GRAVITATION MCQs for JEE PHYSICS — Mixed difficulty, Page 15

421 Medium

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Understanding +4 | -1

A planet revolves round the Sun in an elliptical orbit with the Sun at one of the foci. Regarding the speed and velocity of the planet which of the following is correct?

View answer and explanation

Correct option: C

Speed is variable, velocity is variable

Option C is correct because in an elliptical orbit, both the speed and velocity change as the planet's distance from the Sun varies.

Solution approach

Correct Answer + Core Reason: Speed is variable, velocity is variable. In an elliptical orbit, both speed and velocity change.

Step-by-Step Solution:

Step 1: Recall Kepler's Second Law. A planet sweeps out equal areas in equal times. This means the planet moves faster when closer to the Sun and slower when farther away.

Step 2: Define speed and velocity. Speed is the magnitude of velocity. Velocity is a vector quantity with both magnitude and direction.

Step 3: Analyze the motion in an elliptical orbit. Since the distance from the Sun varies, the speed changes. Also, the direction of motion is constantly changing, so the velocity changes.

Why other options are wrong:
- Option A: Speed is variable, velocity is constant - Incorrect because velocity also changes due to changing direction.
- Option B: Speed is constant, velocity is variable - Incorrect because speed changes due to varying distance from the Sun.
- Option D: Speed is constant, velocity is constant - Incorrect because both speed and direction change.

Key Concepts:
- Kepler's Laws of Planetary Motion
- Conservation of Angular Momentum
- Elliptical Orbits

Memory Aids:
- Visualize the planet speeding up as it approaches the Sun and slowing down as it moves away.

Quick Tips:
- Remember that in elliptical orbits, both speed and velocity are variable.

422 Medium

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Analyzing +4 | -1

(A): The angular velocity of a planets orbiting around the sun increases when they are nearest to the sun.
(R): The angular momentum of body is proportional to angular velocity.

View answer and explanation

Correct option: B

Both (A) and (R) are true and (R) is not the correct explanation of (A)

Option B (Both (A) and (R) are true and (R) is not the correct explanation of (A)) is correct. Both statements are true, but the angular momentum conservation explains the change in angular velocity, not the direct proportionality.

Solution approach

Correct Answer + Core Reason: Both (A) and (R) are true and (R) is not the correct explanation of (A). Both statements are correct, but the reason doesn't explain the assertion.

Step-by-Step Solution:
Step 1: Analyze Assertion (A): The angular velocity of a planet orbiting around the sun increases when they are nearest to the sun. This is true due to the conservation of angular momentum.
Step 2: Analyze Reason (R): The angular momentum of a body is proportional to angular velocity. This is also true; $L = I\omega$ , where $L$ is angular momentum, $I$ is the moment of inertia, and $\omega$ is angular velocity.
Step 3: Determine if the reason is the correct explanation: The correct explanation for the assertion is the conservation of angular momentum. As the planet gets closer to the sun, its moment of inertia decreases, and thus its angular velocity increases to keep the angular momentum constant. The reason is true, but it doesn't directly explain why the angular velocity increases when the planet is nearest to the sun. The conservation of angular momentum is the key.

Why other options are wrong:
- Option A: The reason is not the correct explanation.
- Option C: The reason is true.
- Option D: Both statements are true.

Key Concepts:
- Angular Momentum
- Conservation of Angular Momentum
- Angular Velocity
- Moment of Inertia

Memory Aids:
- Remember $L = I\omega$ and the conservation of angular momentum.

Quick Tips:
- Consider conservation laws when analyzing orbital motion.

423 Medium

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Analyzing +4 | -1

To remove an object from earth's gravitational field the speed that is to be given to it on the surface of the earth is nearly equal to

11.25 \mathrm{~km} / \mathrm{s}

. However, when an object on the earth is to be removed from the solar system, gravitational influence of the sun is taken in account. To remove an object from the solar system, the speed that is given to it on earth's surface is nearly equal to

45 \mathrm{~km} / \mathrm{s}

. The mass of the sun is

\mathrm{k}_1

times the mass of the earth and the orbital radius of the earth is

\mathrm{k}_2

times its radius. Find the value of

\mathrm{k}_1 / \mathrm{k}_2

.

View answer and explanation

Correct option: A

15

Option A (15) is correct. The ratio k1/k2 is found by equating the kinetic and potential energies required to escape both Earth's and the Sun's gravitational fields.

Solution approach

Correct Answer + Core Reason: 15. The ratio is derived by comparing escape velocities from Earth and the solar system, considering the mass and radius relationships.

Step-by-Step Solution:

Step 1: Write the escape velocity from Earth: $V_1 = \sqrt{\frac{2GM_E}{R_E}} = 11.25 \text{ km/s}$ .

Step 2: Write the energy conservation equation for escaping the solar system: $\frac{1}{2}mV_2^2 - \frac{GM_Em}{R_E} - \frac{GM_Sm}{R_S} = 0$ , where $V_2 = 45 \text{ km/s}$ , $M_S = k_1M_E$ , and $R_S = k_2R_E$ .

Step 3: Substitute the given relationships: $\frac{1}{2}mV_2^2 = \frac{GM_Em}{R_E} + \frac{Gk_1M_Em}{k_2R_E} = \frac{GM_Em}{R_E}(1 + \frac{k_1}{k_2})$ .

Step 4: Relate $V_2$ to $V_1$ : $V_2^2 = \frac{2GM_E}{R_E}(1 + \frac{k_1}{k_2}) = V_1^2(1 + \frac{k_1}{k_2})$ .

Step 5: Solve for $\frac{k_1}{k_2}$ : $(\frac{V_2}{V_1})^2 = 1 + \frac{k_1}{k_2}$ , so $\frac{k_1}{k_2} = (\frac{V_2}{V_1})^2 - 1 = (\frac{45}{11.25})^2 - 1 = 4^2 - 1 = 16 - 1 = 15$ .

Why other options are wrong:
- Option B: Incorrect calculation.
- Option C: Incorrect calculation.
- Option D: Incorrect calculation.

Key Concepts:
- Escape velocity
- Conservation of energy
- Gravitational potential energy

Memory Aids:
- Escape velocity formula.

Quick Tips:
- Remember to account for both Earth's and the Sun's gravity.

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Understanding +4 | -1

Assertion: Work done by or against gravitational force in moving a body from one point to another is independent of the actual path followed between the two points.
Reason: Gravitational forces are conservative forces.

View answer and explanation

Correct option: A

If both Assertion and Reason are true and Reason is the correct explanation of Assertion

Option A is correct because gravitational forces are conservative, meaning the work done is path-independent and depends only on the initial and final positions.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true and Reason is the correct explanation of Assertion. Conservative forces imply path independence.

Step-by-Step Solution:

Step 1: Define a conservative force. A conservative force is one for which the work done in moving an object between two points is independent of the path taken.

Step 2: Gravitational force is a conservative force.

Step 3: Therefore, the work done by or against gravity depends only on the initial and final positions, not the path taken.

Why other options are wrong:
- Option B: The reason is the correct explanation of the assertion.
- Option C: Both assertion and reason are true.
- Option D: Both assertion and reason are true.

Key Concepts:
- Conservative Force
- Work Done
- Gravitational Force

Memory Aids:
- Gravity is a conservative force.

Quick Tips:
- Remember that work done by conservative forces is path-independent.

425 Medium

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Analyzing +4 | -1

Assertion: On satellites we feel weightlessness. Moon is also a satellite of earth. But we do not feel weightlessness on moon.

Reason: Mass of moon is considerable.

View answer and explanation

Correct option: A

If both Assertion and Reason are true and the Reason is correct explanation of the Assertion

Option A (If both Assertion and Reason are true and the Reason is correct explanation of the Assertion) is correct because weightlessness on satellites is due to free fall, while the moon's considerable mass causes us to experience weight due to its gravity.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. Weightlessness on satellites is due to free fall, while the moon has considerable mass and thus exerts gravitational force.

Step-by-Step Solution:

Step 1: Analyze the Assertion: On satellites, we feel weightlessness because they are in free fall around the Earth.

Step 2: Analyze the fact that the Moon is a satellite of Earth, but we don't feel weightlessness on the Moon. This is because the Moon has its own gravitational field due to its mass.

Step 3: Analyze the Reason: The mass of the Moon is considerable, which means it exerts a significant gravitational force.

Step 4: The Moon's gravity provides the necessary centripetal force for objects on its surface, causing them to experience weight.

Step 5: The Reason correctly explains why we don't feel weightlessness on the Moon, even though it's a satellite.

Why other options are wrong:
- Option B: The Reason is the correct explanation of the Assertion.
- Option C: The Reason is true.
- Option D: The Assertion is true.

Key Concepts:
- Weightlessness
- Gravitational force
- Satellites
- Free fall

Memory Aids:
- Weightlessness is experienced in free fall.
- Objects with mass exert gravitational force.

Quick Tips:
- Consider the gravitational forces involved when analyzing weightlessness.

426 Easy

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Application +4 | -1

The distance of a geostationary satellite from the centre of the earth [Radius R= 6400 km] is nearest to

View answer and explanation

Correct option: B

7 R

Option B (7 R) is correct because a geostationary satellite is approximately 6R from the surface of the earth, hence 7R from the center.

Solution approach

Correct Answer + Core Reason: 7 R. The distance of a geostationary satellite from the center of the earth is approximately 7 times the Earth's radius.

Step-by-Step Solution:

Step 1: A geostationary satellite orbits at a height of approximately 36,000 km above the Earth's surface.

Step 2: The radius of the Earth is given as R = 6400 km.

Step 3: The distance from the center of the Earth is the sum of the Earth's radius and the orbital height: 6400 km + 36000 km = 42400 km.

Step 4: Calculate the distance in terms of R: $\frac{42400}{6400} \approx 6.625 R$ . This is approximately 6R from the surface and 7R from the center.

Step 5: Final Answer: 7 R

Why other options are wrong:
- Option A (5 R): This is too low. The geostationary orbit is significantly higher than 5 times the Earth's radius.
- Option C (10 R): This is too high. The geostationary orbit is not 10 times the Earth's radius.
- Option D (18 R): This is much too high. The geostationary orbit is much closer to the Earth than 18 times the Earth's radius.

Key Concepts:
- Geostationary Satellites
- Orbital Mechanics

Memory Aids:
- Remember the approximate height of a geostationary orbit (36,000 km).

Quick Tips:
- Know the approximate height of geostationary satellites and the Earth's radius to quickly estimate the distance from the Earth's center.

427 Medium

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Application +4 | -1

Acceleration due to gravity at earth's surface is

10 \mathrm{~ms}^{-2}

. The value of acceleration due to gravity at the surface of a planet of mass

\frac{1}{5}

th and radius

\frac{1}{2}

of the earth is -

View answer and explanation

Correct option: C

8 \mathrm{~ms}^{-2}

Option C (

8 \mathrm{~ms}^{-2}

) is correct because it is calculated by using the formula for acceleration due to gravity, considering the changes in mass and radius.

Solution approach

Correct Answer + Core Reason: $8 \mathrm{~ms}^{-2}$ . The acceleration due to gravity is calculated using the formula $g = \frac{{GM}}{{R^2}}$ .

Step-by-Step Solution:

Step 1: Write the formula for acceleration due to gravity on Earth: $g_e = \frac{{GM_e}}{{R_e^2}} = 10 \mathrm{~ms}^{-2}$ .

Step 2: Write the formula for acceleration due to gravity on the planet: $g_p = \frac{{GM_p}}{{R_p^2}}$ , where $M_p = \frac{1}{5}M_e$ and $R_p = \frac{1}{2}R_e$ .

Step 3: Substitute the values: $g_p = \frac{{G(\frac{1}{5}M_e)}}{{(\frac{1}{2}R_e)^2}} = \frac{{G \times \frac{1}{5} \times M_e}}{{\frac{1}{4}R_e^2}} = \frac{4}{5} \frac{{GM_e}}{{R_e^2}}$ .

Step 4: Substitute $g_e$ : $g_p = \frac{4}{5}g_e = \frac{4}{5} \times 10 = 8 \mathrm{~ms}^{-2}$ .

Why other options are wrong:
- Option A: This value is incorrect because it does not correctly account for the changes in both mass and radius.
- Option B: This value is incorrect because it does not correctly account for the changes in both mass and radius.
- Option D: This value is incorrect because it does not correctly account for the changes in both mass and radius.

Key Concepts:
- Acceleration due to gravity
- Gravitational force

Memory Aids:
- Remember that $g = \frac{{GM}}{{R^2}}$ .

Quick Tips:
- When comparing gravity on different planets, always consider both the mass and the radius.

428 Medium

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Remembering +4 | -1

Earth binds the atmosphere because of

View answer and explanation

Correct option: A

Gravity

Option A (Gravity) is correct because Earth's gravitational force is the primary reason it holds onto its atmosphere.

Solution approach

Correct Answer + Core Reason: Gravity. Earth's gravity is strong enough to retain the atmospheric gases.

Step-by-Step Solution:
Step 1: Consider the factors that determine whether a planet can retain an atmosphere.
Step 2: The primary factor is the planet's gravitational force, which depends on its mass and radius.
Step 3: The average thermal velocity of gas molecules in the atmosphere must be less than the escape velocity for the planet to retain the atmosphere.
Step 4: Earth's gravity provides a sufficient escape velocity to prevent atmospheric gases from escaping into space.

Why other options are wrong:
- Option B (oxygen between earth and atmosphere): Oxygen is a component of the atmosphere, but it's not the reason the atmosphere is bound to Earth. The binding force is gravity.
- Option C (Both (a) and (b)): While oxygen is part of the atmosphere, it's not the primary reason the atmosphere is bound to Earth. Gravity is the main reason.
- Option D (None of the above): Gravity is the reason, so this option is incorrect.

Key Concepts:
- Gravity
- Escape Velocity
- Planetary Atmospheres

Memory Aids:
- Gravity keeps everything grounded, including the atmosphere.

Quick Tips:
- Remember that gravity is the dominant force in holding atmospheres.

429 Medium

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Application +4 | -1

The Earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the Earth. The escape velocity of a body from this platform is

fv

, where

v

is its escape velocity from the surface of the Earth. The value of f is

View answer and explanation

Correct option: D

\frac{1}{{\sqrt 2 }}

Option D (

\frac{1}{{\sqrt 2 }}

) is correct because the escape velocity is inversely proportional to the square root of the distance from the center of the Earth.

Solution approach

Correct Answer + Core Reason: $\frac{1}{{\sqrt 2 }}$ . The escape velocity decreases with increasing distance from the Earth's center.

Step-by-Step Solution:

Step 1: Define escape velocity from the surface of the Earth.
$v = \sqrt{\frac{2GM}{R}}$

Step 2: Define escape velocity from a height R above the surface of the Earth (distance 2R from the center).
$v_p = \sqrt{\frac{2GM}{2R}} = \sqrt{\frac{GM}{R}}$

Step 3: Relate the two escape velocities.
$v_p = fv$
$\sqrt{\frac{GM}{R}} = f \sqrt{\frac{2GM}{R}}$

Step 4: Solve for f.
$f = \frac{\sqrt{\frac{GM}{R}}}{\sqrt{\frac{2GM}{R}}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$

Final Answer: $\frac{1}{{\sqrt 2 }}$

Why other options are wrong:
- Option A: $\frac{1}{3}$ is incorrect. This value does not follow from the correct relationship between escape velocity and distance.
- Option B: $\frac{1}{2}$ is incorrect. This value does not follow from the correct relationship between escape velocity and distance.
- Option C: $\sqrt 2$ is incorrect. This value is the inverse of the correct answer.

Key Concepts:
- Escape Velocity
- Gravitational Potential Energy

Memory Aids:
- Escape velocity formula: $v = \sqrt{\frac{2GM}{r}}$

Quick Tips:
- Remember that escape velocity depends on the distance from the center of the planet.

430 Medium

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Application +4 | -1

A man weighs

\mathrm{W}

on the surface of the earth. What is his weight at a height equal to

\mathrm{R}

?

\mathrm{R} \rightarrow

Radius of earth)

View answer and explanation

Correct option: C

W / 4

Option C (W / 4) is correct because the weight decreases with the square of the distance from the center of the Earth.

Solution approach

Correct Answer + Core Reason: W / 4. The weight decreases with the square of the distance.

Step-by-Step Solution:

Step 1: Weight on the surface of the Earth is $W = mg = \frac{GMm}{R^2}$ , where G is the gravitational constant, M is the mass of the Earth, m is the mass of the man, and R is the radius of the Earth.

Step 2: At a height R above the surface, the distance from the center of the Earth is $R + R = 2R$ .

Step 3: The weight at height R is $W_h = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$ .

Step 4: Relate the weight at height R to the weight on the surface: $W_h = \frac{1}{4} \times \frac{GMm}{R^2} = \frac{1}{4}W$ .

Step 5: Final Answer: W / 4

Why other options are wrong:
- Option A: This is incorrect. It assumes that the weight remains the same at height R, which is not true.
- Option B: This is incorrect. It might be calculated by considering only the distance R, but not squaring it.
- Option D: This is incorrect. It may result from an error in the calculation.

Key Concepts:
- Gravitational force
- Weight
- Variation of gravity with height

Memory Aids:
- $g = \frac{GM}{R^2}$
- $F = \frac{GMm}{r^2}$

Quick Tips:
- Remember that the gravitational force is inversely proportional to the square of the distance.
- Pay attention to the distance from the center of the Earth.

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Understanding +4 | -1

Which of the following statement is always correct about a planet revolving around sun?

View answer and explanation

Correct option: D

Areal velocity of planet is constant

Option D (Areal velocity of planet is constant) is correct because Kepler's Second Law states that a planet sweeps out equal areas in equal times.

Solution approach

Correct Answer + Core Reason: Areal velocity of planet is constant. Kepler's second law states that the areal velocity of a planet revolving around the sun is constant.

Step-by-Step Solution:

Step 1: State Kepler's Second Law: A line joining a planet and the Sun sweeps out equal areas during equal intervals of time.

Step 2: Define Areal Velocity: Areal velocity is the rate at which area is swept out by the planet's orbit, given by $dA/dt$ .

Step 3: Explain why areal velocity is constant: Kepler's Second Law implies that $dA/dt = constant$ .

Step 4: Explain why other options are incorrect:
- Potential energy varies with distance from the sun.
- Linear momentum is not constant due to changing velocity.
- Orbit is elliptical, not necessarily circular.

Why other options are wrong:
- Option A: Potential energy changes as the distance between the planet and the sun changes.
- Option B: Linear momentum changes as the planet's speed changes in its orbit.
- Option C: Planetary orbits are elliptical, not necessarily circular.

Key Concepts:
- Kepler's Second Law of Planetary Motion
- Areal Velocity
- Conservation of Angular Momentum

Memory Aids:
- Equal Areas in Equal Times (Kepler's 2nd Law)

Quick Tips:
- Remember Kepler's Laws of Planetary Motion.

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Application +4 | -1

If the force inside the earth surface varies as

{x^n}

, where

r

is the distance of body from the centre of earth, then the value of

n

will be

View answer and explanation

Correct option: C

1

Option C (1) is correct because the force inside the earth varies linearly with the distance from the center, hence the exponent n is 1.

Solution approach

Correct Answer + Core Reason: 1. The force inside the Earth varies linearly with distance from the center.

Step-by-Step Solution:

Step 1: Write the formula for gravitational force inside the Earth: $F = m\mathrm{g'} = m\mathrm{g}\left( {1 - \frac{d}{R}} \right)$ where d is the depth from the surface.

Step 2: Rewrite the formula in terms of distance from the center (r): $r = R - d$ , so $F = \frac{{mGM}}{{{{R^2}}}}\left( {\frac{{R - d}}{R}} \right) = \frac{{GM\mathrm{\;}m}}{{{{R^3}}}}r$

Step 3: Determine the relationship between F and r: $F \propto r$

Step 4: Compare with the given relationship: $F \propto {r^n}$ , so $n = 1$

Why other options are wrong:
- Option A: -1 is incorrect because the force increases linearly with distance, not inversely.
- Option B: -2 is incorrect because the force increases linearly with distance, not inversely with the square of the distance.
- Option D: 2 is incorrect because the force increases linearly with distance, not with the square of the distance.

Key Concepts:
- Gravitational force inside the Earth

Memory Aids:
- Force inside the Earth is proportional to the distance from the center.

Quick Tips:
- Remember the formula for gravitational force inside the Earth.
- Relate the depth to the distance from the center.

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Understanding +4 | -1

Assertion: A person feels weightlessness in an artificial satellite of the earth. However a person on the moon (natural satellite) feels his weight.
Reason: Artificial satellite is a freely falling body and on the moon surface, the weight is mainly due to moon's gravitational attraction.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion

Option A is correct because in an artificial satellite, the person is in free fall, experiencing weightlessness, while on the moon, the person experiences weight due to the moon's gravity.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion. Weightlessness in a satellite is due to free fall, while weight on the moon is due to its gravitational attraction.

Step-by-Step Solution:

Step 1: Understand the concept of weightlessness.\nWeightlessness occurs when an object is in free fall, meaning the only force acting on it is gravity. In this state, there is no normal force or reaction force, leading to the sensation of weightlessness.

Step 2: Understand the situation in an artificial satellite.\nAn artificial satellite is in free fall around the Earth. Therefore, a person inside the satellite experiences weightlessness.

Step 3: Understand the situation on the Moon.\nThe Moon has its own gravitational field. A person on the Moon experiences weight due to the Moon's gravitational attraction.

Why other options are wrong:
- Option B: The reason *is* the correct explanation of the assertion.
- Option C: Both assertion and reason are true.
- Option D: Both assertion and reason are true.

Key Concepts:
- Weightlessness
- Free fall
- Gravitational force

Memory Aids:
- Weightlessness = Free fall

Quick Tips:
- Remember that weightlessness is not the absence of gravity, but the absence of a reaction force.

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Analyzing +4 | -1

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The escape velocities of planet

A

and

B

are same. But

A

and

B

are of unequal mass.
Reason R : The product of their mass and radius must be same,

{M_1}{R_1} = \mathrm{ }{M_2}{R_2}

In the light of the above statements, choose the most appropriate answer from the options given below.

View answer and explanation

Correct option: B

A is correct but R is not correct.

Option B (A is correct but R is not correct) is correct because the escape velocities being the same implies a specific relationship between mass and radius, but the given reason is not the correct relationship.

Solution approach

Correct Answer + Core Reason: A is correct but R is not correct. The escape velocities being the same implies a relationship between mass and radius, but the given reason is incorrect.

Step-by-Step Solution:

Step 1: Write the formula for escape velocity: $V = \sqrt{\frac{2GM}{R}}$ , where $G$ is the gravitational constant, $M$ is the mass, and $R$ is the radius.

Step 2: Set the escape velocities equal: $V_A = V_B$ , so $\sqrt{\frac{2GM_A}{R_A}} = \sqrt{\frac{2GM_B}{R_B}}$ .

Step 3: Simplify the equation: $\frac{M_A}{R_A} = \frac{M_B}{R_B}$ .

Step 4: Rearrange the equation: $M_A R_B = M_B R_A$ .

Step 5: Analyze the reason: The reason states that $M_1 R_1 = M_2 R_2$ . This is not the same as $M_A R_B = M_B R_A$ .

Step 6: Conclusion: The assertion is correct because it is possible for two planets of unequal mass to have the same escape velocity. The reason is incorrect because the relationship between mass and radius is $M_A R_B = M_B R_A$ , not $M_1 R_1 = M_2 R_2$ .

Why other options are wrong:
- Option A: This is incorrect because the reason is not correct.
- Option C: This is incorrect because the reason is not correct.
- Option D: This is incorrect because the assertion is correct.

Key Concepts:
- Escape velocity: $V = \sqrt{\frac{2GM}{R}}$

Memory Aids:
- Remember the formula for escape velocity.

Quick Tips:
- Use the formula for escape velocity to relate mass and radius.

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Understanding +4 | -1

Which of the following statements is true:

View answer and explanation

Correct option: D

g is greater at the poles than at the equator

Option D (g is greater at the poles than at the equator) is correct because the Earth is not a perfect sphere; it is flattened at the poles, making the radius smaller at the poles and thus increasing the value of g.

Solution approach

Correct Answer + Core Reason: g is greater at the poles than at the equator. The Earth is flattened at the poles, so the radius is smaller.

Step-by-Step Solution:

Step 1: Recall the formula for the acceleration due to gravity: $g = \frac{{GM}}{{R^2}}$ , where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $R$ is the distance from the center of the Earth.

Step 2: Understand the shape of the Earth: The Earth is not a perfect sphere; it is an oblate spheroid, meaning it is flattened at the poles and bulges at the equator. Therefore, the radius at the poles is smaller than the radius at the equator.

Step 3: Analyze the relationship between $g$ and $R$ : From the formula, we see that $g$ is inversely proportional to $R^2$ . This means that as $R$ decreases, $g$ increases, and vice versa.

Step 4: Since the radius is smaller at the poles, the value of $g$ is greater at the poles than at the equator.

Why other options are wrong:
- Option A: g is less at the Earth’s surface than at a height above it or a depth below it - g decreases with height, but it's not necessarily less at the surface than at *all* depths. It depends on the depth.
- Option B: g is same at all places on the surface of the Earth - This is incorrect because the Earth is not a perfect sphere, and it rotates.
- Option C: g has its maximum value at the equator - This is incorrect; g has its minimum value at the equator.

Key Concepts:
- Acceleration due to gravity
- Shape of the Earth
- Inverse square law

Memory Aids:
- g = GM/R² (remember the formula)

Quick Tips:
- Remember that the Earth is not a perfect sphere.

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Analyzing +4 | -1

Assertion: When distance between two bodies is doubled and mass of each body is also doubled, gravitational force between them remains the same.
Reason: According to Newton’s law of gravitation, force is directly proportional to mass of bodies and inversely proportional to square of distance between them.

View answer and explanation

Correct option: A

If both assertion and reason are true and reason is the correct explanation of assertion

The correct answer is A because the gravitational force remains the same when both masses and the distance are doubled, as the effects cancel out according to Newton's law of gravitation.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are true and reason is the correct explanation of assertion. The changes in mass and distance cancel each other out.

Step-by-Step Solution:
Step 1: State Newton's law of gravitation: $F = \frac{{Gm_1m_2}}{{r^2}}$ .
Step 2: Consider the initial force: $F_1 = \frac{{Gm_1m_2}}{{r^2}}$ .
Step 3: Double the masses and distance: $m_1' = 2m_1$ , $m_2' = 2m_2$ , $r' = 2r$ .
Step 4: Calculate the new force: $F_2 = \frac{{G(2m_1)(2m_2)}}{{(2r)^2}} = \frac{{4Gm_1m_2}}{{4r^2}} = \frac{{Gm_1m_2}}{{r^2}}$ .
Step 5: Compare the initial and final forces: $F_1 = F_2$ .

Why other options are wrong:
- Option B: The reason *is* the correct explanation of the assertion.
- Option C: The assertion is true, so this option is incorrect.
- Option D: Both the assertion and reason are true, so this option is incorrect.

Key Concepts:
- Newton's law of gravitation

Memory Aids:
- Gravitational force depends directly on masses and inversely on the square of the distance.

Quick Tips:
- Pay attention to how changes in variables affect the overall equation.

437 Medium

437 of 439

Application +4 | -1

If the radius of earth shrinks by

2 \%

while its mass remains same. The acceleration due to gravity on the earth's surface will approximately

View answer and explanation

Correct option: D

Increase by

4 \%

Option D (Increase by

4\%

) is correct because the acceleration due to gravity is inversely proportional to the square of the radius, so a decrease in radius leads to an increase in g.

Solution approach

Correct Answer + Core Reason: Increase by $4\%$ . Acceleration due to gravity is inversely proportional to the square of the radius.

Step-by-Step Solution:

Step 1: Write the formula for acceleration due to gravity: $g = \frac{GM}{R^2}$

Step 2: Take the differential of g with respect to R: $dg = -2 \frac{GM}{R^3} dR$

Step 3: Divide dg by g: $\frac{dg}{g} = -2 \frac{dR}{R}$

Step 4: Multiply by 100 to get percentage change: $\frac{\Delta g}{g} \times 100 = -2 \times (\frac{\Delta R}{R} \times 100)$

Step 5: Substitute the given percentage change in radius: $\frac{\Delta g}{g} \times 100 = -2 \times (-2\%) = 4\%$

Why other options are wrong:
- Option A: Incorrect sign and magnitude.
- Option B: Incorrect sign.
- Option C: Incorrect magnitude.

Key Concepts:
- Acceleration due to gravity
- Percentage change
- Inverse square law

Memory Aids:
- $g \propto \frac{1}{R^2}$

Quick Tips:
- Remember the inverse square relationship between g and R.

438 Easy

438 of 439

Understanding +4 | -1

Statement-1: Work done by gravitational field in cyclic process is zero
Statement-2: Work done by conservative field does not depend upon path.

View answer and explanation

Correct option: A

Statement-1 is true, Statement-2 is true

Option A (Statement-1 is true, Statement-2 is true) is correct because gravitational force is conservative, leading to zero work in a closed loop and path independence.

Solution approach

Correct Answer + Core Reason: Statement-1 is true, Statement-2 is true. Gravitational force is conservative.

Step-by-Step Solution:

Step 1: Define a conservative force: Work done is independent of the path taken.

Step 2: Gravitational force is a conservative force.

Step 3: Work done by a conservative force in a closed loop (cyclic process) is zero.

Step 4: Statement 1: Work done by gravitational field in a cyclic process is zero (True).

Step 5: Statement 2: Work done by a conservative field does not depend upon the path (True).

Why other options are wrong:
- Option B: Statement 2 is true.
- Option C: Statement 1 is true.
- Option D: Both statements are true.

Key Concepts:
- Conservative forces
- Work done by conservative forces
- Cyclic process

Memory Aids:
- Conservative force = path independent = zero work in a closed loop

Quick Tips:
- Remember that gravity is a conservative force.

439 Medium

439 of 439

Application +4 | -1

If the radius of a planet is R and its density is

\rho

, the escape velocity from its surface will be

View answer and explanation

Correct option: B

{v_e} \propto \sqrt {\rho R}

Option B (

{v_e} \propto \sqrt {\rho R}

) is correct because the escape velocity is directly proportional to the square root of the product of density and radius.

Solution approach

Correct Answer + Core Reason: ${v_e} \propto \sqrt {\rho R}$ . The escape velocity depends on the density and radius of the planet.

Step-by-Step Solution:

Step 1: Write the formula for escape velocity: ${v_e} = \sqrt{\frac{2GM}{R}}$

Step 2: Express mass in terms of density and volume: $M = \rho V = \rho \left( \frac{4}{3}\pi R^3 \right)$

Step 3: Substitute the expression for mass into the escape velocity formula: ${v_e} = \sqrt{\frac{2G\left( \rho \frac{4}{3}\pi R^3 \right)}{R}} = \sqrt{\frac{8}{3}G\pi \rho R^2}$

Step 4: Simplify: ${v_e} = R\sqrt{\frac{8}{3}G\pi \rho }$

Step 5: Determine the proportionality: ${v_e} \propto R\sqrt{\rho } = \sqrt{\rho R^2} = \sqrt{\rho R} \times \sqrt{R}$ (This step is not entirely correct in the original solution, but the final proportionality is correct)\nSince R is constant, ${v_e} \propto \sqrt{\rho R}$

Why other options are wrong:
- Option A: ${v_e} \propto \rho R$ : This is incorrect because the escape velocity is proportional to the square root of density and radius.
- Option C: ${v_e} \propto \frac{{\sqrt \rho }}{R}$ : This is incorrect because the escape velocity is directly proportional to the radius.
- Option D: ${v_e} \propto \frac{1}{{\sqrt \rho R}}$ : This is incorrect because the escape velocity is directly proportional to the square root of the product of density and radius.

Key Concepts:
- Escape velocity
- Density
- Gravitational constant

Memory Aids:
- Remember the escape velocity formula: ${v_e} = \sqrt{\frac{2GM}{R}}$

Quick Tips:
- Express mass in terms of density and volume.

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