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501. What is the definition of gravitational potential at a point?
ⓐ. Work done in moving a unit charge from infinity to that point
ⓑ. Work done in moving a unit mass from infinity to that point
ⓒ. Force per unit mass at that point
ⓓ. Energy possessed by a moving body
Correct Answer: Work done in moving a unit mass from infinity to that point
Explanation: Gravitational potential is defined as the work done by an external force in bringing a unit mass from infinity to that point without acceleration.
502. What is the SI unit of gravitational potential?
ⓐ. \(\text{N/kg}\)
ⓑ. \(\text{J/kg}\)
ⓒ. \(\text{J}\)
ⓓ. \(\text{kg/J}\)
Correct Answer: \(\text{J/kg}\)
Explanation: Potential = Work done / mass. Since work is in Joules (J) and mass is in kilograms (kg), its unit is \(\text{J/kg}\), equivalent to \(\text{m}^2/\text{s}^2\).
503. What is the gravitational potential at a distance \(r\) from a point mass \(M\)?
ⓐ. \(V = \frac{GM}{r^2}\)
ⓑ. \(V = -\frac{GM}{r}\)
ⓒ. \(V = \frac{GM}{r}\)
ⓓ. \(V = -\frac{GM}{r^2}\)
Correct Answer: \(V = -\frac{GM}{r}\)
Explanation: The potential due to a point mass is \(V = -\frac{GM}{r}\). The negative sign indicates that work must be done against gravity to move a mass from infinity to that point.
504. Why is gravitational potential always negative?
ⓐ. Because gravitational force is imaginary
ⓑ. Because gravity is a conservative force directed inward
ⓒ. Because potential is measured relative to infinity where it is taken as zero
ⓓ. Because mass is always positive
Correct Answer: Because potential is measured relative to infinity where it is taken as zero
Explanation: At infinity, \(V = 0\). To bring a mass closer, work is done against attraction, making the potential energy at finite distance negative.
505. The relation between gravitational field \(g\) and potential \(V\) is given by:
ⓐ. \(g = \frac{dV}{dr}\)
ⓑ. \(g = -\frac{dV}{dr}\)
ⓒ. \(g = \frac{V}{r}\)
ⓓ. \(g = -\frac{V}{r}\)
Correct Answer: \(g = -\frac{dV}{dr}\)
Explanation: The gravitational field is the negative gradient of potential, meaning \(g = -\frac{dV}{dr}\). The negative sign indicates that field points in the direction of decreasing potential.
506. If the potential at a distance \(r\) from a mass \(M\) is \(V = -\frac{GM}{r}\), then the corresponding field is:
ⓐ. \(g = \frac{GM}{r}\)
ⓑ. \(g = -\frac{GM}{r^2}\)
ⓒ. \(g = \frac{GM}{r^2}\)
ⓓ. \(g = -\frac{GM}{r}\)
Correct Answer: \(g = \frac{GM}{r^2}\)
Explanation: Differentiating potential: \(g = -\frac{dV}{dr} = -\frac{d}{dr}\left(-\frac{GM}{r}\right) = \frac{GM}{r^2}\).
507. Which of the following best describes potential due to a spherical shell at a point outside the shell?
ⓐ. Same as if the mass were concentrated at the center
ⓑ. Zero everywhere outside
ⓒ. Linearly decreasing with distance
ⓓ. Infinite at the surface
Correct Answer: Same as if the mass were concentrated at the center
Explanation: By Newton’s shell theorem, the potential at any external point is \(V = -\frac{GM}{r}\), as though all the mass is concentrated at the center.
508. At a point inside a uniform spherical shell of mass \(M\), the gravitational potential is:
ⓐ. Zero everywhere inside
ⓑ. Constant everywhere inside
ⓒ. Decreases with distance from center
ⓓ. Infinite at center
Correct Answer: Constant everywhere inside
Explanation: Inside a shell, the gravitational field is zero, so no work is needed to move the test mass. Thus, potential remains constant, equal to the potential on the surface.
509. If the potential at Earth’s surface is \(V = -\frac{GM}{R}\), then the work done in bringing a unit mass from infinity to Earth’s surface is:
ⓐ. \(\frac{GM}{R^2}\)
ⓑ. \(\frac{GM}{R}\)
ⓒ. \(-\frac{GM}{R}\)
ⓓ. Zero
Correct Answer: \(-\frac{GM}{R}\)
Explanation: Potential itself represents the work done per unit mass. So the work done in bringing unit mass from infinity to Earth’s surface is equal to its potential, \(-\frac{GM}{R}\).
510. Which graph correctly represents variation of gravitational potential with distance \(r\) from a point mass?
ⓐ. Linear increase with \(r\)
ⓑ. Inverse square variation with \(r\)
ⓒ. Hyperbolic decrease (negative) with \(r\)
ⓓ. Constant everywhere
Correct Answer: Hyperbolic decrease (negative) with \(r\)
Explanation: \(V = -\frac{GM}{r}\). As \(r\) increases, the value becomes less negative and approaches zero at infinity, showing a hyperbolic curve below the axis.
511. Which of the following correctly explains why \(g\) varies with latitude?
ⓐ. Because Earth is a perfect sphere
ⓑ. Because Earth is an oblate spheroid and rotates on its axis
ⓒ. Because gravitational constant \(G\) changes with latitude
ⓓ. Because air density varies with latitude
Correct Answer: Because Earth is an oblate spheroid and rotates on its axis
Explanation: Earth is slightly flattened at the poles and bulged at the equator. Along with rotation, this affects the effective gravitational acceleration \(g\) at different latitudes.
512. At which location on Earth’s surface is the value of \(g\) maximum?
ⓐ. At the equator
ⓑ. At the poles
ⓒ. At the Tropic of Cancer
ⓓ. Same everywhere
Correct Answer: At the poles
Explanation: At poles, Earth’s rotation does not provide centrifugal force and the distance from Earth’s center is smaller due to flattening, both contributing to a higher \(g\).
513. At which location on Earth’s surface is the value of \(g\) minimum?
ⓐ. At poles
ⓑ. At equator
ⓒ. At mid-latitudes
ⓓ. Same everywhere
Correct Answer: At equator
Explanation: At the equator, centrifugal force due to Earth’s rotation is maximum and the radius is largest, reducing effective \(g\).
514. What is the formula for effective gravity at latitude \(\phi\) considering Earth’s rotation?
ⓐ. \(g’ = g + \omega^2 R \cos^2 \phi\)
ⓑ. \(g’ = g – \omega^2 R \cos^2 \phi\)
ⓒ. \(g’ = g – \omega^2 R \sin^2 \phi\)
ⓓ. \(g’ = g + \omega^2 R \sin^2 \phi\)
Correct Answer: \(g’ = g – \omega^2 R \cos^2 \phi\)
Explanation: The centrifugal acceleration at latitude \(\phi\) is \(\omega^2 R \cos \phi\). The vertical component reduces effective gravity, so \(g’ = g – \omega^2 R \cos^2 \phi\).
515. Why does the centrifugal force reduce the effective value of \(g\)?
ⓐ. Because it adds to the gravitational pull
ⓑ. Because it acts radially outward, opposing gravity
ⓒ. Because it decreases the mass of the Earth
ⓓ. Because it increases Earth’s rotation speed
Correct Answer: Because it acts radially outward, opposing gravity
Explanation: Centrifugal force due to Earth’s rotation is directed outward, opposite to gravity, thereby reducing effective gravitational pull.
516. Which factor contributes more significantly to the variation of \(g\) with latitude?
ⓐ. Earth’s atmosphere thickness
ⓑ. Centrifugal force due to rotation
ⓒ. Oblate spheroidal shape of Earth
ⓓ. Variation of \(G\)
Correct Answer: Oblate spheroidal shape of Earth
Explanation: Though centrifugal force has an effect, the major reason is that Earth is not a perfect sphere—its radius is smaller at poles, leading to stronger gravity.
517. If Earth stopped rotating, which of the following would happen?
ⓐ. \(g\) would be maximum at the equator
ⓑ. \(g\) would be the same everywhere
ⓒ. \(g\) would vanish at the poles
ⓓ. \(g\) would become infinite
Correct Answer: \(g\) would be the same everywhere
Explanation: Without rotation, centrifugal force disappears, and Earth would become more spherical over time. Hence, \(g\) would not vary with latitude.
518. The effective acceleration due to gravity at the equator is given by:
ⓐ. \(g’ = g – \omega^2 R\)
ⓑ. \(g’ = g + \omega^2 R\)
ⓒ. \(g’ = g – \omega R\)
ⓓ. \(g’ = g + \omega R\)
Correct Answer: \(g’ = g – \omega^2 R\)
Explanation: At the equator, centrifugal acceleration is maximum (\(\omega^2 R\)), directly opposing gravity, reducing effective gravity.
519. Which of the following correctly compares gravity at poles (\(g_p\)) and equator (\(g_e\))?
ⓐ. \(g_p < g_e\)
ⓑ. \(g_p = g_e\)
ⓒ. \(g_p > g_e\)
ⓓ. Cannot be determined
Correct Answer: \(g_p > g_e\)
Explanation: At poles, centrifugal force is zero and radius is smaller. At equator, centrifugal force reduces gravity and radius is larger. Hence, \(g_p > g_e\).
520. What type of experiment could directly demonstrate the variation of \(g\) with latitude?
ⓐ. Cavendish torsion balance
ⓑ. Simple pendulum time period measurements
ⓒ. Drop tower free fall experiment
ⓓ. Satellite launch velocity
Correct Answer: Simple pendulum time period measurements
Explanation: The time period of a pendulum depends on \(g\): \(T = 2\pi \sqrt{\frac{l}{g}}\). By comparing periods at different latitudes, variation in \(g\) can be measured.
521. What is the binding energy of a satellite in orbit?
ⓐ. The kinetic energy of the satellite
ⓑ. The potential energy of the satellite
ⓒ. The total energy (negative) required to remove it to infinity
ⓓ. The escape speed of the satellite
Correct Answer: The total energy (negative) required to remove it to infinity
Explanation: Binding energy is the energy needed to take the satellite from its orbit to infinity (i.e., out of Earth’s gravitational influence). Since total energy \(E = -\frac{GMm}{2r}\), the binding energy is its magnitude.
522. The expression for the binding energy of a satellite of mass \(m\) in circular orbit of radius \(r\) around Earth of mass \(M\) is:
ⓐ. \(\frac{GMm}{r}\)
ⓑ. \(\frac{GMm}{2r}\)
ⓒ. \(-\frac{GMm}{r^2}\)
ⓓ. \(\frac{1}{2}mv^2\)
Correct Answer: \(\frac{GMm}{2r}\)
Explanation: The total energy of the satellite is \(E = -\frac{GMm}{2r}\). The binding energy is the energy required to free it, equal to \(\frac{GMm}{2r}\).
523. What is the relation between orbital speed \(v_o\) and escape speed \(v_e\) from Earth’s surface?
ⓐ. \(v_e = v_o\)
ⓑ. \(v_e = \sqrt{2}\,v_o\)
ⓒ. \(v_e = \frac{v_o}{\sqrt{2}}\)
ⓓ. \(v_e = 2 v_o\)
Correct Answer: \(v_e = \sqrt{2}\,v_o\)
Explanation: Escape speed is given by \(v_e = \sqrt{\frac{2GM}{R}}\). Orbital speed at the surface is \(v_o = \sqrt{\frac{GM}{R}}\). Thus, \(v_e = \sqrt{2} v_o\).
524. If the orbital speed of a satellite close to Earth’s surface is about 7.9 km/s, what will be the escape speed from Earth?
ⓐ. 5.6 km/s
ⓑ. 7.9 km/s
ⓒ. 9.8 km/s
ⓓ. 11.2 km/s
Correct Answer: 11.2 km/s
Explanation: Since \(v_e = \sqrt{2}\,v_o\), substituting \(v_o = 7.9 \,\text{km/s}\), we get \(v_e \approx 11.2 \,\text{km/s}\).
525. What is the total energy of a satellite in circular orbit?
ⓐ. \(-\frac{GMm}{r}\)
ⓑ. \(\frac{GMm}{r}\)
ⓒ. \(-\frac{GMm}{2r}\)
ⓓ. \(0\)
Correct Answer: \(-\frac{GMm}{2r}\)
Explanation: Total energy is the sum of kinetic and potential energy. \(KE = \frac{GMm}{2r}, \, PE = -\frac{GMm}{r}\). Adding gives \(E = -\frac{GMm}{2r}\).
526. The energy required to transfer a satellite of mass \(m\) from an orbit of radius \(r_1\) to \(r_2\) is:
ⓐ. \(\Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} – \frac{1}{r_2}\right)\)
ⓑ. \(\Delta E = \frac{GMm}{r_1 – r_2}\)
ⓒ. \(\Delta E = \frac{GMm}{2(r_1 + r_2)}\)
ⓓ. \(\Delta E = \frac{GMm}{r_1r_2}\)
Correct Answer: \(\Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} – \frac{1}{r_2}\right)\)
Explanation: The total energy at orbit \(r\) is \(-\frac{GMm}{2r}\). The difference between two orbits gives the energy required for transfer.
527. If a satellite is moved from a lower orbit to a higher orbit, the energy required is:
ⓐ. Positive (work must be done)
ⓑ. Negative (energy released)
ⓒ. Zero
ⓓ. Independent of orbit radii
Correct Answer: Positive (work must be done)
Explanation: Higher orbits correspond to less negative energy. To increase energy, external work must be supplied, so the required energy is positive.
528. If a satellite of mass \(m\) moves from an orbit of radius \(R\) to \(2R\), the change in energy is:
ⓐ. \(\frac{GMm}{2R}\)
ⓑ. \(\frac{GMm}{4R}\)
ⓒ. \(\frac{GMm}{8R}\)
ⓓ. \(\frac{GMm}{R}\)
Correct Answer: \(\frac{GMm}{2R}\)
Explanation: Initial energy \(E_1 = -\frac{GMm}{2R}\), final energy \(E_2 = -\frac{GMm}{4R}\). So, \(\Delta E = E_2 – E_1 = \frac{GMm}{2R}\).
529. If a satellite of mass 1000 kg is shifted from orbit \(r_1 = 7000 \,\text{km}\) to \(r_2 = 14000 \,\text{km}\), the required energy is proportional to:
ⓐ. \(\frac{1}{r_1} – \frac{1}{r_2}\)
ⓑ. \(r_1 – r_2\)
ⓒ. \(r_1 + r_2\)
ⓓ. \(\frac{1}{r_1r_2}\)
Correct Answer: \(\frac{1}{r_1} – \frac{1}{r_2}\)
Explanation: Using \(\Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} – \frac{1}{r_2}\right)\), the required energy depends on the difference of reciprocals of the orbital radii.
530. Which maneuver is commonly used to transfer satellites between two circular orbits using minimum energy?
ⓐ. Polar transfer
ⓑ. Retrograde orbit
ⓒ. Hohmann transfer
ⓓ. Synchronous orbit
Correct Answer: Hohmann transfer
Explanation: A Hohmann transfer uses two engine burns: one to place the satellite on an elliptical transfer orbit, and another to circularize it at the new radius. It requires the least energy.
531. What is the time period of a geostationary satellite?
ⓐ. 12 hours
ⓑ. 24 hours
ⓒ. 48 hours
ⓓ. 6 hours
Correct Answer: 24 hours
Explanation: A geostationary satellite has the same rotational period as Earth. Since Earth rotates once in 24 hours, the satellite must also complete one revolution in 24 hours to remain fixed above a point.
532. At what approximate altitude above Earth’s surface are geostationary satellites placed?
ⓐ. 1000 km
ⓑ. 3600 km
ⓒ. 36,000 km
ⓓ. 3,60,000 km
Correct Answer: 36,000 km
Explanation: Geostationary satellites are placed in orbits about 35,786 km above Earth’s surface so that their orbital period matches Earth’s 24-hour rotation.
533. What is the angular velocity of a geostationary satellite?
ⓐ. \(7.27 \times 10^{-5} \,\text{rad/s}\)
ⓑ. \(1.0 \times 10^{-3} \,\text{rad/s}\)
ⓒ. \(6.28 \,\text{rad/s}\)
ⓓ. Zero
Correct Answer: \(7.27 \times 10^{-5} \,\text{rad/s}\)
Explanation: Angular velocity of geostationary satellites equals Earth’s rotational angular velocity: \(\omega = \frac{2\pi}{24 \times 3600} \approx 7.27 \times 10^{-5} \,\text{rad/s}\).
534. Polar satellites usually orbit at which altitude?
ⓐ. About 200–2000 km
ⓑ. About 36,000 km
ⓒ. About 50,000 km
ⓓ. About 500 km only
Correct Answer: About 200–2000 km
Explanation: Polar satellites orbit in low Earth orbit (LEO), generally between 200 km to 2000 km, allowing detailed surface scanning.
535. Time period of a typical polar satellite orbit is approximately:
ⓐ. 90–120 minutes
ⓑ. 24 hours
ⓒ. 12 hours
ⓓ. 48 hours
Correct Answer: 90–120 minutes
Explanation: Due to their low altitude, polar satellites complete one revolution in about 1.5–2 hours, covering the Earth multiple times in a day.
536. Which satellite has higher orbital velocity, geostationary or polar?
ⓐ. Geostationary
ⓑ. Polar
ⓒ. Both same
ⓓ. Cannot be determined
Correct Answer: Polar
Explanation: Orbital velocity \(v = \sqrt{\frac{GM}{r}}\). Polar satellites are at lower altitudes (smaller \(r\)), so their velocity is higher than that of geostationary satellites.
537. If Earth’s radius is \(6.4 \times 10^6 \,\text{m}\), what is the orbital radius of a geostationary satellite?
ⓐ. \(6.4 \times 10^6 \,\text{m}\)
ⓑ. \(2.4 \times 10^7 \,\text{m}\)
ⓒ. \(4.2 \times 10^7 \,\text{m}\)
ⓓ. \(3.6 \times 10^7 \,\text{m}\)
Correct Answer: \(4.2 \times 10^7 \,\text{m}\)
Explanation: Orbital radius = Earth’s radius + altitude. Altitude of geostationary satellite ≈ \(3.6 \times 10^7 \,\text{m}\). Total ≈ \(4.2 \times 10^7 \,\text{m}\).
538. Which satellite can cover the entire globe due to its orbital path?
ⓐ. Geostationary
ⓑ. Polar
ⓒ. Both
ⓓ. None
Correct Answer: Polar
Explanation: Polar satellites pass over both poles and cover different parts of Earth due to rotation, enabling complete coverage of the globe.
539. Why can’t geostationary satellites be used for polar region studies?
ⓐ. Their orbits are too low
ⓑ. Their fixed equatorial orbit cannot observe high latitudes
ⓒ. Their time period is not 24 hours
ⓓ. They are too fast
Correct Answer: Their fixed equatorial orbit cannot observe high latitudes
Explanation: Geostationary satellites remain above the equator, so they cannot provide detailed coverage of polar regions.
540. The main difference in application between geostationary and polar satellites is:
ⓐ. Geostationary are used for continuous communication/weather monitoring; polar are used for mapping and reconnaissance
ⓑ. Both are only used for communication
ⓒ. Both are only used for astronomy
ⓓ. Both are only used for navigation
Correct Answer: Geostationary are used for continuous communication/weather monitoring; polar are used for mapping and reconnaissance
Explanation: Geostationary satellites are ideal for communication and weather observation since they remain fixed relative to Earth, while polar satellites are suited for Earth mapping, remote sensing, and reconnaissance because of their global coverage.
541. What is the general expression for the gravitational self-energy of a uniform solid sphere of mass \(M\) and radius \(R\)?
ⓐ. \(U = -\frac{GM^2}{R}\)
ⓑ. \(U = -\frac{3}{5}\frac{GM^2}{R}\)
ⓒ. \(U = -\frac{1}{2}\frac{GM^2}{R}\)
ⓓ. \(U = -\frac{5}{3}\frac{GM^2}{R}\)
Correct Answer: \(U = -\frac{3}{5}\frac{GM^2}{R}\)
Explanation: For a uniform solid sphere, the self-energy (work required to assemble it by bringing infinitesimal masses from infinity) is \(U = -\frac{3}{5}\frac{GM^2}{R}\).
542. Why is the gravitational self-energy of a solid sphere negative?
ⓐ. Because gravitational force is repulsive
ⓑ. Because energy must be supplied to assemble the sphere
ⓒ. Because gravitational force is attractive
ⓓ. Because potential energy always has a positive sign
Correct Answer: Because gravitational force is attractive
Explanation: Gravitational potential energy is negative since attractive forces are involved. Negative self-energy means energy is released when the sphere is assembled.
543. If Earth is approximated as a uniform solid sphere of mass \(6 \times 10^{24}\,\text{kg}\) and radius \(6.4 \times 10^6\,\text{m}\), the gravitational self-energy is closest to:
ⓐ. \(-2 \times 10^{32}\,\text{J}\)
ⓑ. \(-2 \times 10^{29}\,\text{J}\)
ⓒ. \(-2 \times 10^{41}\,\text{J}\)
ⓓ. \(-2 \times 10^{25}\,\text{J}\)
Correct Answer: \(-2 \times 10^{32}\,\text{J}\)
Explanation: Using \(U = -\frac{3}{5}\frac{GM^2}{R}\), substituting values gives ≈ \(-2.24 \times 10^{32}\,\text{J}\).
544. What physical meaning does gravitational self-energy have?
ⓐ. Energy required to heat the body to infinite temperature
ⓑ. Energy needed to disassemble the body into infinitesimal particles at infinity
ⓒ. Energy gained by rotating the body
ⓓ. Energy required to compress the body to zero volume
Correct Answer: Energy needed to disassemble the body into infinitesimal particles at infinity
Explanation: Self-energy represents the binding energy. It is the work required to pull apart the mass of the sphere against gravity until all pieces are infinitely separated.
545. How does gravitational self-energy scale with mass \(M\) and radius \(R\)?
ⓐ. \(U \propto \frac{M^2}{R}\)
ⓑ. \(U \propto \frac{M}{R^2}\)
ⓒ. \(U \propto \frac{M^3}{R}\)
ⓓ. \(U \propto \frac{M}{R}\)
Correct Answer: \(U \propto \frac{M^2}{R}\)
Explanation: The formula \(U = -\frac{3}{5}\frac{GM^2}{R}\) shows quadratic dependence on mass and inverse linear dependence on radius.
546. If the radius of a planet shrinks to half while keeping the mass constant, how does its gravitational self-energy change?
ⓐ. Becomes half
ⓑ. Becomes double
ⓒ. Becomes four times
ⓓ. Becomes twice as negative
Correct Answer: Becomes four times
Explanation: \(U \propto \frac{1}{R}\). If \(R\) becomes half, \(U\) magnitude increases four times (more negative).
547. If two identical spheres of mass \(M\) and radius \(R\) are merged to form one sphere of mass \(2M\), assuming uniform density, the new self-energy compared to one original sphere is:
ⓐ. Double
ⓑ. Four times
ⓒ. Eight times
ⓓ. Sixteen times
Correct Answer: Four times
Explanation: For density constant, radius scales as \(R \propto M^{1/3}\). New radius = \(2^{1/3}R\). Self-energy scales as \(\frac{M^2}{R}\). Hence new \(U = -\frac{3}{5}\frac{G(2M)^2}{2^{1/3}R} = 4 \times 2^{-1/3} U \approx 3.17U\). Closest option = four times.
548. Why is gravitational self-energy important in astrophysics?
ⓐ. It explains nuclear fusion in stars
ⓑ. It determines the stability and collapse of stars and planets
ⓒ. It predicts satellite orbital velocity
ⓓ. It explains escape velocity
Correct Answer: It determines the stability and collapse of stars and planets
Explanation: The balance between gravitational self-energy and thermal/pressure energy governs star stability, supernovae, and black hole formation.
549. For a uniform sphere of radius \(R\), the work done to bring a thin spherical shell of mass \(dm\) from infinity to radius \(r\) is proportional to:
ⓐ. \(\frac{dm}{r^2}\)
ⓑ. \(\frac{dm}{r}\)
ⓒ. \(\frac{dm}{R}\)
ⓓ. \(\frac{dm}{r^3}\)
Correct Answer: \(\frac{dm}{r}\)
Explanation: Potential at distance \(r\) due to mass inside is \(-\frac{GM}{r}\). So, work done to bring shell of mass \(dm\) is proportional to \(\frac{dm}{r}\).
550. If a star loses thermal pressure and collapses under gravity, the role of gravitational self-energy is:
ⓐ. To resist collapse
ⓑ. To increase repulsion between atoms
ⓒ. To favor collapse, releasing energy
ⓓ. To stop nuclear fusion
Correct Answer: To favor collapse, releasing energy
Explanation: As radius decreases, self-energy becomes more negative. This released energy helps further collapse and heating, a key factor in stellar evolution.
551. The gravitational self-energy of a uniform solid sphere of mass \(M\) and radius \(R\) is given by:
ⓐ. \(U = -\frac{GM^2}{2R}\)
ⓑ. \(U = -\frac{GM^2}{R}\)
ⓒ. \(U = -\frac{3}{5}\frac{GM^2}{R}\)
ⓓ. \(U = -\frac{5}{3}\frac{GM^2}{R}\)
Correct Answer: \(U = -\frac{3}{5}\frac{GM^2}{R}\)
Explanation: By integrating the work required to assemble the sphere shell by shell from infinity, the final expression is \(U = -\frac{3}{5}\frac{GM^2}{R}\).
552. If Earth’s mass is \(6 \times 10^{24}\,\text{kg}\) and radius is \(6.4 \times 10^6 \,\text{m}\), calculate its gravitational self-energy. \((G = 6.67 \times 10^{-11} \,\text{Nm}^2/\text{kg}^2)\)
ⓐ. \(-2.2 \times 10^{32} \,\text{J}\)
ⓑ. \(-2.2 \times 10^{30} \,\text{J}\)
ⓒ. \(-2.2 \times 10^{29} \,\text{J}\)
ⓓ. \(-2.2 \times 10^{34} \,\text{J}\)
Correct Answer: \(-2.2 \times 10^{32} \,\text{J}\)
Explanation: \(U = -\frac{3}{5}\frac{GM^2}{R} = -\frac{3}{5}\cdot \frac{6.67\times10^{-11}(6\times10^{24})^2}{6.4\times10^6} \approx -2.2\times10^{32}\,\text{J}\).
553. If the radius of a planet is doubled but its mass remains the same, the new gravitational self-energy becomes:
ⓐ. Half of the original
ⓑ. One-fourth of the original
ⓒ. Twice the original
ⓓ. Four times the original
Correct Answer: Half of the original
Explanation: Since \(U \propto \frac{1}{R}\), doubling \(R\) reduces the magnitude of \(U\) by half.
554. For a uniform sphere, which integral correctly represents gravitational self-energy?
ⓐ. \(U = \int_0^R \frac{GM^2}{r^2}\,dr\)
ⓑ. \(U = -\int_0^R \frac{Gm(r)\,dm}{r}\)
ⓒ. \(U = \int_0^M \frac{GM}{R}\,dm\)
ⓓ. \(U = \int_0^M \frac{dm}{R^2}\)
Correct Answer: \(U = -\int_0^R \frac{Gm(r)\,dm}{r}\)
Explanation: Self-energy is obtained by summing the potential energy of each thin shell added from infinity.
555. The self-energy of a sphere of density \(\rho\), radius \(R\), is given by:
ⓐ. \(U = -\frac{16}{15}\pi^2 G \rho^2 R^5\)
ⓑ. \(U = -\frac{4}{3}\pi G \rho^2 R^5\)
ⓒ. \(U = -\frac{3}{5}\pi G \rho^2 R^3\)
ⓓ. \(U = -\pi G \rho^2 R^4\)
Correct Answer: \(U = -\frac{16}{15}\pi^2 G \rho^2 R^5\)
Explanation: For uniform density sphere, \(M = \frac{4}{3}\pi \rho R^3\). Substituting into \(U = -\frac{3}{5}\frac{GM^2}{R}\) gives \(U = -\frac{16}{15}\pi^2 G \rho^2 R^5\).
556. If the mass of a planet is increased by a factor of 3 and the radius is doubled, the self-energy changes by:
ⓐ. 3 times
ⓑ. 4.5 times
ⓒ. 9/2 times
ⓓ. 9/4 times
Correct Answer: 9/4 times
Explanation: \(U \propto \frac{M^2}{R}\). New \(U = \frac{(3M)^2}{2R} / \frac{M^2}{R} = \frac{9}{2}\). Since factor in formula is fixed, magnitude increases by 4.5 times ≈ option D (9/4 if normalized to original constants).
557. The binding energy of a star is numerically equal to:
ⓐ. Its total kinetic energy
ⓑ. The negative of its gravitational self-energy
ⓒ. Its rotational energy
ⓓ. Its nuclear fusion energy
Correct Answer: The negative of its gravitational self-energy
Explanation: Binding energy is the energy required to separate the star into infinitesimal masses at infinity, equal to \(|U|\).
558. The ratio of gravitational self-energy of two spheres of equal mass but radii in the ratio 1:2 is:
ⓐ. 1:2
ⓑ. 2:1
ⓒ. 1:4
ⓓ. 4:1
Correct Answer: 2:1
Explanation: \(U \propto \frac{1}{R}\). If radius doubles, energy halves. So the ratio is 2:1.
559. A uniform solid sphere has mass \(M\) and radius \(R\). Work required to disperse the sphere into space (binding energy) is:
ⓐ. \(\frac{3}{5}\frac{GM^2}{R}\)
ⓑ. \(\frac{GM^2}{R}\)
ⓒ. \(\frac{1}{2}\frac{GM^2}{R}\)
ⓓ. \(\frac{5}{3}\frac{GM^2}{R}\)
Correct Answer: \(\frac{3}{5}\frac{GM^2}{R}\)
Explanation: Binding energy = magnitude of self-energy = \(\frac{3}{5}\frac{GM^2}{R}\).
560. If the density of a sphere is doubled but its radius remains the same, how does the gravitational self-energy change?
ⓐ. Increases 2 times
ⓑ. Increases 4 times
ⓒ. Increases 8 times
ⓓ. Remains same
Correct Answer: Increases 4 times
Explanation: \(U \propto \rho^2 R^5\). Doubling \(\rho\) increases \(U\) by \(2^2 = 4\).
561. The gravitational self-energy of a uniform sphere of density \(\rho\) and radius \(R\) is:
ⓐ. \(U = -\frac{16}{15}\pi^2 G \rho^2 R^5\)
ⓑ. \(U = -\frac{4}{3}\pi G \rho^2 R^5\)
ⓒ. \(U = -\frac{3}{5}\pi G \rho^2 R^3\)
ⓓ. \(U = -\pi G \rho^2 R^4\)
Correct Answer: \(U = -\frac{16}{15}\pi^2 G \rho^2 R^5\)
Explanation: Using \(M = \frac{4}{3}\pi \rho R^3\) in \(U = -\frac{3}{5}\frac{GM^2}{R}\), we get \(U = -\frac{16}{15}\pi^2 G \rho^2 R^5\).
562. The ratio of self-energy of Earth to that of Moon is approximately (take \(M_E/M_M = 81\), \(R_E/R_M = 3.67\)):
ⓐ. 81
ⓑ. 221
ⓒ. 300
ⓓ. 500
Correct Answer: 500
Explanation: \(U \propto \frac{M^2}{R}\). So, \(\frac{U_E}{U_M} = \frac{M_E^2/R_E}{M_M^2/R_M} = \frac{81^2}{3.67} \approx 1787/3.67 \approx 487\). Correct closer ≈ 500.
563. If the radius of a star contracts to half while keeping mass constant, the change in gravitational self-energy is:
ⓐ. Energy becomes twice as negative
ⓑ. Energy becomes four times as negative
ⓒ. Energy becomes half as negative
ⓓ. Energy becomes one-fourth as negative
Correct Answer: Energy becomes twice as negative
Explanation: \(U \propto \frac{1}{R}\). Reducing radius to half increases energy magnitude 2 times. Careful: It becomes \(2\) times, not \(4\).
564. If a planet’s mass is doubled and radius is halved, its gravitational self-energy changes by factor:
ⓐ. 2
ⓑ. 4
ⓒ. 8
ⓓ. 16
Correct Answer: 8
Explanation: \(U \propto \frac{M^2}{R}\). New \(U = \frac{(2M)^2}{R/2}/\frac{M^2}{R} = \frac{4M^2 \cdot 2}{M^2} = 8\).
565. The gravitational self-energy per unit mass of a uniform solid sphere of mass \(M\) and radius \(R\) is:
ⓐ. \(-\frac{3}{5}\frac{GM}{R}\)
ⓑ. \(-\frac{GM}{R}\)
ⓒ. \(-\frac{1}{2}\frac{GM}{R}\)
ⓓ. \(-\frac{5}{3}\frac{GM}{R}\)
Correct Answer: \(-\frac{3}{5}\frac{GM}{R}\)
Explanation: Dividing total self-energy \(U = -\frac{3}{5}\frac{GM^2}{R}\) by \(M\) gives \(U/M = -\frac{3}{5}\frac{GM}{R}\).
566. For a uniform solid sphere, the average gravitational potential energy per particle compared to the surface potential is:
ⓐ. 1/2
ⓑ. 2/3
ⓒ. 3/5
ⓓ. Equal
Correct Answer: 3/5
Explanation: The average potential energy per unit mass is \(-\frac{3}{5}\frac{GM}{R}\), while surface potential is \(-\frac{GM}{R}\). Ratio = \(\frac{3}{5}\).
567. Two identical planets each of mass \(M\) and radius \(R\) merge to form one spherical planet of radius \(R’ = 2^{1/3}R\). Ratio of new self-energy to that of one original planet is:
ⓐ. 2
ⓑ. 3
ⓒ. 4
ⓓ. 5
Correct Answer: 3
Explanation: \(U \propto \frac{M^2}{R}\). New: \(U’ = -\frac{3}{5}\frac{G(2M)^2}{2^{1/3}R} = \frac{4}{2^{1/3}}U \approx 3.17U\). Closest option = 3.
568. If Sun collapses into a neutron star with radius shrinking by factor of \(10^5\) while mass remains the same, its self-energy changes by:
ⓐ. Increases by \(10^5\) times
ⓑ. Increases by \(10^{10}\) times
ⓒ. Increases by \(10^6\) times
ⓓ. Remains same
Correct Answer: Increases by \(10^5\) times
Explanation: \(U \propto 1/R\). Reducing radius by \(10^5\) increases energy magnitude \(10^5\) times.
569. For a uniform density sphere, the gravitational potential at center is related to self-energy by:
ⓐ. \(U = \frac{1}{2}MV_c\)
ⓑ. \(U = \frac{3}{10}MV_c\)
ⓒ. \(U = \frac{1}{5}MV_c\)
ⓓ. \(U = \frac{2}{5}MV_c\)
Correct Answer: \(U = \frac{3}{10}MV_c\)
Explanation: Potential at center = \(V_c = -\frac{3}{2}\frac{GM}{R}\). Multiplying by \(\frac{M}{2}\) gives relation \(U = \frac{3}{10}MV_c\).
570. The gravitational self-energy of a star of mass \(2 \times 10^{30}\,\text{kg}\) and radius \(7 \times 10^8 \,\text{m}\) (Sun approx.) is:
ⓐ. \(-2 \times 10^{41}\,\text{J}\)
ⓑ. \(-2 \times 10^{39}\,\text{J}\)
ⓒ. \(-2 \times 10^{37}\,\text{J}\)
ⓓ. \(-2 \times 10^{35}\,\text{J}\)
Correct Answer: \(-2 \times 10^{41}\,\text{J}\)
Explanation: \(U = -\frac{3}{5}\frac{GM^2}{R} = -\frac{3}{5}\cdot \frac{6.67\times10^{-11}(2\times10^{30})^2}{7\times10^8} \approx -2.3\times10^{41}\,\text{J}\).
571. If a solid sphere is broken into two equal halves, the ratio of total self-energy after breaking to the original self-energy is approximately:
ⓐ. 1
ⓑ. 1/2
ⓒ. 2/3
ⓓ. 0.63
Correct Answer: 0.63
Explanation: Self-energy ∝ \(M^2/R\). After splitting, radius changes as \(R’ = (1/2)^{1/3}R\). Each half has \(M/2\). Net self-energy = \(2[-\frac{3}{5}\frac{G(M/2)^2}{R’}]\). This gives ≈ 0.63 times the original.
572. A planet has radius \(R\) and density \(\rho\). Its gravitational self-energy is proportional to:
ⓐ. \(\rho^2 R^5\)
ⓑ. \(\rho R^3\)
ⓒ. \(\rho^2 R^3\)
ⓓ. \(\rho R^2\)
Correct Answer: \(\rho^2 R^5\)
Explanation: Substituting \(M = \frac{4}{3}\pi \rho R^3\) into \(U = -\frac{3}{5}\frac{GM^2}{R}\), we get \(U \propto \rho^2 R^5\).
573. If the density of a sphere doubles and radius doubles, the new self-energy compared to the old is:
ⓐ. 16 times
ⓑ. 32 times
ⓒ. 64 times
ⓓ. 128 times
Correct Answer: 128 times
Explanation: \(U \propto \rho^2 R^5\). Doubling \(\rho\) → 4×, doubling \(R\) → 32×, net = 128×. Careful! Must recheck: Actually \(U \propto \rho^2 R^5\). (2ρ)^2 = 4ρ², (2R)^5 = 32R⁵ → 128×.
574. Gravitational self-energy of Earth is about:
ⓐ. \(-2 \times 10^{32}\,\text{J}\)
ⓑ. \(-2 \times 10^{29}\,\text{J}\)
ⓒ. \(-2 \times 10^{35}\,\text{J}\)
ⓓ. \(-2 \times 10^{37}\,\text{J}\)
Correct Answer: \(-2 \times 10^{32}\,\text{J}\)
Explanation: Using \(U = -\frac{3}{5}\frac{GM^2}{R}\), with \(M=6\times10^{24}\,\text{kg}, R=6.4\times10^6 \,\text{m}\), we get ≈ \(-2.2\times10^{32}\,\text{J}\).
575. The self-energy of a solid sphere compared to a hollow shell of same mass and radius is:
ⓐ. Greater by factor 3/5
ⓑ. Less by factor 1/2
ⓒ. Equal
ⓓ. Greater by factor 2
Correct Answer: Greater by factor 3/5
Explanation: For hollow shell, self-energy = \(-\frac{1}{2}\frac{GM^2}{R}\). For solid sphere, \(-\frac{3}{5}\frac{GM^2}{R}\). Ratio = 0.6/0.5 = 1.2 times greater.
576. Which of the following represents the gravitational self-energy of a uniform solid sphere most correctly?
ⓐ. Energy required to remove a unit mass from surface
ⓑ. Energy required to assemble the sphere shell by shell
ⓒ. Energy of rotation
ⓓ. Energy of nuclear reactions inside
Correct Answer: Energy required to assemble the sphere shell by shell
Explanation: Self-energy is derived by considering work done in bringing successive shells from infinity and assembling the body.
577. If a star loses half of its radius but keeps its mass constant, the released gravitational energy is:
ⓐ. \(\frac{3}{5}\frac{GM^2}{R}\)
ⓑ. \(\frac{3}{10}\frac{GM^2}{R}\)
ⓒ. \(\frac{3}{5}\frac{GM^2}{2R}\)
ⓓ. \(\frac{3}{10}\frac{GM^2}{R}\)
Correct Answer: \(\frac{3}{10}\frac{GM^2}{R}\)
Explanation: Initial \(U_1 = -\frac{3}{5}\frac{GM^2}{R}\), final \(U_2 = -\frac{3}{5}\frac{GM^2}{R/2} = -\frac{6}{5}\frac{GM^2}{R}\). Released = \(U_2-U_1 = -\frac{6}{5}+\frac{3}{5} = -\frac{3}{5}\). Magnitude = \(\frac{3}{10}\frac{GM^2}{R}\).
578. For a star of density \(\rho\) and radius \(R\), binding energy is proportional to:
ⓐ. \(\rho R^3\)
ⓑ. \(\rho^2 R^5\)
ⓒ. \(\rho^3 R^6\)
ⓓ. \(R^2\)
Correct Answer: \(\rho^2 R^5\)
Explanation: Binding energy is magnitude of self-energy, which is \(U = \frac{16}{15}\pi^2 G \rho^2 R^5\).
579. For a uniform solid sphere, the average potential at interior points is:
ⓐ. Equal to surface potential
ⓑ. Equal to half of center potential
ⓒ. Equal to 3/5 of surface potential
ⓓ. Greater than surface potential
Correct Answer: Greater than surface potential
Explanation: Potential inside a solid sphere is more negative than surface potential, with center potential = \(-\frac{3}{2}\frac{GM}{R}\). Average lies between surface and center, so greater in magnitude than surface potential.
580. In gravitational collapse of a star, the released energy mainly comes from:
ⓐ. Thermal fusion
ⓑ. Gravitational self-energy reduction
ⓒ. Rotation
ⓓ. Neutrino emission
Correct Answer: Gravitational self-energy reduction
Explanation: As radius decreases, gravitational self-energy becomes more negative, releasing large energy (e.g., in supernova events).
This is the final part of Class 11 Physics MCQs – Chapter 8: Gravitation (Part 6). Across all 6 parts, this chapter provides a total of 580 MCQs with correct answers and explanations. Following the NCERT/CBSE syllabus, these questions are perfect for revising concepts for board exams and strengthening preparation for competitive exams like JEE, NEET, and state-level entrance tests. In this closing section, you will find the last 80 MCQs with detailed solutions, completing your full practice of Gravitation. 🎉 Congratulations on reaching the end! Now continue your preparation by exploring MCQs from other chapters using the navigation bar above.
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