Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The kinetic energy needed to project a body of mass $m$ from earth surface to infinity is $\frac{1}{2} \mathrm{mgR}$, where R is the radius of earth. Reason R: The maximum potential energy of a body is zero when it is projected to infinity from earth surface.
Assertion (A): The kinetic energy needed to project a body of mass \( m \) from the Earth’s surface to infinity is \( \frac{1}{2} mgR \), where \( R \) is the radius of the Earth.
Reason (R): The maximum potential energy of a body is zero when it is projected to infinity from the Earth’s surface.
The gravitational potential energy of a body of mass \( m \) at a distance \( r \) from the center of the Earth is given by:
\[ U = -\frac{GMm}{r} \]where \( G \) is the gravitational constant and \( M \) is the mass of the Earth. At infinity, \( U = 0 \). The minimum kinetic energy required to just escape from Earth's gravitational field is known as the escape energy.
It is given by:
\[ K = \frac{1}{2} m v_e^2 \] \[ \text{where } v_e = \sqrt{\frac{2GM}{R}} \] \[ \therefore K = \frac{GMm}{R} \]Step 1: Compute the correct expression for the required kinetic energy.
\[ K = \frac{GMm}{R} \]Step 2: Relate \( \frac{GM}{R^2} \) to \( g \), the acceleration due to gravity.
\[ g = \frac{GM}{R^2} \Rightarrow GM = gR^2 \] \[ K = \frac{gR^2 m}{R} = mgR \]Step 3: Therefore, the kinetic energy needed to project a body from the Earth's surface to infinity is:
\[ K = mgR \]This shows that the Assertion (A) is false because it states \( \frac{1}{2} mgR \) instead of \( mgR \).
Step 4: Analyze the Reason (R).
The Reason correctly states that the maximum potential energy (at infinity) is zero, since gravitational potential energy becomes zero at infinite separation.
Final Answer: Assertion (A) is false, but Reason (R) is true.
Net gravitational force at the center of a square is found to be \( F_1 \) when four particles having masses \( M, 2M, 3M \) and \( 4M \) are placed at the four corners of the square as shown in figure, and it is \( F_2 \) when the positions of \( 3M \) and \( 4M \) are interchanged. The ratio \( \dfrac{F_1}{F_2} = \dfrac{\alpha}{\sqrt{5}} \). The value of \( \alpha \) is 

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