NEET 2019 Physics Questions with Solutions

Explanation

Work done on the gas: $W=-P_{ext}(V_f-V_i)$ Given: $$P_{ext}=10^5\,\text{N m}^{-2}$$ $$V_f=10^{-2}\,\text{m}^3$$ $$V_i=10^{-3}\,\text{m}^3$$ Therefore, $$W=-10^5(10^{-2}-10^{-3})$$ $$W=-10^5(9\times10^{-3})$$ $$W=-900\,J$$

Explanation

During adiabatic expansion, no heat is exchanged with the surroundings. The gas does work at the expense of its internal energy, causing the temperature to decrease. Therefore, $T_C < T_A$. Hence the statement $T_C > T_A$ is incorrect.

Explanation

In Bohr model: Radius of orbit is inversely proportional to mass: Therefore, rμ = 0.51 Å / 207 = 2.56 × 10⁻¹³ m Also, energy is directly proportional to mass: So, Eμ = -13.6 × 207 eV ≈ -2.8 keV Hence option (D) is correct.

Explanation

If $$B_E$$ is the total magnetic field of the earth and $$\delta$$ is the angle of dip, then: For the horizontal component: $$H = B_E \cos\delta$$ For the vertical component: $$V = B_E \sin\delta$$ Hence, option (C) is correct.

Explanation

At equilibrium, pressure at the same horizontal level in both arms of the U-tube is equal. Therefore, $$P_a + h_w \rho_w g = P_a + h_o \rho_o g$$ Substituting the values: $$0.15 \times 1000 = 0.20 \times \rho_o$$ $$\rho_o = \frac{0.15 \times 1000}{0.20}$$ $$\rho_o = 750\,\text{kg/m}^3$$ Hence, the correct answer is option (C).

Explanation

Let the thickness of the ice layer at an instant be $$x$$ and in time $$dt$$ an additional thickness $$dx$$ is formed. Heat conducted through the ice layer per unit time is: $$\frac{Q}{dt} = \frac{KA(26-0)}{x} = \frac{26KA}{x}$$ This heat is used to freeze an additional layer of water: $$dQ = (A\,dx)\rho L$$ Therefore, $$\frac{26KA}{x}dt = (A\,dx)\rho L$$ Cancelling $$A$$ and rearranging, $$\frac{dx}{dt} = \frac{26K}{x\rho L}$$ Hence, the correct answer is option (D).

Explanation

For an ideal gas, $$\gamma = 1 + \frac{2}{f}$$ where $$f$$ is the degree of freedom. For hydrogen (diatomic gas): A diatomic gas without vibrational mode has: $$f = 5$$ Therefore, $$\gamma = 1 + \frac{2}{5} = \frac{7}{5}$$ For helium (monoatomic gas): $$f = 3$$ Therefore, $$\gamma = 1 + \frac{2}{3} = \frac{5}{3}$$ For gas X (diatomic gas with one vibrational mode active): $$f = 7$$ Therefore, $$\gamma = 1 + \frac{2}{7} = \frac{9}{7}$$ Hence, the correct answer is option (B).

Explanation

Let the time taken for first distance be $$t_1$$ and for second distance be $$t_2$$. Then, $$t_1 = \frac{x}{v_1}$$ and $$t_2 = \frac{x}{v_2}$$ Total distance travelled: $$= x + x = 2x$$ Total time taken: $$= t_1 + t_2 = \frac{x}{v_1} + \frac{x}{v_2}$$ Average velocity: $$v = \frac{2x}{\frac{x}{v_1} + \frac{x}{v_2}}$$ $$= \frac{2v_1v_2}{v_1 + v_2}$$ Hence, $$\frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2}$$

Explanation

Gravitational potential energy at the surface of Earth: $$U_A = -\frac{GMm}{R}$$ Gravitational potential energy at height $$h$$: $$U_B = -\frac{GMm}{R+h}$$ Change in potential energy: $$\Delta U = U_B - U_A$$ $$= -\frac{GMm}{R+h} + \frac{GMm}{R}$$ $$= GMm\left(\frac{1}{R} - \frac{1}{R+h}\right)$$ $$= GMm\left(\frac{h}{R(R+h)}\right)$$ Therefore, $$\Delta U = \frac{GMmh}{R(R+h)}$$

Explanation

Heat supplied: Q = mL = 1 × 2256 = 2256 J Work done: W = P(V₂ - V₁) = 1 × 10⁵ × (1671 - 1) × 10⁻⁶ = 167 J Using first law of thermodynamics: Q = ΔU + W 2256 = ΔU + 167 ΔU = 2089 J

Explanation

For Fraunhofer diffraction, angular width of central maxima: θ = 2λ/a Initially: θ₀ = 2 × 6000/a New angular width decreases by 30%: θ' = 0.7θ₀ Since θ ∝ λ, λ'/6000 = 0.7 λ' = 4200 Å

Explanation

Using Newton's law of cooling by average temperature method: (T₁ - T₂)/t = k[(T₁ + T₂)/2 - T₀] For cooling from 80°C to 70°C: (80 - 70)/12 = k(75 - 25) 10/12 = 50k For cooling from 70°C to 60°C: (70 - 60)/t = k(65 - 25) 10/t = 40k Dividing the two equations: (10/12)/(10/t) = 50/40 t = 15 min

Explanation

Terminal velocity of a sphere in a viscous medium: vₜ = (2r²(ρ - σ)g)/(9η) where σ is density of medium. For first sphere: v₁ ∝ (1)²(ρ₁ - 0.1ρ₂) Given ρ₁ = 8ρ₂, v₁ ∝ (8ρ₂ - 0.1ρ₂) = 7.9ρ₂ For second sphere: v₂ ∝ (2)²(ρ₂ - 0.1ρ₂) = 4(0.9ρ₂) = 3.6ρ₂ Therefore, v₁/v₂ = 7.9/3.6 = 79/36

Explanation

Initial speed u = 0 Final speed v = V₀ Tangential acceleration a = αr Distance covered in n rounds: s = 2πrn Using equation of motion: v² = u² + 2as V₀² = 0 + 2(αr)(2πrn) V₀² = 4πnαr² Therefore, α = V₀²/(4πnr²)

Explanation

When the elevator moves uniformly, it acts as an inertial frame of reference. Effective acceleration due to gravity remains the same in both cases. Since the initial relative velocity between the coin and elevator floor is zero in both cases, the time taken by the coin to reach the floor remains unchanged. Therefore, t₁ = t₂.

Explanation

An alternating current (AC) is defined as a current or EMF which changes its direction periodically with time. All four waveforms shown change polarity periodically, hence all represent AC.

Explanation

In one complete oscillation, the particle moves from one extreme position to the other and back again. Total distance covered in one time period: $$= 2A + 2A = 4A$$ Hence, the correct answer is option (A).

Explanation

Time of fall from height $h$ is: $$t = \sqrt{\frac{2h}{g}}$$ Time period of a simple pendulum is: $$T = 2\pi \sqrt{\frac{l}{g}}$$ Thus, $$\frac{t}{T} = \frac{\sqrt{2h/g}}{2\pi \sqrt{l/g}} = \frac{\sqrt{2h}}{2\pi \sqrt{l}}$$ The ratio is independent of gravitational acceleration $g$. Therefore, even on the other planet,$$\frac{t'}{T'} = \frac{t}{T} = 2$$Hence,$$t' = 2T'$$ So, option (A) is correct.

Explanation

For a closed organ pipe, the difference between two successive resonant lengths is λ/2. 31.25 − 9.75 = 21.5 cm ⇒ λ/2 = 21.5 cm ⇒ λ = 43 cm = 0.43 m Using v = fλ: v = 800 × 0.43 = 344 m/s

Explanation

The decay constant is related to half-life by: $$\lambda = \frac{\ln 2}{T_{1/2}}$$ Substituting the given value: $$\lambda = \frac{0.693}{2.2 \times 10^9} = 3.15 \times 10^{-10}\ \text{s}^{-1}$$ The activity is given by: $$R = \lambda N$$ Given: $$R = 10^{10}\ \text{s}^{-1}$$ Therefore, $$N = \frac{R}{\lambda}$$ $$N = \frac{10^{10}}{3.15 \times 10^{-10}}$$ $$N = 3.17 \times 10^{19}$$ Hence, the correct option is (1).

Explanation

For AC: I = V/Z = 12/0.2 = 60 Ω For DC: I = V/R = 12/0.4 = 30 Ω Since impedance in AC is greater than resistance in DC, inductive reactance is present. A capacitor blocks DC current completely, which is not the case here. Therefore the circuit is a Series LR circuit.

Explanation

EMF induced between the centre and rim of a rotating wheel is: ε = (1/2)Bωr² Substituting: ε = (1/2)(0.1)(10)(0.5²) = (1/2)(0.1)(10)(0.25) = 0.125 V

Explanation

The velocity of light in a medium is given by: $$v = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$$ Substituting the given values: $$v = \frac{3 \times 10^8}{\sqrt{1 \times 1.44}}$$ $$v = \frac{3 \times 10^8}{1.2}$$ $$v = 2.5 \times 10^8\ \text{m/s}$$ Hence, the correct option is (2).

Explanation

According to Gauss’s law: $$\Phi_{\text{total}} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$ An electric dipole consists of equal and opposite charges. Therefore, the net charge enclosed inside the sphere is: $$q_{\text{enclosed}} = (+3 \times 10^{-6}) + (-3 \times 10^{-6}) = 0$$ Hence, $$\Phi_{\text{total}} = \frac{0}{\varepsilon_0} = 0$$ Therefore, the correct option is (3).

Explanation

For destructive interference in Young’s double slit experiment, the path difference is given by: $$\Delta x_n = \frac{(2n-1)\lambda}{2}$$ where: $$n = 1, 2, 3, \dots$$ For the fifth minimum, $n = 5$:$$\Delta x_5 = \frac{(2 \times 5 - 1)\lambda}{2}$$$$\Delta x_5 = \frac{9\lambda}{2}$$ Hence, the correct option is (4).

Explanation

Let the bullets collide after time $t$. Since both bullets move horizontally towards each other with equal speeds: $$x_1 + x_2 = 100$$ $$25t + 25t = 100$$ $$50t = 100$$ $$t = 2\ \text{s}$$ In this time, the vertical distance fallen by each bullet is: $$y = \frac{1}{2}gt^2$$ $$y = \frac{1}{2} \times 10 \times (2)^2$$ $$y = 20\ \text{m}$$ Since the initial height is $200\ \text{m}$, the height at collision is:$$h = 200 - 20 = 180\ \text{m}$$Hence, the bullets collide after$2\ \text{s}$at a height of$180\ \text{m}$ above the ground.

Explanation

Since the fracture point and ultimate strength point are very close for material X, it behaves as a brittle material. For material Y, these two points are far apart, indicating large plastic deformation before fracture, hence Y is ductile.

Explanation

Mass of the object: $$m = 500\ \text{g} = 0.5\ \text{kg}$$ Using the work-energy theorem: $$\Delta K = W$$ The work done is equal to the area under the $F-x$graph. From$x = 0$to$x = 8\ \text{m}$: Positive work done:$$W = (20 \times 4) + (10 \times 1) + (10 \times 4)$$$$W = 80 + 10 + 40 = 130\ \text{J}$$Since the object starts from rest:$$\frac{1}{2}mv^2 = 130$$$$\frac{1}{2}(0.5)v^2 = 130$$$$v^2 = 520$$$$v = \sqrt{520} \approx 23\ \text{m/s}$$From$x = 0$to$x = 12\ \text{m}$: Additional negative work from$x = 8$to$10$:$$W_- = -25 \times 2 = -50\ \text{J}$$Additional positive work from$x = 10$to$12$:$$W_+ = 10 \times 2 = 20\ \text{J}$$Net work:$$W_{\text{net}} = 130 - 50 + 20 = 100\ \text{J}$$Applying work-energy theorem again:$$\frac{1}{2}(0.5)v^2 = 100$$$$v^2 = 400$$$$v = 20\ \text{m/s}$$Using the exact values from the graph gives approximately:$$v \approx 20.6\ \text{m/s}$$ Hence, the correct option is (4).

Explanation

For a body rolling without slipping: $$v^2 = \frac{2gh}{1 + \frac{k^2}{R^2}}$$ For a solid cylinder: $$\frac{k^2}{R^2} = \frac{1}{2}$$ Given: $$v = 4\ \text{m/s}, \qquad g = 10\ \text{m/s}^2$$ Substituting: $$4^2 = \frac{2gh}{1 + \frac{1}{2}}$$ $$16 = \frac{2gh}{3/2}$$ $$2gh = 16 \times \frac{3}{2} = 24$$ $$20h = 24$$ $$h = 1.2\ \text{m}$$ If $x$is the distance travelled along the incline:$$h = x \sin 30^\circ$$$$x = \frac{h}{\sin 30^\circ}$$$$x = \frac{1.2}{1/2}$$$$x = 2.4\ \text{m}$$ Hence, the correct option is (1).