Re-NEET 2026 Question Paper with Solutions

Physics Questions 1 to 45

1. A particle of mass $M$ moves along a horizontal $x$ axis from $x = 0$ to $x = L$. The coefficient of kinetic friction varies as a function of $x$ as $\mu_k(x) = \mu_0 - \alpha x$, where $\mu_0$, $\alpha$ are constants of appropriate dimensions, so that $\mu_k(L) = 0$. The total work done by the frictional force during the motion is $n\mu_0 MgL$, where $g$ is the acceleration due to gravity. The value of $n$ is:

   1. (1)

      3

   2. (2)

      1

   3. (3)

      $\frac{1}{3}$

   4. (4)

      $\frac{1}{2}$

   Answer and solution

   Answer: (4)

   The condition $\mu_k(L) = 0$ fixes $\alpha = \mu_0/L$, so $\mu_k(x) = \mu_0\left(1 - \frac{x}{L}\right)$ falls linearly from $\mu_0$ to zero. Because the force varies with position, the work must be integrated: $|W| = \int_0^L \mu_k(x) Mg\,dx = \mu_0 Mg\left[L - \frac{L}{2}\right] = \frac{1}{2}\mu_0 MgL$. Comparing with $n\mu_0 MgL$ gives $n = 1/2$, which is simply the average of the initial and final friction coefficients since the variation is linear. NCERT Class 11, Work, Energy and Power (work done by a variable force).

2. The mean free path of molecules in an ideal gas $A$ is half that of another ideal gas $B$. The diameter of the spherical molecules of gas $A$ is twice the diameter of the molecules of $B$. If number densities of the gases $A$ and $B$ are $n_A$ and $n_B$, respectively, the correct option is:

   1. (1)

      $n_A = n_B$

   2. (2)

      $n_A = 2n_B$

   3. (3)

      $n_A = \frac{1}{4}n_B$

   4. (4)

      $n_A = \frac{1}{2}n_B$

   Answer and solution

   Answer: (4)

   Mean free path is $\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$, so $\lambda \propto \frac{1}{d^2 n}$. Taking the ratio, $\frac{\lambda_A}{\lambda_B} = \frac{d_B^2 n_B}{d_A^2 n_A}$. Substituting $\lambda_A/\lambda_B = \frac{1}{2}$ and $d_A = 2d_B$ gives $\frac{1}{2} = \frac{n_B}{4n_A}$, hence $n_A = \frac{1}{2}n_B$. NCERT Class 11, Kinetic Theory (mean free path).

3. Two identical inductors are connected in two different configurations $P$ and $Q$, where a time varying current $I(t)$ is flowing, as shown in the figure. The induced emf between points $a$ and $b$ for configuration $P$ is $E_P$ and that for configuration $Q$ is $E_Q$. The ratio $E_P/E_Q$ is: [Neglect the effect of mutual inductance.]

   
   1. (1)

      $\frac{1}{4}$

   2. (2)

      $\frac{1}{2}$

   3. (3)

      1

   4. (4)

      2

   Answer and solution

   Answer: (4)

   In configuration $P$ the two inductors are in series and the segment between $a$ and $b$ spans just ONE of them, carrying the full current, so $E_P = L\frac{dI}{dt}$. In configuration $Q$ the points $a$ and $b$ straddle the parallel pair, whose equivalent inductance is $\frac{L}{2}$, so $E_Q = \frac{L}{2}\frac{dI}{dt}$. The ratio is therefore 2. NCERT Class 12, Electromagnetic Induction (self-inductance).
   
   Why the other options are wrong:
   
   - (1) takes $E_P$ across the series pair as $2L$ and inverts the ratio
   - (2) inverts the correct answer by swapping which configuration sits in the numerator
   - (3) assumes identical inductors must produce identical electromotive forces (emfs) regardless of how $a$ and $b$ are placed

4. For sound waves, if the number of nodes for the 5th harmonic of an open-ended pipe is $n$ and that for the 9th harmonic of the same pipe with one of its ends closed is $m$, the ratio $\frac{n}{m}$ is

   1. (1)

      $\frac{5}{9}$

   2. (2)

      $\frac{9}{5}$

   3. (3)

      1

   4. (4)

      $\frac{3}{5}$

   Answer and solution

   Answer: (3)

   An open-open pipe has antinodes at both ends, and its $p$-th harmonic contains exactly $p$ nodes, so the 5th harmonic gives $n = 5$. A closed pipe supports only odd harmonics, with a node at the closed end and an antinode at the open end; its $q$-th harmonic ($q$ odd) contains $\frac{q+1}{2}$ nodes, so the 9th harmonic gives $m = \frac{10}{2} = 5$. The ratio is therefore 1. NCERT Class 11, Waves (standing waves in air columns).

5. Consider a long solenoid of length $l$ and radius $r$. If $n$ is the number of turns per unit length and $\mu_0$ is the permeability of free space, the inductance of the solenoid is :

   1. (1)

      $\mu_0\pi n^2 r^2 l$

   2. (2)

      $\mu_0 n^2 r^2 l$

   3. (3)

      $(\mu_0/2\pi) n^2 r^2 l$

   4. (4)

      $2\mu_0\pi n^2 r^2 l$

   Answer and solution

   Answer: (1)

   For a long solenoid $L = \mu_0 n^2 A l$, where $n$ is turns per unit length and $A$ the cross-sectional area. Substituting $A = \pi r^2$ gives $L = \mu_0\pi n^2 r^2 l$. NCERT Class 12, Electromagnetic Induction (self-inductance).

6. Consider a particle moving along a straight line, whose position as a function of time is given by $s(t) = \alpha t^2 - \beta t + \gamma$, where $\alpha = 1\ \text{ms}^{-2}$, $\beta = 6\ \text{ms}^{-1}$ and $\gamma = 5\ \text{m}$. The average speed of the particle, in $\text{ms}^{-1}$ from $t = 0$ to $t = 6$ s is:

   1. (1)

      12

   2. (2)

      6

   3. (3)

      3

   4. (4)

      0

   Answer and solution

   Answer: (3)

   Average speed uses total PATH LENGTH, not displacement, so the turning point must be located first. Here $v = \frac{ds}{dt} = 2t - 6$, which vanishes at $t = 3$ s. Positions are $s(0) = 5\,\text{m}$, $s(3) = -4\,\text{m}$, $s(6) = 5\,\text{m}$, giving 9 m travelled in each leg and 18 m in total. Hence average speed $= 18/6 = 3\ \text{ms}^{-1}$. NCERT Class 11, Motion in a Straight Line.

7. Consider the following nuclear reaction $^{238}\mathrm{U} \rightarrow ^{234}\mathrm{Th} + ^{4}\mathrm{He}$. Take masses of $^{238}\mathrm{U}$, $^{234}\mathrm{Th}$ and $^{4}\mathrm{He}$ as 238.050 u, 234.043 u and 4.003 u, respectively. The Q value for the reaction, in keV, is: [Given: 1 u = 931.5 MeV $c^{-2}$]

   1. (1)

      3726

   2. (2)

      3730

   3. (3)

      3736

   4. (4)

      3740

   Answer and solution

   Answer: (1)

   The Q value is the mass defect converted to energy: $\Delta m = 238.050 - (234.043 + 4.003) = 0.004\ \text{u}$. Multiplying by the conversion factor, $Q = 0.004 \times 931.5 = 3.726\ \text{MeV} = 3726\ \text{keV}$. The positive value confirms the decay is spontaneous. NCERT Class 12, Nuclei (mass-energy and nuclear reactions).

8. A beam of light falls on a metal surface such that photo-electrons are generated. If power of the light source starts to decrease linearly with time $t$, then variation of the photocurrent $I$ and magnitude of the stopping potential $|V|$ with time is best represented by:

   1. (1)
      
   2. (2)
      
   3. (3)
      
   4. (4)
      
   Answer and solution

   Answer: (1)

   Photocurrent is set by the NUMBER of photons striking the surface per second, so it is proportional to the source power and falls linearly with $t$. Stopping potential, by contrast, depends only on the maximum kinetic energy of the emitted electrons, $eV_0 = h\nu - \phi_0$, which is fixed by the FREQUENCY and the work function - neither of which changes here. So $|V|$ stays flat while $I$ decays. NCERT Class 12, Dual Nature of Radiation and Matter.
   
   Why the other options are wrong:
   
   - (2) treats stopping potential as intensity-dependent, the single most common photoelectric misconception
   - (3) swaps the roles entirely, tying current to frequency and stopping potential to intensity
   - (4) assumes neither quantity responds to the source, ignoring that current tracks photon flux

9. Three identical capacitors, $P$, $Q$ and $S$, each of the capacitance $C$, are connected to a battery of voltage $V$, as shown in the figure. If the energy stored in the capacitor $P$ and total energy stored in the system are $U_P$ and $U_T$, respectively, then the ratio $\frac{U_P}{U_T}$ is:

   
   1. (1)

      2/3

   2. (2)

      1/3

   3. (3)

      1/2

   4. (4)

      1/6

   Answer and solution

   Answer: (4)

   $P$ and $Q$ form a series branch of equivalent capacitance $C/2$, and that branch sits in parallel with $S$ across the battery, so $C_{eq} = \frac{C}{2} + C = \frac{3C}{2}$ and $U_T = \frac{1}{2}\left(\frac{3C}{2}\right)V^2 = \frac{3}{4}CV^2$. Identical capacitors in series split the voltage equally, so $P$ holds $V/2$ and $U_P = \frac{1}{2}C\left(\frac{V}{2}\right)^2 = \frac{CV^2}{8}$. The ratio is $\frac{CV^2/8}{3CV^2/4} = \frac{1}{6}$. NCERT Class 12, Electrostatic Potential and Capacitance.
   
   Why the other options are wrong:
   
   - (1) treats all three capacitors as being in parallel, so each stores an equal third and $P$ plus $Q$ gives 2/3
   - (2) assumes the three capacitors share the energy equally, ignoring that $P$ sees only half the battery voltage
   - (3) computes the ratio of $U_P$ to the energy of the series branch alone, forgetting to include $S$ in the total

10. Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between $P$ and $Q$ is 15 $\text{Nm}^{-2}$. The area of cross-section at $P$ and $Q$ are 40 $\text{cm}^2$ and 20 $\text{cm}^2$, respectively. The rate of flow of water through the pipe, in $\text{cm}^3\text{s}^{-1}$, is: [Take density of water = 1000 $\text{kg m}^{-3}$]

    
    1. (1)

       100

    2. (2)

       200

    3. (3)

       300

    4. (4)

       400

    Answer and solution

    Answer: (4)

    Continuity gives $A_P v_P = A_Q v_Q$, so halving the area doubles the speed: $v_Q = 2v_P$. For a horizontal pipe Bernoulli's equation reduces to $\Delta P = \frac{1}{2}\rho(v_Q^2 - v_P^2) = \frac{3}{2}\rho v_P^2$. Substituting $15 = 1500\,v_P^2$ gives $v_P = 0.1\ \text{ms}^{-1} = 10\ \text{cm s}^{-1}$, and the volume flow rate is $A_P v_P = 40 \times 10 = 400\ \text{cm}^3\text{s}^{-1}$. NCERT Class 11, Mechanical Properties of Fluids.
    
    Why the other options are wrong:
    
    - (1) uses the narrow area $A_Q$ with the wide-section speed $v_P$, mismatching the pair
    - (2) drops the factor 3 from $v_Q^2 - v_P^2$, effectively using $\Delta P = \frac{1}{2}\rho v_Q^2$ alone
    - (3) mixes SI and centimetre-gram-second (CGS) units partway through, most often by leaving $\rho$ in $\text{kg m}^{-3}$ alongside areas in $\text{cm}^2$

11. A current $I_0$ flows through a metallic circular loop of radius $r$ as shown in the figure. Resistance of the segment $ABC$ is half that of $ADC$. Magnitude of magnetic field at the centre $O$ of the loop is :

    
    1. (1)

       $\frac{\mu_0 I_0}{12r}$

    2. (2)

       $\frac{\mu_0 I_0}{4r}$

    3. (3)

       $\frac{\mu_0 I_0}{2r}$

    4. (4)

       $\frac{\mu_0 I_0}{2\pi r}$

    Answer and solution

    Answer: (1)

    The current entering at $A$ divides between the two arcs in INVERSE proportion to their resistances, so with $R_{ABC} = \frac{1}{2}R_{ADC}$ the currents are $\frac{2}{3}I_0$ and $\frac{1}{3}I_0$. Each arc is a semicircle ($A$ at the top, $C$ at the bottom), contributing $\frac{\mu_0 I}{4r}$ but in OPPOSITE directions at $O$. The net field is $\frac{\mu_0}{4r}\left(\frac{2}{3}I_0 - \frac{1}{3}I_0\right) = \frac{\mu_0 I_0}{12r}$. NCERT Class 12, Moving Charges and Magnetism (Biot-Savart law).
    
    Why the other options are wrong:
    
    - (2) sends the whole current $I_0$ round one semicircle and ignores the opposing contribution from the other arc
    - (3) uses the full-loop result $\frac{\mu_0 I}{2r}$ without splitting the current or accounting for cancellation
    - (4) applies the formula for a long straight wire, $\frac{\mu_0 I}{2\pi r}$, instead of an arc

12. In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used. The variation of the terminal velocity ($v$) with the ratio of density of spherical ball ($\sigma$) to density of the liquid ($\rho$), is best represented by:

    1. (1)
       
    2. (2)
       
    3. (3)
       
    4. (4)
       
    Answer and solution

    Answer: (1)

    Stokes' law gives $v = \frac{2r^2 g}{9\eta}(\sigma - \rho)$, which can be rewritten as $v = \frac{2r^2 g\rho}{9\eta}\left(\frac{\sigma}{\rho} - 1\right)$. With $r$ fixed this is a STRAIGHT LINE in $\sigma/\rho$ whose intercept on the horizontal axis is exactly at $\sigma/\rho = 1$. Below that value the ball is lighter than the liquid and does not sink at all, so the downward terminal velocity stays zero. NCERT Class 11, Mechanical Properties of Fluids (Stokes' law).
    
    Why the other options are wrong:
    
    - (2) takes $v \propto \sigma$ and lets the line pass through the origin, ignoring the buoyant term $-\rho$
    - (3) places the zero-velocity crossing at some arbitrary density ratio rather than at the neutral-buoyancy point $\sigma = \rho$
    - (4) treats terminal velocity as a fixed property of the ball once it sinks, missing its linear dependence on excess density

13. Two planets $P_1$ and $P_2$ with equal mass have radii $R_1$ and $R_2$, respectively, where $R_2 = \frac{R_1}{2}$. The escape speeds of $P_1$ and $P_2$ are $v_1$ and $v_2$, respectively. Then $\frac{v_2}{v_1}$ is:

    1. (1)

       $\frac{1}{\sqrt{2}}$

    2. (2)

       1

    3. (3)

       $\sqrt{2}$

    4. (4)

       2

    Answer and solution

    Answer: (3)

    Escape speed is $v_e = \sqrt{\frac{2GM}{R}}$. With the masses equal, $v_e \propto R^{-1/2}$, so $\frac{v_2}{v_1} = \sqrt{\frac{R_1}{R_2}} = \sqrt{2}$. The denser, smaller planet is harder to escape from, which is the physical sense check. NCERT Class 11, Gravitation (escape speed).

14. In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius $R$ is proportional to:

    1. (1)

       $R^{1/2}$

    2. (2)

       $R^{3/2}$

    3. (3)

       $R^{2}$

    4. (4)

       $R^{3}$

    Answer and solution

    Answer: (2)

    Kepler's third law states $T^2 \propto R^3$ for orbits about a common central mass, so $T \propto R^{3/2}$. It also follows directly from equating gravitational and centripetal force: $\frac{GMm}{R^2} = m\omega^2 R$ with $\omega = 2\pi/T$. NCERT Class 11, Gravitation (Kepler's laws).

15. Two infinitely long parallel conducting wires $A$ and $B$ carry currents $I$ and $2I$, respectively, in the same direction. The wire $A$ has uniform mass per unit length $\lambda$ and lies on an insulated floor. The wire $B$ is kept fixed at a height $h$ above the floor. The minimum magnitude of $h$ so that the wire $A$ does not rise from the floor is: [$g$ is the acceleration due to gravity and $\mu_0$ is the permeability of free space.]

    1. (1)

       $\frac{\mu_0 I^2}{2\pi\lambda g}$

    2. (2)

       $\frac{\mu_0 I^2}{\pi\lambda g}$

    3. (3)

       $\frac{2\mu_0 I^2}{\pi\lambda g}$

    4. (4)

       $\frac{4\mu_0 I^2}{\pi\lambda g}$

    Answer and solution

    Answer: (2)

    Parallel currents in the SAME direction attract, so wire $B$ pulls wire $A$ upwards with force per unit length $\frac{\mu_0 (I)(2I)}{2\pi h} = \frac{\mu_0 I^2}{\pi h}$. Wire $A$ stays on the floor as long as this does not exceed its weight per unit length $\lambda g$, so $\frac{\mu_0 I^2}{\pi h} \le \lambda g$ and $h_{min} = \frac{\mu_0 I^2}{\pi\lambda g}$. NCERT Class 12, Moving Charges and Magnetism (force between two parallel currents).

16. An ideal Zener diode with breakdown voltage of -3 V is reverse biased with a negative input voltage $V_i = -5\ \text{V}$. The magnitude of voltage difference between point $B$ and $A$ is:

    
    1. (1)

       3 V

    2. (2)

       2 V

    3. (3)

       1 V

    4. (4)

       0 V

    Answer and solution

    Answer: (2)

    Once an ideal Zener diode is driven past its breakdown voltage it CLAMPS the voltage across itself at 3 V and behaves like a constant-voltage source. The remaining $5 - 3 = 2\ \text{V}$ of the supply must therefore appear across the series resistor, which is the element between $B$ and $A$. NCERT Class 12, Semiconductor Electronics (Zener diode as a voltage regulator).
    
    Why the other options are wrong:
    
    - (1) reports the voltage across the ZENER rather than across the resistor between $B$ and $A$
    - (3) subtracts twice, or assumes the diode drops an extra 1 V forward-bias contribution
    - (4) assumes the reverse-biased diode blocks all current, so no drop appears across the resistor - but breakdown means current does flow

17. In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas ($\gamma = 5/3$) decreases from 60 K to 50 K. The work done by the gas in the process is: (Take the universal gas constant as $R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}$)

    1. (1)

       41.5 J

    2. (2)

       83 J

    3. (3)

       124.5 J

    4. (4)

       166 J

    Answer and solution

    Answer: (3)

    In an adiabatic process $Q = 0$, so the work done by the gas comes entirely out of its internal energy: $W = -\Delta U = nC_V(T_1 - T_2)$. For a monatomic gas $C_V = \frac{3}{2}R$, giving $W = 1 \times \frac{3}{2}(8.3)(10) = 124.5\ \text{J}$. The same follows from $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$ with $\gamma - 1 = \frac{2}{3}$. NCERT Class 11, Thermodynamics.

18. A ray of light with wavelength $\lambda$ is incident on three different photoelectric cells namely 1, 2 and 3. The threshold wavelength of these photoelectric cells are $\lambda_1$, $\lambda_2$, and $\lambda_3$, respectively and the magnitude of stopping potentials of these cells are $V_1$, $V_2$ and $V_3$, respectively. The relation between $\lambda$ and threshold wavelengths are $\lambda_1 < \lambda$, $\lambda_2 > \lambda$ and $\lambda_3 \gg \lambda$. The correct option is:

    1. (1)

       $V_1 = 0,\ V_2 < V_3$

    2. (2)

       $V_1 = 0,\ V_2 > V_3$

    3. (3)

       $V_1 > V_2,\ V_3 = 0$

    4. (4)

       $V_1 < V_2,\ V_3 = 0$

    Answer and solution

    Answer: (1)

    Emission occurs only when the incident wavelength is SHORTER than the threshold wavelength. Cell 1 has $\lambda_1 < \lambda$, so the photons are too weak and $V_1 = 0$. For the other two, $eV_0 = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_{th}}\right)$; a LARGER threshold wavelength means a smaller work function and hence a larger stopping potential. Since $\lambda_3 \gg \lambda_2$, the term $1/\lambda_3$ is nearly zero and $V_3 > V_2$. NCERT Class 12, Dual Nature of Radiation and Matter (Einstein's photoelectric equation).

19. A photon and an electron, each of 20 eV energy, move in free space. The ratio of linear momentum of electron $p_e$ to that of photon $p_{Ph}$, $\frac{p_e}{p_{Ph}}$ is: [Take speed of light = $3 \times 10^8\ \text{ms}^{-1}$, charge of electron = $-1.6 \times 10^{-19}$ C and mass of electron = $9 \times 10^{-31}$ kg]

    1. (1)

       $\frac{2}{450}$

    2. (2)

       $\frac{1}{250}$

    3. (3)

       225

    4. (4)

       275

    Answer and solution

    Answer: (3)

    A photon's momentum is $p = \frac{E}{c} = \frac{20 \times 1.6 \times 10^{-19}}{3 \times 10^8} \approx 1.07 \times 10^{-26}\ \text{kg m s}^{-1}$. A non-relativistic electron of the same kinetic energy has $p = \sqrt{2mE} = \sqrt{2(9 \times 10^{-31})(3.2 \times 10^{-18})} = 2.4 \times 10^{-24}\ \text{kg m s}^{-1}$. The ratio is therefore about 225 - a massive particle carries far more momentum than a photon of the same energy, because $p \propto \sqrt{E}$ for matter but $p \propto E$ for light. NCERT Class 12, Dual Nature of Radiation and Matter.

20. Which of the following measurements require 'index correction'?

    1. (1)

       Measurement of resistance of a wire using meter bridge

    2. (2)

       Measurement of gravitational acceleration using simple pendulum

    3. (3)

       Measurement of focal length of lenses using optical bench

    4. (4)

       Measurement of speed of sound using resonance tube

    Answer and solution

    Answer: (3)

    On an optical bench the positions read on the scale are those of the UPRIGHTS, not of the optical centres of the lens, mirror or object pin. The fixed offset between an upright's index mark and the actual optical element is removed by the index (or bench) correction, applied to every measured object and image distance before computing the focal length. NCERT Class 12 practical work, Ray Optics and Optical Instruments.

21. A unit positive point charge is taken slowly through an infinitesimally thin tube that is inside a charged dielectric sphere of radius $R$, having uniform positive charge density $\rho$, as shown in the figure. The initial and final positions of the charge are marked by $A$ and $B$ at distance $2R$ and $3R$ respectively, from the centre of the sphere. In this process, the magnitude of the total work done on the point charge is $\frac{\rho R^2}{n\varepsilon_0}$. The value of $n$ is : ($\varepsilon_0$ is the permittivity of vacuum)

    
    1. (1)

       2

    2. (2)

       6

    3. (3)

       9

    4. (4)

       18

    Answer and solution

    Answer: (4)

    Work done in moving a charge slowly is path-independent and depends only on the endpoint potentials, so the detour through the tube is a red herring. Both $A$ and $B$ lie OUTSIDE the sphere, where $V(r) = \frac{Q}{4\pi\varepsilon_0 r}$ with $Q = \frac{4}{3}\pi R^3\rho$. Then $|W| = |V_B - V_A| = \frac{Q}{4\pi\varepsilon_0 R}\left|\frac{1}{3} - \frac{1}{2}\right| = \frac{\rho R^2}{18\varepsilon_0}$, giving $n = 18$. NCERT Class 12, Electrostatic Potential and Capacitance.
    
    Why the other options are wrong:
    
    - (1) uses only the $\frac{1}{2}$ from position $A$ and drops the $\frac{4}{3}\pi R^3$ volume factor
    - (2) computes $\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ but forgets the $\frac{1}{3}$ carried by the sphere's total charge
    - (3) integrates the internal field along the tube as if the endpoints were inside the sphere

22. Consider three media $P$, $Q$ and $R$ with refractive indices 1, 1.25, and 1.5 respectively. The medium $Q$ having a thickness of 5 cm is placed between extended media $P$ and $R$ as shown in the figure. An object $O$ is placed at the centre of medium $Q$. If viewed from medium $P$ near the normal direction, the apparent depth of $O$ is $h_1$. For similar observation from medium $R$, the apparent depth is $h_2$. The value of $|h_1 - h_2|$, in cm, is:

    
    1. (1)

       0

    2. (2)

       1

    3. (3)

       2

    4. (4)

       3

    Answer and solution

    Answer: (2)

    For near-normal viewing across a plane boundary, apparent depth $=$ real depth $\times \frac{n_{\text{observer}}}{n_{\text{object}}}$, and the object sits 2.5 cm from each face. Viewed from $P$: $h_1 = 2.5 \times \frac{1}{1.25} = 2\ \text{cm}$. Viewed from $R$: $h_2 = 2.5 \times \frac{1.5}{1.25} = 3\ \text{cm}$. Hence $|h_1 - h_2| = 1\ \text{cm}$. Looking in from the DENSER side makes the object appear farther away, which is the sense check here. NCERT Class 12, Ray Optics and Optical Instruments (refraction at plane surfaces).
    
    Why the other options are wrong:
    
    - (1) assumes the symmetric placement of the object makes the two apparent depths equal, ignoring the different observer media
    - (3) uses the full 5 cm thickness instead of the 2.5 cm from the object to each face
    - (4) computes $h_2$ alone and reports it as the difference

23. A frictionless circular wire of unit radius is fixed on the horizontal plane. Two-point particles of unit mass start moving simultaneously from point $A\left(\theta = \frac{\pi}{2}\right)$ with identical uniform angular speeds in opposite directions, and meet again at point $B\left(\theta = -\frac{\pi}{2}\right)$. During this time, which of the following figures schematically represent the magnitude of the total linear momentum $P$ of the system, as a function of $\theta$?

    
    1. (1)
       
    2. (2)
       
    3. (3)
       
    4. (4)
       
    Answer and solution

    Answer: (3)

    By symmetry the two particles sit at mirror-image angles about the vertical, so their tangential velocities have equal and opposite horizontal components while their vertical components add. Writing the particles at $\theta$ and $\pi - \theta$, the resultant works out to $P = 2m\omega r|\cos\theta|$, which is zero at the start ($\theta = \pi/2$), peaks when the particles are diametrically opposite the meeting point ($\theta = 0$), and returns to zero as they collide at $B$. That single smooth hump is option (3). NCERT Class 11, System of Particles and Rotational Motion (motion of the centre of mass).
    
    Why the other options are wrong:
    
    - (1) assumes momentum is conserved as a constant because the SPEEDS are constant, forgetting that momentum is a vector whose direction keeps changing
    - (2) inverts the phase - takes $P \propto |\sin\theta|$, i.e. maximum where the particles start rather than where they are farthest apart
    - (4) assumes $P$ falls off linearly with angle instead of following the cosine of the half-separation

24. The temperature of a metallic sphere of radius $R$ is increased by a small amount $\Delta T$. If the linear coefficient of thermal expansion of the metal is $\alpha$, the approximate increase in the volume of the sphere is:

    1. (1)

       $2\pi R^3\alpha\Delta T$

    2. (2)

       $3\pi R^3\alpha\Delta T$

    3. (3)

       $4\pi R^3\alpha\Delta T$

    4. (4)

       $6\pi R^3\alpha\Delta T$

    Answer and solution

    Answer: (3)

    For small expansions the coefficient of volume expansion is $\gamma = 3\alpha$. Hence $\Delta V = V\gamma\Delta T = \frac{4}{3}\pi R^3 \times 3\alpha \times \Delta T = 4\pi R^3\alpha\Delta T$. Equivalently, differentiating $V = \frac{4}{3}\pi R^3$ gives $\Delta V = 4\pi R^2 \Delta R$ with $\Delta R = R\alpha\Delta T$. NCERT Class 11, Thermal Properties of Matter.

25. A cylindrical cork of uniform density floats in a liquid of density $\rho_1$. If the cork is depressed slightly and released, it oscillates harmonically with time period $T$. If the same cork floats in another liquid of density $\rho_2$, then the similar oscillation has time period $2T$. The value of $\rho_2/\rho_1$ is:

    1. (1)

       4

    2. (2)

       2

    3. (3)

       1/2

    4. (4)

       1/4

    Answer and solution

    Answer: (4)

    Depressing the cork by $x$ creates an extra buoyant force $A\rho g x$ directed back towards equilibrium, so the effective spring constant is $k = A\rho g$ and $T = 2\pi\sqrt{\frac{m}{A\rho g}}$. With $m$ and $A$ unchanged, $T \propto \rho^{-1/2}$. Doubling the period therefore requires the density to fall by a factor of four: $\frac{\rho_2}{\rho_1} = \frac{1}{4}$. NCERT Class 11, Oscillations (simple harmonic motion (SHM) examples).

26. One main scale division of a Vernier calliper is equal to 1 mm and the number of divisions on the Vernier scale is 10. When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that 4th Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of the wire to be 1 cm, the actual length of the wire is :

    1. (1)

       0.60 cm

    2. (2)

       0.96 cm

    3. (3)

       1.00 cm

    4. (4)

       1.06 cm

    Answer and solution

    Answer: (4)

    NTA dropped this question because no printed option matched the correct answer; option (4) is corrected here to 1.06 cm. Working: least count $= \frac{1\,\text{MSD}}{N} = \frac{1\,\text{mm}}{10} = 0.1\,\text{mm}$. The Vernier zero lies to the left of the main-scale zero, so the zero error is negative. With the 4th Vernier division coinciding, zero error $= -(N-n)\times LC = -(10-4)(0.1) = -0.6\,\text{mm}$. Therefore true length $= \text{observed reading} - \text{zero error} = 10.0 - (-0.6) = 10.6\,\text{mm} = 1.06\,\text{cm}$. NCERT Class 11, Units and Measurements (least count and instrumental error).

27. A solid sphere $A$ of radius $R$ and mass $M$ is attached at a point to a smaller solid sphere $B$ of radius $r < R$ and mass $m < M$. Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of $A$ is $I_A$ and that calculated about a vertical axis passing through the centre of $B$ is $I_B$. The difference $I_A - I_B$ is:

    
    1. (1)

       $(M - m)(R + r)^2$

    2. (2)

       $(m - M)(R + r)^2$

    3. (3)

       $(m - M)(R - r)^2$

    4. (4)

       0

    Answer and solution

    Answer: (2)

    The spheres touch, so their centres are separated by $d = R + r$. About the axis through $A$: $I_A = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + m d^2$; about the axis through $B$: $I_B = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + M d^2$. The two spin terms are common to both and cancel, leaving $I_A - I_B = (m - M)(R + r)^2$, a negative quantity as expected since the heavier sphere sits off-axis in the second case. NCERT Class 11, System of Particles and Rotational Motion (parallel axis theorem).
    
    Why the other options are wrong:
    
    - (1) gets the separation right but flips the sign - attaches the parallel-axis term to the wrong body in each case
    - (3) uses $R - r$ as the centre separation, as if the small sphere were embedded in the large one rather than attached externally
    - (4) assumes the moment of inertia is independent of the choice of axis, ignoring the parallel axis theorem altogether

28. Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m$ kg and the spring constant is $k\ \text{Nm}^{-1}$. At a given instant, the extension of the spring is $x$-meter and the speed of the particle is $v\ \text{ms}^{-1}$. On the $x$-$v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is:

    1. (1)

       $k = \frac{1}{m}$

    2. (2)

       $k = m$

    3. (3)

       $k = m^2$

    4. (4)

       $k = \sqrt{m}$

    Answer and solution

    Answer: (2)

    Conservation of energy gives $\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = E$, which rearranges to $\frac{v^2}{2E/m} + \frac{x^2}{2E/k} = 1$ - an ellipse in the $x\!-\!v$ plane. An ellipse becomes a CIRCLE only when the two semi-axes are equal, i.e. $\frac{2E}{m} = \frac{2E}{k}$, which requires $k = m$ numerically. NCERT Class 11, Oscillations (energy in simple harmonic motion).

29. The lens combination as shown in the figure, consists of two lenses, $L_1$ and $L_2$, of the focal lengths +10 cm and -10 cm, respectively. The position of the image formed is:

    
    1. (1)

       20 cm to the left of the concave lens

    2. (2)

       60 cm to the left of the concave lens

    3. (3)

       30 cm to the right of the concave lens

    4. (4)

       60 cm to the right of the concave lens

    Answer and solution

    Answer: (2)

    For the convex lens, $u = -30\ \text{cm}$ and $f = +10\ \text{cm}$ give $\frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}$, so its image lies 15 cm to its right. The lenses are 3 cm apart, so that image sits 12 cm BEYOND the concave lens and acts as a VIRTUAL OBJECT with $u = +12\ \text{cm}$. Then $\frac{1}{v} = -\frac{1}{10} + \frac{1}{12} = -\frac{1}{60}$, giving $v = -60\ \text{cm}$: 60 cm to the left of the concave lens. NCERT Class 12, Ray Optics and Optical Instruments (combination of thin lenses).
    
    Why the other options are wrong:
    
    - (1) uses the thin-lens-in-contact formula $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$, which fails for a 3 cm separation
    - (3) keeps the sign of $v$ positive, missing that a negative $v$ places a virtual image on the incoming side
    - (4) gets the magnitude right but assigns the image to the wrong side of the concave lens

30. An ac voltage $V = 220\sin(2 \times 10^3 t)$ Volt is applied to a series LCR circuit. Then the current amplitude in this circuit is: (Given : $L = 10\ \text{mH}$, $C = 25\ \mu\text{F}$, $R = 100\ \Omega$)

    1. (1)

       2.2 A

    2. (2)

       5.5 A

    3. (3)

       11.0 A

    4. (4)

       22.0 A

    Answer and solution

    Answer: (1)

    With $\omega = 2 \times 10^3\ \text{rad s}^{-1}$, the reactances are $X_L = \omega L = 20\ \Omega$ and $X_C = \frac{1}{\omega C} = 20\ \Omega$. Since $X_L = X_C$ the circuit is at RESONANCE, the reactances cancel and the impedance reduces to $Z = R = 100\ \Omega$. The current amplitude is then $I_0 = \frac{V_0}{Z} = \frac{220}{100} = 2.2\ \text{A}$. NCERT Class 12, Alternating Current (LCR series circuit and resonance).

31. A thin horizontal disc is rotating about a vertical axis passing through its fixed centre $O$. Its angular momentum is $L_A$ and $L_B$ computed about points $A$ and $B$, respectively, with $OB = 2 \times OA$. The value of $\frac{L_A}{L_B}$ is:

    
    1. (1)

       $\frac{1}{4}$

    2. (2)

       $\frac{1}{2}$

    3. (3)

       1

    4. (4)

       2

    Answer and solution

    Answer: (3)

    The angular momentum of a rigid body about any point splits as $\vec{L} = \vec{L}_{cm} + \vec{r} \times M\vec{v}_{cm}$. Here the centre is FIXED, so $\vec{v}_{cm} = 0$ and the orbital term vanishes for every reference point. Only the spin term $I_{cm}\omega$ survives, which is the same whether you compute about $A$ or $B$, so the ratio is 1 regardless of the distances. NCERT Class 11, System of Particles and Rotational Motion (angular momentum of a rotating rigid body).
    
    Why the other options are wrong:
    
    - (1) assumes $L \propto 1/r^2$, importing an inverse-square intuition that does not apply
    - (2) assumes $L$ scales inversely with the distance to the reference point
    - (4) assumes $L$ grows in proportion to the distance, as it would for a single particle moving in a straight line

32. A conducting loop of finite resistance lies on the $x - y$ plane. There is a constant magnetic field in the $z$ direction. The area of the loop varies with time $t$, as $A = A_0(1 + \sin t)$ in appropriate units. The figure that correctly indicates the qualitative behaviour of the power $P$ dissipated in the loop as a function of time is:

    1. (1)
       
    2. (2)
       
    3. (3)
       
    4. (4)
       
    Answer and solution

    Answer: (2)

    Faraday's law gives $\varepsilon = -\frac{d\Phi}{dt} = -BA_0\cos t$, so the dissipated power is $P = \frac{\varepsilon^2}{R} = \frac{B^2A_0^2}{R}\cos^2 t$. Because $\cos^2 t$ is MAXIMUM at $t = 0$ and touches zero at $t = \pi/2$, the correct graph starts at a peak and dips to zero, unlike a $\sin^2 t$ curve which would start at zero. $P$ is never negative because the electromotive force (emf) is squared. NCERT Class 12, Electromagnetic Induction (Faraday's law).
    
    Why the other options are wrong:
    
    - (1) differentiates $\sin t$ to $\sin t$ rather than $\cos t$, producing a $\sin^2 t$ curve that starts at zero
    - (3) assumes the constant field means a constant emf, forgetting that it is the CHANGING area that drives induction
    - (4) confuses the growing area with growing power, ignoring that the flux change is periodic and bounded

33. A point charge $Q$ is placed inside a cavity within a solid isolated conducting sphere. Consider points $A$, $B$ and $C$ as shown in the figure, where the magnitudes of the electric fields are $E_A$, $E_B$, $E_C$, respectively. The points $B$ and $C$ are at the same distance from the center of the solid sphere. The correct option is :

    
    1. (1)

       $E_A = 0,\ E_B = E_C$

    2. (2)

       $E_A \neq 0,\ E_B = E_C$

    3. (3)

       $E_A = 0,\ E_B > E_C$

    4. (4)

       $E_A \neq 0,\ E_B < E_C$

    Answer and solution

    Answer: (2)

    Point $A$ lies in the empty cavity, not in the conducting material, so the field of $Q$ is fully present there and $E_A \neq 0$. The induced $-Q$ on the cavity wall and $+Q$ on the outer surface arrange themselves so that the outer surface charge is UNIFORM regardless of where the cavity sits, so outside the sphere the field is exactly that of a point charge $Q$ at the centre. Since $B$ and $C$ are equidistant from that centre, $E_B = E_C$. NCERT Class 12, Electric Charges and Fields (electrostatics of conductors).
    
    Why the other options are wrong:
    
    - (1) confuses the cavity with the conducting material - the field vanishes inside the METAL, not inside the hollow
    - (3) combines the cavity error with the idea that the off-centre cavity biases the external field towards $B$
    - (4) correctly sees $E_A \neq 0$ but wrongly assumes the external field is skewed away from the cavity side

34. Consider a fixed uniformly charged insulating sphere with radius $R$ and total charge $+Q$. A point charge $-q$ ($q \ll Q$) with mass $m$ is released from rest at a distance of $3R$ from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is: ($\varepsilon_0$ is the permittivity of vacuum, neglect gravitational forces).

    1. (1)

       $\sqrt{\frac{3Qq}{4\pi\varepsilon_0 mR}}$

    2. (2)

       $\sqrt{\frac{2Qq}{3\pi\varepsilon_0 mR}}$

    3. (3)

       $\sqrt{\frac{Qq}{3\pi\varepsilon_0 mR}}$

    4. (4)

       $\sqrt{\frac{Qq}{4\pi\varepsilon_0 mR}}$

    Answer and solution

    Answer: (3)

    Outside the sphere the potential energy of the attracting pair is $U(r) = -\frac{Qq}{4\pi\varepsilon_0 r}$. Energy conservation from $r = 3R$ (at rest) to $r = R$ gives $\frac{1}{2}mv^2 = \frac{Qq}{4\pi\varepsilon_0 R}\left(1 - \frac{1}{3}\right) = \frac{Qq}{6\pi\varepsilon_0 R}$. Solving, $v = \sqrt{\frac{Qq}{3\pi\varepsilon_0 mR}}$. NCERT Class 12, Electrostatic Potential and Capacitance.

35. In Geiger-Marsden experiment, the number of scattered $\alpha$-particles $N(\theta)$ is plotted as a function of scattering angle $\theta$. Which of the following options represents the correct plot?

    1. (1)
       
    2. (2)
       
    3. (3)
       
    4. (4)
       
    Answer and solution

    Answer: (3)

    Rutherford's scattering formula gives $N(\theta) \propto \frac{1}{\sin^4(\theta/2)}$, so the count is enormous for small deflections and drops by several orders of magnitude as $\theta$ increases - which is exactly why the plots are drawn on a logarithmic vertical axis. The experimental finding was that the overwhelming majority of $\alpha$-particles pass almost undeviated, with only a tiny fraction turned through large angles by the nucleus. NCERT Class 12, Atoms (alpha-particle scattering).
    
    Why the other options are wrong:
    
    - (1) reverses the physics, implying most particles suffer large deflections and the atom is uniformly dense
    - (2) captures the falling trend but far too weakly - it misses that the drop spans many orders of magnitude
    - (4) invents a preferred scattering angle, as if the nucleus deflected particles by a characteristic amount

36. Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, $+\alpha$, while the other one has a negative temperature of coefficient, $-\alpha$, as shown in the figure. The current through these circuits are denoted by $I_A$ and $I_B$. At initial temperature, the resistance of the two resistors is $R_0$. As the temperature is increased, the correct option that describes the variation of current in these circuits is:

    
    1. (1)

       $I_A$ remains constant while $I_B$ increases

    2. (2)

       $I_A$ decreases while $I_B$ increases

    3. (3)

       $I_A$ increases while $I_B$ decreases

    4. (4)

       Both $I_A$ and $I_B$ remain constant

    Answer and solution

    Answer: (1)

    In circuit (A) the resistors are in SERIES, so $R_A = R_0(1-\alpha\Delta T) + R_0(1+\alpha\Delta T) = 2R_0$ - the two temperature terms cancel exactly and $I_A$ is unchanged. In circuit (B) they are in PARALLEL, giving $R_B = \frac{R_0(1-\alpha^2\Delta T^2)}{2}$, which decreases as $\Delta T$ grows because of the $-\alpha^2\Delta T^2$ term, so $I_B$ increases. NCERT Class 12, Current Electricity (temperature dependence of resistivity).
    
    Why the other options are wrong:
    
    - (2) assumes the series combination must heat up and resist more, missing the exact cancellation between $+\alpha$ and $-\alpha$
    - (3) swaps the two circuits, applying the parallel result to (A) and the series result to (B)
    - (4) spots the cancellation in series and wrongly assumes the same cancellation holds for reciprocals in parallel

37. Consider that $\sigma_s$, $k_B$, $b$ represents Stefan-Boltzmann constant, Boltzmann constant and Wien's displacement law constant, respectively. The dimension of $\sigma_s k_B^{-1} b$ is

    1. (1)

       $[L^{-1}T^{-1}K^{-2}]$

    2. (2)

       $[L^{-1}K^{-2}]$

    3. (3)

       $[L^{-1}T^{-1}K^{-3}]$

    4. (4)

       $[L^{-1}T^{-1}K^{-4}]$

    Answer and solution

    Answer: (1)

    Take each constant from its defining equation. Stefan's law $E = \sigma_s T^4$ makes $\sigma_s$ a power per unit area per $K^4$, so $[\sigma_s] = [MT^{-3}K^{-4}]$. Boltzmann's constant is an energy per kelvin, $[k_B] = [ML^2T^{-2}K^{-1}]$, and Wien's constant is a wavelength times a temperature, $[b] = [LK]$. Multiplying, $[MT^{-3}K^{-4}][M^{-1}L^{-2}T^{2}K][LK] = [L^{-1}T^{-1}K^{-2}]$. NCERT Class 11, Units and Measurements; the constants themselves come from Thermal Properties of Matter.

38. An electromagnetic wave travelling in a lossless dielectric medium having a dielectric constant, $\varepsilon_r = 9$, has the electric field, $E_x = E_0\sin(kz - 2\pi \times 10^6 t)\ \text{Vm}^{-1}$ where $E_0$ is the amplitude and $k$ is the wave vector. Among the following options, the incorrect choice is

    1. (1)

       The speed of the electromagnetic wave inside the medium is $10^8\ \text{ms}^{-1}$

    2. (2)

       The wavelength of the electromagnetic wave inside the medium is 300 m

    3. (3)

       The magnetic field is given by the relation $B_y = \frac{B_0}{v}\sin(kz - 2\pi \times 10^6 t)$ where $v$ is the speed of the electromagnetic wave inside the medium

    4. (4)

       The direction of propagation of the electromagnetic wave is along $+z$

    Answer and solution

    Answer: (2) or (3) NTA accepted more than one option.

    NTA accepted both option (2) and option (3) as answers. The refractive index is $n = \sqrt{\varepsilon_r} = 3$, so $v = \frac{c}{n} = 10^8\ \text{m s}^{-1}$, making option (1) correct. From $\omega = 2\pi \times 10^6$, the frequency is $10^6\ \text{Hz}$, hence $\lambda = \frac{v}{f} = 100\ \text{m}$, not 300 m; therefore option (2) is incorrect. Option (3) is also incorrect as printed: the magnetic field should be $B_y = \frac{E_0}{v}\sin(kz-\omega t)$, whereas the option uses $B_0/v$. Option (4) is correct because $kz-\omega t$ represents propagation along $+z$. NCERT Class 12, Electromagnetic Waves.

39. One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. The total heat supplied to the gas is:

    
    1. (1)

       400 J

    2. (2)

       500 J

    3. (3)

       600 J

    4. (4)

       800 J

    Answer and solution

    Answer: (3)

    Over a complete cycle the gas returns to its initial state, so $\Delta U = 0$ and the first law reduces to $Q = W$. The cycle is a rectangle on the $P\!-\!V$ diagram between $P = 100$ and $300\ \text{N m}^{-2}$ and $V = 2$ and $5\ \text{m}^3$, traversed clockwise, so the work done by the gas is the enclosed area $= (300-100)(5-2) = 600\ \text{J}$. Hence 600 J of heat is supplied. NCERT Class 11, Thermodynamics (cyclic process).
    
    Why the other options are wrong:
    
    - (1) miscounts one side of the rectangle, typically taking $\Delta V = 2\ \text{m}^3$
    - (2) computes only the net heat of the two isobaric legs and omits the isochoric contributions to the balance
    - (4) adds the areas under the two isobaric legs instead of taking their difference, double-counting the return path

40. Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6} \times 10^5\ \text{ms}^{-1}$. The radius of the orbit is $x \times 10^{-9}$ m. The value of $x$ is: [Take the mass of electron to the $9 \times 10^{-31}$ kg, charge of electron = $-1.6 \times 10^{-19}$ C and $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \text{Nm}^2\text{C}^{-2}$]

    1. (1)

       4

    2. (2)

       3

    3. (3)

       2

    4. (4)

       1

    Answer and solution

    Answer: (4)

    In a Bohr orbit the Coulomb attraction supplies the centripetal force: $\frac{ke^2}{r^2} = \frac{mv^2}{r}$, which rearranges to $r = \frac{ke^2}{mv^2}$. Substituting $v^2 = 25.6 \times 10^{10} = 2.56 \times 10^{11}$, the numerator is $9 \times 10^9 \times (1.6 \times 10^{-19})^2 = 2.304 \times 10^{-28}$ and the denominator is $9 \times 10^{-31} \times 2.56 \times 10^{11} = 2.304 \times 10^{-19}$. Their ratio is $1 \times 10^{-9}\ \text{m}$, so $x = 1$. NCERT Class 12, Atoms (Bohr model).

41. A car travels on a circular racetrack of radius 50 m, which is banked at an angle $\theta$. If the car travels at a speed 10 $\text{ms}^{-1}$, then the wear and tear on its tyres is minimum. Taking the acceleration due to gravity to be 10 $\text{ms}^{-2}$, the value of $\theta$ is:

    1. (1)

       $\tan^{-1}\left(\frac{1}{5}\right)$

    2. (2)

       $\tan^{-1}\left(\frac{2}{5}\right)$

    3. (3)

       $\tan^{-1}(\sqrt{3}/2)$

    4. (4)

       $\tan^{-1}(2\sqrt{3})$

    Answer and solution

    Answer: (1)

    Minimum wear and tear is the exam's way of saying no sideways friction is needed, i.e. the car is at the optimum (design) speed of the banked track. Then the horizontal component of the normal reaction alone supplies the centripetal force, giving $\tan\theta = \frac{v^2}{rg} = \frac{10^2}{50 \times 10} = \frac{1}{5}$, so $\theta = \tan^{-1}(1/5)$. NCERT Class 11, Laws of Motion (motion on a banked road).

42. Three identical p-n junction diodes $D_1$, $D_2$, and $D_3$ are connected across a battery as shown in the figure. If the width of the depletion regions of $D_1$, $D_2$ and $D_3$ are $W_1$, $W_2$ and $W_3$, respectively, then the correct option is:

    
    1. (1)

       $W_1 > W_2 > W_3$

    2. (2)

       $W_3 = W_1 > W_2$

    3. (3)

       $W_3 > W_2 > W_1$

    4. (4)

       $W_2 > W_1 = W_3$

    Answer and solution

    Answer: (3)

    Depletion width is set by the net potential barrier at the junction. Forward bias lowers the barrier and NARROWS the depletion layer; reverse bias raises it and WIDENS the layer; an unbiased junction sits at its equilibrium width in between. In this circuit $D_1$ is forward biased (narrowest), $D_2$'s branch carries no bias so it keeps its equilibrium width, and $D_3$ is reverse biased (widest), giving $W_3 > W_2 > W_1$. NCERT Class 12, Semiconductor Electronics (p-n junction).
    
    Why the other options are wrong:
    
    - (1) reverses the bias effect, assuming forward bias widens and reverse bias narrows the depletion region
    - (2) assumes forward and reverse bias shift the width by equal and opposite amounts so that $W_1$ and $W_3$ coincide
    - (4) treats the unbiased diode as the extreme case rather than the intermediate one

43. The following table presents the part of the electromagnetic spectrum and their corresponding major applications. The correct option is:

    List I	List II
    P) Microwave	I) For purifying the water
    Q) UV rays	II) For warming the food
    R) Gamma rays	III) For AM and FM communication systems
    S) Radio wave	IV) For treating the Cancer cells

    1. (1)

       P-I, Q-II, R-III, S-IV

    2. (2)

       P-I, Q-IV, R-II, S-III

    3. (3)

       P-II, Q-I, R-IV, S-III

    4. (4)

       P-II, Q-IV, R-III, S-I

    Answer and solution

    Answer: (3)

    Microwaves are absorbed by water molecules in food and heat it, so P-II. Ultraviolet radiation is germicidal and is used in water sterilisation, so Q-I. Gamma rays carry enough energy to destroy malignant tissue and are used in radiotherapy, so R-IV. Radio waves are the carriers in amplitude modulated (AM) and frequency modulated (FM) broadcasting, so S-III. NCERT Class 12, Electromagnetic Waves (electromagnetic spectrum).

44. Bob $B$ of mass $m$ at rest is hanging vertically from the ceiling via a massless string of length 10 m, as shown in the figure. Point mass $A$ of mass $m$ travelling horizontally with speed 10 $\text{ms}^{-1}$ hits bob $B$ elastically. The bob $B$ rises $h$ meter after the collision. Taking the acceleration due to gravity $g = 10\ \text{ms}^{-2}$ and neglecting the size of the bob, the value of $h$ is:

    
    1. (1)

       8

    2. (2)

       7

    3. (3)

       5

    4. (4)

       2.5

    Answer and solution

    Answer: (3)

    In a head-on ELASTIC collision between equal masses the velocities are simply exchanged, so $A$ stops dead and $B$ leaves the collision at the full 10 $\text{ms}^{-1}$. Energy conservation on the swing then gives $h = \frac{v^2}{2g} = \frac{100}{20} = 5\ \text{m}$. The bob rises only 5 m, less than the 10 m string length, so it stays below the horizontal through the point of suspension and the string never goes slack. NCERT Class 11, Work, Energy and Power (collisions).
    
    Why the other options are wrong:
    
    - Option (1) (8 m) and option (2) (7 m) are higher than 5 m. Even if $B$ took all the kinetic energy of $A$, it could rise only $\frac{v^2}{2g} = 5\ \text{m}$. A greater height needs energy the collision never supplied.
    - Option (4) (2.5 m) gives $B$ only half the kinetic energy of $A$, as if the energy were shared equally between the two masses. That halves the height. In an elastic collision of equal masses, all of it passes to $B$.

45. An ideal gas is made of polyatomic molecules. Each of the molecules has three translational, three rotational and $f$ number of vibrational modes. If the ratio of heat capacities $C_P/C_V$ of the gas is 8/7, then the value of $f$ is:

    1. (1)

       4

    2. (2)

       3

    3. (3)

       2

    4. (4)

       1

    Answer and solution

    Answer: (1)

    Equipartition assigns $\frac{1}{2}k_BT$ to each translational and rotational mode but a full $k_BT$ to each vibrational mode, so $C_V = \frac{R}{2}(3 + 3 + 2f) = (3+f)R$ and $C_P = C_V + R = (4+f)R$. Setting $\frac{4+f}{3+f} = \frac{8}{7}$ gives $28 + 7f = 24 + 8f$, so $f = 4$. NCERT Class 11, Kinetic Theory (law of equipartition of energy).