Physics MCQs for NEET — Practice Questions with Answers

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.

According to Newton's second law, F(t) = ma(t). For SHM, acceleration a(t) = -ω²x(t). Substituting this into Newton's second law gives F(t) = -mω²x(t). The text also defines k = mω², so F(t) = -kx(t). Therefore, all the given options are correct representations of the force law for SHM. (NCERT, page 267)

The context states: 'Like acceleration, force is always directed towards the mean position—hence it is sometimes called the restoring force in SHM.' This is a fundamental characteristic of SHM. (NCERT, page 267)

The context states: 'Note that the force in Eq. (13.13) is linearly proportional to x(t). A particle oscillating under such a force is, therefore, calling a linear harmonic oscillator.' (NCERT, page 267)

The text explains: 'In the real world, the force may contain small additional terms proportional to x², x³, etc. These then are called non-linear oscillators.' (NCERT, page 267)

The context states: 'To summarise the discussion so far, simple harmonic motion can be defined in two equivalent ways, either by Eq. (13.4) for displacement or by Eq. (13.13) that gives its force law.' Eq. (13.13) is F(t) = -kx(t). (NCERT, page 267)

The context mentions: 'Going from Eq. (13.4) to Eq. (13.13) required us to differentiate two times. Likewise, by integrating the force law Eq. (13.13) two times, we can get back Eq. (13.4).' (NCERT, page 267)

From the formula ω = √(k/m), given in the text (Eq. 13.14b), if 'm' increases and 'k' remains constant, ω will decrease. (NCERT, page 267)

In Example 13.6, when two identical springs (spring constant k) are attached to a mass 'm' with fixed supports such that one spring elongates and the other compresses by 'x', the forces are F₁ = -kx and F₂ = -kx. The net force F = F₁ + F₂ = -2kx. (NCERT, page 268)