JEE Main 2026 Jan 23 shift 1

We need to evaluate the Assertion and Reason about ferromagnetic materials. Assertion (A): "The individual atoms in a ferromagnetic material possess a magnetic dipole moment and interact with one another in such a way that they spontaneously align themselves forming domains." Analysis: In ferromagnetic materials (like iron, cobalt, nickel), each atom has a permanent magnetic dipole moment due to unpaired electrons. These atomic magnetic moments interact through a quantum mechanical exchange interaction, which causes neighboring moments to align parallel to each other spontaneously, forming regions called magnetic domains. Assertion (A) is TRUE. Reason (R): "At high enough temperature, the domain structure of ferromagnetic material disintegrates. Thus, magnetization will disappear at high enough temperature known as Curie temperature." Analysis: Above the Curie temperature, thermal energy overcomes the exchange interaction energy, destroying the long-range ordering of magnetic moments. The material transitions from ferromagnetic to paramagnetic behavior. Reason (R) is TRUE. Relationship: While both statements are true, Reason (R) describes what happens at high temperatures (loss of domain structure), which does not explain why domains form in the first place (Assertion A). The formation of domains is due to exchange interactions at lower temperatures, not related to their destruction at the Curie temperature. R does NOT explain A. The correct answer is Option 1 : Both (A) and (R) are true but (R) is not the correct explanation of (A).

The Zener diode voltage regulator maintains $V_o = V_z = 5V$.
The voltage drop across $R_s$ at maximum input voltage is $V_{Rs} = V_{\in,max} - V_z = 25V - 5V = 20V$.
The problem implies that the total current through $R_s$ ($I_s$) is chosen as four times the load current ($I_L$) for design purposes.
So, $I_s = 4 \times I_L = 4 \times 5mA = 20mA$.
Therefore, $R_s = \frac{V_{Rs}}{I_s} = \frac{20V}{20mA} = \frac{20V}{0.02A} = 1000 \Omega$.
At this resistance, the Zener current is $I_z = I_s - I_L = 20mA - 5mA = 15mA$, which is less than the $4 \times I_L = 20mA$ it can withstand.

We need to find the angular velocity of a system of two balls (masses $m$ and $2m$) attached to a rod of length $d$, given angular momentum $L$ about the axis through the center of mass. Find the center of mass. Taking the position of mass $m$ as origin: $x_{cm} = \frac{m \cdot 0 + 2m \cdot d}{m + 2m} = \frac{2d}{3}$ So mass $m$ is at distance $\frac{2d}{3}$ from the CM, and mass $2m$ is at distance $d - \frac{2d}{3} = \frac{d}{3}$ from the CM. Calculate the moment of inertia about the CM: $I = m\left(\frac{2d}{3}\right)^2 + 2m\left(\frac{d}{3}\right)^2 = m \cdot \frac{4d^2}{9} + 2m \cdot \frac{d^2}{9} = \frac{4md^2 + 2md^2}{9} = \frac{6md^2}{9} = \frac{2md^2}{3}$ Use $L = I\omega$ to find $\omega$: $\omega = \frac{L}{I} = \frac{L}{2md^2/3} = \frac{3L}{2md^2}$ The correct answer is Option (2): $\frac{3}{2}\frac{L}{md^2}$.

Let the mass of each cylinder be $m = \frac{M}{4}$. The moment of inertia of a solid cylinder of mass $m$, radius $R$ and length $L$ about a diameter of its circular cross section (axis through centre perpendicular to its length) is given by the formula $I_{\diameter} = \frac{1}{4}mR^2 + \frac{1}{12}mL^2$ $-(1)$ The two vertical rods (left and right sides) have their midpoints on the rotation axis and for each the axis is a diameter of its cross section. Moment of inertia of one vertical rod about the given axis is $I_{\text{v,1}} = \frac{1}{4}mR^2 + \frac{1}{12}mL^2$. For two vertical rods, $I_{\text{vertical}} = 2\,I_{\text{v,1}} = \frac{1}{2}mR^2 + \frac{1}{6}mL^2$ $-(2)$ The moment of inertia of a solid cylinder of mass $m$ and radius $R$ about its central (longitudinal) axis is $I_{\text{long}} = \frac{1}{2}mR^2$ $-(3)$ The top and bottom rods are horizontal, parallel to the rotation axis, and at a perpendicular distance $\frac{L}{2}$ from it. Using the parallel axis theorem for one such rod: $I_{\text{h,1}} = I_{\text{long}} + m\Bigl(\frac{L}{2}\Bigr)^2 = \frac{1}{2}mR^2 + \frac{1}{4}mL^2$ $-(4)$ For both horizontal rods, $I_{\text{horizontal}} = 2\,I_{\text{h,1}} = mR^2 + \frac{1}{2}mL^2$ $-(5)$ The total moment of inertia of the square loop about the given axis is $I = I_{\text{vertical}} + I_{\text{horizontal}} = \Bigl(\frac{1}{2}mR^2 + \frac{1}{6}mL^2\Bigr) + \Bigl(mR^2 + \frac{1}{2}mL^2\Bigr)\,.$ Substituting $m = \frac{M}{4}$ gives $I = \Bigl(\frac{1}{2}\cdot\frac{M}{4}R^2 + \frac{1}{6}\cdot\frac{M}{4}L^2\Bigr) + \Bigl(\frac{M}{4}R^2 + \frac{1}{2}\cdot\frac{M}{4}L^2\Bigr)$ $= \frac{M}{8}R^2 + \frac{M}{24}L^2 + \frac{M}{4}R^2 + \frac{M}{8}L^2 = \frac{3M}{8}R^2 + \frac{M}{6}L^2\,.$ Therefore, the required moment of inertia is $\displaystyle \frac{3}{8}MR^2 + \frac{1}{6}ML^2\,,$ which corresponds to Option B.

We need to verify which statements about the hydrogen atom spectrum are correct. Key Formula: The wavelength of spectral lines is given by: $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ where $R$ is Rydberg's constant, $n_1$ is the lower level, and $n_2$ is the upper level. Statement A: The maximum wavelength of Lyman series is $\frac{4}{3R}$. The Lyman series has $n_1 = 1$. Maximum wavelength corresponds to minimum energy transition ($n_2 = 2$): $\frac{1}{\lambda_{max}} = R\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R \times \frac{3}{4}$ $\lambda_{max} = \frac{4}{3R}$ Statement A is CORRECT. Statement B: The Balmer series lies \in the visible region of the spectrum. The Balmer series has $n_1 = 2$. Its wavelengths range from 364.6 nm (series limit) to 656.3 nm, which falls \in the visible region (380-700 nm). Statement B is CORRECT. Statement C: The minimum wavelength of Paschen series is $\frac{9}{R}$. The Paschen series has $n_1 = 3$. Minimum wavelength corresponds to the series limit ($n_2 \to \infty$): $\frac{1}{\lambda_{min}} = R\left(\frac{1}{3^2} - 0\right) = \frac{R}{9}$ $\lambda_{min} = \frac{9}{R}$ Statement C is CORRECT. Statement D: The minimum wavelength of Lyman series is $\frac{5}{4R}$. The minimum wavelength of Lyman series corresponds to $n_2 \to \infty$: $\frac{1}{\lambda_{min}} = R\left(\frac{1}{1^2} - 0\right) = R$ $\lambda_{min} = \frac{1}{R}$ This is NOT $\frac{5}{4R}$. Statement D is INCORRECT. The correct statements are A, B, and C. The correct answer is Option 3 : A, B and C Only.

We need to find the de Broglie wavelength of an O$_2$ molecule at 27°C. The de Broglie wavelength of a gas molecule at temperature T is: $\lambda = \frac{h}{\sqrt{3mk_BT}}$ where $h$ is Planck's constant, $m$ is the mass of one molecule, $k_B$ is Boltzmann's constant, and $T$ is the absolute temperature. $h = 6.63 \times 10^{-34} \, \text{J·s}$ $m = 32 \times 1.66 \times 10^{-27} \, \text{kg} = 53.12 \times 10^{-27} \, \text{kg}$ (O$_2$ has molar mass 32 g/mol) $k_B = 1.38 \times 10^{-23} \, \text{J/K}$ $T = 27°C = 300 \, \text{K}$ $3mk_BT = 3 \times 53.12 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300$ $= 3 \times 53.12 \times 1.38 \times 300 \times 10^{-50}$ $= 3 \times 21992.64 \times 10^{-50} = 65977.9 \times 10^{-50}$ $= 6.598 \times 10^{-46}$ $\sqrt{3mk_BT} = \sqrt{6.598 \times 10^{-46}} = 2.569 \times 10^{-23}$ $\lambda = \frac{6.63 \times 10^{-34}}{2.569 \times 10^{-23}} = 2.581 \times 10^{-11} \, \text{m} = 25.81 \times 10^{-12} \, \text{m} \approx 26 \times 10^{-12} \, \text{m}$ So $x = 26$. The correct answer is Option (4): 26.

$T = 2\pi\sqrt{L/g}$. $T_1 = 10/20 = 0.5$s. $T_2 = 10/40 = 0.25$s. $T_2/T_1 = 1/2 = \sqrt{L_2/L_1}$. $L_2 = L_1/4 = 30/4 = 7.5$ cm. The answer is Option 4 : 7.5 cm.

Bob A's speed before impact: $v_A = \sqrt{2gL(1-\cos 60^\circ)} = \sqrt{2g(1)(1-0.5)} = \sqrt{g}$ m/s.
Due to an elastic collision between identical masses, bob B acquires $v_B = v_A = \sqrt{g}$ m/s.
For bob B to just complete a vertical circular path of radius R, its speed at the highest point (height $2R$) must be $v_{top} = \sqrt{gR}$.
By conservation of energy for bob B: $\frac{1}{2}mv_B^2 = \frac{1}{2}mv_{top}^2 + mg(2R)$.
Substituting values: $\frac{1}{2}m(g) = \frac{1}{2}m(gR) + mg(2R) \implies \frac{1}{2} = \frac{1}{2}R + 2R \implies \frac{1}{2} = \frac{5}{2}R \implies R = \frac{1}{5}$ m.

step 1: resistance of circular wire $Total\ length=2\pi r$ $Total\ resis\tan ce=2\pi r\lambda$ Between A and B, the circle splits into two arcs: \bullet smaller arc (quarter circle): $length=\frac{\pi r}{2}$ $\to resis\tan ce=\frac{\pi r\lambda}{2}$ \bullet larger arc (remaining): $length=\frac{3\pi r}{2}$ $\to resis\tan ce=\frac{3\pi r\lambda}{2}$ These two are \in parallel: $R_{\text{circle}}=\frac{\left(\frac{\pi r\lambda}{2}\right)\left(\frac{3\pi r\lambda}{2}\right)}{\frac{\pi r\lambda}{2}+\frac{3\pi r\lambda}{2}}=\frac{3\pi^2r^2\lambda^2/4}{2\pi r\lambda}=\frac{3\pi r\lambda}{8}$ step 2: resistance of straight wire AB Length = 2r RAB=2r\lambda step 3: both paths are \in parallel $R_{eq}=\frac{\left(\frac{3\pi r\lambda}{8}\right)(2r\lambda)}{\frac{3\pi r\lambda}{8}+2r\lambda}$ Factor \lambda : $R_{eq}=r\lambda\cdot\frac{(3\pi/8)(2)}{(3\pi/8)+2}=r\lambda\cdot\frac{3\pi/4}{(3\pi+16)/8}$ $=r\lambda\cdot\frac{3\pi/4\times8}{3\pi+16}=r\lambda\cdot\frac{6\pi}{3\pi+16}$

Use Gauss law. Charge at center of face CDEF: It lies exactly on surface of cube. A charge on a face contributes half its flux through the cube. So contribution: $\Phi_1=\frac{q}{2\epsilon_0}$ Charge 2q is at vertex A. A vertex is shared by 8 identical cubes. So this cube gets one-eighth contribution: $\Phi_2=\frac{2q}{8\epsilon_0}=\frac{q}{4\epsilon_0}$ Total flux: $\Phi=\Phi_1+\Phi_2$ $=\frac{q}{2\epsilon_0}+\frac{q}{4\epsilon_0}$ $=\frac{3q}{4\epsilon_0}$

We are given a screw gauge with a zero error where the zero of the circular scale lies 3 divisions above the horizontal pitch line when the studs are in contact. The least count (LC) is $0.01$ mm. When the zero of the circular scale lies 3 divisions above the reference line, the circular scale has not reached its zero mark. This means extra rotation is needed to bring it to zero, which represents a negative zero error: $\text{Zero error} = -3 \times \text{LC} = -3 \times 0.01 = -0.03 \text{ mm}$ The observed reading is given by: $\text{Observed reading} = \text{PSR} + (\text{CSR} \times \text{LC}) = 1 + (51 \times 0.01) = 1.51 \text{ mm}$ The zero correction is the negative of the zero error, so the corrected reading is: $\text{Correct thickness} = \text{Observed reading} - \text{Zero error} = 1.51 - (-0.03) = 1.54 \text{ mm}$ The correct answer is Option B) 1.54 mm .

We need to find the relative error in the measurement of the average length of a rod measured by four persons as 20.00 cm, 19.75 cm, 17.01 cm, and 18.25 cm. Calculate the average (mean) length: $\bar{x} = \frac{20.00 + 19.75 + 17.01 + 18.25}{4} = \frac{75.01}{4} = 18.7525 \, \text{cm}$ Calculate the absolute deviations from the mean: $|20.00 - 18.7525| = 1.2475$ $|19.75 - 18.7525| = 0.9975$ $|17.01 - 18.7525| = 1.7425$ $|18.25 - 18.7525| = 0.5025$ Calculate the mean absolute deviation (mean error): $\Delta\bar{x} = \frac{1.2475 + 0.9975 + 1.7425 + 0.5025}{4} = \frac{4.49}{4} = 1.1225 \, \text{cm}$ Calculate the relative error: $\text{Relative error} = \frac{\Delta\bar{x}}{\bar{x}} = \frac{1.1225}{18.7525} = 0.0599 \approx 0.06$ The correct answer is Option (4): 0.06.

Refractive index of a medium is defined by the formula $n = \frac{c}{v}$, where $c$ is the speed of light \in vacuum and $v$ is the speed of light \in the medium. -(1) For medium A: $n_{1} = \frac{c}{2.4\times10^{8}}$ For medium B: $n_{2} = \frac{c}{2.7\times10^{8}}$ Since $2.4\times10^{8}\;\&lt\;2.7\times10^{8}$, it follows that $n_{1}\;\&gt\;n_{2}$, so total internal reflection is possible when light travels from A to B. The critical angle $\theta_{c}$ is given by Snell's law at the limit of refraction: $\sin\theta_{c} = \frac{n_{2}}{n_{1}}$ -(2) Substituting the values of $n_{1}$ and $n_{2}$ from (1), we get $\sin\theta_{c} = \frac{\frac{c}{2.7\times10^{8}}}{\frac{c}{2.4\times10^{8}}} = \frac{2.4\times10^{8}}{2.7\times10^{8}} = \frac{8}{9}$ -(3) Therefore, $\theta_{c} = \sin^{-1}\!\Bigl(\frac{8}{9}\Bigr)$ -(4) To express $\theta_{c}$ \in the form involving an inverse tangent, use the identity $\tan\theta = \frac{\sin\theta}{\sqrt{1-\sin^{2}\theta}}$. Substituting $\sin\theta_{c} = \frac{8}{9}$ gives $\tan\theta_{c} = \frac{\tfrac{8}{9}}{\sqrt{1-\bigl(\tfrac{8}{9}\bigr)^{2}}} = \frac{\tfrac{8}{9}}{\tfrac{\sqrt{17}}{9}} = \frac{8}{\sqrt{17}}$ -(5) Hence, $\theta_{c} = \tan^{-1}\!\Bigl(\frac{8}{\sqrt{17}}\Bigr)\,.\!$ Comparing with the given options, this corresponds to Option C.

An object is projected from point A with kinetic energy KKK at an angle of 60∘60^\circ60∘. Since there is no air resistance, mechanical energy is conserved throughout the motion. At point A, the total energy is purely kinetic. $K_A\ =\ K=\ \ \frac{\ 1}{2}\times \ m\times \ v^2$ = $\ \frac{\ mv^2}{2}$ As the object moves upward to point B, it gains height. Due to this increase \in height, some of the kinetic energy is converted into potential energy, so the kinetic energy at B decreases. $K_A\ =\ K_B+m\times \ g\times \ h$ At the highest point B, the vertical component of velocity becomes zero, and only the horizontal component remains. Therefore, kinetic energy at B depends only on the horizontal velocity. $v_x=v\cos\ 60^{\circ\ \ }=\ \frac{\ v}{2}$ So kinetic energy at B: $K_B=\ \ \frac{\ 1}{2}\times \ m\times \left(v\cos\ 60^{\circ\ }\right)^2\ =\ \ \frac{\ 1}{2}\times \ m\times \ \left(\ \frac{\ v}{2}\right)^2\ =\frac{\ mv^2}{8}$ = $\ \frac{\ K}{4}\$ Now, as the object moves from B to C, it comes back to the same vertical level as A. Hence, the potential energy at C is the same as at A, and thus the kinetic energy at C becomes equal to the initial kinetic energy. $K_A=K_C=K$ Now consider the difference \in kinetic energies between C and B , and kinetic energy at A $K_C-K_B=K-\ \frac{\ K}{4}\ =\ \ \frac{\ 3K}{4}$ $K_A=K$ Finally, take the ratio of these differences. $\ \frac{\ diff\ between\ B\ and\ C}{Kinetic\ energy\ at\ A}$ = $\ \frac{\ \frac{\ 3K}{4}}{K}$ = $\ \frac{\ 3}{4}$ Thus, the required ratio is 3:4

We need to match Maxwell's equations (List I) with their corresponding laws (List II). $\oint \vec{E} \cdot \vec{dl} = -\frac{d}{dt}\oint \vec{B} \cdot \vec{da}$ This is the integral form of Faraday's law of electromagnetic induction - a changing magnetic flux induces an electric field. \to II (Faraday's law) $\oint \vec{B} \cdot \vec{dl} = \mu_0\left(I + \epsilon_0\frac{d\phi_E}{dt}\right)$ This is Ampere's law with Maxwell's displacement current correction - magnetic fields are produced by both conduction current and changing electric flux. \to III (Ampere-Maxwell law) $\oint \vec{E} \cdot \vec{da} = \frac{1}{\epsilon_0}\int_V \rho \, dv$ This is Gauss's law for electricity - the electric flux through a closed surface equals the enclosed charge divided by $\epsilon_0$. \to IV (Gauss's law of electrostatics) $\oint \vec{B} \cdot \vec{dl} = \mu_0 I$ This is the original Ampere's circuital law (without the displacement current term). → I (Ampere's circuital law) The correct answer is Option (3): A-II, B-III, C-IV, D-I.

A conductor of length $L$ moving with velocity $v$ perpendicular to a magnetic field of strength $B$ has an induced EMF given by the formula: $\mathcal{E} = B\,L\,v\quad-(2)$ First, find the speed $v$ of the wire after falling through a vertical distance $h=200\text{ m}$ under gravity $g=10\text{ m/s}^2$. Using the kinematic relation with initial speed zero: $v^2 = u^2 + 2\,g\,h = 0 + 2\times 10\times 200 = 4000\quad-(1)$ Hence $v = \sqrt{4000} = 20\sqrt{10}\text{ m/s}$ Convert the magnetic field from gauss to tesla: 1 Gauss = $10^{-4}$ T. Thus $B = 0.5\text{ Gauss} = 0.5\times 10^{-4}\text{ T} = 5\times 10^{-5}\text{ T}$ Substitute $B=5\times 10^{-5}\text{ T}$, $L=20\text{ m}$ and $v=20\sqrt{10}\text{ m/s}$ into equation $(2)$: $\mathcal{E} = (5\times 10^{-5})\times 20\times 20\sqrt{10} = 0.02\sqrt{10}\text{ V}$ Convert volts to millivolts: $0.02\sqrt{10}\text{ V} = 20\sqrt{10}\text{ mV}$ Therefore, the induced EMF is $20\sqrt{10}\text{ mV}$, which is Option C.

Let extension in spring A = $x_A$ and extension \in spring B = $x_B$ Energy stored \in spring A = $E_A$ = E and energy stored \in spring B = $E_B$ From Force balance \in spring A, $K_Ax_A\ =\ mg\ =\ 0.1\ \times \ 10\ =\ 1N$ $\therefore\ x_A\ =\ \frac{1}{K_A}$ $E_A\ =\ \frac{1}{2}K_Ax_A^2\ =\ \frac{1}{2}K_A\times \ \frac{1}{K_A^2}=\frac{0.5}{K_A}$ Similarly, from spring B, $K_Bx_B\ =\ 0.2\ \times \ 10\ =\ 2$ $\therefore\ x_B\ =\ \frac{2}{K_B}$ $E_B\ =\ \frac{1}{2}K_Bx_B^2\ =\ \frac{1}{2}K_B\times \ \frac{4}{K_B^2}\ =\ \frac{2}{K_B}$ Since, $K_B\ =\ \frac{4}{3}K_A$ $\therefore\ E_B\ =\ \frac{2}{\frac{4}{3}K_A}=\frac{1.5}{K_A}=3\ \times \ \frac{0.5}{K_A}=3E$

We need to find the angle of the second prism when two thin prisms are combined to produce dispersion without deviation. We first state the deviation formula for a thin prism. For a thin prism with small angle $A$ and refractive index $n$, the deviation is given by: $\delta = (n - 1)A$ This formula is derived from Snell's law under the small-angle approximation ($\sin\theta \approx \theta$). Next, we set up the condition for zero net deviation. When two prisms are combined \in opposite orientations, their deviations act \in opposite directions. For the net deviation to be zero: $\delta_1 + \delta_2 = 0$ This implies: $(n_1 - 1)A_1 = (n_2 - 1)A_2$ (The deviations must be equal \in magnitude but opposite \in direction.) Substituting the given values $n_1 = 1.72$, $A_1 = 5°$, and $n_2 = 1.9$ into this equation gives: $(1.72 - 1) \times 5 = (1.9 - 1) \times A_2$ $0.72 \times 5 = 0.9 \times A_2$ $3.6 = 0.9 \times A_2$ Solving for $A_2$ yields: $A_2 = \frac{3.6}{0.9} = 4°$ The correct answer is Option B: $4°$.

Let the masses be $m_1 = 15\ \text{kg}$ and $m_2 = 25\ \text{kg}$, and their velocities before collision be $v_1 = +10\ \text{m/s}$ and $v_2 = -30\ \text{m/s}$ (opposite directions). In a perfectly inelastic collision, momentum is conserved. The formula is: $m_1v_1 + m_2v_2 = (m_1 + m_2)\,v_f$ $-(1)$ Substitute the given values into $(1)$: $15\times 10 + 25\times (-30) = 40\,v_f$ $150 - 750 = 40\,v_f$ $-600 = 40\,v_f$ $v_f = -\frac{600}{40} = -15\ \text{m/s}$ $-(2)$ The heat produced equals the loss \in kinetic energy. First, compute the initial kinetic energy: $K_i = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$ $-(3)$ Substitute values: $K_i = \frac{1}{2}\times 15\times 10^2 + \frac{1}{2}\times 25\times 30^2$ $K_i = 750 + 11250 = 12000\ \text{J}$ $-(4)$ Next, compute the final kinetic energy of the combined mass: $K_f = \frac{1}{2}(m_1 + m_2)\,v_f^2$ $-(5)$ Substitute values: $K_f = \frac{1}{2}\times 40\times (-15)^2 = 20\times 225 = 4500\ \text{J}$ $-(6)$ Thus, the heat generated is: $Q = K_i - K_f = 12000 - 4500 = 7500\ \text{J}$ $-(7)$ Given specific heat \in calories: $c = 31\ \text{cal/kg}^{\circ}\text{C}$ and $1\ \text{cal} = 4.2\ \text{J}$. Converting to SI units: $c = 31\times 4.2 = 130.2\ \text{J/kg}^{\circ}\text{C}$ $-(8)$ The total mass is: $M = m_1 + m_2 = 15 + 25 = 40\ \text{kg}$ $-(9)$ The temperature rise is: $\Delta T = \frac{Q}{M\,c} = \frac{7500}{40\times 130.2} \approx 1.44\ ^{\circ}\text{C}$ $-(10)$ Final Answer: The rise \in temperature is $1.44\ ^{\circ}\text{C}$ (Option D).

Young's modulus, $Y = \frac{\text{Stress}}{\text{Strain}}$ In the given plot, strain is on the $y$-axis and stress is on the $x$-axis. $\text{Slope} = \frac{\text{Strain}}{\text{Stress}} = \frac{1}{Y}$ Therefore, the material with the least slope will have the largest Young's modulus. Observing the plot, material C has the minimum slope. Therefore, it possesses the largest Young's modulus.

The pendulum consists of a metallic wire of length $L = 10 \text{ cm} = 0.1 \text{ m}$ and a bob of mass $m = 10 \text{ g} = 0.01 \text{ kg}$, oscillating \in a uniform magnetic field $B = 2 \text{ T}$ perpendicular to the plane of oscillation. The maximum induced EMF occurs when the angular velocity is maximum, which is at the lowest point of the swing. The induced EMF \in a rotating conductor of length $L$ about one end \in a uniform magnetic field $B$ perpendicular to the plane of rotation is given by: $\mathcal{E} = \frac{1}{2} B \omega L^2$ where $\omega$ is the angular velocity. To find the maximum angular velocity $\omega_{\text{max}}$, use conservation of energy. The initial angle is $\theta_0 = 60^\circ$. The height $h$ above the lowest point is: $h = L (1 - \cos \theta_0)$ Substitute $\cos 60^\circ = 0.5$: $h = 0.1 \times (1 - 0.5) = 0.1 \times 0.5 = 0.05 \text{ m}$ The potential energy at the initial position converts to kinetic energy at the lowest point: $m g h = \frac{1}{2} m v_{\text{max}}^2$ where $g = 10 \text{ m/s}^2$. Solve for $v_{\text{max}}$: $v_{\text{max}} = \sqrt{2 g h} = \sqrt{2 \times 10 \times 0.05} = \sqrt{1} = 1 \text{ m/s}$ The maximum angular velocity $\omega_{\text{max}}$ is related to $v_{\text{max}}$ by: $\omega_{\text{max}} = \frac{v_{\text{max}}}{L} = \frac{1}{0.1} = 10 \text{ rad/s}$ Now substitute into the EMF formula: $\mathcal{E}_{\text{max}} = \frac{1}{2} B \omega_{\text{max}} L^2 = \frac{1}{2} \times 2 \times 10 \times (0.1)^2$ Calculate step by step: $(0.1)^2 = 0.01$ $\mathcal{E}_{\text{max}} = \frac{1}{2} \times 2 \times 10 \times 0.01 = 1 \times 10 \times 0.01 = 0.1 \text{ V}$ Convert to millivolts: $0.1 \text{ V} = 100 \text{ mV}$. Thus, the maximum induced EMF is $100 \text{ mV}$.

We are given the electric field $E = \sqrt{377} \sin(6.27 \times 10^3 t - 2.09 \times 10^{-5} x)$ N/C and must determine $\alpha$ such that the time‐averaged power density equals $\frac{1}{\alpha}$ W/m$^2$. For a plane electromagnetic wave, the instantaneous power per unit area is described by the Poynting vector, and its time‐averaged value (or intensity) is $\langle S \rangle = \frac{E_0^2}{2Z_0}$ In this expression, $E_0$ denotes the amplitude of the electric field, while $Z_0 = \sqrt{\mu_0/\epsilon_0}$ represents the intrinsic impedance of free space. The factor of $\tfrac{1}{2}$ arises because $\langle \sin^2(\omega t)\rangle = \tfrac{1}{2}$ over a full cycle. From the given wave, one reads off the amplitude $E_0 = \sqrt{377}$ N/C, and since $Z_0 = \sqrt{\mu_0/\epsilon_0}$ takes the value $377\ \Omega$ \in SI units, substitution yields $\langle S \rangle = \frac{E_0^2}{2Z_0} = \frac{(\sqrt{377})^2}{2 \times 377} = \frac{377}{2 \times 377} = \frac{1}{2} \text{ W/m}^2$ Because this result must equal $\frac{1}{\alpha}$, we set $\frac{1}{\alpha} = \frac{1}{2}$, which leads directly to $\alpha = 2$. The answer is 2.

We need to find the ratio $D_1/D_2$ given that two Young's double-slit setups produce fringes of equal width. The fringe width \in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$ Here $\lambda$ is the wavelength, $D$ is the distance from slits to screen, and $d$ is the slit separation. Since the fringe widths are equal, $\beta_1 = \beta_2$, which implies $\frac{\lambda_1 D_1}{d_1} = \frac{\lambda_2 D_2}{d_2}$ Substituting the given ratios $d_1 : d_2 = 2 : 1$ and $\lambda_1 : \lambda_2 = 1 : 2$, we rearrange for $D_1/D_2$: $\frac{D_1}{D_2} = \frac{\lambda_2}{\lambda_1} \times \frac{d_1}{d_2}$ Evaluating this gives $\frac{D_1}{D_2} = \frac{2}{1} \times \frac{2}{1} = 4$ The answer is 4.

Let original capacitance (no dielectric): $C=\frac{\varepsilon_0A}{d}$ step 1: understand geometry \bullet Top half (thickness d/2): dielectric $K_1=2$, area A \bullet Bottom half (thickness d/2): split into two equal areas: left: $K_2=3$, area A/2 right: $K_3=5$, area A/2 step 2: bottom part (parallel combination) Each has thickness d/2: $C_2=\frac{3\varepsilon_0(A/2)}{d/2}=\frac{3\varepsilon_0A}{d}$ $C_3=\frac{5\varepsilon_0(A/2)}{d/2}=\frac{5\varepsilon_0A}{d}$ Parallel: $C_{bottom}=C_2+C_3=\frac{8\varepsilon_0A}{d}$ step 3: top part $C_{top}=\frac{2\varepsilon_0A}{d/2}=\frac{4\varepsilon_0A}{d}$ step 4: series combination Top and bottom are \in series: $\frac{1}{C'}=\frac{1}{C_{top}}+\frac{1}{C_{bottom}}=\frac{d}{4\varepsilon_0A}+\frac{d}{8\varepsilon_0A}$ $=\frac{3d}{8\varepsilon_0A}$ $C'=\frac{8\varepsilon_0A}{3d}$

At maximum current, the LCR circuit is at resonance, implying $X_L = X_C$.
From $V = 5 \sin(100t)$, the angular frequency is $\omega = 100$ rad/s.
The resonance condition is $\omega L = 1/(\omega C)$, so $C = 1/(\omega^2 L)$.
Substituting the values: $C = 1/((100 \text{ rad/s})^2 \times 2 \text{ H}) = 1/(10000 \times 2) = 1/20000$ F.
Converting to microfarads: $C = 5 \times 10^{-5}$ F $= 50 \times 10^{-6}$ F $= 50$ $\mu$F.

[CORRECT_OPTION: 50]

A. Baeyer's test for unsaturation involves decolourization of pink $KMnO_4$. So, A-IV.
B. The Ceric ammonium nitrate test for alcohols produces a red-colored complex. So, B-I.
C. Tollen's reagent test for aldehydes results in the formation of a silver mirror. So, C-II.
D. The $FeCl_3$ test for phenolic groups typically yields a violet coloration. So, D-III.
[CORRECT_OPTION: B]

We need to identify the incorrect statements about the nickel(II) complex of dimethylglyoxime (DMG), which is $[Ni(DMG)_2]$. Statement A: It is red \in colour. The nickel-DMG complex is indeed bright red (also called strawberry red). Statement A is CORRECT. Statement B: It has high solubility \in water at pH = 9. The nickel-DMG complex is insoluble \in water. It precipitates out at pH 5-9, which is actually how nickel is detected gravimetrically. Statement B is INCORRECT. Statement C: The Ni ion has two unpaired d-electrons. In the square planar nickel-DMG complex, $Ni^{2+}$ has a $d^8$ configuration. In a square planar field, all 8 electrons are paired ($d_{xy}^2, d_{yz}^2, d_{xz}^2, d_{z^2}^2$, empty $d_{x^{2-}y^2}$). So there are zero unpaired electrons. Statement C is INCORRECT. Statement D: The N-Ni-N bond angle is almost close to 90 degrees. In the square planar geometry of the complex, the N-Ni-N bond angle between adjacent nitrogen atoms is approximately 90 degrees. Statement D is CORRECT. Statement E: The complex contains four five-membered metallacycles. Each DMG molecule forms one five-membered chelate ring with Ni (Ni-N-C-C-N). With two DMG ligands, there are only two five-membered metallacycles, not four. Statement E is INCORRECT. The incorrect statements are B, C, and E. The answer is Option D) B, C and E Only .

A. As we move down group 13, the principal quantum number increases, causing the ionic radii of $M^{3+}$ ions to increase. Thus, statement A is false.
B. Electronegativity does not strictly decrease due to the poor shielding effects of $d$ and $f$ electrons (lanthanide contraction), making statement B false.
C. Boron possesses the smallest atomic radius and highest effective nuclear charge in the group, resulting in the highest first ionisation enthalpy. Thus, statement C is true.
D. According to Fajans' rule, triiodides exhibit strong covalent character due to the high polarizability of the large iodide ion by the $M^{3+}$ cation. Thus, statement D is true.

Reactivity towards electrophilic aromatic substitution, such as nitration, depends on the electron density of the benzene ring. Electron-donating groups increase reactivity, while electron-withdrawing groups decrease it.

1. Compound b (Nitrobenzene): The $-NO_2$ group exerts strong $-I$ and $-M$ effects, making the ring electron-deficient and the least reactive.
2. Compound a (Chlorobenzene): The $-Cl$ group has a $-I$ effect that outweighs its weak $+M$ effect, making it deactivating, but less so than the nitro group.
3. Compound c (Anisole): The $-OCH_3$ group has a strong $+M$ effect that dominates its $-I$ effect, strongly activating the ring.