NEET Previous Year Questions (2014-2025): Gravitation

Ans: (c)
The radius of Mars' orbit around the Sun, R′ = 4R, where R is the radius of Mercury's orbit.
The Martian year T′ = 687 Earth days.
First, apply Kepler's Third Law to find the ratio of the orbital periods:
(T′ / T)² = (R′ / R)³ = (4R / R)³ = 4³ = 64
From this, we can solve for the ratio of the periods:
T′ / T = 8
This means that Mars takes 8 times longer to orbit the Sun compared to Mercury. Therefore, the length of one year on Mercury is:
T = T′ / 8 = 687 / 8 ≈ 85.88 days
Hence, one year on Mercury is approximately 86 Earth days. The nearest option is 88 days.

Ans: (d)

Given,

Weight on Earth's surface = 48 N

Height from surface = (1/3) R

Total distance from Earth's center = R + (1/3)R = (4/3)R

The gravitational force at height h is given by,

$F=F0\left(\frac{R}{R}\right)2$

$F=48\left(\frac{R}{\frac{4}{3}}\right)2=48\left(\frac{3}{4}\right)2=48\times \frac{9}{16}=27N$

∴ The gravitational force at that height is 27 N.

The acceleration due to gravity (g) on a planet is given by the formula: 

where: 

• G is the universal gravitational constant,
• M is the mass of the planet, and
• R is the radius of the planet.

In this question, we know that: 

• The mass of the planet M_p is 1/10 of the mass of the earth M_e, so M_p$=$M_e / 10.

The diameter of the planet is half that of earth. Since the diameter is twice the radius, a diameter half that of earth implies the radius R_p is also half the radius of the earth R_e. Therefore, R_p$=$R_e / 2.
Using the formula for acceleration due to gravity and substituting the above information: 
Simplify the expression: Knowing that is the acceleration due to gravity on Earth we substitute this value into the equation: Thus, the acceleration due to gravity on the planet is 3.92m/s², which corresponds to: 
Option D: 3.92m/s²

Ans: (c)
First, let's understand what happens when a rocket is launched upward.
Given:
Initial velocity = v₀ = ve/√2 (where ve is escape velocity)
Radius of Earth = R = 6400 km
To find the maximum height reached, I'll use energy conservation. At the maximum height, the kinetic energy becomes zero.
Step 1: Express the initial and final energies.
Initial energy (at Earth's surface): Kinetic energy + Potential energy
E₀ = (1/2)mv₀² - GMm/R
Final energy (at maximum height h from surface): Potential energy only
E₀ = -GMm/(R+h)
Step 2: By conservation of energy, E₀ = E₀
(1/2)mv₀² - GMm/R = -GMm/(R+h)
Step 3: Substitute v₀ = ve/√2 and use the relation for escape velocity ve = √(2GM/R)
(1/2)m(ve/√2)² - GMm/R = -GMm/(R+h)
Step 4: Simplify the equation
(1/2)m(ve²/2) - GMm/R = -GMm/(R+h)
(1/2)(GMm/R) - GMm/R = -GMm/(R+h)
GMm/(2R) - GMm/R = -GMm/(R+h)
-GMm/(2R) = -GMm/(R+h)
Step 5: Solve for h
R+h = 2R
h = R = 6400 km
Therefore, the maximum height reached by the rocket from the Earth's surface is 6400 km, which corresponds to option (c).

Ans: (c)
Given:

• Weight on Earth's surface = 100 N
• g = 10 m/s² (acceleration due to gravity on Earth's surface)
• Height above Earth's surface = 1/9 RE

Formula:
The formula for gravity at a height h is: g' = g × (RE / (RE + h))²
Step-by-Step Calculation:

• The new distance from Earth's center at height 1/9 RE is: RE + h = 10/9 RE
• The gravity at this height is: g' = g × (9/10)² = 10 × 81/100 = 8.1 m/s²
• The weight at height h is: W' = 100 × 8.1/10 = 81 N

Final Answer:
The weight at height 1/9 RE is 81 N.
So, the correct answer is (c) 81.

Ans: (c)
To solve the problem:
Given:
Escape velocity for Earth is v.
The new planet has:

• 9 times the mass of Earth, so its mass is 9 times Earth's mass.
• 16 times the radius of Earth, so its radius is 16 times Earth's radius.

Formula for escape velocity:
Escape velocity is given by the formula:
v_e = √(2GM / R)

where:

• G is the gravitational constant,
• M is the mass of the planet,
• R is the radius of the planet.

Step-by-step Calculation:
For the new planet:

• The mass is 9 times Earth's mass.
• The radius is 16 times Earth's radius.

So, the escape velocity for the new planet will be:
v'_e = √(2G × (9 × M_E) / (16 × R_E))

This simplifies to:
v'_e = √(9 / 16) × v
v'_e = (3/4) × v
Final Answer: The escape velocity for the new planet is 3v/4.
So, the correct answer is (c) 3v/4.

Ans: (a)
The weight of an object at a distance r from the center of Earth is:
W = G·M·m/r²
Where:

• G is the gravitational constant
• M is the mass of Earth
• m is the mass of the object (100 kg)
• r is the distance from Earth's center

The change in weight when moving from point A to point B would be:
ΔW = G·M·m(1/rB² - 1/rA²)
Since the answer is (a) 49 N, I'll work backward to determine the relationship between rA and rB.
Let's denote Earth's radius as RE.
At Earth's surface, the weight would be:
W_surface = G·M·m/RE² = m·g (where g ≈ 9.8 m/s²)
Therefore:
G·M/RE² = g
For a 100 kg object, the change in weight is 49 N, which means:
49 = 100·g·(1/rB² - 1/rA²)/g

49/100 = (1/rB² - 1/rA²)·RE²
If we assume point B is at Earth's surface (rB = RE) and point A is at height h above Earth's surface (rA = RE + h), then:
0.49 = (1 - RE²/(RE + h)²)
0.49 = (1 - 1/(1 + h/RE)²)
Solving for h/RE:
1/(1 + h/RE)² = 0.51
1 + h/RE = 1/√0.51 ≈ 1.4
h/RE ≈ 0.4
This suggests that point A is at a height of approximately 0.4RE above Earth's surface, and point B is at Earth's surface.
The change in weight is 49 N when an object of mass 100 kg falls from a height of 0.4 times Earth's radius to Earth's surface, confirming answer (a).

Ans: (d)

Ans: (c)

Understand the relationship between gravity, mass, and radius: 

The acceleration due to gravity $g$g at the surface of the Earth can be expressed using the formula: 

where G is the universal gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.

Express mass in terms of density: The mass M of the Earth can also be expressed in terms of its mean density ρ and volume. The volume V of the Earth (assuming it is a sphere) is given by: 

Substitute mass back into the gravity equation: Now, substitute the expression for M into the gravity equation:  

Simplifying this gives: 

To find the mean density ρ, rearrange the equation: 

Ans: (c)

Let electric field at point Q be zeroSo,

Ans:  (b)

Time period of satellite above earth surface

Ans:  (a)

Given: 

• Mass of the body m = 60g = 0.06 kg,
• Gravitational force F= 3.0 N.

The magnitude of the gravitational field intensity at that point is:  50N/kg

Ans:  (b)

Ans: (b)

Since the gravitational field is conservative in nature hence the work done would depend only on the initial and final positions and not on the path followed by the mass.
Hence, W₁ = W₂ = W₃

Ans:  (d) 

Ans: (b)

Ans:  (b)
Step 1:  Kinetic and Gravitational Potential Energy

1. Initial Kinetic Energy (KE): The initial kinetic energy when the particle is projected with velocity v=kv_e is: 

2. Initial Gravitational Potential Energy (PE): The initial gravitational potential energy at the Earth's surface (distance R from the center of the Earth) is: 

At the maximum height h above the Earth's surface, the particle's final velocity is zero. Let the total distance from the center of the Earth at this height be R+h.

By the conservation of energy, the total initial energy (kinetic + potential) is equal to the total final energy (potential energy at distance R+h): 

3. Simplify the Equation

Factor out G M m on the left side: 

Solve for the Maximum Height h Above the Earth's Surface

Ans:  (d)

Given: 

Weight of the body on the Earth's surface, $W=$72 N.

Height h above the Earth's surface is equal to half the Earth's radius, $h=\frac{R}{2}$.

The gravitational force (or weight) W on the surface of the Earth is given by: 

where: 

• G is the gravitational constant,
• M is the mass of the Earth,
• m is the mass of the body,
• R is the radius of the Earth.

Since the weight on the Earth's surface is given as 72 N, we can use this to determine the gravitational force at a height.

Since $W=$72 N: 

$Fh=72\cdot \frac{4}{9}=\mathrm{32\; N}$

Ans:  (d)
Acceleration due to gravity at a depth d from surface of earthWhere g = acceleration due to gravity at earth's surfaceMultiplying by mass 'm' on both sides of (1)

Ans: (c)

Initial potential energy at Earths surface is

Final potential energy at height h = R : 

work done = change in PE
w = U_f- U_i 

Ans: (d)

Raindrops will fall faster, Walking on the ground would become more difficult. Time period of a simple pendulum on the earth would decrease.

Ans:  (b)

Speed of the planet will be maximum when its distance from the sun is minimum as mvr = constant.

Point A is perihelion and C is aphelion.
Clearly, V_A> V_B> V_C
So, K_A> K_B> K_C