Physics MCQs for NEET — Practice Questions with Answers

The NCERT text explicitly states: 'Thus, for a varying force the work done can be expressed as a definite integral of force over displacement: $W = \int_{x_i}^{x_f} F(x) dx$.'

For a varying force, the work done is represented by the area under the force-displacement curve. If the curve forms a triangle, calculating the area of the triangle($1/2 \times base \times height$) would yield the work done.

The work-energy theorem states that the net work done on an object equals the change in its kinetic energy ($\Delta K = W_{net}$). This holds true for both constant and variable forces, as explained under 'THE WORK-ENERGY THEOREM FOR A VARIABLE FORCE'.

Given $m = 0.5 \text{ kg}$, $v = ax^{3/2}$, $a = 5 \text{ m}^{-1/2} \text{ s}^{-1}$. Initial velocity at $x=0$, $v_i = a(0)^{3/2} = 0$. Final velocity at $x=2 \text{ m}$, $v_f = a(2)^{3/2} = 5 \times (2^{3/2}) = 5 \times (2 \sqrt{2}) = 10 \sqrt{2} \text{ m/s}$. Work done by net force (Work-Energy Theorem) is $W = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$. $W = \frac{1}{2} (0.5 \text{ kg}) (10 \sqrt{2} \text{ m/s})^2 - 0$ $W = \frac{1}{2} (0.5) (100 \times 2) = \frac{1}{2} (0.5) (200) = 0.5 \times 100 = 50 \text{ J}$. NOTE: Re-calculating. The given answer for similar problem 5.20 results in 125 J. Let's recheck the calculation of $v_f$: $v_f = 5 \times (2^{3/2}) = 5 \times (2 \cdot \sqrt{2}) = 10\sqrt{2}$. Then $v_f^2 = (10\sqrt{2})^2 = 100 \times 2 = 200$. So, $W = \frac{1}{2} \times 0.5 \times 200 = 50 \text{ J}$. Let's assume there's a misunderstanding of the problem from the textbook. The physics is about applying the work-energy theorem. Given the exact problem from NCERT (5.20), let's ensure the calculation is accurate. $v = 5 x^{3/2}$ $v_i = 0$ at $x=0$ $v_f = 5 (2)^{3/2} = 5 \times 2 \sqrt{2} = 10 \sqrt{2} \text{ m/s}$ at $x=2 \text{ m}$ $K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} (0.5) (10\sqrt{2})^2 = \frac{1}{2} (0.5) (100 \times 2) = \frac{1}{2} (0.5) (200) = 50 \text{ J}$ $W = K_f - K_i = 50 - 0 = 50 \text{ J}$. However, if we are to derive the given solution from NCERT (which yields 125 J in the solution part of text related to similar problems), there must be a mismatch somewhere. Let's re-read the context. Ah, wait, this problem is actually part of the 'Additional Exercises' (Question 5.20) in the NCERT, for which the solution is not explicitly provided in the excerpt. My calculation gives 50 J. Let me ensure if there was any mistake in my understanding of the problem that could lead to 125 J. No, the calculation follows the work-energy theorem correctly. So 50 J is the correct value. Since it's an MCQ, let's assume the options are based on possible values, and the calculation of 50J is solid. Let's re-evaluate in case the question was implicitly asking for the work done by a force $F = ma = m \frac{dv}{dt}$. $v = ax^{3/2} \implies \frac{dv}{dt} = \frac{d}{dt} (ax^{3/2}) = a \frac{3}{2} x^{1/2} \frac{dx}{dt} = a \frac{3}{2} x^{1/2} v = a \frac{3}{2} x^{1/2} (ax^{3/2}) = \frac{3}{2} a^2 x^2$ $F = m \frac{dv}{dt} = m \frac{3}{2} a^2 x^2$ $W = \int F dx = \int_0^2 m \frac{3}{2} a^2 x^2 dx = m \frac{3}{2} a^2 \left[\frac{x^3}{3}\right]_0^2 = m \frac{3}{2} a^2 \frac{8}{3} = 4 m a^2$ Substitute values: $m = 0.5 \text{ kg}$, $a = 5 \text{ m}^{-1/2} \text{ s}^{-1}$ $W = 4 \times 0.5 \times (5)^2 = 2 \times 25 = 50 \text{ J}$. Both methods yield 50 J. So, the correct option should reflect 50 J. If 125 J was expected, the 'a' or velocity function might be different implicitly. Sticking to my calculation from the problem statement, 50 J is correct.

The context states: 'This is not possible according to the Heisenberg uncertainty principle. Bohr model of the hydrogen atom, therefore, not only ignores the dual behaviour of electron but also contradicts Heisenberg uncertainty principle.'

The context mentions: 'It was also unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or an electric field (Stark effect).'

The context states: 'However it fails when applied to microscopic objects like electrons, atoms, molecules etc. This is mainly because of the fact that classical mechanics ignores the concept of dual behaviour of matter especially for sub-atomic particles and the uncertainty principle.'

The context states: 'In fact an insight into the structure of the atom was needed which could account for wave-particle duality of matter and be consistent with Heisenberg uncertainty principle. This came with the advent of quantum mechanics.'

The context notes: 'An orbit is a clearly defined path and this path can completely be defined only if both the exact position and the exact velocity of the electron at the same time are known. This is not possible according to the Heisenberg uncertainty principle.'

The context defines it: '*Hydrogenic atoms are the atoms consisting of a nucleus with positive charge +Ze and a single electron, where Z is the proton number. Examples are hydrogen atom, singly ionised helium, doubly ionised lithium, and so forth. In these atoms more complex electron-electron interactions are nonexistent.'