Gravitation Previous Year Questions — JEE Main Physics
22 past-year Gravitation questions from JEE Main (Physics) — page 2.
Q6 (2025)
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time and thus areal velocity of planet is constant. [cite: 55]
Reason (R) For a central force field the angular momentum is a constant. [cite: 56]
In the light of the above statements, choose the most appropriate answer from the options given below: [cite: 57]
- Both (A) and (R) are correct and (R) is the correct explanation of (A) [cite: 63]
- Both (A) and (R) are correct but (R) is not the correct explanation of (A) [cite: 65]
- A is correct but R is not correct [cite: 72]
- A is not correct but R is correct [cite: 73]
### Related Formula
$$\frac{dA}{dt} = \frac{L}{2m}$$ [cite: 684]
### Core Logic
Kepler's Second Law state that the areal velocity $\frac{dA}{dt}$ is directly proportional to the angular momentum $L$ of the planet[cite: 55, 684]. Because the gravitational force between the Sun and the planet acts strictly along the line joining their centers (a central force field), its torque $\vec{\tau} = \vec{r} \times \vec{F} = 0$[cite: 56, 686]. Since torque is zero, the angular momentum $L$ remains completely constant over time[cite: 686]. As a consequence, $\frac{dA}{dt} = \text{constant}$[cite: 55, 684]. Both statements are true and (R) is the correct explanation of (A)[cite: 63].
### Pattern Recognition
Areal velocity constancy is a direct geometric manifestation of the conservation of angular momentum under any central force field[cite: 55, 56, 686].
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q25 (2025)
Acceleration due to gravity on the surface of earth is 'g'. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is ____ g.
### Related Formula
Surface gravitational acceleration:
$$g = \frac{GM}{R^2}$$
### Core Logic
Since diameter drops to $1/3$, the radius $R'$ also scales down to $1/3$ its original value ($R' = R/3$), while mass $M$ remains constant.
New acceleration value $g'$ calculation:
$$g' = \frac{GM}{(R/3)^2} = 9 \cdot \frac{GM}{R^2} = 9g$$
The scaling factor is 9.
### Pattern Recognition
Gravity follows an inverse-square law with respect to radius. Shrinking the radius by a factor of $n$ increases surface gravity by $n^2$ if the mass is unchanged.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q1139 (2025)
A satellite is launched into a circular orbit of radius 'R' around the earth. A second satellite is launched into an orbit of radius 1.03 R. The time period of revolution of the second satellite is larger than the first one approximately by :-
- 3%
- 4.5%
- 9%
- 2.5%
### Related Formula
By Kepler's Third Law of Planetary Motion, the square of the orbital period $T$ is proportional to the cube of the orbital radius $R$:
$$T^{2} = K \cdot R^{3}$$
### Core Logic
Taking logs and differentiating to find fractional errors for small changes:
$$2 \frac{\Delta T}{T} = 3 \frac{\Delta R}{R}$$
$$\frac{\Delta T}{T} = \frac{3}{2} \left(\frac{\Delta R}{R}\right)$$
### Step 1: Computing Percentage Change
The change in radius is $\Delta R = 1.03R - R = 0.03R$, which means $\frac{\Delta R}{R} = 0.03$ or $3\%$.
Substitute this into our fraction scaling relation:
$$\frac{\Delta T}{T} = \frac{3}{2} \times 0.03 = 0.045 = 4.5\%$$
### Pattern Recognition
Power factors act as direct linear multipliers for small percentage shifts. Here, the scaling factor is simply $\frac{3}{2}$ times the radius change.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q11 (2025)
Earth has mass 8 \times and radius 2 \times that of a planet. If the escape velocity from the earth is $11.2 \, \text{km/s}$ , the escape velocity in $\text{km/s}$ from the planet will be:
- 11.2
- 5.6
- 2.8
- 8.4
### Related Formula
The expression for escape velocity from a spherical planetary body is given by:
$$v_{\text{escape}} = \sqrt{\frac{2GM}{R}}$$
### Core Logic
Let the planet's mass be $M_P$ and its radius be $R_P$.
According to the problem statement :
* Earth's mass, $M_E = 8 M_P \implies \frac{M_P}{M_E} = \frac{1}{8}$
* Earth's radius, $R_E = 2 R_P \implies \frac{R_E}{R_P} = 2$
Taking the ratio of escape velocities :
$$\frac{v_{P}}{v_{E}} = \sqrt{\left(\frac{M_P}{M_E}\right) \times \left(\frac{R_E}{R_P}\right)}$$
$$\frac{v_{P}}{v_{E}} = \sqrt{\frac{1}{8} \times 2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$
Given that $v_E = 11.2 \text{ km/s}$:
$$v_P = \frac{1}{2} \times 11.2 = 5.6 \text{ km/s}$$
### Pattern Recognition
Setting up quick ratios prevents substitution mistakes. For any planetary variant, notice how scaling properties scale inside the root operator directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q32 (2024)
If $R$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g = \pi^2 \mathrm{~m/s^2}$, then the length of the second's pendulum at a height $h = 2R$ from the surface of earth will be:
- $\frac{2}{9}\mathrm{~m}$
- $\frac{1}{9}\mathrm{~m}$
- $\frac{4}{9}\mathrm{~m}$
- $\frac{8}{9}\mathrm{~m}$
### Related Formula
Variation of $g$ with height:
$$g' = g\left(\frac{R}{R+h}\right)^2$$
Time period of a simple pendulum:
$$T = 2\pi\sqrt{\frac{l}{g'}}$$
### Core Logic
Given height $h = 2R$, the effective acceleration due to gravity becomes:
$$g' = g\left(\frac{R}{R+2R}\right)^2 = \frac{g}{9}$$
For a second's pendulum, the time period is defined exactly as $T = 2\mathrm{~s}$.
### Step 1: Calculate Length
Substitute $T = 2\mathrm{~s}$ and $g' = \frac{g}{9}$ into the time period formula:
$$2 = 2\pi\sqrt{\frac{l}{g/9}}$$
$$1 = \pi\sqrt{\frac{9l}{g}}$$
Squaring both sides:
$$1 = \pi^2 \cdot \frac{9l}{g}$$
Since $g = \pi^2 \mathrm{~m/s^2}$:
$$1 = 9l \implies l = \frac{1}{9}\mathrm{~m}$$
### Pattern Recognition
Second's pendulum always has $T = 2\mathrm{~s}$. Height $2R$ from the surface means a total distance of $3R$ from the center, which yields a $\frac{1}{9}$ drop in gravity.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Class 11 Physics: Oscillations
