NEET Practice Questions (MCQs) with Answers & Solutions

First, calculate the acceleration (retardation) using Newton's Second Law: $F = ma \Rightarrow a = F/m = -50 N / 20 kg = -2.5 m/s^2$. (Negative sign indicates retardation). Then, use the kinematic equation $v = u + at$, where $v = 0$ (final velocity when stopped), $u = 15 m/s$, and $a = -2.5 m/s^2$. So, $0 = 15 + (-2.5)t \Rightarrow 2.5t = 15 \Rightarrow t = 15 / 2.5 = 6$ seconds.

The NCERT text states: 'T aken together, the product of mass and velocity, that is momentum, is evidently a relevant variable of motion. The greater the change in the momentum in a given time, the greater is the force that needs to be applied.' This directly relates the change in momentum to the applied force, which is fundamental to Newton's Second Law.

The NCERT text notes: 'The second law of motion given by Eq. (4.5) is applicable to a single point particle... It turns out, however, that the law in the same form applies to a rigid body or, even more generally, to a system of particles. In that case, F refers to the total external force on the system and a refers to the acceleration of the system as a whole. Any internal forces in the system are not to be included in F.'

The net upward force must overcome gravity and provide the upward acceleration. Let T be the thrust. F_net = T - mg. According to Newton's Second Law, F_net = ma. So, T - mg = ma \Rightarrow T = ma + mg = m(a + g). Given m = 20,000 kg, a = 5.0 m/s², and g = 10 m/s². T = 20,000 kg (5.0 m/s² + 10 m/s²) = 20,000 kg * 15 m/s² = 300,000 N.

The NCERT text states: 'If necessary, follow the same procedure for another choice of the system. In doing so, employ Newton’s third law. That is, if in the free-body diagram of A, the force on A due to B is shown as F, then in the free-body diagram of B, the force on B due to A should be shown as –F.'

Before being dropped, the stone has the same velocity as the truck. The truck starts from rest ($u_t = 0$) and accelerates at $a_t = 2.0 m/s^2$. At $t = 10 s$, the velocity of the truck is $v_t = u_t + a_t t = 0 + (2.0 m/s^2)(10 s) = 20 m/s$. When the stone is dropped, it retains this horizontal velocity due to inertia, neglecting air resistance. The vertical motion starts from rest relative to the truck, but horizontally it continues with the truck's velocity at that instant.

The NCERT text states: 'A rigid body is in mechanical equilibrium if (1) it is in translational equilibrium, i.e., the total external force on it is zero : $\Sigma F_i = 0$, and (2) it is in rotational equilibrium, i.e. the total external torque on it is zero : $\Sigma \tau_i = \Sigma (r_i \times F_i) = 0$.'

The NCERT text specifies: 'The second law of motion is a local relation which means that force F at a point in space (location of the particle) at a certain instant of time is related to a at that point at that instant. Acceleration here and now is determined by the force here and now, not by any history of the motion of the particle (See Fig. 4.5).'

In uniform circular motion on a smooth horizontal surface, the only horizontal force acting on the particle towards the center is the tension in the string. This tension provides the necessary centripetal force. Therefore, the net force directed towards the center is T. This directly corresponds to problem 4.4 in the provided context.

First, calculate the acceleration: $a = (v - u) / t = (3.5 m/s - 2.0 m/s) / 25 s = 1.5 m/s / 25 s = 0.06 m/s^2$. Then, use Newton's Second Law: $F = ma = (3.0 kg)(0.06 m/s^2) = 0.18 N$. This aligns with problem 4.6 in the provided context.