NEET 2018 Physics Questions with Solutions

Explanation

For a resonance tube closed at one end, the difference between two successive resonant lengths is equal to half the wavelength: $$L_2 - L_1 = \frac{\lambda}{2}$$ Given: $$L_1 = 20\text{ cm}, \quad L_2 = 73\text{ cm}$$ $$L_2 - L_1 = 53\text{ cm} = 0.53\text{ m}$$ Hence, $$\lambda = 2 \times 0.53 = 1.06\text{ m}$$ Velocity of sound: $$v = f\lambda$$ $$v = 320 \times 1.06 = 339.2\text{ m/s}$$ Therefore, the velocity of sound in air is approximately $$339\text{ m/s}$$.

Explanation

For motion under constant acceleration: $$h = \frac{1}{2}at^2$$ In an electric field: $$a = \frac{eE}{m}$$ Therefore, $$t = \sqrt{\frac{2hm}{eE}}$$ Thus, $$t \propto \sqrt{m}$$ Since the mass of the electron is much smaller than the mass of the proton, the electron takes less time to fall. Hence, the correct answer is: $$\boxed{\text{Smaller}}$$

Explanation

When the current in the electromagnet is switched on, energy is supplied by the current source to establish the magnetic field. The diamagnetic rod is pushed upward and gains gravitational potential energy. This energy ultimately comes from the electrical energy supplied by the current source. Hence, option (A) is correct.

Explanation

The angular fringe width in Young's double slit experiment is given by: $$\theta = \frac{\lambda}{d}$$ Initially: $$0.20^\circ = \frac{\lambda}{2\text{ mm}}$$ Let the new slit separation be $$d$$ when the angular width becomes $$0.21^\circ$$. $$0.21^\circ = \frac{\lambda}{d}$$ Dividing the two equations: $$\frac{0.20}{0.21} = \frac{d}{2}$$ $$d = \frac{0.20}{0.21} \times 2$$ $$d \approx 1.9\text{ mm}$$ Hence, option (B) is correct.

Explanation

When the reflected and refracted rays are perpendicular to each other, the angle of incidence is called Brewster angle. At Brewster angle: $$\tan i = \mu$$ The reflected light becomes completely plane polarised with its electric vector perpendicular to the plane of incidence. Therefore, option (B) is correct.

Explanation

For an electromagnetic wave, the propagation direction is given by: $$\vec{E} \times \vec{B} = \vec{V}$$ Given: $$\vec{V} = V\hat{i}$$ and electric field is along +y direction: $$\vec{E} = E\hat{j}$$ Using the vector product: $$\hat{j} \times \hat{k} = \hat{i}$$ Therefore, the magnetic field must be along +z direction. Hence, option (B) is correct.

Explanation

Energy stored in an inductor is given by: $$U = \frac{1}{2}LI^2$$ Given: $$U = 25\text{ mJ} = 25 \times 10^{-3}\text{ J}$$ $$I = 60\text{ mA} = 60 \times 10^{-3}\text{ A}$$ Substituting: $$25 \times 10^{-3} = \frac{1}{2}L(60 \times 10^{-3})^2$$ $$L = \frac{2 \times 25 \times 10^{-3}}{3600 \times 10^{-6}}$$ $$L = \frac{500}{36} \approx 13.89\text{ H}$$ Hence, option (C) is correct.

Explanation

Initially, the number of nuclei is: $$N_0 = 600$$ If 450 nuclei disintegrate, the remaining nuclei are: $$N = 600 - 450 = 150$$ Radioactive decay law: $$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{t/t_{1/2}}$$ Substituting values: $$\frac{150}{600} = \left(\frac{1}{2}\right)^{t/10}$$ $$\frac{1}{4} = \left(\frac{1}{2}\right)^{t/10}$$ $$\left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^{t/10}$$ Therefore: $$\frac{t}{10} = 2$$ $$t = 20\text{ minutes}$$ Hence, option (A) is correct.

Explanation

For an electron in a Bohr orbit: $$\text{Total Energy} = -\text{Kinetic Energy}$$ Thus, $$KE : TE = 1 : -1$$ Hence, option (B) is correct.

Explanation

The viscous force on a sphere moving through a liquid is given by Stokes' law: $$F = 6\pi\eta rv_t$$ The rate of production of heat equals the power dissipated: $$P = Fv_t$$ $$P = 6\pi\eta r v_t^2$$ The terminal velocity of a sphere in a viscous liquid is: $$v_t \propto r^2$$ Therefore: $$P \propto r \times (r^2)^2$$ $$P \propto r^5$$ Hence, option (D) is correct.

Explanation

Extension in a wire is given by: $$\Delta l = \frac{Fl}{AY}$$ where $$Y$$ is Young's modulus. Since both wires have the same volume: For wire 1: $$A_1 = A, \quad l_1 = 3l$$ So, $$\Delta l = \frac{F(3l)}{AY} \quad ...(1)$$ For wire 2: $$A_2 = 3A, \quad l_2 = l$$ If force required is $$F'$$: $$\Delta l = \frac{F'l}{3AY} \quad ...(2)$$ Equating (1) and (2): $$\frac{3Fl}{AY} = \frac{F'l}{3AY}$$ $$F' = 9F$$ Hence, option (A) is correct.

Explanation

For oxygen molecules to escape Earth's atmosphere, their rms speed must equal the escape velocity. Escape velocity of Earth: $$v_e = 11200\text{ m/s}$$ RMS speed is given by: $$v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}}$$ Equating: $$11200 = \sqrt{\frac{3k_BT}{m}}$$ Squaring both sides: $$11200^2 = \frac{3k_BT}{m}$$ Therefore: $$T = \frac{m(11200)^2}{3k_B}$$ Substituting values: $$T = \frac{2.76 \times 10^{-26} \times (11200)^2}{3 \times 1.38 \times 10^{-23}}$$ $$T \approx 8.360 \times 10^4\text{ K}$$ Hence, option (B) is correct.

Explanation

For a closed organ pipe, the third harmonic frequency is: $$f_c = \frac{3v}{4l}$$ For an open organ pipe, the fundamental frequency is: $$f_o = \frac{v}{2L}$$ Given: $$f_o = f_c$$ Therefore: $$\frac{v}{2L} = \frac{3v}{4l}$$ Cancelling $$v$$: $$\frac{1}{2L} = \frac{3}{4l}$$ $$L = \frac{2l}{3}$$ Given closed pipe length: $$l = 20\text{ cm}$$ $$L = \frac{2 \times 20}{3} = 13.33\text{ cm}$$ Hence, option (A) is correct.

Explanation

Since the track is frictionless, mechanical energy is conserved. At the lowest point A: $$mgh = \frac{1}{2}mv^2$$ Thus: $$h = \frac{v^2}{2g}$$ For just completing a vertical circle, the minimum speed at the lowest point is: $$v = \sqrt{5gR}$$ Substituting: $$h = \frac{5gR}{2g} = \frac{5R}{2}$$ Given diameter: $$D = 2R \Rightarrow R = \frac{D}{2}$$ Therefore: $$h = \frac{5}{2} \times \frac{D}{2}$$ $$h = \frac{5D}{4}$$ Hence, option (C) is correct.

Explanation

Let the lighter block of mass $$m$$ move initially with velocity $$v$$ and the heavier block of mass $$4m$$ be at rest. After collision, the lighter block comes to rest. Using conservation of linear momentum: $$mv + 4m(0) = m(0) + 4mV$$ where $$V$$ is the final velocity of the heavier block. Thus: $$V = \frac{v}{4}$$ Coefficient of restitution: $$e = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$$ $$e = \frac{\frac{v}{4} - 0}{v - 0}$$ $$e = \frac{1}{4} = 0.25$$ Hence, option (B) is correct.

Explanation

Initial acceleration: $$a = \frac{6 - 0}{1} = 6\text{ m/s}^2$$ From $$t = 0$$ to $$1\text{ s}$$: $$s_1 = \frac{1}{2}at^2 = \frac{1}{2}(6)(1)^2 = 3\text{ m}$$ Velocity at $$t = 1\text{ s}$$ is $$6\text{ m/s}$$. The electric field is then reversed, so acceleration becomes $$-6\text{ m/s}^2$$. From $$t = 1$$ to $$2\text{ s}$$: $$s_2 = ut + \frac{1}{2}at^2$$ $$s_2 = 6(1) + \frac{1}{2}(-6)(1)^2 = 3\text{ m}$$ Velocity becomes zero at $$t = 2\text{ s}$$. From $$t = 2$$ to $$3\text{ s}$$: $$s_3 = \frac{1}{2}(-6)(1)^2 = -3\text{ m}$$ Total displacement: $$S = 3 + 3 - 3 = 3\text{ m}$$ Average velocity: $$\bar{v} = \frac{S}{t} = \frac{3}{3} = 1\text{ m/s}$$ Total distance travelled: $$= 3 + 3 + 3 = 9\text{ m}$$ Average speed: $$= \frac{9}{3} = 3\text{ m/s}$$ Hence, option (B) is correct.