A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to…

Year	2026	2025	2024	Total
Questions	5	9	8	22

Year

2026

2025

2024

Total

Questions

Solution & Explanation

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

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

A planet takes $200 days$ $200\text{ days}$ 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?

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate the New Time Period

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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A planet takes 200 days$200\text{ days}$ 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?

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate the New Time Period

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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A planet takes $200 days$ $200\text{ days}$ 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?