JEE Main · Physics ↓ Falling

Gravitation appeared 22 times across 3 years — 2.5% of Physics. This question is from Gravitational Force and Superposition.

Year	2026	2025	2024	Total
Questions	5	9	8	22

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Net gravitational force at the centre of a square is found to be $F₁$ $F_{1}$ when four particles having mass M, 2M, 3M and 4M are placed at the four corners of the square as shown in figure and it is $F₂$ $F_{2}$ when the positions of 3M and 4M are interchanged. The ratio $F₁F₂$ $\frac{F_{1}}{F_{2}}$ is $α\sqrt(5)$ $\frac{\alpha}{\sqrt{5}}$. The value of $α$ $\alpha$ is \_\_\_\_.
Four point masses placed at the corners of a square.

A. 2
B. 3
C. 1
D. $2\sqrt(5)$ $2\sqrt{5}$

Solution & Explanation

Related Formula

F = (G m₁ m₂)/(r²)

Core Logic

Solution diagram for Q35 - JEE Main 2026 Morning — Four point masses placed at the corners of a square.

Initial configuration:

$F₁ = 2\sqrt(2) (Gmm₀)/(r²)$|$

New configuration (after interchanging 3M and 4M):

F_2 = \sqrt{10} \frac{Gmm_0}{r^2}

Ratio:

\frac{F_1}{F_2} = \frac{2\sqrt{2}}{\sqrt{10}} = \frac{2}{\sqrt{5}} \implies \alpha = 2$$

Pattern Recognition

Sees: Gravitational force vector sum at center of square. Shortcut: Compute net vector resultants for both mass configurations and take ratio. Check: Matches option (1). ✓

Chapter Mix

Class 11 Physics: Gravitation

Solution diagram for Q35 - JEE Main 2026 Morning — Four point masses placed at the corners of a square.

Solution diagram for Q35 - JEE Main 2026 Morning — Four point masses placed at the corners of a square.

Solution diagram for Q35 - JEE Main 2026 Morning — Four point masses placed at the corners of a square.

Reference Study Guides

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

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