Gravitation Previous Year Questions — JEE Main Physics
22 past-year Gravitation questions from JEE Main (Physics) — page 3.
Q44 (2024)
A planet takes $200\text{ days}$ to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution?
- $25$
- $50$
- $100$
- $20$
### Related Formula
According to Kepler's Third Law (Law of Periods):
$T^2 \propto r^3$
where:
* $T$ is the time period of revolution.
* $r$ is the orbital radius of the planet.
### Core Logic
Using the proportionality relationship for two states:
$$\frac{T_2^2}{T_1^2} = \left( \frac{r_2}{r_1} \right)^3$$
Given:
* $T_1 = 200\text{ days}$
* $r_2 = \frac{r_1}{4}$
### Step 1: Calculate the New Time Period
Substitute the values into the proportionality relation:
$$\frac{T_2^2}{(200)^2} = \left( \frac{r_1 / 4}{r_1} \right)^3 = \left( \frac{1}{4} \right)^3 = \frac{1}{64}$$
Taking the square root on both sides:
$$\frac{T_2}{200} = \sqrt{\frac{1}{64}} = \frac{1}{8}$$
$$T_2 = \frac{200}{8} = 25\text{ days}$$
### Pattern Recognition
If orbital distance scales by $x$, the period scales by $x^{3/2}$. Here, distance scales by $\frac{1}{4}$, so the period scales by $\left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}$. Thus, $200 \times \frac{1}{8} = 25\text{ days}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q35 (2024)
The acceleration due to gravity on the surface of earth is $g$.
If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be:
- $\frac{g}{4}$
- $2g$
- $\frac{g}{2}$
- $4g$
### Related Formula
$$g = \frac{GM}{R^2} \implies g \propto \frac{1}{R^2}$$
### Core Logic
Since diameter reduces to half, the radius $R_{2}$ also reduces to half of its initial value $R_{1}$:
$$R_{2} = \frac{R_{1}}{2}$$
Setting up the ratio:
$$\frac{g_{2}}{g_{1}} = \left(\frac{R_{1}}{R_{2}}\right)^{2} = \left(\frac{R_{1}}{R_{1}/2}\right)^{2} = 4$$
### Step 1: Final Calculation
$$g_{2} = 4g_{1} = 4g$$
### Pattern Recognition
Inverse square dependence means halving the distance scale amplifies the surface field metric by a factor of $2^2 = 4$ matching constant mass bounds.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q1637 (2024)
At what distance above and below the surface of the earth a body will have same weight, (take radius of earth as R.)
- $\sqrt{5} \mathrm{R} - \mathrm{R}$
- $\frac{\sqrt{3} \mathrm{R} - \mathrm{R}}{2}$
- $\frac{R}{2}$
- $\frac{\sqrt{5} \mathrm{R} - \mathrm{R}}{2}$
### Related Formula
Acceleration due to gravity at a height $h$ above the Earth's surface:
$$g_p = \frac{g R^2}{(R + h)^2}$$
Acceleration due to gravity at a depth $h$ below the Earth's surface:
$$g_q = g \left(1 - \frac{h}{R}\right)$$
{{SOL_IMG1}}
### Core Logic
We need the weights to be identical, meaning $g_p = g_q$ at the exact same distance value $h$:
$$\frac{g R^2}{(R + h)^2} = g \left(1 - \frac{h}{R}\right)$$
Dividing by $g$ and simplifying the left side denominator fraction:
$$\frac{1}{\left(1 + \frac{h}{R}\right)^2} = 1 - \frac{h}{R}$$
$$\left(1 - \frac{h}{R}\right)\left(1 + \frac{h}{R}\right)^2 = 1$$
### Step 1: Set Up Algebraic Equation
Let $\frac{h}{R} = x$. Then:
$$(1 - x)(1 + x)^2 = 1$$
$$(1 - x)(1 + 2x + x^2) = 1$$
$$1 + 2x + x^2 - x - 2x^2 - x^3 = 1$$
$$x - x^2 - x^3 = 0$$
### Step 2: Solve for x
Since $x \neq 0$ (distance cannot be zero), divide by $x$:
$$1 - x - x^2 = 0 \implies x^2 + x - 1 = 0$$
Solving via quadratic formula:
$$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$$
Since distance parameter $x \gt 0$, we take the positive root:
$$x = \frac{\sqrt{5} - 1}{2}$$
### Step 3: Find Height h
Substitute back $x = \frac{h}{R}$:
$$\frac{h}{R} = \frac{\sqrt{5} - 1}{2} \implies h = \frac{R}{2} (\sqrt{5} - 1) = \frac{\sqrt{5}R - R}{2}$$
Therefore, the required distance is $\frac{\sqrt{5}R - R}{2}$.
### Pattern Recognition
Do not use the linear approximation formula $g_h \approx g(1 - \frac{2h}{R})$ unless the problem explicitly states $h \ll R$. Equating the approximated form to depth gives $h_{\text{height}} = \frac{1}{2} h_{\text{depth}}$, which fails when looking for a single unified distance value $h$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q42 (2024)
Escape velocity of a body from earth is $11.2 \,\mathrm{km/s}$. If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is:
- $11.2 \mathrm{~km / s}$
- $8.4 \mathrm{~km / s}$
- $4.2 \mathrm{~km / s}$
- $7.9 \mathrm{~km / s}$
### Related Formula
$$V_e = \sqrt{\frac{2GM}{R}}$$
### Core Logic
For Earth: $V_e = \sqrt{\frac{2GM_E}{R_E}} = 11.2 \mathrm{~km/s}$
For the planet:
$\mathrm{R}_{\mathrm{P}} = \frac{\mathrm{R}_{\mathrm{E}}}{3}$ and $\mathrm{M}_{\mathrm{P}} = \frac{\mathrm{M}_{\mathrm{E}}}{6}$
We can express the escape velocity of the planet $V_p$ as a ratio of the Earth's escape velocity.
### Step 1: Ratio of Velocities
$$\frac{V_p}{V_e} = \sqrt{\frac{M_p}{M_E} \times \frac{R_E}{R_p}}$$
$$\frac{V_p}{V_e} = \sqrt{\left(\frac{1}{6}\right) \times \left(\frac{3}{1}\right)} = \sqrt{\frac{1}{2}}$$
### Step 2: Calculate Escape Velocity
$$V_p = \frac{V_e}{\sqrt{2}}$$
$$V_p = \frac{11.2}{1.414} \approx 7.92 \mathrm{~km/s}$$
### Pattern Recognition
Any scaling of a planet's mass by factor $\alpha$ and radius by factor $\beta$ scales the escape velocity by a factor of $\sqrt{\alpha / \beta}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
Q42 (2024)
The gravitational potential at a point above the surface of earth is $-5.12 \times 10^{7} \mathrm{~J / kg}$ and the acceleration due to gravity at that point is $6.4 \mathrm{~m/s^2}$. Assume that the mean radius of earth to be $6400 \mathrm{~km}$. The height of this point above the earth's surface is:
- $1600 \mathrm{\,km}$
- $540 \mathrm{\,km}$
- $1200 \mathrm{\,km}$
- $1000 \mathrm{\,km}$
### Related Formula
$$V = -\frac{GM_E}{R_E + h}$$
$$g' = \frac{GM_E}{(R_E + h)^2}$$
### Core Logic
The gravitational potential ($V$) and acceleration due to gravity ($g'$) at a distance $r = R_E + h$ from the center of the earth can be related by dividing their magnitudes: $|V| / g' = r$.
### Step 1: Set Up Equations
From the given data:
$$-\frac{GM_E}{R_E + h} = -5.12 \times 10^7 \quad \dots (i)$$
$$\frac{GM_E}{(R_E + h)^2} = 6.4 \quad \dots (ii)$$
### Step 2: Isolate Variable
Divide equation (i) by (ii) (taking magnitudes):
$$R_E + h = \frac{5.12 \times 10^7}{6.4}$$
$$R_E + h = 0.8 \times 10^7 \mathrm{~m} = 8000 \mathrm{~km}$$
### Step 3: Solve for h
Given mean radius of the earth $R_E = 6400 \mathrm{~km}$:
$6400 + h = 8000$
$$h = 1600 \mathrm{~km}$$
### Pattern Recognition
Always exploit the $V/g = r$ relationship to extract distances cleanly without having to substitute large values for $G$ or $M_E$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
