Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length a. When released…

Year	2026	2025	2024	Total
Questions	5	9	8	22

Q44 jee_main_2026_21_jan_morning Satellite Energy

Initially a satellite of 100 kg is in a circular orbit of radius

1.5RE

. This satellite can be moved to a circular orbit of radius

3RE

by supplying

α \times 10⁶J

of energy. The value of

α

is ____. (Take Radius of Earth

RE = 6 \times 10⁶ m

and

g = 10 m/s²

)

A. 150
B. 500
C. 100
D. 1000

Solution

Related Formula

E = (-GME m)/(2r)

Δ E = Ef - Eᵢ

g = (GME)/(RE²) GME = g RE²

Core Logic

Energy of a satellite in a circular orbit is given as $E = (-GME m)/(2r)$ $E = \frac{-GM_E m}{2r}$ where r is the radius of the circular orbit. Required energy to be supplied $Δ E = Ef - Eᵢ$ $\Delta E = E_f - E_i$:

Δ E = ( (-GME m)/(2(3RE)) ) - ( (-GME m)/(2(1.5RE)) )

Δ E = (-GME m)/(6RE) + (GME m)/(3RE) = (GME m)/(6RE)

Step 1: Evaluate Delta E

Substitute $GME = g RE²$ $GM_E = g R_E^2$ into the expression:

Δ E = ((g RE²) m)/(6RE) = (1)/(6) m g RE

= (1)/(6) \times 100 \times 10 \times (6 \times 10⁶)

= 1000 \times 10⁶ J

Comparing with $α \times 10⁶ J$ $\alpha \times 10^6\text{ J}$, we get $α = 1000$ $\alpha = 1000$.

Pattern Recognition

Orbital transition energy is always $Δ E = (GMm)/(2) ((1)/(rᵢ) - (1)/(rf))$ . Never forget the factor of 2 in the denominator (which accounts for kinetic energy contribution in orbit).

Chapter Mix

Class 11 Physics: Gravitation

Q30 jee_main_2026_22_january_morning Escape Velocity

The escape velocity from a spherical planet A is 10 km/s. The escape velocity from another planet B whose density and radius are 10% of those of planet A, is \_\_\_\_ m/s.

A. 1000
B. $200\sqrt(5)$ $200\sqrt{5}$
C. $100\sqrt(10)$ $100\sqrt{10}$
D. $1000\sqrt(2)$ $1000\sqrt{2}$

Solution

Related Formula

Vₑ = \sqrt((2GM)/(R)) = \sqrt((2G ρ (4π R³)/(3))/(R)) Vₑ \propto \sqrt(ρ) \times R

Core Logic

Given density and radius of B are 10% ( $0.1$ $0.1$) of planet A:

((Vₑ)B)/((Vₑ)A) = \sqrt((ρB)/(ρA)) \times (RB)/(RA) = \sqrt(0.1) \times 0.1 = 110\sqrt(10)

(Vₑ)B = 10 \times 100010\sqrt(10) = 100\sqrt(10) m/sec

Pattern Recognition

Sees: Escape velocity scaling with density and radius. Shortcut: Express escape velocity in terms of density $ρ$ $\rho$ and radius $R$ $R$, then take ratios. Check: Matches option (3). ✓

Chapter Mix

Class 11 Physics: Gravitation

Q35 jee_main_2026_22_january_morning Gravitational Force and Superposition

Net gravitational force at the centre of a square is found to be

F₁

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

when the positions of 3M and 4M are interchanged. The ratio

F₁F₂

is

α\sqrt(5)

. The value of

α

is \_\_\_\_.

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

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

Solution

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

Q39 jee_main_2026_22_january_evening Satellite Motion and Time Period

Given below are two statements : Statement I : A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth. Statement II: The time period of revolution of the satellite is

T = 2π Rₑg

(for satellite very close to the earth surface), where

Rₑ

radius of earth and g acceleration due to gravity. In the light of the above statements, choose the correct answer from the options given below :

A. Both Statement I and Statement II are false
B. Both Statement I and Statement II are true
C. Statement I is true but Statement II is false
D. Statement I is false but Statement II is true

Solution

Related Formula

T = 2π \sqrt((R³)/(GM))

M = ρ \cdot (4)/(3)π R³

g = (GM)/(R²)

Core Logic

Evaluating Statement I: Substituting total mass $M = ρ \cdot (4)/(3)π R³$ $M = \rho \cdot \frac{4}{3}\pi R^3$ into time period formula:

T = 2π \sqrt((R³)/(G (ρ \cdot (4)/(3)π R³))) = \sqrt((3π)/(G ρ))

Hence, time period depends inversely on the square root of Earth's density $ρ$ $\rho$. Statement I is true.

Evaluating Statement II: Substituting $g = (GM)/(Rₑ²)$ $g = \frac{GM}{R_e^2}$ into $T = 2π \sqrt((Rₑ³)/(GM))$ $T = 2\pi \sqrt{\frac{R_e^3}{GM}}$:

T = 2π \sqrt((Rₑ)/(g))

Statement II is also true.

Step 1: Final Conclusion

Both Statement I and Statement II are true.

Pattern Recognition

Near-surface satellite shortcut: $T = \sqrt(3π / (Gρ)) \approx 84.6 min$ . Shows direct dependence on Earth's density $ρ$ $\rho$ and formula $T = 2π \sqrt(Rₑ / g)$ $T = 2\pi \sqrt{R_e / g}$.

Chapter Mix

Class 11 Physics: Gravitation

Q45 jee_main_2026_24_january_morning Gravitational Potential Energy

Three masses 200 kg, 300 kg and 400 kg are placed at the vertices of an equilateral triangle with sides 20 m. They are rearranged on the vertices of a bigger triangle of side 25 m and with the same centre. The work done in this process ____ J. (Gravitational constant

G = 6.7 \times 10⁻¹¹ N m² / kg²

)

A. $9.86 \times 10⁻⁶$ $9.86 \times 10^{-6}$
B. $2.85 \times 10⁻⁷$ $2.85 \times 10^{-7}$
C. $1.74 \times 10⁻⁷$ $1.74 \times 10^{-7}$
D. $4.77 \times 10⁻⁷$ $4.77 \times 10^{-7}$

Solution

Related Formula

U = -(Gm₁ m₂)/(r)

Wₑₓₜ = Δ U = Uf - Uᵢ

Core Logic

Equilateral triangle mass configuration scaling

Initial potential energy of the system ( $rᵢ = 20 m$ $r_i = 20 \text{ m}$):

Uᵢ = -(G)/(rᵢ) (m₁ m₂ + m₂ m₃ + m₁ m₃)

Uᵢ = -6.67 \times 10⁻¹¹20 [200 \times 300 + 300 \times 400 + 200 \times 400]

Uᵢ = -6.67 \times 10⁻¹¹20 [60000 + 120000 + 80000]

Uᵢ = -6.67 \times 10⁻¹¹20 \times 260000 = -86.71 \times 10⁻⁸ J

Final potential energy of the system ( $rf = 25 m$ $r_f = 25 \text{ m}$):

Uf = -(G)/(rf) (260000)

Uf = -6.67 \times 10⁻¹¹25 \times 260000

Uf = -69.36 \times 10⁻⁸ J

Step 1: Calculate Work Done

The work done by an external agent is equal to the change in potential energy:

Δ U = Uf - Uᵢ

Δ U = (-69.36 \times 10⁻⁸) - (-86.71 \times 10⁻⁸)

Δ U = (86.71 - 69.36) \times 10⁻⁸ = 17.35 \times 10⁻⁸ J \approx 1.74 \times 10⁻⁷ J

Pattern Recognition

When an entire geometric configuration is isotropically scaled (e.g., side $a$ $a$ to $b$ $b$), the total potential energy scales by the ratio of their reciprocals. $Uf = Uᵢ (rᵢ/rf)$ $U_f = U_i (r_i/r_f)$.

Chapter Mix

Class 11 Physics: Gravitation Class 11 Physics: Work, Energy and Power

Solution & Explanation

Related Formula

Core Logic

Step 1: Scaling Formula of Time

Step 2: Calculating Final Time

Pattern Recognition

Chapter Mix

Reference Study Guides

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