Weight is the force exerted by gravity on an object. The weight of a body is given by the formula: \[ W = mg \] where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity. However, when a body is in free fall, it experiences a state of weightlessness. This is because the force of gravity acting on the body and the acceleration due to gravity are balanced, resulting in no normal force acting on the body.
Therefore, the effective weight of the body is zero while in free fall.
The correct option is (E) : zero
Answer: mg
Explanation:
The weight of a body is defined as the force with which it is attracted towards the Earth due to gravity. It is given by the formula: $$ W = mg $$ where:
- $W$ is the weight of the body
- $m$ is the mass of the body
- $g$ is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\) near the Earth's surface)
Even during free fall (neglecting air resistance), the weight remains mg. What changes is the apparent weight if the body is in a reference frame that's also falling.
Match the LIST-I with LIST-II
\[ \begin{array}{|l|l|} \hline \text{LIST-I} & \text{LIST-II} \\ \hline \text{A. Gravitational constant} & \text{I. } [LT^{-2}] \\ \hline \text{B. Gravitational potential energy} & \text{II. } [L^2T^{-2}] \\ \hline \text{C. Gravitational potential} & \text{III. } [ML^2T^{-2}] \\ \hline \text{D. Acceleration due to gravity} & \text{IV. } [M^{-1}L^3T^{-2}] \\ \hline \end{array} \]
Choose the correct answer from the options given below:
A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is: 