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 (alpha/sqrt(5)). The value of alpha is ____.

Initial configuration: F2 = sqrt(10) (Gmm0/r2)Ratio:(F1/F2) = 2sqrt(2)sqrt(10) = (2/sqrt(5)) alpha = 2$ 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). ✓ Class 11 Physics: Gravitation

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 t

Initial configuration: F2 = sqrt(10) (Gmm0/r2)Ratio:(F1/F2) = 2sqrt(2)sqrt(10) = (2/sqrt(5)) alpha = 2$ 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). ✓ Class 11 Physics: Gravitation

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 (alpha/sqrt(5)). The value of alpha is ____.

Initial configuration: F2 = sqrt(10) (Gmm0/r2)Ratio:(F1/F2) = 2sqrt(2)sqrt(10) = (2/sqrt(5)) alpha = 2$$ 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). ✓ Class 11 Physics: Gravitation

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 t

Initial configuration: F2 = sqrt(10) (Gmm0/r2)Ratio:(F1/F2) = 2sqrt(2)sqrt(10) = (2/sqrt(5)) alpha = 2$$ 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). ✓ Class 11 Physics: Gravitation

Net gravitational force at the centre of a square is found to be F1 when four particles having mass M, 2M, 3M and 4M…

Year	2026	2025	2024	Total
Questions	5	9	8	22

Q6 jee_main_2025_07_april_evening Kepler's Laws of Planetary Motion

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]

A. Both (A) and (R) are correct and (R) is the correct explanation of (A) [cite: 63]
B. Both (A) and (R) are correct but (R) is not the correct explanation of (A) [cite: 65]
C. A is correct but R is not correct [cite: 72]
D. A is not correct but R is correct [cite: 73]

Solution

Related Formula

$(dA)/(dt) = (L)/(2m)$ $$\frac{dA}{dt} = \frac{L}{2m}$$ [cite: 684]

Core Logic

Kepler's Second Law state that the areal velocity $(dA)/(dt)$ $\frac{dA}{dt}$ is directly proportional to the angular momentum $L$ $L$ 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 $τ = r \times F = 0$ [cite: 56, 686]. Since torque is zero, the angular momentum $L$ $L$ remains completely constant over time[cite: 686]. As a consequence, $(dA)/(dt) = constant$ $\frac{dA}{dt} = \text{constant}$[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].

Chapter Mix

Class 11 Physics: Gravitation

Q25 jee_main_2025_24_jan_evening Acceleration due to Gravity

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.

Numerical Answer. Answer: 9 to 9

Solution

Related Formula

Surface gravitational acceleration:

g = (GM)/(R²)

Core Logic

Since diameter drops to $1/3$ $1/3$, the radius $R'$ $R'$ also scales down to $1/3$ $1/3$ its original value ( $R' = R/3$ $R' = R/3$), while mass $M$ $M$ remains constant.

New acceleration value $g'$ $g'$ calculation:

g' = (GM)/((R/3)²) = 9 \cdot (GM)/(R²) = 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 $n$ $n$ increases surface gravity by $n²$ $n^2$ if the mass is unchanged.

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Class 11 Physics: Gravitation

Q jee_main_2025_24_jan_morning Kepler's Laws of Planetary Motion

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 :-

A. 3%
B. 4.5%
C. 9%
D. 2.5%

Solution

Related Formula

By Kepler's Third Law of Planetary Motion, the square of the orbital period $T$ $T$ is proportional to the cube of the orbital radius $R$ $R$:

T² = K \cdot R³

Core Logic

Taking logs and differentiating to find fractional errors for small changes:

2 (Δ T)/(T) = 3 (Δ R)/(R)

(Δ T)/(T) = (3)/(2) ((Δ R)/(R))

Step 1: Computing Percentage Change

The change in radius is $Δ R = 1.03R - R = 0.03R$ $\Delta R = 1.03R - R = 0.03R$, which means $(Δ R)/(R) = 0.03$ $\frac{\Delta R}{R} = 0.03$ or $3%$ $3\%$.

Substitute this into our fraction scaling relation:

(Δ T)/(T) = (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 $(3)/(2)$ $\frac{3}{2}$ times the radius change.

Chapter Mix

Class 11 Physics: Gravitation

Q11 jee_main_2025_28_jan_evening Escape Velocity

Earth has mass 8 \times and radius 2 \times that of a planet. If the escape velocity from the earth is

11.2 km/s

, the escape velocity in

km/s

from the planet will be:

A. 11.2
B. 5.6
C. 2.8
D. 8.4

Solution

Related Formula

The expression for escape velocity from a spherical planetary body is given by:

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

Core Logic

Let the planet's mass be $MP$ $M_P$ and its radius be $RP$ $R_P$. According to the problem statement :

Earth's mass, $ME = 8 MP (MP)/(ME) = (1)/(8)$
Earth's radius, $RE = 2 RP (RE)/(RP) = 2$

Taking the ratio of escape velocities :

vPvE = \sqrt(((MP)/(ME)) \times ((RE)/(RP)))

vPvE = \sqrt((1)/(8) \times 2) = \sqrt((1)/(4)) = (1)/(2)

Given that $vE = 11.2 km/s$ $v_E = 11.2 \text{ km/s}$:

vP = (1)/(2) \times 11.2 = 5.6 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.

Chapter Mix

Class 11 Physics: Gravitation

Q32 jee_main_2024_01_february_morning Acceleration Due to Gravity

If

R

is the radius of the earth and the acceleration due to gravity on the surface of earth is

g = π² ~m/s²

, then the length of the second's pendulum at a height

h = 2R

from the surface of earth will be:

A. $(2)/(9)~m$ $\frac{2}{9}\mathrm{~m}$
B. $(1)/(9)~m$ $\frac{1}{9}\mathrm{~m}$
C. $(4)/(9)~m$ $\frac{4}{9}\mathrm{~m}$
D. $(8)/(9)~m$ $\frac{8}{9}\mathrm{~m}$

Solution

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

Solution & Explanation

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Core Logic

Pattern Recognition

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