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

Q44 jee_main_2024_29_january_evening Kepler's Laws of Planetary Motion

A planet takes

200 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?

A. $25$ $25$
B. $50$ $50$
C. $100$ $100$
D. $20$ $20$

Solution

Related Formula

According to Kepler's Third Law (Law of Periods):

$T² \propto r³$ $T^2 \propto r^3$

where:

$T$ $T$ is the time period of revolution.
$r$ $r$ is the orbital radius of the planet.

Core Logic

Using the proportionality relationship for two states:

(T₂²)/(T₁²) = ( (r₂)/(r₁) )³

Given:

$T₁ = 200 days$ $T_1 = 200\text{ days}$
$r₂ = (r₁)/(4)$ $r_2 = \frac{r_1}{4}$

Step 1: Calculate the New Time Period

Substitute the values into the proportionality relation:

(T₂²)/((200)²) = ( (r₁ / 4)/(r₁) )³ = ( (1)/(4) )³ = (1)/(64)

Taking the square root on both sides:

(T₂)/(200) = \sqrt((1)/(64)) = (1)/(8)

T₂ = (200)/(8) = 25 days

Pattern Recognition

If orbital distance scales by $x$ $x$, the period scales by $x3/2$ $x^{3/2}$. Here, distance scales by $(1)/(4)$ $\frac{1}{4}$, so the period scales by $((1)/(4))3/2 = (1)/(8)$ . Thus, $200 \times (1)/(8) = 25 days$ .

Chapter Mix

Class 11 Physics: Gravitation

Q35 jee_main_2024_27_jan_morning Acceleration due to Gravity

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:

A. $(g)/(4)$ $\frac{g}{4}$
B. $2g$ $2g$
C. $(g)/(2)$ $\frac{g}{2}$
D. $4g$ $4g$

Solution

Related Formula

g = (GM)/(R²) g \propto (1)/(R²)

Core Logic

Since diameter reduces to half, the radius $R₂$ $R_{2}$ also reduces to half of its initial value $R₁$ $R_{1}$:

R₂ = R₁2

Setting up the ratio:

g₂g₁ = ( R₁R₂)² = ( R₁R₁/2)² = 4

Step 1: Final Calculation

g₂ = 4g₁ = 4g

Pattern Recognition

Inverse square dependence means halving the distance scale amplifies the surface field metric by a factor of $2² = 4$ $2^2 = 4$ matching constant mass bounds.

Chapter Mix

Class 11 Physics: Gravitation

Q jee_main_2024_29_jan_morning Acceleration due to Gravity

At what distance above and below the surface of the earth a body will have same weight, (take radius of earth as R.)

A. $\sqrt(5) R - R$ $\sqrt{5} \mathrm{R} - \mathrm{R}$
B. $\sqrt(3) R - R2$
C. $(R)/(2)$ $\frac{R}{2}$
D. $\sqrt(5) R - R2$

Solution

Related Formula

Acceleration due to gravity at a height $h$ $h$ above the Earth's surface:

gₚ = (g R²)/((R + h)²)

Acceleration due to gravity at a depth $h$ $h$ below the Earth's surface:

gq = g (1 - (h)/(R))

Visual representation of points p and q showing positions above and below Earth's surface for Q47

Core Logic

We need the weights to be identical, meaning $gₚ = gq$ $g_p = g_q$ at the exact same distance value $h$ $h$:

(g R²)/((R + h)²) = g (1 - (h)/(R))

Dividing by $g$ $g$ and simplifying the left side denominator fraction:

(1)/((1 + (h)/(R))²) = 1 - (h)/(R)

(1 - (h)/(R))(1 + (h)/(R))² = 1

Step 1: Set Up Algebraic Equation

Let $(h)/(R) = x$ $\frac{h}{R} = x$. Then:

(1 - x)(1 + x)² = 1

(1 - x)(1 + 2x + x²) = 1

1 + 2x + x² - x - 2x² - x³ = 1

x - x² - x³ = 0

Step 2: Solve for x

Since $x \neq 0$ $x \neq 0$ (distance cannot be zero), divide by $x$ $x$:

1 - x - x² = 0 x² + x - 1 = 0

Solving via quadratic formula:

x = -1 \pm \sqrt(1² - 4(1)(-1))2 = -1 \pm \sqrt(5)2

Since distance parameter $x gt 0$ $x \gt 0$, we take the positive root:

x = \sqrt(5) - 12

Step 3: Find Height h

Substitute back $x = (h)/(R)$ $x = \frac{h}{R}$:

(h)/(R) = \sqrt(5) - 12 h = (R)/(2) (\sqrt(5) - 1) = \sqrt(5)R - R2

Therefore, the required distance is $\sqrt(5)R - R2$ $\frac{\sqrt{5}R - R}{2}$.

Pattern Recognition

Do not use the linear approximation formula $gh \approx g(1 - (2h)/(R))$ $g_h \approx g(1 - \frac{2h}{R})$ unless the problem explicitly states $h ll R$ $h \ll R$. Equating the approximated form to depth gives $hheight = (1)/(2) hdepth$ , which fails when looking for a single unified distance value $h$ $h$.

Chapter Mix

Class 11 Physics: Gravitation

Q42 jee_main_2024_30_january_evening Escape Velocity

Escape velocity of a body from earth is

11.2 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:

A. $11.2 ~km / s$ $11.2 \mathrm{~km / s}$
B. $8.4 ~km / s$ $8.4 \mathrm{~km / s}$
C. $4.2 ~km / s$ $4.2 \mathrm{~km / s}$
D. $7.9 ~km / s$ $7.9 \mathrm{~km / s}$

Solution

Related Formula

Vₑ = \sqrt((2GM)/(R))

Core Logic

For Earth: $Vₑ = \sqrt((2GME)/(RE)) = 11.2 ~km/s$ For the planet: $RP = RE3$ and $MP = ME6$ We can express the escape velocity of the planet $Vₚ$ $V_p$ as a ratio of the Earth's escape velocity.

Step 1: Ratio of Velocities

(Vₚ)/(Vₑ) = \sqrt((Mₚ)/(ME) \times (RE)/(Rₚ))

(Vₚ)/(Vₑ) = \sqrt(((1)/(6)) \times ((3)/(1))) = \sqrt((1)/(2))

Step 2: Calculate Escape Velocity

Vₚ = Vₑ\sqrt(2)

Vₚ = (11.2)/(1.414) \approx 7.92 ~km/s

Pattern Recognition

Any scaling of a planet's mass by factor $α$ $\alpha$ and radius by factor $β$ $\beta$ scales the escape velocity by a factor of $\sqrt(α / β)$ $\sqrt{\alpha / \beta}$.

Chapter Mix

Class 11 Physics: Gravitation

Q42 jee_main_2024_30_jan_morning Gravitational Potential and Field

The gravitational potential at a point above the surface of earth is

-5.12 \times 10⁷ ~J / kg

and the acceleration due to gravity at that point is

6.4 ~m/s²

. Assume that the mean radius of earth to be

6400 ~km

. The height of this point above the earth's surface is:

A. $1600 km$ $1600 \mathrm{\,km}$
B. $540 km$ $540 \mathrm{\,km}$
C. $1200 km$ $1200 \mathrm{\,km}$
D. $1000 km$ $1000 \mathrm{\,km}$

Solution

Related Formula

V = -(GME)/(RE + h)

g' = (GME)/((RE + h)²)

Core Logic

The gravitational potential ( $V$ $V$) and acceleration due to gravity ( $g'$ $g'$) at a distance $r = RE + h$ $r = R_E + h$ from the center of the earth can be related by dividing their magnitudes: $|V| / g' = r$ $|V| / g' = r$.

Step 1: Set Up Equations

From the given data:

-(GME)/(RE + h) = -5.12 \times 10⁷ (i)

(GME)/((RE + h)²) = 6.4 (ii)

Step 2: Isolate Variable

Divide equation (i) by (ii) (taking magnitudes):

RE + h = (5.12 \times 10⁷)/(6.4)

RE + h = 0.8 \times 10⁷ ~m = 8000 ~km

Step 3: Solve for h

Given mean radius of the earth $RE = 6400 ~km$ $R_E = 6400 \mathrm{~km}$: $6400 + h = 8000$ $6400 + h = 8000$

h = 1600 ~km

Pattern Recognition

Always exploit the $V/g = r$ $V/g = r$ relationship to extract distances cleanly without having to substitute large values for $G$ $G$ or $ME$ $M_E$.

Chapter Mix

Class 11 Physics: Gravitation

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Gravitation Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Calculate the New Time Period

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set Up Algebraic Equation

Step 2: Solve for x

Step 3: Find Height h

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Ratio of Velocities

Step 2: Calculate Escape Velocity

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set Up Equations

Step 2: Isolate Variable

Step 3: Solve for h

Pattern Recognition

Chapter Mix

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