\).
177. What is the formula for the orbital period of a planet around the Sun using \(G\)?
ⓐ. \(T = \sqrt{\frac{2GM}{a}}\)
ⓑ. \(T^2 = \frac{4\pi^2 a^3}{GM}\)
ⓒ. \(T = \frac{GM}{a^2}\)
ⓓ. \(T^2 = \frac{GM}{a^3}\)
Correct Answer: \(T^2 = \frac{4\pi^2 a^3}{GM}\)
Explanation: This is Newton’s derivation of Kepler’s Third Law. \(T\) is the orbital period, \(a\) the semi-major axis, \(M\) the mass of the Sun, and \(G\) the gravitational constant. It is a cornerstone equation in celestial mechanics.
178. How is \(G\) involved in calculating the gravitational potential at a point?
ⓐ. \(V = \frac{GM}{r}\)
ⓑ. \(V = -\frac{GM}{r}\)
ⓒ. \(V = -\frac{G}{Mr}\)
ⓓ. \(V = \frac{GM}{r^2}\)
Correct Answer: \(V = -\frac{GM}{r}\)
Explanation: The potential at a distance \(r\) from a mass \(M\) is given by \(V = -\frac{GM}{r}\). The negative sign shows work must be done against gravity to bring a unit mass from infinity to that point.
179. Which equation shows how \(G\) is used to calculate the escape velocity of a satellite from Mars (mass \(M\), radius \(R\))?
ⓐ. \(v_e = \sqrt{\frac{2GM}{R}}\)
ⓑ. \(v_e = \sqrt{\frac{GM}{R^2}}\)
ⓒ. \(v_e = \frac{GM}{R}\)
ⓓ. \(v_e = \frac{G}{MR}\)
Correct Answer: \(v_e = \sqrt{\frac{2GM}{R}}\)
Explanation: The escape velocity depends only on \(G\), the planet’s mass, and its radius. For Mars, inserting values of \(M\) and \(R\) with \(G\) gives the escape velocity required to overcome its gravitational pull.
180. Why is \(G\) considered fundamental in all gravitational calculations?
ⓐ. Because it is used only for Earth-based experiments
ⓑ. Because it provides the link between measurable quantities (force, mass, distance) and cosmic phenomena (planetary motion, satellite orbits)
ⓒ. Because it determines the value of \(c\) (speed of light)
ⓓ. Because it eliminates the effect of distance
Correct Answer: Because it provides the link between measurable quantities (force, mass, distance) and cosmic phenomena (planetary motion, satellite orbits)
Explanation: The gravitational constant \(G\) connects terrestrial experiments with celestial observations. It allows us to move from simple weight measurements on Earth to predicting orbits of satellites, planets, and galaxies, making it one of the most important constants in physics.
181. What is the definition of acceleration due to gravity (\(g\)) at the surface of Earth?
ⓐ. The acceleration of an object due to air resistance
ⓑ. The acceleration produced when Earth rotates
ⓒ. The acceleration experienced by a body freely falling under Earth’s gravitational force
ⓓ. The acceleration caused only by centrifugal force
Correct Answer: The acceleration experienced by a body freely falling under Earth’s gravitational force
Explanation: Acceleration due to gravity (\(g\)) is the acceleration imparted to objects due to Earth’s gravitational pull. Near the Earth’s surface, its average value is \(9.8 \, \text{m/s}^2\).
182. Which equation relates \(g\) to the universal gravitational constant \(G\)?
ⓐ. \(g = \frac{GM}{R^2}\)
ⓑ. \(g = \frac{GR^2}{M}\)
ⓒ. \(g = GM R^2\)
ⓓ. \(g = \sqrt{\frac{GM}{R}}\)
Correct Answer: \(g = \frac{GM}{R^2}\)
Explanation: By applying Newton’s law of gravitation with Earth’s mass \(M\) and radius \(R\), the acceleration due to gravity at Earth’s surface is given by \(g = \frac{GM}{R^2}\).
183. What is the SI unit of \(g\)?
ⓐ. \(\text{m/s}\)
ⓑ. \(\text{m/s}^2\)
ⓒ. \(\text{N/kg}\)
ⓓ. Both B and C
Correct Answer: Both B and C
Explanation: Since \(g\) is an acceleration, its SI unit is \(\text{m/s}^2\). But because weight is \(W = mg\), the unit can also be written as \(\text{N/kg}\), showing the equivalence of acceleration and field strength.
184. What is the average value of \(g\) on Earth’s surface?
ⓐ. \(10.8 \, \text{m/s}^2\)
ⓑ. \(8.9 \, \text{m/s}^2\)
ⓒ. \(9.8 \, \text{m/s}^2\)
ⓓ. \(7.8 \, \text{m/s}^2\)
Correct Answer: \(9.8 \, \text{m/s}^2\)
Explanation: The standard value of \(g\) at Earth’s surface is \(9.8 \, \text{m/s}^2\). This value is an average, as actual \(g\) varies slightly with altitude, latitude, and local geology.
185. Which of the following is the correct dimensional formula for \(g\)?
ⓐ. \([M L^0 T^{-1}]\)
ⓑ. \([M^0 L T^{-2}]\)
ⓒ. \([M^0 L^0 T^{-2}]\)
ⓓ. \([M L T^{-2}]\)
Correct Answer: \([M^0 L T^{-2}]\)
Explanation: Since \(g\) is acceleration, it is derived from velocity over time: \(L/T^2\). It has no dependence on mass, so the dimensional formula is \([M^0 L T^{-2}]\).
186. Why is \(g\) called “acceleration” and not “force”?
ⓐ. Because it depends on Earth’s shape
ⓑ. Because it is independent of the mass of the falling body
ⓒ. Because it depends on the gravitational constant
ⓓ. Because it measures direction of fall only
Correct Answer: Because it is independent of the mass of the falling body
Explanation: According to Newton’s laws, all bodies fall with the same acceleration in the absence of resistance. This universality means \(g\) is an acceleration, not a force, though weight (force) is derived from it.
187. If the radius of Earth increases while its mass remains the same, what happens to \(g\)?
ⓐ. It increases
ⓑ. It decreases
ⓒ. It remains unchanged
ⓓ. It becomes infinite
Correct Answer: It decreases
Explanation: From \(g = \frac{GM}{R^2}\), if \(R\) increases while \(M\) is constant, \(g\) decreases since it is inversely proportional to the square of the radius.
188. At Earth’s center, what would be the value of \(g\)?
ⓐ. \(g = 9.8 \, \text{m/s}^2\)
ⓑ. \(g = \infty\)
ⓒ. \(g = 0\)
ⓓ. \(g = 4.9 \, \text{m/s}^2\)
Correct Answer: \(g = 0\)
Explanation: At Earth’s center, the gravitational pulls from all sides cancel out, resulting in zero net force. Thus, acceleration due to gravity is zero at the exact center of Earth.
189. Why do all bodies fall at the same rate in the absence of air resistance?
ⓐ. Because Earth pulls harder on heavier bodies
ⓑ. Because \(g\) is independent of mass
ⓒ. Because weight equals mass
ⓓ. Because distance does not matter
Correct Answer: Because \(g\) is independent of mass
Explanation: From \(F = ma\) and \(F = G \frac{Mm}{R^2}\), the \(m\) cancels out, leaving \(a = g = \frac{GM}{R^2}\). Hence, free-fall acceleration is the same for all masses.
190. Why is the definition of \(g\) significant in gravitational studies?
ⓐ. It explains electromagnetic effects
ⓑ. It helps measure universal gravitational constant
ⓒ. It connects planetary properties with local gravity and free-fall motion
ⓓ. It defines weight as mass
Correct Answer: It connects planetary properties with local gravity and free-fall motion
Explanation: The formula \(g = \frac{GM}{R^2}\) links Earth’s mass and radius (cosmic properties) to everyday free-fall acceleration. This makes \(g\) the bridge between universal gravitation and observable motion on Earth.
191. Which formula gives the value of acceleration due to gravity at a height \(h\) above Earth’s surface?
ⓐ. \(g_h = g \left( \frac{R}{R+h} \right)^2\)
ⓑ. \(g_h = g \left( \frac{R+h}{R} \right)^2\)
ⓒ. \(g_h = g \frac{R}{R-h}\)
ⓓ. \(g_h = g \left( \frac{R}{R-h} \right)^2\)
Correct Answer: \(g_h = g \left( \frac{R}{R+h} \right)^2\)
Explanation: At a height \(h\), the distance from Earth’s center is \(R+h\). Substituting into \(g = \frac{GM}{r^2}\) gives \(g_h = g \left( \frac{R}{R+h} \right)^2\). This shows that \(g\) decreases with increasing altitude.
192. If altitude \(h \ll R\), what approximate formula can be used for \(g\)?
ⓐ. \(g_h \approx g \left(1 – \frac{2h}{R}\right)\)
ⓑ. \(g_h \approx g \left(1 + \frac{h}{R}\right)\)
ⓒ. \(g_h \approx g \left(1 – \frac{h}{R}\right)^2\)
ⓓ. \(g_h \approx g \left(\frac{h}{R}\right)^2\)
Correct Answer: \(g_h \approx g \left(1 – \frac{2h}{R}\right)\)
Explanation: For small heights compared to Earth’s radius, the binomial expansion of \((R/(R+h))^2\) gives \(g_h \approx g \left(1 – \frac{2h}{R}\right)\). This linear approximation simplifies calculations for small altitudes.
193. By what percentage does \(g\) decrease at an altitude of 1000 km above Earth’s surface? (Take \(R = 6400 \, \text{km}\)).
ⓐ. \~10%
ⓑ. \~20%
ⓒ. \~25%
ⓓ. \~30%
Correct Answer: \~25%
Explanation: \(g_h = g \left( \frac{R}{R+h} \right)^2 = g \left( \frac{6400}{7400} \right)^2 \approx 0.75 g\). This is about 75% of surface gravity, so \(g\) decreases by \~25%, not 10%. Wait carefully: \((6400/7400)^2 \approx 0.75\). Hence decrease is \~25%. Correct option is C. At 1000 km altitude, \(g_h \approx 0.75 g\). Therefore, \(g\) decreases by about 25% compared to the surface value.
194. Which formula gives the value of \(g\) at a depth \(d\) below Earth’s surface (assuming uniform density)?
ⓐ. \(g_d = g \left(1 – \frac{d}{R}\right)\)
ⓑ. \(g_d = g \left(1 – \frac{d}{R}\right)^2\)
ⓒ. \(g_d = g \left(\frac{d}{R}\right)\)
ⓓ. \(g_d = g \frac{R}{R-d}\)
Correct Answer: \(g_d = g \left(1 – \frac{d}{R}\right)\)
Explanation: At depth \(d\), the effective mass contributing to gravity is reduced. The result from Newton’s shell theorem is \(g_d = g \left(1 – \frac{d}{R}\right)\). Thus, \(g\) decreases linearly with depth and becomes zero at Earth’s center.
195. At a depth equal to half of Earth’s radius, the value of \(g\) is:
ⓐ. \(g/2\)
ⓑ. \(g/4\)
ⓒ. Zero
ⓓ. \(2g\)
Correct Answer: \(g/2\)
Explanation: Using \(g_d = g \left(1 – \frac{d}{R}\right)\), for \(d = R/2\), we get \(g_d = g \left(1 – \frac{1}{2}\right) = g/2\). Gravity reduces linearly with depth inside Earth.
196. At Earth’s center, what is the value of \(g\)?
ⓐ. Maximum
ⓑ. Zero
ⓒ. Half of surface value
ⓓ. Twice surface value
Correct Answer: Zero
Explanation: At the center, all gravitational forces from different layers cancel out due to symmetry. Therefore, the net acceleration due to gravity at Earth’s center is zero.
197. Which physical principle is used to derive the variation of \(g\) with depth?
ⓐ. Archimedes’ principle
ⓑ. Newton’s shell theorem
ⓒ. Pascal’s law
ⓓ. Law of inertia
Correct Answer: Newton’s shell theorem
Explanation: Newton’s shell theorem states that only the mass enclosed within radius \(r\) contributes to gravity at that point, while outer shells exert no net force. This explains why \(g\) decreases linearly with depth.
198. Why does \(g\) decrease with altitude?
ⓐ. Because Earth’s mass decreases with height
ⓑ. Because the distance from Earth’s center increases
ⓒ. Because Earth rotates faster at high altitudes
ⓓ. Because gravitational constant decreases
Correct Answer: Because the distance from Earth’s center increases
Explanation: As altitude increases, the distance between the object and Earth’s center grows. Since \(g \propto 1/r^2\), this increase in \(r\) causes a decrease in gravitational acceleration.
199. Why does \(g\) decrease linearly with depth inside Earth?
ⓐ. Because density increases with depth
ⓑ. Because effective mass contributing decreases proportionally with radius
ⓒ. Because Earth’s shape is not spherical
ⓓ. Because \(G\) decreases underground
Correct Answer: Because effective mass contributing decreases proportionally with radius
Explanation: At depth \(d\), only the mass enclosed within \(R-d\) contributes. This effective mass decreases linearly, so \(g\) also decreases linearly until it becomes zero at the center.
200. At what approximate altitude does \(g\) become half of its surface value? (Take \(R = 6400 \, \text{km}\)).
ⓐ. 6400 km
ⓑ. 2600 km
ⓒ. 3600 km
ⓓ. 12800 km
Correct Answer: 2600 km
Explanation: From \(g_h = g \left( \frac{R}{R+h} \right)^2\), setting \(g_h = g/2\) gives \((R/(R+h))^2 = 1/2\). Solving: \(R+h = R\sqrt{2} \Rightarrow h = R(\sqrt{2}-1)\). Substituting \(R=6400 \, km\) gives \(h \approx 2600 \, km\).
The chapter Gravitation in Class 11 Physics (NCERT/CBSE syllabus) is a fundamental topic that explains concepts of gravitational force, motion of planets, escape velocity, and geostationary satellites. These topics are frequently asked in board exams and are extremely important for competitive exams like JEE, NEET, and state entrance tests. To simplify preparation, we provide a total of 580 MCQs with solutions, divided into 6 systematic parts. This section (Part 2) contains the next 100 MCQs with answers, helping you master concepts step by step.
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