Physics MCQs for NEET — Practice Questions with Answers

In Example 13.6, the net restoring force is F = -2kx. Comparing this with F = -K_eff x, we find the effective spring constant K_eff = 2k. The period of oscillation T = 2π√(m/K_eff). Substituting K_eff = 2k gives T = 2π√(m/2k). (NCERT, page 268)

The context states: 'The total mechanical energy of a harmonic oscillator is thus independent of time as expected for motion under any conservative force.' (NCERT, page 269)

The text clarifies: 'For x = 0, the energy is kinetic; at the extremes x = ± A, it is all potential energy.' At extreme positions, the velocity is zero, hence kinetic energy is zero, and the potential energy is maximum. (NCERT, page 269)

The text clearly states the relationship as ω = k/m (13.14b). This is a crucial formula for SHM. (NCERT, page 267)

The force law is F(t) = -kx(t). If x is positive, then F = -k(positive value), which results in a negative force. This indicates the force acts in the opposite direction to the displacement, towards the mean position. (NCERT, page 267-268)

According to the provided text and the derived expression in Example 2.6, the stopping distance is given by $d_s = \frac{-v_0^2}{2a}$. This shows that the stopping distance is proportional to the square of the initial velocity ($v_0^2$). The context explicitly states, 'Thus, the stopping distance is proportional to the square of the initial velocity.'

The context states, 'Doubling the initial velocity increases the stopping distance by a factor of 4 (for the same deceleration).' This is because stopping distance ($d_s$) is proportional to the square of the initial velocity ($v_0^2$). If $v_0$ becomes $2v_0$, then $d_s'$ will be proportional to $(2v_0)^2 = 4v_0^2$, making the new stopping distance $4x$.

Example 2.7 clearly defines reaction time: 'Reaction time is the time a person takes to observe, think and act.' It encompasses the entire cognitive and motor process from stimulus recognition to initial response.

From Example 2.7, an object under free fall (like the ruler) starts with an initial velocity $v_0 = 0$. The distance travelled ($d$) is related to time ($t_r$) by the kinematic equation $d = v_0 t_r + \frac{1}{2} a t_r^2$. Since $v_0 = 0$ and acceleration $a = g$, the equation simplifies to $d = \frac{1}{2} g t_r^2$. Note: The solution in the text uses $a = -g$ when defining the direction of motion relative to the coordinate system, but for the magnitude of distance traveled downwards, 'g' is used positively.

During the reaction time, the car continues to move at its initial speed before deceleration begins. The distance covered during reaction time can be calculated as distance = speed × time. So, distance = $20 \text{ m/s} \times 0.5 \text{ s} = 10 \text{ m}$.