If R is the radius of the earth and the acceleration due to gravity on the surface of earth is g = pi2 ~m/s2, then the…

Year	2026	2025	2024	Total
Questions	5	9	8	22

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Variation of $g$ $g$ with height:

g' = g((R)/(R+h))²

Time period of a simple pendulum:

T = 2π\sqrt((l)/(g'))

Core Logic

Given height $h = 2R$ $h = 2R$, the effective acceleration due to gravity becomes:

g' = g((R)/(R+2R))² = (g)/(9)

For a second's pendulum, the time period is defined exactly as $T = 2~s$ $T = 2\mathrm{~s}$.

Step 1: Calculate Length

Substitute $T = 2~s$ $T = 2\mathrm{~s}$ and $g' = (g)/(9)$ $g' = \frac{g}{9}$ into the time period formula:

2 = 2π\sqrt((l)/(g/9))

1 = π\sqrt((9l)/(g))

Squaring both sides:

1 = π² \cdot (9l)/(g)

Since $g = π² ~m/s²$ $g = \pi^2 \mathrm{~m/s^2}$:

1 = 9l l = (1)/(9)~m

Pattern Recognition

Second's pendulum always has $T = 2~s$ $T = 2\mathrm{~s}$. Height $2R$ $2R$ from the surface means a total distance of $3R$ $3R$ from the center, which yields a $(1)/(9)$ $\frac{1}{9}$ drop in gravity.

Chapter Mix

Class 11 Physics: Gravitation Class 11 Physics: Oscillations

More Gravitation Previous-Year Questions — Page 5

Q46 jee_main_2024_31_jan_evening Escape Velocity

The mass of the moon is

1/144

times the mass of a planet and its diameter

1/16

times the diameter of a planet. If the escape velocity on the planet is v, the escape velocity on the moon will be:

A. $(v)/(3)$ $\frac{v}{3}$
B. $(v)/(4)$ $\frac{v}{4}$
C. $(v)/(12)$ $\frac{v}{12}$
D. $(v)/(6)$ $\frac{v}{6}$

Solution

Related Formula

vescape = \sqrt((2GM)/(R))

Core Logic

For the planet: $v = \sqrt((2GMₚ)/(Rₚ))$ $v = \sqrt{\frac{2GM_p}{R_p}}$ For the moon: $Mm = (Mₚ)/(144)$ $M_m = \frac{M_p}{144}$ and $Rm = (Rₚ)/(16)$ $R_m = \frac{R_p}{16}$.

Step 1: Setup the Ratio

vm = \sqrt((2G Mm)/(Rm))

vm = \sqrt((2G ((Mₚ)/(144)))/(((Rₚ)/(16))))

vm = \sqrt((2G Mₚ)/(Rₚ) \times (16)/(144))

Step 2: Simplification

vm = \sqrt((2G Mₚ)/(Rₚ)) \times \sqrt((1)/(9))

vm = v \times (1)/(3) = (v)/(3)

Pattern Recognition

Escape velocity scales as $\sqrt(M/R)$ $\sqrt{M/R}$. If $M$ $M$ scales by $x$ $x$ and $R$ $R$ scales by $y$ $y$, velocity scales by $\sqrt(x/y)$ $\sqrt{x/y}$. Here, $\sqrt((1/144)/(1/16)) = \sqrt(16/144) = \sqrt(1/9) = 1/3$ .

Chapter Mix

Class 11 Physics: Gravitation

Q jee_main_2024_31_jan_morning Superposition Principle

Four identical particles of mass

m

are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is

( 2\sqrt(2) + 132) Gm²L²

, the length of the sides of the square is

A. $L2$ $\frac{\mathrm{L}}{2}$
B. $4 L$ $4 L$
C. $3L$ $3L$
D. $2 L$ $2 L$

Solution

Related Formula

F = (G m₁ m₂)/(r²)

Core Logic

Superposition Principle diagram for Q37 - JEE Main 2024 Morning

Let the side length of the square be $a$ $a$. Considering one corner mass, it experiences forces from the adjacent two masses (distance $a$ $a$) and the diagonally opposite mass (distance $\sqrt(2)a$ $\sqrt{2}a$).

The forces from the two adjacent masses are at $90^°$ $90^\circ$ to each other:

F = (Gm²)/(a²)

The resultant of these two is $\sqrt(2)F = \sqrt(2) (Gm²)/(a²)$ $\sqrt{2}F = \sqrt{2} \frac{Gm^2}{a^2}$, directed along the diagonal.

Step 2: Total Force Equation

The force from the diagonal mass is:

F' = Gm²(\sqrt(2)a)² = (Gm²)/(2a²)

Total resultant force $Fₙₑₜ = \sqrt(2)F + F'$ $F_{\text{net}} = \sqrt{2}F + F'$:

Fₙₑₜ = \sqrt(2) (Gm²)/(a²) + (Gm²)/(2a²) = (Gm²)/(a²) ( \sqrt(2) + (1)/(2) )

Fₙₑₜ = (Gm²)/(a²) ( 2\sqrt(2) + 12 )

Equating this to the given force value:

( 2\sqrt(2) + 132)(Gm²)/(L²) = (Gm²)/(a²) ( 2\sqrt(2) + 12 )

(1)/(32 L²) = (1)/(2 a²)

$a² = 16 L²$ $a^2 = 16 L^2$ $a = 4L$ $a = 4L$

Chapter Mix

Class 11 Physics: Gravitation

If $R$ $R$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g = π² ~m/s²$ $g = \pi^2 \mathrm{~m/s^2}$, then the length of the second's pendulum at a height $h = 2R$ $h = 2R$ from the surface of earth will be:

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Length

Pattern Recognition

Chapter Mix

Reference Study Guides

More Gravitation Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Setup the Ratio

Step 2: Simplification

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Total Force Equation

Chapter Mix

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If R$R$ is the radius of the earth and the acceleration due to gravity on the surface of earth is g = π² ~m/s²$g = \pi^2 \mathrm{~m/s^2}$, then the length of the second's pendulum at a height h = 2R$h = 2R$ from the surface of earth will be:

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Length

Pattern Recognition

Chapter Mix

Reference Study Guides

More Gravitation Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Setup the Ratio

Step 2: Simplification

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Total Force Equation

Chapter Mix

More Gravitation Questions — jee_main_2024_01_february_morning

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If $R$ $R$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g = π² ~m/s²$ $g = \pi^2 \mathrm{~m/s^2}$, then the length of the second's pendulum at a height $h = 2R$ $h = 2R$ from the surface of earth will be: