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 required to project a body of mass \( m \) from Earth's surface to infinity is \( \frac{1}{2} mgR \), where \( R \) is Earth's radius.
Reason (R): The maximum potential energy of a body is zero when projected to infinity from Earth's surface.
Gravitational potential energy of a body of mass \( m \) at distance \( r \) from Earth's center is \( U = -\frac{GMm}{r} \), where \( G \) is the gravitational constant and \( M \) is Earth's mass. At infinity, \( U = 0 \). The minimum kinetic energy to escape Earth's gravity is 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: Calculate 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 Earth's surface to infinity is:
\[K = mgR\]This indicates that Assertion (A) is false as it states \( \frac{1}{2} mgR \) instead of \( mgR \).
Step 4: Analyze Reason (R).
Reason (R) is correct, stating that maximum potential energy (at infinity) is zero, as gravitational potential energy is 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 
