QJEEM24Phfe0f: JEE Main 2024 Physics - Class 11 JEEM24Phfe0f Solution: A body placed at distance from surface of earth experiences…

The gravitational force $F$ that an object of mass $m$ placed at a distance $r$ (from the center of Earth) experiences can be calculated using the universal law of gravitation, given by:

$F = \frac{G M m}{r^2}$

Where:

• $G$ is the gravitational constant ( $6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2$ )
• $M$ is the mass of the Earth (approximately $5.972 \times 10^{24} \text{kg}$ )
• $m$ is the mass of the object
• $r$ is the distance from the center of the Earth to the object

However, when the object is at a distance $2R$ from the surface of the Earth, the total distance from the center of the Earth $r'$ becomes $R + 2R = 3R$ . This is because the radius of the Earth $R$ is the distance from the Earth's center to its surface, so if the object is $2R$ above the surface, the total distance from the center is $3R$ .

At the surface of the Earth, the gravitational force ( $F_{earth}$ ) that acts on an object is given by its weight, which can be calculated using the formula $F_{earth} = m \cdot g$ , where $g$ is the acceleration due to gravity on the surface of the Earth. Given that $g = 10 \text{m/s}^2$ and the mass of the body $m = 90 \text{kg}$ , we get:

$F_{earth} = 90 \text{kg} \cdot 10 \text{m/s}^2 = 900 \text{N}$

To find the gravitational pull at a distance $2R$ from the Earth's surface, we use the fact that gravitational force varies inversely as the square of the distance from the center of the Earth. Since the distance triples ( $3R$ from $R$ ), the gravitational force becomes $\frac{1}{3^2} = \frac{1}{9}$ of the force at the surface.

Therefore, the gravitational pull $F_{2R}$ on the body when placed at $2R$ from the Earth's surface is:

$F_{2R} = \frac{F_{earth}}{9} = \frac{900 \text{N}}{9} = 100 \text{N}$

Thus, the gravitational pull on the body when it is placed at a distance of $2R$ from the Earth's surface is 100 N. So, the correct option is:

Option C: 100 N