Test: Universal Law of Gravitation - NEET MCQ

The correct answer is Option C - Directly proportional to the product of their masses and inversely proportional to the square of the distance between them

The magnitude of the gravitational force between two point masses m₁ and m₂ separated by a distance r is given by F = G m₁ m₂ / r².

Here G is the gravitational constant with approximate value 6.67 × 10^-11 N m² kg^-2.

From the formula, the force is directly proportional to the product of the masses, so F ∝ m₁ m₂; for example, doubling one mass doubles the force.

The force is inversely proportional to the square of the distance, so F ∝ 1 / r²; for example, doubling r reduces the force to one quarter.

In vector form the force is attractive along the line joining the masses: ⃗F = -G m₁ m₂ / r² r̂, the negative sign indicating attraction toward the other mass.

These proportionalities and the formula confirm that Option C correctly states the dependence of gravitational force on the masses and separation.

Topic in NCERT: Universal law of gravitation

Line in NCERT: "every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them."

The correct answer is Option A - 6.67 × 10⁻¹¹ N m² kg⁻²

F = G m₁ m₂ / r² is the mathematical form of the universal law of gravitation, where G is the constant of proportionality.

Rearranging gives G = F r² / (m₁ m₂), so the SI units of G follow from the units of force, distance and mass.

Thus [G] = N · m² · kg^-2, which matches the unit shown in the correct option.

The experimentally determined numerical value in SI units is approximately 6.67 × 10⁻¹¹, so the full SI expression is 6.67 × 10⁻¹¹ N m² kg^-2.

Topic in NCERT: The gravitational constant

Line in NCERT: "the currently accepted value is g = 6.67 × 10^-11 n m²/kg²."

The correct answer is Option D - 6.67 × 10⁻⁵ N

F = G \t m1 \t m2 / r²

Given G = 6.67 × 10^-11 N m² kg^-2, m₁ = 100 kg, m₂ = 10000 kg, r = 1 m.

m₁ · m₂ = 100 × 10000 = 1000000

Substituting, F = 6.67 × 10^-11 × 1000000 / 1² = 6.67 × 10^-5 N

Hence the magnitude of the gravitational force is 6.67 × 10⁻⁵ N, which matches the stated option.

Topic in NCERT: Universal law of gravitation

Line in NCERT: "newton's law of universal gravitation states that the gravitational force of attraction between any two particles of masses m₁ and m₂ separated by a distance r has the magnitude f = g(m₁m₂/r²)."

The correct answer is Option C - One-fourth of the original value

According to Newton's law of universal gravitation, the magnitude of the force between two point masses is given by

F = G m₁ m₂ / r²

For a new separation r' = 2r, the gravitational force becomes

F' = G m₁ m₂ / (r')²

F' = G m₁ m₂ / (2r)² = G m₁ m₂ / (4 r²)

Since F = G m₁ m₂ / r², it follows that F' = F / 4.

Therefore the gravitational force becomes one-fourth of the original value, confirming Option C.

Topic in NCERT: Universal law of gravitation

Line in NCERT: "every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them."

The correct answer is Option A - A, C and D only

Statement A is true because gravitational force between two masses is always attractive; two masses always attract each other under classical gravity.

Statement B is false because gravitational interaction depends on the masses involved and their separation, not on the material medium; gravity acts through vacuum as well as through matter.

Statement C is true since a central force is one that acts along the line joining two bodies and depends only on their separation; gravitational force meets both conditions.

Statement D is true because the magnitude of gravitational force follows an inverse-square law.

F = G m₁ m₂ / r²

Statement E is false; gravity is the weakest of the fundamental interactions, much weaker than the electromagnetic and strong nuclear forces.

Topic in NCERT: Gravitation

Line in NCERT: "the gravitational force is attractive, i.e., the force f is along - r."

The correct answer is Option C - M⁻¹ L³ T⁻²

Using the law of gravitation between two point masses, F = G m₁ m₂ / r².

The dimensional symbol of force is [F] = M L T^-2.

The dimensional symbols of mass and distance are [m] = M and [r] = L, respectively.

Rearranging the relation to find G gives [G] = [F][r]² / [m]².

Substituting dimensions: [G] = (M L T^-2)(L²)/(M²).

Simplifying yields [G] = M^-1 L³ T^-2, which matches Option C.

Topic in NCERT: Dimensions unit

Line in NCERT: "dimensions unit [m l³ t⁻²]"

The correct answer is Option B - (A)-(II), (B)-(I), (C)-(III), (D)-(IV)

F = G m₁ m₂ / r²

The correct answer is Option C - 16 times the original force

According to Newton's law of gravitation, the force is given by F = G m₁ m₂ / r², where G is the gravitational constant.

Both masses are doubled, so they become 2 m₁ and 2 m₂, and the distance is halved to r/2.

The new force is F' = G (2 m₁)(2 m₂) / (r/2)².

Simplifying, the numerator gives 4 m₁ m₂ and the denominator is r²/4, so F' = G (4 m₁ m₂) / (r²/4).

Thus F' = 4 × 4 × (G m₁ m₂ / r²) = 16 F, i.e., the force becomes 16 times the original value.

Topic in NCERT: Universal law of gravitation

Line in NCERT: "every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them."

The correct answer is Option B - Both statements are correct but Statement II does not explain Statement I

Statement I is true because, in classical gravity, the force depends only on the masses involved and their separation; the intervening material does not change the gravitational interaction between the masses.

F = G m₁ m₂ / r²

In the formula above, G is the gravitational constant, and m₁, m₂ and r are the two masses and the distance between them; none of these quantities is altered by placing a medium between the masses, and classical gravity has no shielding, so the force is independent of the medium.

Statement II is also true because the gravitational force between two point masses is directed along the line of the relative position vector; for extended bodies with spherical symmetry, the shell theorem shows the net force is as if the entire mass were concentrated at the centre, so the direction is along the line joining their centres.

Statement II gives the direction of the force but does not address the dependence (or independence) on the intervening medium; independence follows from the dependence on masses, distance and G, not from the force's direction.

Statement II does not explain Statement I. Therefore Option B is correct.

Topic in NCERT: Gravitational forces and shielding

Line in NCERT: "gravitational shielding is not possible."

The correct answer is Option C - √3 × G × m² / a²

Each of the other two masses exerts a gravitational force of magnitude Gm²/a² on the chosen mass.

The two forces are symmetrically placed with an angle of 60° between them.

For two equal forces F with angle θ between them, the resultant magnitude is R = 2F cos(θ/2).

Here F = Gm²/a² and θ = 60°, so cos(θ/2) = cos30° = √3/2.

R = 2 × (Gm²/a²) × (√3/2)

R = √3 × Gm²/a²

Thus the magnitude of the resultant gravitational force equals √3 × G × m² / a², which corresponds to Option C.

Topic in NCERT: Gravitation

Line in NCERT: "example 7.2 three equal masses of mkg each are fixed at the vertices of an equilateral triangle abc."

The correct answer is Option D - 8F / 3

Using Newton's law of gravitation, the force is given by F = G m₁ m₂ / r².

One mass is increased by 50%, so it becomes 1.5 m₁ (= 3/2 m₁), and the separation is reduced to 75% of original, so r' = 0.75 r (= 3/4 r).

The new force is F' = G (1.5 m₁) m₂ / (0.75 r)².

Taking the ratio with the original force, F'/F = (3/2) / (3/4)².

Evaluating, (3/2) / (9/16) = (3/2) × (16/9) = 8/3, so F' = (8/3) F.

Therefore the new force equals 8F / 3, which matches Option D.

The correct answer is Option A - A, B and D only

Topic in NCERT: Gravitational forces and shielding

Line in NCERT: "gravitational shielding is not possible."

The correct answer is Option C - 10R

Because there is no external force, the position of the centre of mass is fixed; hence the displacements of the two masses about their initial positions satisfy m₁Δx₁ = m₂Δx₂.

The reduction in separation between the centres from initial to contact is 12R - 3R = 9R, so the magnitudes of the displacements satisfy Δx₁ + Δx₂ = 9R.

With m₁ = M and m₂ = 5M, the centre-of-mass relation gives Δx₁ = 5Δx_{2. Substituting into the sum, 5Δx2 + Δx2 = 9R, so 6Δx2 = 9R and Δx2 = 1.5R.}

Therefore Δx₁ = 5×1.5R = 7.5R, so the smaller body travels 7.5R before collision.

Since 7.5R is not among the listed options, none of the given choices is correct.

The correct answer is Option B - Statement I is incorrect but Statement II is correct

The law of universal gravitation gives the force between two point masses as F = G m₁ m₂ / r², directed along the line joining them, where G is the gravitational constant.

Statement I is incorrect because the law itself is universal and applies to every pair of mass elements; the simple closed form F = G m₁ m₂ / r² is valid directly only for point masses or when an extended body is spherically symmetric and the force on an external point can be treated as if the entire mass were concentrated at the centre.

Statement II is correct because, for an extended body, the total gravitational force is obtained by the principle of superposition by summing (integrating) contributions of all mass elements. For a point mass m interacting with a continuous mass distribution with density ρ(r'), a differential force is dF = G m·dm / r² (with vector direction along the line joining the elements), and the net force is found by integrating over the body's volume.

In vector form, for a point at position r acted on by a continuous distribution ρ(r'), the force can be written as F = G m ∫ ρ(r') (r' - r) / |r' - r|³ dV', which implements the required integration over all mass elements.

Therefore, the correct assessment is that the first statement is false while the second statement is true, so Option B is correct.

Topic in NCERT: Gravitation

Line in NCERT: "before we can apply eq. (7.5) to objects under consideration, we have to be careful since the law refers to point masses whereas we deal with extended objects which have finite size."

The correct answer is Option D - 4 : 1

According to Newton's law of gravitation, the gravitational force between two point masses is given by F = G M m / r².

Let the force at the first distance be F₁ = G M m / r².

At twice the distance the force is F₂ = G M m / (2r)² = G M m / 4 r².

Therefore, the ratio F₁ : F₂ = (G M m / r²) : (G M m / 4 r²) which simplifies to 4 : 1.

Topic in NCERT: Gravitation

Line in NCERT: "the force due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre."