NEET 2023 Paper — NEET Complete Yearwise Papers MCQs with Solutions

Solution: For a solid sphere about its own axis, $I = \dfrac{2}{5}MR^2$, so $K_{solid} = R\sqrt{\dfrac{2}{5}}$. For a thin hollow sphere, $I = \dfrac{2}{3}MR^2$, so $K_{hollow} = R\sqrt{\dfrac{2}{3}}$. $\dfrac{K_{solid}}{K_{hollow}} = \sqrt{\dfrac{2/5}{2/3}} = \sqrt{\dfrac{3}{5}}$, i.e. $\sqrt{3} : \sqrt{5}$. Note: the options as printed in this paper have lost their square-root signs, and the printed key marks (d). Read the intended answer as $\sqrt{3} : \sqrt{5}$.

Solution: Secondary current: $i_s = \dfrac{P}{V_s} = \dfrac{60}{12} = 5\ \text{A}$. For an ideal transformer, $\dfrac{N_s}{N_p} = \dfrac{V_s}{V_p} = \dfrac{i_p}{i_s}$, so $i_p = \dfrac{i_s V_s}{V_p} = \dfrac{5 \times 12}{220} = 0.27\ \text{A}$

Solution: No current flows through the galvanometer, so the potential drop across $R$ must equal the $2\ \text{V}$ of the right-hand cell. That leaves $10 - 2 = 8\ \text{V}$ across the $400\ \Omega$ resistor: $i = \dfrac{8}{400} = \dfrac{1}{50}\ \text{A}$ The same current flows through $R$: $2 = \dfrac{1}{50} \times R \Rightarrow R = 100\ \Omega$

Solution: The capacitor acts as the filter. It charges while the rectified voltage rises and discharges through the load while it falls, smoothing out the ripple and leaving a steadier dc output.

Solution: For the photoelectric effect the incident energy must be at least equal to the work function, $E \geq W$. With $E = 2.20\ \text{eV}$: only caesium qualifies, since $2.20 > 2.14$. Potassium ($2.30\ \text{eV}$) and sodium ($2.75\ \text{eV}$) both need more energy than is supplied.

Solution: For an open pipe, $f_{open} = \dfrac{v}{2L}$. For a closed pipe, $f_{closed} = \dfrac{v}{4L}$. $\dfrac{f_{open}}{f_{closed}} = \dfrac{v/2L}{v/4L} = \dfrac{2}{1}$

Solution: A soap bubble has TWO surfaces, so the work done is $W = T \times \Delta A = T \times 2 \times 4\pi R^2 = 8\pi R^2 T$. $W = 0.03 \times 4\pi \times (2 \times 10^{-2})^2 \times 2 = 3.01 \times 10^{-4}\ \text{J}$

Solution: Longitudinal stress is the restoring force per unit area of cross-section. The whole weight $W$ acts along the wire, so the stress at any point is $\text{Stress} = \dfrac{W}{A}$

Solution: When equal DISTANCES are covered at two speeds, the average speed is their harmonic mean: $\bar{v} = \dfrac{2v_1v_2}{v_1 + v_2} = \dfrac{2 \times v \times 2v}{v + 2v} = \dfrac{4v^2}{3v} = \dfrac{4v}{3}$

Solution: Statement I is true: the angular separation is $\theta = \dfrac{\lambda}{d}$, which does not involve the screen distance $D$ at all. Moving the screen changes the fringe WIDTH, not the angular separation. Statement II is false: since $\theta \propto \lambda$, a larger wavelength INCREASES the angular separation rather than decreasing it.

Solution: Speed in air $v_1 = \dfrac{x}{t_1}$; speed in the other medium $v_2 = \dfrac{10x}{t_2}$. At the critical angle, $\sin C = \dfrac{\mu_{rarer}}{\mu_{denser}} = \dfrac{v_2}{v_1}$ $\sin C = \dfrac{10x/t_2}{x/t_1} = \dfrac{10t_1}{t_2} \Rightarrow C = \sin^{-1}\left(\dfrac{10t_1}{t_2}\right)$ Note: the answer letter printed beside this question in the source reads (a), but the paper's own worked solution uses exactly the formula above, which gives option (b).

Solution: Capacitive reactance $X_C = \dfrac{1}{2\pi fC}$, so $X_C \propto \dfrac{1}{f}$. As the frequency falls, $X_C$ RISES. The current is $I_D = \dfrac{V}{X_C}$, so as $X_C$ rises the displacement current falls.

Solution: The two $3\ \mu F$ capacitors on the right are in parallel: $3 + 3 = 6\ \mu F$. That combination is in series with the remaining $3\ \mu F$: $C_{eff} = \dfrac{6 \times 3}{6 + 3} = \dfrac{18}{9} = 2\ \mu F$

Solution: For an electromagnetic wave in free space, $E_0 = B_0 c$. $B_0 = \dfrac{E_0}{c} = \dfrac{48}{3 \times 10^8} = 1.6 \times 10^{-7}\ \text{T}$ (The frequency is not needed.)

Solution: The shortest wavelength of a series is the series limit, $n_2 \to \infty$. Balmer ($n_1 = 2$): $\dfrac{1}{\lambda} = R\left[\dfrac{1}{2^2} - 0\right] = \dfrac{R}{4}$ Brackett ($n_1 = 4$): $\dfrac{1}{\lambda'} = R\left[\dfrac{1}{4^2} - 0\right] = \dfrac{R}{16}$ $\dfrac{\lambda}{\lambda'} = \dfrac{1/16}{1/4} = \dfrac{1}{4} \Rightarrow \lambda' = 4\lambda$

Solution: Density $\rho = \dfrac{m}{A \times l} = \dfrac{m}{\pi r^2 l}$, so $\dfrac{\Delta\rho}{\rho} = \dfrac{\Delta m}{m} + 2\dfrac{\Delta r}{r} + \dfrac{\Delta l}{l}$ $= \dfrac{0.002}{0.4} + 2 \times \dfrac{0.001}{0.3} + \dfrac{0.02}{5} = 0.005 + 0.0067 + 0.004 \approx 1.6\%$

Solution: Force is the rate of change of MOMENTUM, and momentum is a vector. $\vec{F} = \dfrac{m(\vec{v}_f - \vec{v}_i)}{\Delta t} = \dfrac{m\left[v\hat{i} - v(-\hat{j})\right]}{\Delta t} = \dfrac{mv(\hat{i} + \hat{j})}{\Delta t}$ Taking $\hat{i}$ as east and $\hat{j}$ as north, the resultant points north-east.

Solution: $v_{rms} \propto \sqrt{T}$. "Increased by 3 times" means the new speed is $v + 3v = 4v$. $T_1 = -50 + 273 = 223\ \text{K}$ $\dfrac{v_2}{v_1} = \sqrt{\dfrac{T_2}{T_1}} \Rightarrow 4 = \sqrt{\dfrac{T_2}{223}} \Rightarrow T_2 = 16 \times 223 = 3568\ \text{K}$ $t_2 = 3568 - 273 = 3295^\circ C$

Solution: $22000\ \Omega = 22 \times 10^3\ \Omega$. The first two bands give the digits 2 and 2 (red, red), and the THIRD band is the multiplier — here $10^3$, which is orange.

Solution: $H = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{(280)^2 \times (0.5)^2}{2 \times 9.8}$ $H = \dfrac{78400}{19.6} \times \dfrac{1}{4} = 1000\ \text{m}$

Solution: $\eta = 1 - \dfrac{T_2}{T_1}$, with $T_1 = 327 + 273 = 600\ \text{K}$. $\dfrac{1}{2} = 1 - \dfrac{T_2}{600} \Rightarrow \dfrac{T_2}{600} = \dfrac{1}{2} \Rightarrow T_2 = 300\ \text{K}$ $300 - 273 = 27^\circ C$

Solution: At resonance, $f = \dfrac{1}{2\pi\sqrt{LC}}$. $f = \dfrac{1}{2\pi\sqrt{10 \times 10^{-3} \times 10^{-6}}} = \dfrac{10^4}{2\pi} \approx 1.59 \times 10^3\ \text{Hz} = 1.59\ \text{kHz}$

Solution: Zero net flux means the surface encloses no NET charge, so whatever flux enters must also leave — the number of flux lines entering equals the number leaving. It does not follow that the field is zero, uniform or constant on the surface, nor that all charges lie inside it.

Solution: The null point sits at $x$ from the mass $m$, where $\dfrac{Gm}{x^2} = \dfrac{9Gm}{(R-x)^2}$, giving $x = \dfrac{R}{\sqrt{9} + 1} = \dfrac{R}{4}$, so the distances are $\dfrac{R}{4}$ and $\dfrac{3R}{4}$. $V = -\dfrac{Gm}{R/4} - \dfrac{G(9m)}{3R/4} = -\dfrac{4Gm}{R} - \dfrac{12Gm}{R} = -\dfrac{16Gm}{R}$

Solution: The two cells oppose each other, so the net emf is $10 - 5 = 5\ \text{V}$, and the total resistance is $2 + 1 + 7 = 10\ \Omega$. $I = \dfrac{E_1 - E_2}{R} = \dfrac{5}{10} = 0.5\ \text{A}$ The larger cell drives the current, so it flows from $A$ to $B$ through $E$.

Solution: All the kinetic energy of the electron goes into a single photon at the cut-off: $eV = \dfrac{hc}{\lambda_{min}} \Rightarrow \lambda_{min} = \dfrac{hc}{eV}$ So $\lambda_{min} \propto \dfrac{1}{V}$.

Solution: Angular acceleration $\vec{\alpha}$ is an axial vector: it is related to the linear acceleration by $\vec{a} = \vec{\alpha} \times \vec{r}$. Like angular velocity, its direction lies along the AXIS OF ROTATION, given by the right-hand rule.

Solution: $U = \dfrac{1}{2}Li^2 = \dfrac{1}{2} \times 4 \times 10^{-6} \times (2)^2$ $U = \dfrac{1}{2} \times 4 \times 10^{-6} \times 4 = 8 \times 10^{-6}\ \text{J} = 8\ \mu J$

Solution: $\dfrac{1}{16} = \left(\dfrac{1}{2}\right)^4$, so 4 half lives must pass. $t = n \times T_{1/2} = 4 \times 20 = 80\ \text{minutes}$

Solution: $U = \dfrac{1}{2}kx^2$, so $U \propto x^2$. $\dfrac{U_2}{U_1} = \left(\dfrac{x_2}{x_1}\right)^2 = \left(\dfrac{8}{2}\right)^2 = 16$ So $U_2 = 16U$.