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201. Which of the following factors does NOT affect the value of \(g\) on Earth?
ⓐ. Altitude
ⓑ. Depth
ⓒ. Latitude
ⓓ. Mass of the falling object
Correct Answer: Mass of the falling object
Explanation: The value of \(g\) is independent of the falling body’s mass. It only depends on Earth’s mass, radius, rotation, and the body’s position (altitude, depth, latitude). This is why all objects fall with the same acceleration in vacuum.
202. Why does \(g\) vary with latitude?
ⓐ. Because Earth is spherical and perfectly uniform
ⓑ. Because Earth is slightly flattened at the poles and rotates about its axis
ⓒ. Because gravity weakens near mountains
ⓓ. Because gravitational constant \(G\) changes with location
Correct Answer: Because Earth is slightly flattened at the poles and rotates about its axis
Explanation: Due to Earth’s rotation, centrifugal force reduces effective gravity at the equator. Also, Earth’s polar flattening means the radius is smaller at the poles, increasing \(g\) there.
203. At which location is \(g\) maximum?
ⓐ. Equator
ⓑ. Poles
ⓒ. Tropic of Cancer
ⓓ. Equator and poles equally
Correct Answer: Poles
Explanation: At the poles, centrifugal force due to Earth’s rotation is zero, and the radius is smallest. Both effects make \(g\) larger at the poles than at the equator.
204. At which location is \(g\) minimum?
ⓐ. Poles
ⓑ. Equator
ⓒ. Tropic of Capricorn
ⓓ. Mid-latitudes
Correct Answer: Equator
Explanation: At the equator, Earth’s radius is largest and centrifugal force is maximum. Both factors reduce the effective gravitational acceleration, making \(g\) minimum at the equator.
205. How does Earth’s rotation affect the value of \(g\)?
ⓐ. It increases \(g\) everywhere
ⓑ. It decreases \(g\) due to centrifugal force, most strongly at the equator
ⓒ. It increases \(g\) near the equator and decreases at poles
ⓓ. It has no effect on \(g\)
Correct Answer: It decreases \(g\) due to centrifugal force, most strongly at the equator
Explanation: The outward centrifugal force reduces effective gravity. At the poles, its effect is zero, but at the equator it is maximum, reducing \(g\) by about \(0.034 \, \text{m/s}^2\).
206. Which equation shows the effective value of \(g\) at latitude \(\phi\) considering Earth’s rotation?
ⓐ. \(g’ = g – R\omega^2\)
ⓑ. \(g’ = g – R\omega^2 \cos^2\phi\)
ⓒ. \(g’ = g – R\omega^2 \sin^2\phi\)
ⓓ. \(g’ = g + R\omega^2 \sin\phi\)
Correct Answer: \(g’ = g – R\omega^2 \cos^2\phi\)
Explanation: The centrifugal acceleration component along the radial direction is \(R\omega^2 \cos^2\phi\). Subtracting this from \(g\) gives the effective gravity at latitude \(\phi\).
207. Which geological factor can cause local variations in \(g\)?
ⓐ. Distribution of mountains and valleys
ⓑ. Presence of mineral deposits and density variations
ⓒ. Tectonic plate differences
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Local variations in Earth’s density (mountains, ocean trenches, mineral-rich regions) slightly alter gravitational pull. Gravimeters are used in geophysics to detect such anomalies.
208. If Earth stopped rotating, what would happen to the value of \(g\) at the equator?
ⓐ. It would decrease
ⓑ. It would increase
ⓒ. It would remain unchanged
ⓓ. It would become zero
Correct Answer: It would increase
Explanation: At the equator, centrifugal force reduces \(g\). If Earth stopped rotating, this reduction would disappear, making \(g\) larger at the equator, equal to polar gravity.
209. How does altitude affect \(g\)?
ⓐ. \(g\) increases with altitude
ⓑ. \(g\) decreases with altitude
ⓒ. \(g\) first increases then decreases
ⓓ. Altitude has no effect
Correct Answer: \(g\) decreases with altitude
Explanation: As altitude increases, distance from Earth’s center grows. Since \(g \propto 1/r^2\), the gravitational acceleration decreases with increasing altitude.
210. How does depth inside Earth affect \(g\)?
ⓐ. \(g\) increases with depth
ⓑ. \(g\) decreases linearly with depth, reaching zero at the center
ⓒ. \(g\) remains constant with depth
ⓓ. \(g\) fluctuates randomly with depth
Correct Answer: \(g\) decreases linearly with depth, reaching zero at the center
Explanation: From Newton’s shell theorem, only the inner spherical mass contributes to gravity at depth \(d\). Hence, \(g_d = g\left(1 – \frac{d}{R}\right)\), making \(g = 0\) at Earth’s center.
211. Which experiment is commonly used to measure the value of \(g\) near Earth’s surface?
ⓐ. Cavendish torsion balance experiment
ⓑ. Simple pendulum experiment
ⓒ. Photoelectric effect experiment
ⓓ. Michelson–Morley experiment
Correct Answer: Simple pendulum experiment
Explanation: The time period of a simple pendulum is given by \(T = 2\pi \sqrt{\frac{l}{g}}\). By measuring the pendulum length \(l\) and period \(T\), the value of \(g\) can be calculated as \(g = \frac{4\pi^2 l}{T^2}\).
212. In a simple pendulum experiment, if length \(l\) is doubled, how does the calculated value of \(g\) change?
ⓐ. It doubles
ⓑ. It halves
ⓒ. It remains unchanged
ⓓ. It becomes zero
Correct Answer: It remains unchanged
Explanation: The formula \(g = \frac{4\pi^2 l}{T^2}\) ensures that as \(l\) increases, \(T\) also increases proportionally. Thus, \(g\) remains constant regardless of pendulum length if measured correctly.
213. Which formula is used to calculate \(g\) using a simple pendulum of length \(l\) and time period \(T\)?
ⓐ. \(g = \frac{2\pi l}{T}\)
ⓑ. \(g = \frac{4\pi^2 l}{T^2}\)
ⓒ. \(g = \frac{T^2}{4\pi^2 l}\)
ⓓ. \(g = \frac{2l}{T^2}\)
Correct Answer: \(g = \frac{4\pi^2 l}{T^2}\)
Explanation: From the equation of oscillation of a pendulum, \(T = 2\pi \sqrt{\frac{l}{g}}\). Rearranging gives \(g = \frac{4\pi^2 l}{T^2}\).
214. How is \(g\) measured in modern laboratories with high precision?
ⓐ. Using atomic clocks and laser interferometry in free-fall experiments
ⓑ. By observing planetary motion
ⓒ. By calculating tides
ⓓ. By using barometers
Correct Answer: Using atomic clocks and laser interferometry in free-fall experiments
Explanation: Modern techniques involve measuring the time taken for a test mass to fall a known distance in vacuum using laser interferometry. These provide very precise values of \(g\).
215. Which method can be used to measure small variations of \(g\) at different locations on Earth?
ⓐ. Torsion balance
ⓑ. Gravimeter
ⓒ. Pendulum only
ⓓ. Vernier calipers
Correct Answer: Gravimeter
Explanation: Gravimeters are sensitive instruments used in geophysics to detect tiny local variations in \(g\) due to changes in density, mountains, or mineral deposits underground.
216. If the time period of a pendulum is measured as 2 seconds with a length of 1 m, what is the approximate value of \(g\)?
ⓐ. \(9.8 \, \text{m/s}^2\)
ⓑ. \(10.2 \, \text{m/s}^2\)
ⓒ. \(8.9 \, \text{m/s}^2\)
ⓓ. \(12.5 \, \text{m/s}^2\)
Correct Answer: \(9.8 \, \text{m/s}^2\)
Explanation: Using \(g = \frac{4\pi^2 l}{T^2}\), with \(l = 1 \, m, T = 2 \, s\), we get \(g = \frac{4\pi^2}{4} = \pi^2 \approx 9.87 \, m/s^2\), close to 9.8.
217. Which physical principle allows the use of free-fall experiments to calculate \(g\)?
ⓐ. Newton’s first law
ⓑ. Uniform acceleration under constant force
ⓒ. Conservation of momentum
ⓓ. Hooke’s law
Correct Answer: Uniform acceleration under constant force
Explanation: In free fall, the only force is gravity, producing constant acceleration. By measuring distance fallen and time, one can use \(s = \frac{1}{2} g t^2\) to determine \(g\).
218. What is the main limitation of using a simple pendulum for measuring \(g\)?
ⓐ. It requires very heavy masses
ⓑ. Air resistance and finite amplitude of oscillation introduce errors
ⓒ. It does not depend on length of pendulum
ⓓ. It works only at poles
Correct Answer: Air resistance and finite amplitude of oscillation introduce errors
Explanation: The pendulum method assumes small amplitude oscillations and no air resistance. In practice, damping, friction, and large amplitudes cause errors in calculating \(g\).
219. Which relation is used in a free-fall experiment to calculate \(g\) if a body falls a distance \(s\) in time \(t\)?
ⓐ. \(g = \frac{s}{t}\)
ⓑ. \(g = \frac{2s}{t^2}\)
ⓒ. \(g = \frac{t^2}{2s}\)
ⓓ. \(g = \frac{s^2}{t}\)
Correct Answer: \(g = \frac{2s}{t^2}\)
Explanation: For uniformly accelerated motion, \(s = \frac{1}{2} g t^2\). Rearranging gives \(g = \frac{2s}{t^2}\). This is a common formula in free-fall experiments to compute \(g\).
220. Why is accurate measurement of \(g\) significant in physics and engineering?
ⓐ. It only helps in pendulum clocks
ⓑ. It is necessary for structural designs, geophysics, satellite launching, and navigation
ⓒ. It helps calculate speed of light
ⓓ. It measures density of gases
Correct Answer: It is necessary for structural designs, geophysics, satellite launching, and navigation
Explanation: Accurate knowledge of \(g\) is required in civil engineering (load calculations), mining (gravity surveys), satellite orbits, and navigation systems. Hence, the measurement of \(g\) is fundamental in both science and engineering.
221. Which formula gives the acceleration due to gravity at a depth \(d\) below Earth’s surface (assuming uniform density)?
ⓐ. \(g_d = g \left( \frac{R}{R+d} \right)^2\)
ⓑ. \(g_d = g \left(1 – \frac{d}{R}\right)\)
ⓒ. \(g_d = g \left(\frac{d}{R}\right)^2\)
ⓓ. \(g_d = g \left(1 + \frac{d}{R}\right)\)
Correct Answer: \(g_d = g \left(1 – \frac{d}{R}\right)\)
Explanation: By Newton’s shell theorem, only the mass inside radius \(R-d\) contributes to gravity. Thus, \(g\) decreases linearly with depth: \(g_d = g \left(1 – \frac{d}{R}\right)\).
222. At what depth below Earth’s surface will \(g\) become half its surface value?
ⓐ. \(R/2\)
ⓑ. \(R/3\)
ⓒ. \(2R/3\)
ⓓ. \(R\)
Correct Answer: \(R/2\)
Explanation: Using \(g_d = g \left(1 – \frac{d}{R}\right)\), set \(g_d = g/2\). Then \(1 – d/R = 1/2 \Rightarrow d = R/2\).
223. What is the value of \(g\) at Earth’s center?
ⓐ. \(g\) maximum
ⓑ. \(g/2\)
ⓒ. \(g = 0\)
ⓓ. \(g = \infty\)
Correct Answer: \(g = 0\)
Explanation: At the center, gravitational pulls from all directions cancel, making net acceleration due to gravity zero.
224. Why does \(g\) decrease linearly with depth?
ⓐ. Because Earth’s density increases with depth
ⓑ. Because outer spherical shells exert no net gravitational force inside them
ⓒ. Because \(G\) changes below the surface
ⓓ. Because centrifugal force increases
Correct Answer: Because outer spherical shells exert no net gravitational force inside them
Explanation: Newton’s shell theorem proves that only the mass within radius \(r\) contributes to gravity. As depth increases, effective radius decreases, leading to a linear reduction of \(g\).
225. Which principle is applied in deriving the expression for \(g\) below Earth’s surface?
ⓐ. Bernoulli’s theorem
ⓑ. Newton’s shell theorem
ⓒ. Archimedes’ principle
ⓓ. Pascal’s law
Correct Answer: Newton’s shell theorem
Explanation: The shell theorem states that gravitational force inside a hollow spherical shell is zero. Thus, only the inner mass contributes, leading to the linear relation for \(g\) with depth.
226. At depth \(d = R\), what is the value of \(g\)?
ⓐ. \(g\)
ⓑ. \(g/2\)
ⓒ. Zero
ⓓ. Infinite
Correct Answer: Zero
Explanation: At depth equal to Earth’s radius (center), the effective mass contributing to gravity is zero, making \(g = 0\).
227. At depth \(d = R/4\), what is the value of \(g\)?
ⓐ. \(3g/4\)
ⓑ. \(g/2\)
ⓒ. \(g/4\)
ⓓ. Zero
Correct Answer: \(3g/4\)
Explanation: Using \(g_d = g \left(1 – \frac{d}{R}\right)\), substituting \(d = R/4\) gives \(g_d = g(1 – 1/4) = 3g/4\).
228. How does the variation of \(g\) above Earth’s surface differ from that below the surface?
ⓐ. Above: linear decrease; Below: inverse square decrease
ⓑ. Above: inverse square decrease; Below: linear decrease
ⓒ. Both decrease linearly
ⓓ. Both decrease inversely with square
Correct Answer: Above: inverse square decrease; Below: linear decrease
Explanation: Above the surface, \(g_h = g\left(\frac{R}{R+h}\right)^2\) decreases as inverse square of distance. Below, \(g_d = g\left(1 – \frac{d}{R}\right)\) decreases linearly.
229. If \(g = 9.8 \, m/s^2\) on the surface, what is its value at a depth of \(1600 \, km\) (Earth’s radius \(R = 6400 \, km\))?
ⓐ. \(9.8 \, m/s^2\)
ⓑ. \(7.35 \, m/s^2\)
ⓒ. \(4.9 \, m/s^2\)
ⓓ. Zero
Correct Answer: \(7.35 \, m/s^2\)
Explanation: Using \(g_d = g \left(1 – \frac{d}{R}\right)\), \(g_d = 9.8 \left(1 – \frac{1600}{6400}\right) = 9.8 \times 0.75 = 7.35 \, m/s^2\).
230. Which statement is correct about \(g\) inside and outside Earth?
ⓐ. Both increase with distance from center
ⓑ. Outside, \(g\) decreases as \(1/r^2\); inside, \(g\) decreases linearly with depth
ⓒ. Both are constant everywhere
ⓓ. Outside, \(g\) increases with height
Correct Answer: Outside, \(g\) decreases as \(1/r^2\); inside, \(g\) decreases linearly with depth
Explanation: Beyond Earth’s surface, \(g \propto 1/r^2\). Inside, effective mass decreases proportionally with radius, making \(g\) decrease linearly until it becomes zero at the center.
231. Which formula gives the value of acceleration due to gravity at a height \(h\) above Earth’s surface?
ⓐ. \(g_h = g \left(1 – \frac{h}{R}\right)\)
ⓑ. \(g_h = g \left(\frac{R}{R+h}\right)^2\)
ⓒ. \(g_h = g \left(\frac{h}{R}\right)^2\)
ⓓ. \(g_h = g \left(\frac{R+h}{R}\right)^2\)
Correct Answer: \(g_h = g \left(\frac{R}{R+h}\right)^2\)
Explanation: Above Earth’s surface, the distance from the center is \(R+h\). Since \(g \propto \frac{1}{r^2}\), the formula becomes \(g_h = g \left(\frac{R}{R+h}\right)^2\).
232. At what altitude will the value of \(g\) reduce to half its surface value?
ⓐ. \(h = R\)
ⓑ. \(h = R/2\)
ⓒ. \(h = 2R\)
ⓓ. \(h = \infty\)
Correct Answer: \(h = R\)
Explanation: For \(g_h = g/2\), we have \(\left(\frac{R}{R+h}\right)^2 = 1/2\). Solving gives \(R+h = \sqrt{2}R\), hence \(h = (\sqrt{2} – 1)R \approx R\).
233. What happens to the value of \(g\) as altitude \(h\) becomes very large compared to Earth’s radius?
ⓐ. It increases infinitely
ⓑ. It remains constant at \(9.8 \, m/s^2\)
ⓒ. It decreases towards zero
ⓓ. It oscillates
Correct Answer: It decreases towards zero
Explanation: Since \(g_h = g \left(\frac{R}{R+h}\right)^2\), as \(h \to \infty\), the denominator dominates, making \(g_h \to 0\).
234. If the value of \(g\) at Earth’s surface is \(9.8 \, m/s^2\), what will it be at a height equal to Earth’s radius (\(h = R\))?
ⓐ. \(9.8 \, m/s^2\)
ⓑ. \(4.9 \, m/s^2\)
ⓒ. \(2.45 \, m/s^2\)
ⓓ. Zero
Correct Answer: \(4.9 \, m/s^2\)
Explanation: At \(h = R\), \(g_h = g \left(\frac{R}{2R}\right)^2 = g \times \frac{1}{4} = 9.8/2 = 4.9 \, m/s^2\).
235. If altitude is very small compared to Earth’s radius (\(h \ll R\)), what approximation can be used for \(g_h\)?
ⓐ. \(g_h \approx g \left(1 – \frac{2h}{R}\right)\)
ⓑ. \(g_h \approx g \left(1 – \frac{h}{R}\right)\)
ⓒ. \(g_h \approx g \left(1 – \frac{h^2}{R^2}\right)\)
ⓓ. \(g_h \approx g\) (no change)
Correct Answer: \(g_h \approx g \left(1 – \frac{2h}{R}\right)\)
Explanation: Using binomial expansion for small \(h\), \(\left(\frac{R}{R+h}\right)^2 \approx 1 – \frac{2h}{R}\). Thus, \(g_h \approx g\left(1 – \frac{2h}{R}\right)\).
236. At what height above Earth’s surface will \(g\) become \(\frac{1}{4}\) of its surface value?
ⓐ. \(h = R\)
ⓑ. \(h = 2R\)
ⓒ. \(h = 3R\)
ⓓ. \(h = 4R\)
Correct Answer: \(h = R\)
Explanation: For \(g_h = g/4\), \(\left(\frac{R}{R+h}\right)^2 = 1/4\). This gives \(R+h = 2R\), hence \(h = R\). Actually, correcting: \(R+h = 2R \Rightarrow h = R\) gives \(g/4\). So answer is A. \(h = R\).
237. If an astronaut orbits at 400 km above Earth’s surface (Earth’s radius \(6400 \, km\)), what fraction of \(g\) does he experience?
ⓐ. \(1\) (same as surface)
ⓑ. \(0.95\)
ⓒ. \(0.90\)
ⓓ. \(0.50\)
Correct Answer: \(0.95\)
Explanation: At \(h = 400 \, km\), \(g_h = g \left(\frac{6400}{6400+400}\right)^2 = g \left(\frac{6400}{6800}\right)^2 \approx g(0.941)^2 \approx 0.95g\).
238. Why is it incorrect to say astronauts are “weightless” because there is no gravity in orbit?
ⓐ. Because \(g\) is zero at 400 km altitude
ⓑ. Because \(g\) is still about 90–95% of its surface value in orbit
ⓒ. Because mass changes in orbit
ⓓ. Because Newton’s law does not apply in space
Correct Answer: Because \(g\) is still about 90–95% of its surface value in orbit
Explanation: Astronauts experience microgravity not due to lack of \(g\), but because they are in free fall around Earth. Gravity remains nearly the same as on the surface at low Earth orbit.
239. At what approximate altitude does \(g\) reduce to 75% of its surface value?
ⓐ. \(1000 \, km\)
ⓑ. \(1600 \, km\)
ⓒ. \(3200 \, km\)
ⓓ. \(4800 \, km\)
Correct Answer: \(1600 \, km\)
Explanation: For \(g_h = 0.75g\), \(\left(\frac{R}{R+h}\right)^2 = 0.75\). Solving gives \(h \approx 1600 \, km\) for Earth’s radius \(6400 \, km\).
240. Which statement is true regarding \(g\) above Earth’s surface?
ⓐ. It decreases linearly with altitude
ⓑ. It decreases inversely with square of the distance from Earth’s center
ⓒ. It remains constant
ⓓ. It becomes negative after a certain height
Correct Answer: It decreases inversely with square of the distance from Earth’s center
Explanation: From Newton’s law, \(g_h = \frac{GM}{(R+h)^2}\). Thus, \(g\) decreases as \(1/(R+h)^2\) with altitude, unlike the linear decrease seen below Earth’s surface.
241. Which of the following practical applications depends on knowing the variation of \(g\) with altitude?
ⓐ. Designing satellite orbits
ⓑ. Constructing bridges
ⓒ. Measuring atmospheric pressure
ⓓ. Designing musical instruments
Correct Answer: Designing satellite orbits
Explanation: The orbital velocity and period of satellites depend directly on the value of \(g\) (or equivalently, gravitational force) at their altitude. Accurate knowledge of variation of \(g\) with altitude ensures stable orbits.
242. Why does a pendulum clock gain or lose time when moved from equator to poles?
ⓐ. Length of pendulum changes
ⓑ. Variation of \(g\) with latitude changes its time period
ⓒ. Earth’s magnetic field interferes
ⓓ. Air density increases at poles
Correct Answer: Variation of \(g\) with latitude changes its time period
Explanation: At poles, \(g\) is slightly larger, reducing the pendulum’s period so the clock gains time. At the equator, \(g\) is smaller, increasing the period, so the clock loses time.
243. Which application of \(g\) variation is used in mineral and oil exploration?
ⓐ. Pendulum clock corrections
ⓑ. Gravimetric surveys
ⓒ. Free-fall experiments
ⓓ. Torsion balance method
Correct Answer: Gravimetric surveys
Explanation: Geophysicists measure small variations in \(g\) at different points on Earth’s surface. Anomalies in \(g\) indicate underground structures, mineral ores, or oil reserves.
244. At high mountains, why does water boil at a lower temperature?
ⓐ. Because \(g\) is greater at higher altitudes
ⓑ. Because air pressure is lower at higher altitudes
ⓒ. Because \(g\) is smaller at higher altitudes
ⓓ. Both B and C
Correct Answer: Both B and C
Explanation: At high altitudes, atmospheric pressure decreases and \(g\) is slightly less, both contributing to water boiling at lower temperatures than at sea level.
245. Why is the knowledge of \(g\) variation with depth important for tunnel construction?
ⓐ. To calculate pressure of rocks only
ⓑ. To calculate decrease in effective weight and stresses inside Earth
ⓒ. To determine sound speed in rocks
ⓓ. To prevent earthquakes
Correct Answer: To calculate decrease in effective weight and stresses inside Earth
Explanation: As depth increases, \(g\) decreases linearly. Engineers account for this reduction when designing deep tunnels and underground structures to ensure safety and stability.
246. Which of the following uses the concept of variation of \(g\) with latitude?
ⓐ. GPS satellite navigation
ⓑ. Adjustment of weighing machines
ⓒ. Measurement of pressure in fluids
ⓓ. Solar energy calculations
Correct Answer: Adjustment of weighing machines
Explanation: Weight depends on local \(g\). Weighing machines are calibrated for specific latitudes. At equator, \(g\) is slightly less, so apparent weight reduces compared to poles.
247. Why does a satellite in low Earth orbit still experience nearly the same gravitational acceleration as on Earth’s surface?
ⓐ. Because \(g\) does not decrease with altitude
ⓑ. Because \(g\) decreases slowly for small altitudes compared to Earth’s radius
ⓒ. Because Earth’s mass increases at altitude
ⓓ. Because of centrifugal force only
Correct Answer: Because \(g\) decreases slowly for small altitudes compared to Earth’s radius
Explanation: Since \(h \ll R\), \(g_h \approx g(1 – 2h/R)\). At 400 km altitude, \(g\) is still about 90–95% of its surface value, so astronauts experience microgravity due to free fall, not absence of gravity.
248. How does variation of \(g\) affect aviation and space travel?
ⓐ. It changes fuel composition
ⓑ. It determines lift of airplanes and trajectory of rockets
ⓒ. It only affects time zones
ⓓ. It changes shape of Earth
Correct Answer: It determines lift of airplanes and trajectory of rockets
Explanation: The value of \(g\) influences effective weight, thrust requirements, and trajectory calculations. Space agencies must account for variation of \(g\) with altitude when designing launch systems.
249. Why do geologists study the variation of \(g\) across different latitudes?
ⓐ. To map Earth’s density distribution and internal structure
ⓑ. To locate volcanic eruptions
ⓒ. To measure rainfall
ⓓ. To measure Earth’s temperature
Correct Answer: To map Earth’s density distribution and internal structure
Explanation: Changes in \(g\) reveal variations in mass distribution. This data helps geologists study tectonic activity, mountains, and sub-surface density anomalies.
250. Which of the following scientific fields directly benefits from studying variation of \(g\)?
ⓐ. Astronomy and Geophysics
ⓑ. Botany and Zoology
ⓒ. Economics and History
ⓓ. Literature and Linguistics
Correct Answer: Astronomy and Geophysics
Explanation: Astronomy uses \(g\) variations for satellite orbits, while geophysics applies them in mineral exploration, seismic studies, and understanding Earth’s interior. Thus, variation of \(g\) has critical applications in both fields.
251. What is the expression for gravitational potential energy of a body of mass \(m\) at height \(h\) near Earth’s surface?
ⓐ. \(U = \frac{GMm}{R}\)
ⓑ. \(U = mgh\)
ⓒ. \(U = \frac{1}{2}mv^2\)
ⓓ. \(U = \frac{GMm}{h}\)
Correct Answer: \(U = mgh\)
Explanation: Near Earth’s surface, gravitational force is nearly constant, so work done in raising a body of mass \(m\) to height \(h\) is \(U = mgh\). Other options represent kinetic or incorrect forms of potential energy.
252. Which of the following best defines gravitational potential energy (GPE)?
ⓐ. Energy a body possesses due to its speed
ⓑ. Energy a body possesses due to its position in a gravitational field
ⓒ. Energy released only during nuclear reactions
ⓓ. Energy stored in bonds of molecules
Correct Answer: Energy a body possesses due to its position in a gravitational field
Explanation: Gravitational potential energy is the energy associated with a body’s position relative to a source of gravity, typically Earth or another massive body.
253. What is the SI unit of gravitational potential energy?
ⓐ. Newton (N)
ⓑ. Joule (J)
ⓒ. Watt (W)
ⓓ. Pascal (Pa)
Correct Answer: Joule (J)
Explanation: Gravitational potential energy is a form of mechanical energy, measured in joules (J), where \(1 \, J = 1 \, N \cdot m = 1 \, kg \, m^2/s^2\).
254. When a body is lifted against gravity, its gravitational potential energy:
ⓐ. Decreases
ⓑ. Increases
ⓒ. Remains constant
ⓓ. Becomes zero
Correct Answer: Increases
Explanation: Work done against gravity is stored as potential energy. The higher the body is lifted, the greater its gravitational potential energy.
255. If gravitational potential energy is considered zero at Earth’s surface, then at a height \(h\), its value is:
ⓐ. Negative
ⓑ. Positive
ⓒ. Zero
ⓓ. Infinite
Correct Answer: Positive
Explanation: By convention, if \(U = 0\) at Earth’s surface, then lifting the object increases energy, so \(U = mgh\) is positive.
256. What happens to gravitational potential energy of an object when it falls freely under gravity?
ⓐ. It remains constant
ⓑ. It decreases while kinetic energy increases
ⓒ. It increases while kinetic energy decreases
ⓓ. It becomes zero immediately
Correct Answer: It decreases while kinetic energy increases
Explanation: As the object falls, height decreases, so GPE decreases. That lost energy is converted into kinetic energy, keeping total mechanical energy conserved.
257. What is the general formula of gravitational potential energy between two masses \(M\) and \(m\) separated by distance \(r\)?
ⓐ. \(U = \frac{GMm}{r}\)
ⓑ. \(U = -\frac{GMm}{r}\)
ⓒ. \(U = GMmr\)
ⓓ. \(U = \frac{GM}{r^2}\)
Correct Answer: \(U = -\frac{GMm}{r}\)
Explanation: By convention, gravitational potential energy is negative because work must be done to separate two masses against gravitational attraction, with zero at infinite separation.
258. Why is gravitational potential energy taken as negative for two masses at a finite separation?
ⓐ. To indicate instability
ⓑ. Because gravity is repulsive
ⓒ. Because infinite work is required to bring them from infinity to a finite distance
ⓓ. Because gravitational force is attractive, requiring work against it to separate masses
Correct Answer: Because gravitational force is attractive, requiring work against it to separate masses
Explanation: By convention, potential energy is zero at infinity. Bringing masses closer requires work against attraction, so the energy is negative at finite separation.
259. If the gravitational potential energy of a system is \(-200 \, J\), what does it indicate?
ⓐ. The system has no energy
ⓑ. The masses are infinitely apart
ⓒ. The system is bound by gravitational attraction
ⓓ. The masses repel each other
Correct Answer: The system is bound by gravitational attraction
Explanation: Negative potential energy indicates a bound system, meaning the bodies are held together by gravity. Work must be done to separate them to infinity.
260. Which situation has higher gravitational potential energy?
ⓐ. A stone lying on the ground
ⓑ. A stone held at a height of 10 m
ⓒ. Both have equal energy
ⓓ. Cannot be determined
Correct Answer: A stone held at a height of 10 m
Explanation: Potential energy is proportional to height. A stone elevated at 10 m has more stored gravitational energy than one lying on the ground (assuming ground as reference \(U=0\)).
261. What is the correct formula for gravitational potential energy between two point masses \(m_1\) and \(m_2\) separated by distance \(r\)?
ⓐ. \(U = \frac{Gm_1m_2}{r}\)
ⓑ. \(U = -\frac{Gm_1m_2}{r}\)
ⓒ. \(U = \frac{Gm_1m_2}{r^2}\)
ⓓ. \(U = -\frac{Gm_1m_2}{r^2}\)
Correct Answer: \(U = -\frac{Gm_1m_2}{r}\)
Explanation: Gravitational potential energy is defined relative to infinity. At finite separation, it is negative because gravity is attractive. The correct expression is \(U = -\frac{Gm_1m_2}{r}\).
262. Why is gravitational potential energy negative in the formula \(U = -\frac{Gm_1m_2}{r}\)?
ⓐ. Because masses repel each other
ⓑ. Because potential energy at infinity is taken as zero
ⓒ. Because energy is imaginary
ⓓ. Because \(G\) is negative
Correct Answer: Because potential energy at infinity is taken as zero
Explanation: By convention, \(U = 0\) at \(r \to \infty\). Bringing masses closer requires negative work, making potential energy negative at finite distances.
263. If two bodies each of mass \(10 \, kg\) are separated by \(1 \, m\), what is their gravitational potential energy?
ⓐ. \(-6.67 \times 10^{-9} \, J\)
ⓑ. \(-6.67 \times 10^{-10} \, J\)
ⓒ. \(-6.67 \times 10^{-8} \, J\)
ⓓ. \(-6.67 \times 10^{-11} \, J\)
Correct Answer: \(-6.67 \times 10^{-9} \, J\)
Explanation: Using \(U = -\frac{Gm_1m_2}{r}\), with \(G = 6.67 \times 10^{-11}, m_1 = m_2 = 10, r = 1\), we get \(U = -6.67 \times 10^{-9} \, J\).
264. If the distance between two masses is doubled, how does gravitational potential energy change?
ⓐ. It becomes twice
ⓑ. It becomes half (less negative)
ⓒ. It remains same
ⓓ. It becomes zero
Correct Answer: It becomes half (less negative)
Explanation: Since \(U \propto -1/r\), doubling \(r\) makes the energy half of its original value (closer to zero).
265. For two bodies of mass \(m_1\) and \(m_2\), if their separation is reduced to half, gravitational potential energy becomes:
ⓐ. Half of original value
ⓑ. Twice of original value (more negative)
ⓒ. Same as before
ⓓ. Zero
Correct Answer: Twice of original value (more negative)
Explanation: Since \(U \propto -1/r\), halving \(r\) doubles the magnitude of negative potential energy, making the system more strongly bound.
266. What is the gravitational potential energy of a \(1000 \, kg\) satellite at a distance of \(7000 \, km\) from Earth’s center (Earth’s mass \(M = 6 \times 10^{24} \, kg\))? (\(G = 6.67 \times 10^{-11}\))
ⓐ. \(-5.7 \times 10^{10} \, J\)
ⓑ. \(-9.5 \times 10^{10} \, J\)
ⓒ. \(-6.67 \times 10^{7} \, J\)
ⓓ. \(-3.0 \times 10^{11} \, J\)
Correct Answer: \(-5.7 \times 10^{10} \, J\)
Explanation: \(U = -\frac{GMm}{r} = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{7 \times 10^6} \approx -5.7 \times 10^{10} \, J\).
267. When separation \(r \to \infty\), the gravitational potential energy is:
ⓐ. Infinite
ⓑ. Zero
ⓒ. Negative
ⓓ. Positive
Correct Answer: Zero
Explanation: At infinite separation, gravitational attraction vanishes. By convention, potential energy is taken as zero at infinity.
268. What does a more negative gravitational potential energy indicate about the system of two masses?
ⓐ. They are less bound
ⓑ. They are strongly bound
ⓒ. They repel each other
ⓓ. They are free of gravitational influence
Correct Answer: They are strongly bound
Explanation: The greater the magnitude of negative potential energy, the stronger the gravitational bond, requiring more work to separate the masses to infinity.
269. If gravitational potential energy of Earth-satellite system is \(-2 \times 10^{10} \, J\), how much energy is required to move the satellite to infinity?
ⓐ. \(2 \times 10^{10} \, J\)
ⓑ. \(1 \times 10^{10} \, J\)
ⓒ. Zero
ⓓ. Infinite
Correct Answer: \(2 \times 10^{10} \, J\)
Explanation: To take the system from \(U = -2 \times 10^{10} \, J\) to \(U = 0\), energy equal in magnitude to potential energy is required, i.e., \(2 \times 10^{10} \, J\).
270. Why is the expression \(U = -\frac{Gm_1m_2}{r}\) more general than \(U = mgh\)?
ⓐ. Because it applies only on Earth’s surface
ⓑ. Because it is valid everywhere, regardless of distance between masses
ⓒ. Because it is easier to calculate
ⓓ. Because it ignores \(G\)
Correct Answer: Because it is valid everywhere, regardless of distance between masses
Explanation: \(U = mgh\) is an approximation near Earth’s surface where \(g\) is nearly constant. The universal formula \(U = -\frac{Gm_1m_2}{r}\) is valid for any separation in space.
271. What is the relationship between work done by gravity and change in gravitational potential energy?
ⓐ. Work done = Change in kinetic energy
ⓑ. Work done = Negative change in potential energy
ⓒ. Work done = Potential energy itself
ⓓ. Work done = Zero always
Correct Answer: Work done = Negative change in potential energy
Explanation: Work-energy theorem states that work done by conservative forces (like gravity) equals the negative of the change in potential energy: \(W = -\Delta U\).
272. If a body of mass \(m\) is lifted to a height \(h\), how much work is done against gravity?
ⓐ. \(W = mgh\)
ⓑ. \(W = -mgh\)
ⓒ. \(W = 0\)
ⓓ. \(W = \frac{1}{2} mgh\)
Correct Answer: \(W = mgh\)
Explanation: Work done against gravity is stored as potential energy. Since \(U = mgh\), the work done in lifting is equal to the gain in gravitational potential energy.
273. Which statement correctly describes the work done by gravity when an object falls freely?
ⓐ. Work done is positive and equal to decrease in potential energy
ⓑ. Work done is negative and equal to increase in potential energy
ⓒ. Work done is zero
ⓓ. Work done equals kinetic energy only when height is zero
Correct Answer: Work done is positive and equal to decrease in potential energy
Explanation: When an object falls, gravitational force acts in the direction of motion. Hence, gravity does positive work equal to the loss of potential energy.
274. When an object is moved upward by external force at constant velocity, the work done by gravity is:
ⓐ. Positive
ⓑ. Negative
ⓒ. Zero
ⓓ. Infinite
Correct Answer: Negative
Explanation: During upward motion, displacement is opposite to gravitational force. Thus, work done by gravity is negative, equal to \(-mgh\).
275. How is total mechanical energy (kinetic + potential) affected when only gravitational force acts on a body?
ⓐ. It increases
ⓑ. It decreases
ⓒ. It remains constant
ⓓ. It becomes zero
Correct Answer: It remains constant
Explanation: Gravity is a conservative force, so total mechanical energy is conserved. Work done by gravity only converts potential energy into kinetic energy or vice versa.
276. If work done by gravity on a body is zero, what can be inferred about displacement?
ⓐ. Displacement was perpendicular to force
ⓑ. Displacement was zero
ⓒ. The object was lifted up
ⓓ. The object was dropped down
Correct Answer: Displacement was perpendicular to force
Explanation: If motion is perpendicular to gravitational force (e.g., horizontal displacement), then no work is done by gravity since \(W = Fd \cos 90^\circ = 0\).
277. Which formula directly relates work done by gravitational force to potential energy?
ⓐ. \(W = U\)
ⓑ. \(W = \Delta U\)
ⓒ. \(W = -\Delta U\)
ⓓ. \(W = 2U\)
Correct Answer: \(W = -\Delta U\)
Explanation: The definition of potential energy comes from this relation: work done by gravity is the negative of change in potential energy.
278. If a ball of mass \(2 \, kg\) falls from a height of \(5 \, m\), how much work is done by gravity? (\(g = 10 \, m/s^2\))
ⓐ. 50 J
ⓑ. 100 J
ⓒ. 25 J
ⓓ. 200 J
Correct Answer: 100 J
Explanation: Work done by gravity = loss in potential energy = \(mgh = 2 \times 10 \times 5 = 100 \, J\).
279. Why is gravitational potential energy defined as the negative of work done by gravity?
ⓐ. To make energy always positive
ⓑ. To maintain consistency with conservation of energy
ⓒ. Because gravity is a repulsive force
ⓓ. Because infinity has negative energy
Correct Answer: To maintain consistency with conservation of energy
Explanation: By defining potential energy as \(U = -W\), conservation of total energy is ensured: as gravity does positive work, potential energy decreases accordingly.
280. When an object is brought closer to Earth from infinity, the work done by gravity is:
ⓐ. Positive, equal to decrease in potential energy
ⓑ. Negative, equal to increase in potential energy
ⓒ. Zero always
ⓓ. Independent of distance
Correct Answer: Positive, equal to decrease in potential energy
Explanation: Gravity pulls the object closer, doing positive work. Meanwhile, the system’s potential energy decreases (becomes more negative). Thus, \(W = -\Delta U\).
281. Which force provides the necessary centripetal force for a satellite to remain in circular orbit around Earth?
ⓐ. Normal force
ⓑ. Gravitational force
ⓒ. Magnetic force
ⓓ. Centrifugal force
Correct Answer: Gravitational force
Explanation: The gravitational force between Earth and the satellite acts as the centripetal force, keeping the satellite in orbit. No other external force is required in ideal conditions.
282. What is the expression for orbital velocity of a satellite near Earth’s surface?
ⓐ. \(v = \sqrt{gR}\)
ⓑ. \(v = \sqrt{\frac{GM}{R}}\)
ⓒ. \(v = \sqrt{\frac{GM}{R+h}}\)
ⓓ. Both B and C depending on altitude
Correct Answer: Both B and C depending on altitude
Explanation: For satellites at Earth’s surface approximation: \(v = \sqrt{\frac{GM}{R}}\). At altitude \(h\), it becomes \(v = \sqrt{\frac{GM}{R+h}}\).
283. Why do satellites not fall back to Earth despite the gravitational pull?
ⓐ. They are weightless
ⓑ. Their tangential velocity balances gravitational pull
ⓒ. Gravity is absent in space
ⓓ. Centrifugal force cancels gravity completely
Correct Answer: Their tangential velocity balances gravitational pull
Explanation: Satellites are in continuous free fall towards Earth, but their high tangential velocity keeps them moving in orbit, preventing them from crashing.
284. What is the expression for total energy of a satellite in circular orbit of radius \(r\)?
ⓐ. \(E = -\frac{GMm}{2r}\)
ⓑ. \(E = -\frac{GMm}{r}\)
ⓒ. \(E = \frac{GMm}{r}\)
ⓓ. \(E = 0\)
Correct Answer: \(E = -\frac{GMm}{2r}\)
Explanation: Total energy of a satellite is the sum of kinetic and potential energies: \(E = KE + PE = \frac{GMm}{2r} – \frac{GMm}{r} = -\frac{GMm}{2r}\).
285. Which of the following correctly represents escape velocity from Earth’s surface?
ⓐ. \(v_e = \sqrt{\frac{2GM}{R}}\)
ⓑ. \(v_e = \sqrt{\frac{GM}{R}}\)
ⓒ. \(v_e = \sqrt{gR}\)
ⓓ. Both A and C are correct
Correct Answer: Both A and C are correct
Explanation: Escape velocity is \(v_e = \sqrt{\frac{2GM}{R}}\). Since \(g = \frac{GM}{R^2}\), it can also be written as \(v_e = \sqrt{2gR}\).
286. Why is the energy of a satellite always negative?
ⓐ. Because it has no kinetic energy
ⓑ. Because gravitational potential energy dominates
ⓒ. Because it is bound to Earth by gravitational attraction
ⓓ. Because velocity is imaginary
Correct Answer: Because it is bound to Earth by gravitational attraction
Explanation: A negative total energy indicates a bound system. To escape Earth, the satellite must gain enough energy to make its total energy zero or positive.
287. If the orbital radius of a satellite is increased, its orbital speed:
ⓐ. Increases
ⓑ. Decreases
ⓒ. Remains constant
ⓓ. Becomes infinite
Correct Answer: Decreases
Explanation: Orbital velocity is \(v = \sqrt{\frac{GM}{r}}\). As \(r\) increases, velocity decreases because gravity weakens with distance.
288. What is the significance of geostationary orbit?
ⓐ. Satellite revolves around Earth in 12 hours
ⓑ. Satellite remains fixed above one point on equator
ⓒ. Satellite travels at escape velocity
ⓓ. Satellite experiences no gravity
Correct Answer: Satellite remains fixed above one point on equator
Explanation: A geostationary satellite has a time period equal to Earth’s rotation (24 hours), so it appears stationary relative to Earth, making it useful for communication and broadcasting.
289. Why are low Earth orbit satellites preferred for remote sensing?
ⓐ. They move slower
ⓑ. They move faster and provide high resolution images
ⓒ. They require no fuel
ⓓ. They do not experience gravity
Correct Answer: They move faster and provide high resolution images
Explanation: Low Earth orbit (LEO) satellites are closer to Earth, so they move faster and provide better resolution images, making them ideal for remote sensing and mapping.
290. Which condition must be satisfied for a satellite to escape Earth’s gravitational field completely?
ⓐ. Kinetic energy = Potential energy
ⓑ. Kinetic energy > Potential energy
ⓒ. Total energy ≥ 0
ⓓ. Momentum = 0
Correct Answer: Total energy ≥ 0
Explanation: For escape, the satellite’s total energy (kinetic + potential) must reach zero or positive. This ensures it can move to infinity without being pulled back by gravity.
291. What is the definition of escape speed from a planet’s surface?
ⓐ. The minimum speed required to move in a circular orbit
ⓑ. The speed at which a body becomes weightless
ⓒ. The minimum speed required to escape from the gravitational field without further propulsion
ⓓ. The maximum speed in free fall motion
Correct Answer: The minimum speed required to escape from the gravitational field without further propulsion
Explanation: Escape speed is the least speed a body must have to reach infinity, overcoming gravitational attraction, without any additional energy supply.
292. Why is escape speed independent of the mass of the escaping object?
ⓐ. Because gravity does not act on mass
ⓑ. Because gravitational and inertial mass cancel out in the formula
ⓒ. Because mass of object is negligible compared to Earth
ⓓ. Because escape speed is purely a constant for all objects
Correct Answer: Because gravitational and inertial mass cancel out in the formula
Explanation: Escape speed depends only on planet’s mass \(M\) and radius \(R\): \(v_e = \sqrt{\frac{2GM}{R}}\). Mass of the escaping body cancels out, making it independent of the object.
293. Which equation correctly gives escape speed from the surface of a planet?
ⓐ. \(v_e = \sqrt{\frac{GM}{R}}\)
ⓑ. \(v_e = \sqrt{\frac{2GM}{R}}\)
ⓒ. \(v_e = \sqrt{gR}\)
ⓓ. Both B and C are correct
Correct Answer: Both B and C are correct
Explanation: Escape speed formula is \(v_e = \sqrt{\frac{2GM}{R}}\). Since \(g = \frac{GM}{R^2}\), substituting gives \(v_e = \sqrt{2gR}\).
294. What is the approximate escape speed from Earth’s surface?
ⓐ. \(7.9 \, km/s\)
ⓑ. \(9.8 \, km/s\)
ⓒ. \(11.2 \, km/s\)
ⓓ. \(15 \, km/s\)
Correct Answer: \(11.2 \, km/s\)
Explanation: For Earth, \(v_e = \sqrt{\frac{2GM}{R}} \approx 11.2 \, km/s\). This is higher than orbital speed (7.9 km/s).
295. Escape speed of an object on Earth does not depend on:
ⓐ. Radius of Earth
ⓑ. Mass of Earth
ⓒ. Mass of object
ⓓ. Value of \(G\)
Correct Answer: Mass of object
Explanation: Escape speed depends on planet’s mass \(M\) and radius \(R\), not the mass of the body. Hence, all objects require the same escape speed.
296. If Earth’s radius were reduced to half while keeping its mass constant, escape speed would:
ⓐ. Double
ⓑ. Become half
ⓒ. Remain unchanged
ⓓ. Increase by factor \(\sqrt{2}\)
Correct Answer: Double
Explanation: \(v_e = \sqrt{\frac{2GM}{R}}\). If \(R\) becomes half, denominator decreases, making escape velocity increase by \(\sqrt{2/0.5} = 2\) times.
297. On which two factors does escape speed of a planet primarily depend?
ⓐ. Mass of object and height
ⓑ. Mass of planet and radius of planet
ⓒ. Velocity of rotation of planet and altitude
ⓓ. Gravitational constant and orbital velocity
Correct Answer: Mass of planet and radius of planet
Explanation: Escape speed is determined by gravitational field strength, hence depends only on mass \(M\) of the planet and its radius \(R\).
298. Why is escape speed on the Moon lower than on Earth?
ⓐ. Moon has smaller radius and lower density
ⓑ. Moon has smaller mass compared to Earth
ⓒ. Moon has no atmosphere
ⓓ. Moon rotates faster than Earth
Correct Answer: Moon has smaller mass compared to Earth
Explanation: Escape speed depends on \(\sqrt{\frac{M}{R}}\). The Moon’s much smaller mass results in lower escape speed (\~2.38 km/s).
299. If escape speed from Earth is \(11.2 \, km/s\), what will be the escape speed from a planet of same radius but 9 times Earth’s mass?
ⓐ. \(11.2 \, km/s\)
ⓑ. \(22.4 \, km/s\)
ⓒ. \(33.6 \, km/s\)
ⓓ. \(100.8 \, km/s\)
Correct Answer: \(33.6 \, km/s\)
Explanation: \(v_e \propto \sqrt{M}\). If mass is 9 times, \(v_e\) increases by factor of 3: \(11.2 \times 3 = 33.6 \, km/s\).
300. Escape speed is derived using which principle?
ⓐ. Conservation of linear momentum
ⓑ. Conservation of angular momentum
ⓒ. Conservation of energy
ⓓ. Newton’s first law of motion
Correct Answer: Conservation of energy
Explanation: Escape speed is obtained by equating kinetic energy \(\frac{1}{2}mv^2\) with gravitational potential energy \(\frac{GMm}{R}\). This ensures the object just reaches infinity with zero velocity.
In Class 11 Physics, the chapter Gravitation plays a central role in mechanics and is widely covered in NCERT/CBSE syllabus as well as in competitive exams like JEE, NEET, and state-level entrance tests. This chapter covers essential concepts such as orbital velocity, satellite motion, weightlessness, and variation of g with height and depth. Across all 6 parts, we provide a total of 580 MCQs with answers. In this section (Part 3), you will practice another 100 multiple-choice questions with detailed solutions to test your conceptual clarity and problem-solving skills.
- 👉 Total MCQs in this chapter: 580.
- 👉 This page contains: Third set of 100 solved MCQs with answers.
- 👉 Important for board exams and competitive tests (JEE/NEET).
- 👉 To explore more chapters and subjects, click the top navigation bar.
- 👉 To move ahead, click the Part 4 button above.