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JEE Main 2026 Jan 23 shift 2

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Physics25 questions

Q1JEE Main 2026MCQ4MThermodynamics

One mole of an ideal diatomic gas expands from volume $V$ to $2 V$ isothermally at a temperature $27^{o}C$ and does W joule of work. lf the gas undergoes same magnitude of expansion adiabatically from $27^{o}C$ doing the same amount of work $W$ , then its final temperature will be (close to) ____ $^{\circ}C.$ $(\log_{e}2 = 0.693)$

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📖 Explanation

We have one mole of an ideal diatomic gas. We need to compare isothermal and adiabatic expansions from volume $V$ to $2V$ , both starting at $27^\circ C$ (i.e., $T = 300$ K). For isothermal expansion of an ideal gas: $W = nRT \ln\left(\frac{V_f}{V_i}\right) = 1 \times R \times 300 \times \ln 2 = 300R \times 0.693$ For an adiabatic process, the work done by the gas is: $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$ For a diatomic gas, $\gamma = \frac{7}{5}$ , so $\gamma - 1 = \frac{2}{5}$ . Since both processes do the same work $W$ : $300R \times 0.693 = \frac{R(300 - T_2)}{2/5}$ $300 \times 0.693 = \frac{5}{2}(300 - T_2)$ $207.9 = 2.5(300 - T_2)$ $300 - T_2 = \frac{207.9}{2.5} = 83.16$ $T_2 = 300 - 83.16 = 216.84 \text{ K}$ $T_2 = 216.84 - 273 \approx -56^\circ C$ The correct answer is Option 4 : $-56^\circ C$ .

Q2JEE Main 2026MCQ4MProperties of Matter

A small metallic sphere of diameter 2 mm and density $10.5 g/cm ^{3}$ is dropped \in glycerine having viscosity 10 Poise and density $1.5 g/cm^{3}$ respectively. The terminal velocity attained by the sphere is __ $cm/s$ . $(\pi=\frac{22}{7} \text { and } g=m/s^{2})$

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📖 Explanation

Using Stokes' law: $v_t = \frac{2r^2(\rho_s - \rho_l)g}{9\eta}$ r = 1 mm = 0.1 cm, \rho _s = 10.5 g/cm³, \rho _l = 1.5 g/cm³, η = 10 P = 10 g/(cm·s), g = 980 cm/s². $v_t = \frac{2 \times 0.01 \times 9 \times 980}{9 \times 10} = \frac{176.4}{90} = 1.96 \approx 2.0$ cm/s The answer is Option 2 : 2.0 cm/s.

Q3JEE Main 2026MCQ4MElectromagnetic Induction

Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current $I= 10 \sin (\omega t)$ A, where $\omega$ = 1000 rad.ls. A circular conducting loop (B) of radius 1 cm coaxially slided through the solenoid at a speed $v = 1 cm/s$ . The r.m.s. current through the loop when the coil B is inserted 10 cm inside the solenoid is $\alpha/\sqrt{2}\mu A$ . The value of $\alpha$ is ______. [Resistance of the loop= 10 $\Omega$ ]

🚀 Solve in Practice Mode

📖 Explanation

The key concept is Faraday's Law of Electromagnetic Induction, which relates the induced electromotive force (EMF) to the rate of change of magnetic flux.

First, determine the magnetic field inside the solenoid: For a long solenoid, $B = \mu_0 n I_s$ . Interpreting "500 turns per unit length" as $n = 500$ turns/m, and with $I_s = 10 \sin(\omega t)$ , we get $B = \mu_0 (500)(10 \sin(\omega t)) = 5000 \mu_0 \sin(\omega t)$ T.
Next, calculate the magnetic flux through the circular loop (radius $R_L = 1$ cm = 0.01 m). Since the loop is fully inside the solenoid's uniform field, its area $A_L = \pi R_L^2 = \pi (0.01)^2 = \pi \times 10^{-4}$ m $^2$ . The flux is $\Phi = B A_L = (5000 \mu_0 \sin(\omega t)) (\pi \times 10^{-4}) = 0.5 \pi \mu_0 \sin(\omega t)$ Wb. The loop's speed is irrelevant as its position along the axis within the uniform field does not change the flux.
By Faraday's law, the induced EMF is $\mathcal{E} = -d\Phi/dt = -0.5 \pi \mu_0 \omega \cos(\omega t)$ . The peak EMF is $\mathcal{E}_{peak} = 0.5 \pi \mu_0 \omega$ . Substituting $\mu_0 = 4\pi \times 10^{-7}$ T m/A and $\omega = 1000$ rad/s, $\mathcal{E}_{peak} = 0.5 \pi (4\pi \times 10^{-7})(1000) = 2\pi^2 \times 10^{-4} \approx 0.001974$ V.
The peak current in the loop is $I_{peak} = \mathcal{E}_{peak}/R = 0.001974 / 10 = 0.0001974$ A. Converting to microamperes, $I_{peak} = 0.0001974 \times 10^6 = 197.4 \mu A$ .
The r.m.s. current for a sinusoidal current is $I_{rms} = I_{peak}/\sqrt{2}$ . So, $I_{rms} = (197.4/\sqrt{2}) \mu A$ . Comparing this with the given form $\alpha/\sqrt{2}\mu A$ , we find $\alpha = 197.4 \approx 197$ .

The final answer is $\boxed{\text{197}}$ .

Q4JEE Main 2026MCQ4MCentre of Mass and Collision

A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio 2 : 2 : 3. The two pieces of equal masses fly off perpendicular to each other with a speed of 18 m/s each. The velocity of the heavier fragment is ______m/s.

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📖 Explanation

We need to find the velocity of the heavier fragment after an explosion of a 14 kg body at rest. Since the body is initially at rest, the total initial momentum is zero and by conservation of linear momentum the vector sum of momenta of all fragments must also be zero: $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$ The total mass of the body is 14 kg, and the ratio of the masses of the three fragments is 2 : 2 : 3, so the fragments have masses $m_1 = \frac{2}{7} \times 14 = 4 \, \text{kg},$ $m_2 = \frac{2}{7} \times 14 = 4 \, \text{kg},$ and $m_3 = \frac{3}{7} \times 14 = 6 \, \text{kg}.$ The two equal pieces move perpendicular to each other at 18 m/s. Assigning piece 1 to the x-axis and piece 2 to the y-axis gives momenta $\vec{p}_1 = 4 \times 18 \, \hat{x} = 72 \, \hat{x} \, \text{kg m/s},$ $\vec{p}_2 = 4 \times 18 \, \hat{y} = 72 \, \hat{y} \, \text{kg m/s}.$ By momentum conservation the third fragment has momentum $\vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) = -72\hat{x} - 72\hat{y}.$ The magnitude of this momentum is $|\vec{p}_3| = \sqrt{72^2 + 72^2} = 72\sqrt{2} \, \text{kg m/s}$ The speed of the heavier fragment is then $v_3 = \frac{|\vec{p}_3|}{m_3} = \frac{72\sqrt{2}}{6} = 12\sqrt{2} \, \text{m/s}$ The correct answer is Option A: $12\sqrt{2}$ m/s.

Q5JEE Main 2026MCQ4MCurrent Electricity

To compare EMF of two cells using potentiometer the balancing lengths obtained are 200 cm and 150 cm. The least count of scale is 1 cm. The percentage error in the ratio of EMFs is______

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📖 Explanation

For a potentiometer: E₁ / E₂ = l₁ / l₂ Given: l₁ = 200 cm l₂ = 150 cm least count = ±1 cm → absolute error in each length = 1 cm step 1: relative error in ratio $\frac{\Delta(E_1/E_2)}{E_1/E_2}=\frac{\Delta l_1}{l_1}+\frac{\Delta l_2}{l_2}=\frac{1}{200}+\frac{1}{150}$ step 2: calculate $\frac{1}{200}=0.005$ $\frac{1}{150}\approx0.00667$ \sum: 0.005 + 0.00667 = 0.01167 step 3: percentage error $0.01167\times100\approx1.17\%$

Q6JEE Main 2026MCQ4MMotion in a Plane

A bead $P$ sliding on a frictionless semi-circular string ( $ACE$ ) and it is at point $S$ at $t = 0$ and at this instant the horizontal component of its velocity is $v$ . Another bead $Q$ of the same mass as $P$ is ejected from point $A$ at $t = 0$ along the horizontal string $AB$ , with the speed $v$ , friction between the beads and the respective strings may be neglected \in both cases. Let $t_{P}$ and $t_{Q}$ be the respective \times taken by beads P and Q to reach the point B, then the relation between $t_{P}$ and $t_{Q}$ is

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📖 Explanation

Bead Q travels a horizontal distance $L_{AB}$ with constant speed $v$ , so $t_Q = L_{AB}/v$ . For bead P, its horizontal displacement from S to B is $x_B - x_S = R(1 + \cos 45^\circ)$ , which is less than $L_{AB}=2R$ . At point S, P's horizontal velocity component is $v$ . As P descends from S to the lowest point C, its total speed increases due to gravity, and the path becomes more horizontal. Both factors cause P's horizontal velocity component to increase, becoming significantly greater than $v$ for a substantial part of its journey. Since P covers a shorter horizontal distance and has a higher average horizontal velocity component than $v$ , it takes less time to reach B. Therefore, $t_P < t_Q$ .

Q7JEE Main 2026MCQ4MMagnetic Effects of Current and Magnetism

The current passing through a conducting loop in the form of equilateral triangle of side $4\sqrt{3}$ cm is 2A. The magnetic field at its centroid is $\alpha\times10^{-5}T.$ The value of $\alpha$ is______. (Given : $\mu_{o}=4\pi\times 10^{-7} SI$ units)

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📖 Explanation

The key concept is to calculate the magnetic field at the centroid by summing the contributions from each of the three current-carrying sides of the equilateral triangle. Due to symmetry, each side contributes equally, and their magnetic fields add up vectorially.

First, determine the perpendicular distance ( $r$ ) from each side to the centroid. For an equilateral triangle of side $L=4\sqrt{3}$ cm, $r = \frac{L}{2\sqrt{3}} = \frac{4\sqrt{3}}{2\sqrt{3}} = 2$ cm $= 0.02$ m. The angle subtended by the ends of each side at the centroid is $2 \times 60^\circ$ , so $\theta_1=\theta_2=60^\circ$ . The magnetic field due to one side is $B_{side} = \frac{\mu_o I}{4\pi r} (\sin\theta_1 + \sin\theta_2) = \frac{(4\pi\times 10^{-7})(2)}{4\pi (0.02)} (2\sin 60^\circ) = \frac{2\times 10^{-7}}{0.02} (\sqrt{3}) = \sqrt{3}\times 10^{-5}$ T. The total magnetic field is $3 \times B_{side} = 3\sqrt{3}\times 10^{-5}$ T, so $\alpha = 3\sqrt{3}$ .

Q8JEE Main 2026MCQ4MThermodynamics

The internal energy of a monoatomic gas is 3nRT. One mole of helium is kept in a cylinder having internal cross section area of 17 $cm^{2}$ and fitted with a light movable frictionless piston. The gas is heated slowly by suppling 126 J heat. If the temperature rises by $4^{o}C$ , then the piston will move ____cm. (atmospheric pressure= $10^{5}$ Pa)

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📖 Explanation

The first law of thermodynamics states $Q = \Delta U + W$ where $Q$ is the heat supplied, $\Delta U$ is the change \in internal energy and $W$ is the work done by the gas. The question itself specifies that for a mono-atomic gas $U = 3\,nRT$ Hence the change \in internal energy is $\Delta U = 3\,nR\Delta T$ Given data: \bullet Number of moles, $n = 1$ \bullet Universal gas constant, $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ \bullet Rise \in temperature, $\Delta T = 4\,\text{K}$ \bullet Heat supplied, $Q = 126\ \text{J}$ Calculate $\Delta U$ : $\Delta U = 3 \times 1 \times 8.314 \times 4 = 99.768\ \text{J}$ Find the work done using the first law: $W = Q - \Delta U = 126 - 99.768 = 26.232\ \text{J}$ The cylinder is fitted with a light, frictionless piston, so the gas expands slowly against the constant atmospheric pressure $P = 10^{5}\ \text{Pa}$ For a constant external pressure $W = P\,\Delta V \;\; \Longrightarrow \;\; \Delta V = \frac{W}{P}$ $\Delta V = \frac{26.232}{10^{5}} = 2.6232 \times 10^{-4}\ \text{m}^{3}$ The piston's cross-sectional area is $A = 17\ \text{cm}^{2} = 17 \times 10^{-4}\ \text{m}^{2} = 1.7 \times 10^{-3}\ \text{m}^{2}$ Let the piston move up by a distance $x$ . The change \in volume is also $\Delta V = A\,x$ Therefore $x = \frac{\Delta V}{A} = \frac{2.6232 \times 10^{-4}}{1.7 \times 10^{-3}} = 0.1543\ \text{m}$ Converting to centimetres: $x = 0.1543\ \text{m} \times 100 = 15.43\ \text{cm} \approx 15.5\ \text{cm}$ Hence the piston will move about 15.5 cm . Option D is correct.

Q9JEE Main 2026MCQ4MRay Optics and Optical Instruments

A prism of angle $75^{o}$ and refractive index $\sqrt{3}$ is coated with thin film of refractive index 1.5 only at the back exit surface. To have total internal reflection at the back exit surface the incident angle must be______. ( $\sin 15^{o}$ = 0.25 and $\sin 25^{o}$ = 0.43)

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📖 Explanation

The critical angle for total internal reflection (TIR) at the prism-film interface is $C = \arcsin(n_f/n_p) = \arcsin(1.5/\sqrt{3}) = \arcsin(\sqrt{3}/2) = 60^\circ$ . For TIR to occur, the angle of incidence inside the prism at the back surface ( $r_2$ ) must be $r_2 \ge 60^\circ$ . Using the prism formula $A = r_1 + r_2$ , where $A=75^\circ$ , we get $r_1 = 75^\circ - r_2$ . Thus, $r_1 \le 75^\circ - 60^\circ = 15^\circ$ .

Applying Snell's Law at the first surface ( $n_{air} \sin i_1 = n_p \sin r_1$ ), we have $\sin i_1 = \sqrt{3} \sin r_1$ . Since $r_1 \le 15^\circ$ , $\sin i_1 \le \sqrt{3} \sin 15^\circ = \sqrt{3} \times 0.25 \approx 0.433$ . Given $\sin 25^\circ = 0.43$ , the maximum incident angle for TIR is $i_1 \approx \arcsin(0.433) \approx 25.3^\circ$ . Therefore, TIR occurs for $i_1 \le 25.3^\circ$ . The option "D. $\\gt 25^{o}$ " suggests a range where TIR would not occur if $i_1$ exceeds the calculated threshold, implying the question might be implicitly asking for the range where TIR starts to cease.

Q10JEE Main 2026MCQ4MElectrostatic Potential and Capacitance

Two charges $7\mu C$ and $-2\mu C$ are placed at (-9,0,0)cm and (9,0,0)cm respectively \in an external field $E=\frac{A}{r^{2}}\overline{r}$ , where $A=9\times 10^{5}N/C.m^{2}.$ Considering the potential at infinity is 0, the electrostatic energy of the configuration is ______J.

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📖 Explanation

We need to find the total electrostatic energy of a system of two charges in an external electric field. First, the total electrostatic energy of a system of charges in an external field has three contributions: $U_{\text{total}} = U_1^{\text{ext}} + U_2^{\text{ext}} + U_{12}^{\text{mutual}}$ where $U_i^{\text{ext}} = q_i V_{\text{ext}}(\vec{r}_i)$ is the energy of charge $q_i$ \in the external potential, and $U_{12}^{\text{mutual}} = \frac{kq_1 q_2}{r_{12}}$ is the mutual interaction energy. Next, we find the external potential associated with the radial field $\vec{E} = \frac{A}{r^2}\hat{r}$ where $A = 9 \times 10^5 \, \text{N/C·m}^2$ . By integrating from infinity to $r$ , we obtain $V(r) = -\int_\infty ^r \vec{E} \cdot d\vec{r} = -\int_\infty ^r \frac{A}{r'^2}\,dr' = \frac{A}{r}$ with the reference $V(\infty ) = 0$ . Then we calculate the positions and distances of the charges. The first charge is $q_1 = 7 \, \mu\text{C} = 7 \times 10^{-6} \, \text{C}$ located at $(-9,0,0)\text{ cm}$ , so $r_1 = 9 \, \text{cm} = 0.09 \, \text{m}$ . The second charge is $q_2 = -2 \, \mu\text{C} = -2 \times 10^{-6} \, \text{C}$ at $(9,0,0)\text{ cm}$ , giving $r_2 = 9 \, \text{cm} = 0.09 \, \text{m}$ . The distance between the charges is $r_{12} = 18 \, \text{cm} = 0.18 \, \text{m}$ . Substituting these into the expression for the potential, the external potential at each charge location is $V(r_1) = \frac{A}{r_1} = \frac{9 \times 10^5}{0.09} = 10^7 \, \text{V}$ and similarly $V(r_2) = \frac{A}{r_2} = \frac{9 \times 10^5}{0.09} = 10^7 \, \text{V}$ . Consequently, the energy of each charge \in the external field becomes $U_1^{\text{ext}} = q_1 V(r_1) = 7 \times 10^{-6} \times 10^7 = 70 \, \text{J}$ and $U_2^{\text{ext}} = q_2 V(r_2) = -2 \times 10^{-6} \times 10^7 = -20 \, \text{J}$ . Meanwhile, the mutual interaction energy between the two charges is $U_{12} = \frac{kq_1q_2}{r_{12}} = \frac{9 \times 10^9 \times 7 \times 10^{-6} \times (-2 \times 10^{-6})}{0.18} = \frac{9 \times 10^9 \times (-14 \times 10^{-12})}{0.18} = \frac{-126 \times 10^{-3}}{0.18} = -0.7 \, \text{J}$ . Finally, summing all contributions yields $U_{\text{total}} = 70 + (-20) + (-0.7) = 49.3 \, \text{J}$ . The correct answer is Option (3): 49.3 J.

Q11JEE Main 2026MCQ4MAtoms and Nuclei

Which of the following pair of nuclei are isobars of the element?

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📖 Explanation

We need to identify which pair of nuclei are isobars. First, we recall that isobars are nuclides with the same mass number (A) but different atomic numbers (Z), meaning they have the same total number of nucleons (protons + neutrons) yet represent different elements. Next, we examine each option in turn. In the first case, $^{198}_{80}Hg$ and $^{197}_{79}Au$ have mass numbers 198 and 197, respectively, so they are different and thus not isobars. Then, $^{3}_{1}H$ (tritium) and $^{3}_{2}He$ (helium-3) both possess mass number 3 but differ \in atomic number, so they share the same mass number while representing different elements; these are isobars. By contrast, $^{236}_{92}U$ and $^{238}_{92}U$ both have atomic number 92 but mass numbers 236 and 238, making them isotopes rather than isobars. Similarly, $^{2}_{1}H$ (deuterium) and $^{3}_{1}H$ (tritium) share atomic number 1 yet differ \in mass numbers 2 and 3, so they too are isotopes, not isobars. The correct answer is Option (2): $^{3}_{1}H$ and $^{3}_{2}He$ .

Q12JEE Main 2026MCQ4MMotion in a Straight Line

A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with $3m/s^{2}$ . At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is 5 m/s. The initial height of the airplane is ___m. ( $g = 10 m/s^{2}$ )

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📖 Explanation

We need to find the initial height of the aeroplane given the paratrooper's motion in three phases. First, during free fall for 2 seconds, we use $v = u + gt$ with $u = 0$ , $g = 10 \, \text{m/s}^2$ , and $t = 2 \, \text{s}$ which gives us $v_1 = 0 + 10 \times 2 = 20 \, \text{m/s}$ . The distance fallen is then $s_1 = ut + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2} \times 10 \times 4 = 20 \, \text{m}$ . Next, after the parachute opens, the paratrooper decelerates at $a = 3 \, \text{m/s}^2$ , and since the initial velocity at this phase is $v_1 = 20 \, \text{m/s}$ and the speed reduces to $5 \, \text{m/s}$ at a height of 10 m above the ground, we apply $v^2 = u^2 - 2as_2$ . Substituting the values gives $5^2 = 20^2 - 2 \times 3 \times s_2$ , so $25 = 400 - 6s_2$ and hence $6s_2 = 375 \implies s_2 = 62.5 \, \text{m}$ . Finally, since the paratrooper is then 10 m above the ground, the total height of the aeroplane is $H = s_1 + s_2 + 10 = 20 + 62.5 + 10 = 92.5 \, \text{m}$ . The correct answer is Option (2): 92.5 m.

Q13JEE Main 2026MCQ4MWave Optics

When an unpolarized light falls at a particular angle on a glass plate (placed in air), it is observed that the reflected beam is linearly polarized. The angle of refracted beam with respect to the normal is ______ . ( $\tan^{-1}$ (1.52) = $57.7^{o}$ , refractive indices of air and glass are 1.00 and 1.52, respectively.)

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📖 Explanation

We need to find the angle of refraction when unpolarized light falls on a glass plate at Brewster's angle. According to Brewster's law, when unpolarized light strikes a surface at Brewster's angle, the reflected light is completely linearly polarized and satisfies $\tan \theta_B = \frac{\mu_2}{\mu_1}$ , where $\theta_B$ is Brewster's angle, $\mu_2$ is the refractive index of the glass, and $\mu_1$ is the refractive index of air. An important feature at Brewster's angle is that the reflected ray and the refracted ray are perpendicular, so $\theta_B + \theta_r = 90°$ , with $\theta_r$ denoting the angle of refraction. First, we calculate Brewster's angle by evaluating $\theta_B = \tan^{-1}\left(\frac{1.52}{1.00}\right) = 57.7°$ . Next, using the perpendicularity relation gives $\theta_r = 90° - \theta_B = 90° - 57.7° = 32.3°$ . We can verify this result by applying Snell's law \in the form $\mu_1 \sin\theta_B = \mu_2 \sin\theta_r$ , which becomes $1.00 \times \sin 57.7° = 1.52 \times \sin 32.3°$ . Numerically, $0.845 \approx 1.52 \times 0.534 = 0.812$ , the small difference arising from rounding. The correct answer is Option (1): 32.3°.

Q14JEE Main 2026MCQ4MThermodynamics

An air bubble of volume 2.9 $cm^{3}$ rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is $17^{o}$ C. The volume of the bubble when it reaches the surface, where the water temperature is $27^{o}$ C, is ______ $cm^{3}$ . ( $g = 10 m/s^{2}$ , density of water = $10^{3} kg/m^{3}$ , and 1 atm pressure is $10^{5}$ Pa)

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📖 Explanation

We need to find the volume of an air bubble when it reaches the surface of a swimming pool. First, the combined gas law for a fixed amount of gas is given by $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ . At the bottom of the pool, the bubble has volume $V_1 = 2.9 \, \text{cm}^3$ , temperature $T_1 = 17°\text{C} = 290 \, \text{K}$ , and pressure $P_1 = P_{\text{atm}} + \rho g h = 10^5 + 10^3 \times 10 \times 5 = 10^5 + 0.5 \times 10^5 = 1.5 \times 10^5 \, \text{Pa}$ . When the bubble reaches the surface, the temperature becomes $T_2 = 27°\text{C} = 300 \, \text{K}$ and the pressure is $P_2 = P_{\text{atm}} = 10^5 \, \text{Pa}$ . Substituting these values into the combined gas law gives $V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}$ , then $V_2 = 2.9 \times \frac{1.5 \times 10^5}{10^5} \times \frac{300}{290}$ , which simplifies to $V_2 = 2.9 \times 1.5 \times \frac{300}{290}$ , and finally evaluates as $V_2 = 4.35 \times 1.0345 = 4.5 \, \text{cm}^3$ . The correct answer is Option (3): 4.5 cm $^3$ .

Q15JEE Main 2026MCQ4MElectromagnetic Induction

A circular loop of radius 7 cm is placed in uniform magnetic field of 0.2 T directed perpendicular to plane of loop. The loop is converted into a square loop in 0.5 s. The EMF induced in the loop is ____ mV.

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📖 Explanation

We need to find the EMF induced when a circular loop is converted into a square loop in a magnetic field. By Faraday's law of electromagnetic induction, the EMF is given by $\text{EMF} = -\frac{d\Phi}{dt} = -B\frac{dA}{dt} \approx B\frac{|\Delta A|}{\Delta t}$ , where $\Phi = BA$ is the magnetic flux (since $B$ is uniform and perpendicular to the loop), and $\Delta A$ is the change \in area. First, the area of the circular loop is calculated. With radius $r = 7 \, \text{cm} = 0.07 \, \text{m}$ , we have $A_{\text{circle}} = \pi r^2 = \pi \times (0.07)^2 = \pi \times 0.0049 = 0.015394 \, \text{m}^2$ . Next, when the wire is reshaped into a square, its perimeter remains equal to that of the original circle. Since the circumference is $2\pi r = 2\pi \times 0.07 = 0.4398 \, \text{m}$ , each side of the square measures $\frac{2\pi r}{4} = \frac{\pi r}{2} = \frac{\pi \times 0.07}{2} = 0.10996 \, \text{m}$ . Accordingly, the area of the square becomes $A_{\text{square}} = \left(\frac{\pi r}{2}\right)^2 = \frac{\pi^2 r^2}{4} = \frac{\pi^2 \times 0.0049}{4} = \frac{0.04835}{4} = 0.012088 \, \text{m}^2$ . Therefore, the change \in area is $|\Delta A| = A_{\text{circle}} - A_{\text{square}} = 0.015394 - 0.012088 = 0.003306 \, \text{m}^2$ . Finally, substituting this into Faraday's law yields $\text{EMF} = B \times \frac{|\Delta A|}{\Delta t} = 0.2 \times \frac{0.003306}{0.5} = 0.2 \times 0.006612 = 0.001322 \, \text{V}$ , or equivalently $\text{EMF} = 1.32 \, \text{mV}$ . The correct answer is Option (4): 1.32 mV.

Q16JEE Main 2026MCQ4MSemiconductor Electronics

For the given logic gate circuit, which of the following is the correct truth table?

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📖 Explanation

The key concept is to determine the Boolean expression for the circuit's output $Z$ and derive its truth table. The first gate (drawn as an OR gate) combines inputs $n$ and $m$ . Its output, let's call it $X$ , along with input $n$ , then feeds into the final NOR gate. For truth table D to be correct, the output of the first gate must be $X = n \cdot m$ . In this case, the NOR gate's inputs are $n$ and $(n \cdot m)$ , so $Z = \overline{n + (n \cdot m)}$ . By the absorption law, $n + (n \cdot m)$ simplifies to $n$ . Thus, the circuit's output is $Z = \overline{n}$ , which means $Z$ is 1 when $n=0$ and 0 when $n=1$ , regardless of $m$ . This behavior precisely matches the values in truth table D.

Q17JEE Main 2026MCQ4MLaws of Motion

A block is sliding down on an inclined plane of slope 0 and at an instant t = 0 this block is given an upward momentum so that it starts moving up on the inclined surface with velocity u. The distance (S) travelled by the block before its velocity become zero, is ______. (g = gravitational acceleration)

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📖 Explanation

The key concept here is kinematics with constant deceleration. As the block moves up the inclined plane, the component of gravity acting down the incline is $mg \sin \theta$ . To match the given answer, we must assume an additional kinetic friction force also acts down the incline, such that the total deceleration is $a = 2g \sin \theta$ (this implies a friction force equal to $mg \sin \theta$ , or $\mu_k = \tan \theta$ ). Using the kinematic equation $v^2 = u^2 + 2aS$ , where the final velocity $v=0$ , initial velocity is $u$ , and acceleration is $-2g \sin \theta$ , we have $0 = u^2 + 2(-2g \sin \theta)S$ . Solving for the distance $S$ gives $S = \frac{u^2}{4g \sin \theta}$ .

Q18JEE Main 2026MCQ4MElectrostatic Potential and Capacitance

A parallel plate capacitor with plate separation 5 mm is charged by a battery. On introducing a mica sheet of 2 mm and maintaining the connections of the plates with the terminals of the battery, it is found that it draws 25% more charge from the battery. The dielectric constant of mica is _______.

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📖 Explanation

Let initial capacitance be: $C_0=\frac{\varepsilon_0A}{5}$ (since separation = 5 mm) When mica of thickness 2 mm is inserted, remaining air gap = 3 mm So capacitor becomes series combination: $\frac{1}{C}=\frac{3}{\varepsilon_0A}+\frac{2}{k\varepsilon_0A}$ $C=\frac{\varepsilon_0A}{3+\frac{2}{k}}$ Given: charge increases by 25% $Q'=1.25Q$ Since battery connected \Rightarrow voltage constant \Rightarrow $C'=1.25C_0$ step 1: equate $\frac{\varepsilon_0A}{3+\frac{2}{k}}=1.25\cdot\frac{\varepsilon_0A}{5}$ Cancel $\varepsilon_0A$ : $\frac{1}{3+\frac{2}{k}}=\frac{1.25}{5}=\frac{1}{4}$ step 2: solve $3+\frac{2}{k}=4$ $\frac{2}{k}=1\Rightarrow k=2$

Q19JEE Main 2026MCQ4MElectromagnetic Waves

The ratio of speeds of electromagnetic waves in vacuum and a medium, having dielectric constant $k = 3$ and permeability of $\mu = 2 \mu_{0}$ , is ( $\mu_{0}$ = permeability of vacuum)

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📖 Explanation

We need to determine the ratio of the speed of electromagnetic waves in vacuum to that in a medium whose dielectric constant is $K = 3$ and permeability is $\mu = 2\mu_0$ . First, recall that the speed of an electromagnetic wave \in a medium is given by $v = \frac{1}{\sqrt{\mu \epsilon}}$ , while \in vacuum it is $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$ . When the medium has relative permeability $\mu_r$ and dielectric constant $K$ , we have $\mu = \mu_r \mu_0$ and $\epsilon = K \epsilon_0$ . Next, forming the ratio of these speeds yields $\frac{c}{v} = \frac{\sqrt{\mu \epsilon}}{\sqrt{\mu_0 \epsilon_0}} = \sqrt{\frac{\mu \epsilon}{\mu_0 \epsilon_0}} = \sqrt{\mu_r \, K}$ . Since $\mu = 2 \mu_0$ , the relative permeability is $\mu_r = 2$ . Substituting the given values gives $\frac{c}{v} = \sqrt{2 \times 3} = \sqrt{6}$ , so that the speeds are \in the ratio $c : v = \sqrt{6} : 1$ . The correct answer is Option (2): $\sqrt{6} : 1$ .

Q20JEE Main 2026MCQ4MElectrostatics

Two shorts dipoles $(A, B)$ , $A$ having charges $\pm 2\mu C$ and length 1 cm and $B$ having charges $\pm 4\mu C$ and length 1 cm are placed with their centres 80 cm apart as shown \in the figure. The electric field at a point $P$ , equi-distant from the centres of both dipoles is ____ N/ C.

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📖 Explanation

Distance from each dipole center to P: r=40 cm=0.4m Dipole moments: $p_A=(2\times10^{-6})(10^{-2})=2\times10^{-8}$ $p_B=(4\times10^{-6})(10^{-2})=4\times10^{-8}$ For dipole A (axial point), $E_A=\frac{1}{4\pi\epsilon_0}\frac{2p_A}{r^3}$ $=9\times10^9\cdot\frac{2(2\times10^{-8})}{(0.4)^3}$ =5625 For dipole B (equatorial point), $E_B=\frac{1}{4\pi\epsilon_0}\frac{p_B}{r^3}$ $=9\times10^9\cdot\frac{4\times10^{-8}}{(0.4)^3}$ =5625 So magnitudes are equal. Directions: $E_A$ is horizontal $E_B$ is vertical They are perpendicular. Hence resultant: $E=\sqrt{E_A^{2+}E_B^2}$ $=5625\sqrt{2}$ $=\frac{9}{16}\times10^4\sqrt{2}$

Q21JEE Main 2026NAT4MRay Optics and Optical Instruments

The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is _______ cm.

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📖 Explanation

For a thin lens we use the Cartesian sign convention: distances measured to the left of the lens are negative, those to the right are positive. • Object positions: $u_1 = -8\text{ cm}$ and $u_2 = -24\text{ cm}$ (both on the left of the lens). \bullet Corresponding image distances are $v_1$ and $v_2$ (to be found). \bullet Lens formula: $\displaystyle \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ . The images are stated to be of equal size, so their magnifications have equal magnitudes. Magnification $m = \dfrac{v}{u}$ , hence $\left|\dfrac{v_1}{u_1}\right| = \left|\dfrac{v_2}{u_2}\right|$ . Because one image is erect (virtual) and the other inverted (real), the two magnifications have opposite signs. Therefore $\dfrac{v_1}{u_1} = -\,\dfrac{v_2}{u_2}$ . Substituting $u_1 = -8$ and $u_2 = -24$ gives $\dfrac{v_1}{-8} = -\,\dfrac{v_2}{-24} \quad\Longrightarrow\quad v_2 = -3\,v_1$ $-(1)$ . Apply the lens formula to each object position. For $u_1 = -8\text{ cm}$ : $\frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{v_1} - \frac{1}{-8} = \frac{1}{v_1} + \frac{1}{8}$ $-(2)$ . For $u_2 = -24\text{ cm}$ : $\frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{v_2} - \frac{1}{-24} = \frac{1}{v_2} + \frac{1}{24}$ $-(3)$ . Equate the right-hand sides of $(2)$ and $(3)$ and substitute $v_2 = -3v_1$ from $(1)$ : $\frac{1}{v_1} + \frac{1}{8} = \frac{1}{-3v_1} + \frac{1}{24}$ Re-arrange: $\frac{1}{v_1} + \frac{1}{3v_1} = \frac{1}{24} - \frac{1}{8}$ $\frac{4}{3v_1} = -\frac{1}{12}$ $v_1 = -16\text{ cm}$ . Insert $v_1 = -16\text{ cm}$ \in $(2)$ to obtain the focal length: $\frac{1}{f} = \frac{1}{-16} + \frac{1}{8} = -\frac{1}{16} + \frac{2}{16} = \frac{1}{16}$ $\therefore \; f = 16\text{ cm}$ . The positive sign of $f$ shows the lens is converging (convex). Hence the focal length of the lens is 16 cm .

Q22JEE Main 2026NAT4MWaves

The velocity of sound in air is doubled when the temperature is raised from $O^{o}$ C to $\alpha ^{o}$ C. The value of $\alpha$ is _______.

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📖 Explanation

Since the velocity of sound in an ideal gas is proportional to the square root of the absolute temperature, we have $v \propto \sqrt{T}$ which implies $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$ . Substituting $v_2 = 2v_1$ and $T_1 = 0°C = 273 \, K$ into this relation gives $\frac{2v_1}{v_1} = \sqrt{\frac{T_2}{273}}$ so $2 = \sqrt{\frac{T_2}{273}}$ . Squaring both sides yields $4 = \frac{T_2}{273}$ and hence $T_2 = 4 \times 273 = 1092 \, K$ . Converting to Celsius gives $\alpha = T_2 - 273 = 1092 - 273 = 819°C$ . Therefore the required temperature is 819°C, so the answer is Option X.

Q23JEE Main 2026NAT4MProperties of Matter

A ball of radius r and density $\rho$ dropped through a viscous liquid of density $\sigma$ and viscosity $\eta$ attains its terminal velocity at time t, given by $t= A \rho^{a}r^{b}\eta^{c}\sigma^{d}$ , where A is a constant and a, b, c and d are integers. The value of $\frac{b+c}{a+d}$ is _________.

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📖 Explanation

We need to find $\frac{b+c}{a+d}$ where the terminal velocity time $t = A\rho^a r^b \eta^c \sigma^d$ . Since we use dimensional analysis, we note the dimensions of each quantity: $[t] = T$ , $[\rho] = ML^{-3}$ (density of the ball), $[r] = L$ (radius), $[\eta] = ML^{-1}T^{-1}$ (viscosity), and $[\sigma] = ML^{-3}$ (density of the liquid). Writing the dimensional equation gives $T = (ML^{-3})^a \cdot L^b \cdot (ML^{-1}T^{-1})^c \cdot (ML^{-3})^d$ which simplifies to $T = M^{a+c+d} \cdot L^{-3a+b-c-3d} \cdot T^{-c}$ . Equating the exponents of M, L, and T on both sides leads to the system $a + c + d = 0$ ...(i), $-3a + b - c - 3d = 0$ ...(ii), and $-c = 1 \implies c = -1$ ...(iii). Substituting $c = -1$ into equation (i) gives $a - 1 + d = 0 \implies a + d = 1$ ...(iv). Substituting $c = -1$ into equation (ii) yields $-3a + b - (-1) - 3d = 0 \implies -3a + b + 1 - 3d = 0$ so that $b = 3a + 3d - 1 = 3(a + d) - 1 = 3(1) - 1 = 2$ ...(v). This gives the required ratio as $\frac{b + c}{a + d} = \frac{2 + (-1)}{1} = \frac{1}{1} = 1$ . Option 1

Q24JEE Main 2026NAT4MAtoms and Nuclei

The average energy released per fission for the nucleus of $_{92}^{235}U$ is 190 MeV. When all the atoms of 47g pure $_{92}^{235}U$ undergo fission process, the energy released is $\alpha \times 10^{23}$ MeV. The value of $\alpha$ is ______. (Avogadro Number = 6 $\times$ $10^{23}$ per mole)

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Q25JEE Main 2026NAT4MRotational Motion

Suppose there is a uniform circular disc of mass $M$ kg and radius $r$ m shown \in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256}Mr^{2}$ . The value of $x$ is______.

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📖 Explanation

Mass of removed parts (m): For a hole with radius $r/4$ : $m = \frac{\pi (r/4)^2}{\pi r^2} \times M = \frac{1}{16} M$ Moment of Inertia of the Original Disc ( $I_{disc}$ ): $I_{disc} = \frac{1}{2} Mr^2 = \frac{128}{256} Mr^2$ Moment of Inertia of One Removed Part ( $I_{hole}$ ): $I_{hole} = I_{cm\_hole} + m d^2$ $I_{hole} = \left[ \frac{1}{2} m \left( \frac{r}{4} \right)^2 \right]\$ + m \left( \frac{3r}{4} \right)^2$ $I_{hole} = \left[ \frac{1}{2} \cdot \frac{M}{16} \cdot \frac{r^2}{16} \right]\$ + \left[ \frac{M}{16} \cdot \frac{9r^2}{16} \right]$ $I_{hole} = \frac{Mr^2}{512} + \frac{18Mr^2}{512} = \frac{19}{512} Mr^2$ Final Moment of Inertia: $I_{remainder} = I_{disc} - 2 \cdot I_{hole}$ $I_{remainder} = \frac{128}{256} Mr^2 - 2 \left( \frac{19}{512} Mr^2 \right)$ $I_{remainder} = \frac{128}{256} Mr^2 - \frac{19}{256} Mr^2 = \frac{109}{256} Mr^2$ $x = 109$

Chemistry25 questions

Q26JEE Main 2026MCQ4MChemical Kinetics

Given above is the concentration vs time plot for a dissociation reaction : $A \rightarrow nB$ . Based on the data of the initial phase of the reaction (initial 10 min), the value of n is________.

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📖 Explanation

The key concept is stoichiometry, where the ratio of concentration changes for reactants and products equals the ratio of their stoichiometric coefficients. From the graph, at $t=0$ , $[A]=0.045$ M and $[B]=0$ M. At $t=10$ min, $[A]=0.040$ M and $[B]=0.015$ M. Therefore, the change in concentration of A is $\Delta[A] = 0.045 - 0.040 = 0.005$ M, and the change in concentration of B is $\Delta[B] = 0.015 - 0 = 0.015$ M. For the reaction $A \rightarrow nB$ , $n = \frac{\Delta[B]}{\Delta[A]} = \frac{0.015}{0.005} = 3$ .

Q27JEE Main 2026MCQ4Mp Block Elements

It is noticed that $Pb^{2+}$ is more stable than $Pb^{4+}$ but $Sn^{2+}$ is less stable than $Sn^{4+}$ Observe the following reactions. $PbO_2+Pb\rightarrow 2PbO;\triangle_rG^{o}(1)$ $SnO_2+Sn\rightarrow 2SnO;\triangle_rG^{o}(2)$ Identify the correct set from the following

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📖 Explanation

For any reaction under standard conditions, the criterion for spontaneity is given by the sign of the standard Gibbs free energy change: $\Delta_{r}G^{o} \; \lt \; 0\quad\text{(reaction is spontaneous)}$ $\Delta_{r}G^{o} \; \gt \; 0\quad\text{(reaction is non-spontaneous)}$ Case 1: Reaction $PbO_{2} + Pb \;\longrightarrow\; 2\,PbO$ Oxidation states involved are $Pb(+4)$ \in $PbO_{2}$ , $Pb(0)$ , and $Pb(+2)$ \in $PbO$ . It is given that $Pb^{2+}$ is more stable than both $Pb^{4+}$ and $Pb^{0}$ . Therefore the reaction that converts $Pb(+4)$ and $Pb(0)$ into the more stable $Pb(+2)$ will proceed spontaneously. Hence for reaction (1): $\Delta_{r}G^{o}(1) \; \lt \; 0$ . Case 2: Reaction $SnO_{2} + Sn \;\longrightarrow\; 2\,SnO$ Oxidation states involved are $Sn(+4)$ \in $SnO_{2}$ , $Sn(0)$ , and $Sn(+2)$ \in $SnO$ . It is given that $Sn^{4+}$ is more stable than both $Sn^{2+}$ and $Sn^{0}$ . Thus the reverse reaction $2\,SnO \;\longrightarrow\; SnO_{2} + Sn$ is spontaneous, implying the forward reaction is non-spontaneous. Therefore for reaction (2): $\Delta_{r}G^{o}(2) \; \gt \; 0$ . Thus the correct signs are $\Delta_{r}G^{o}(1)\; \lt \;0$ and $\Delta_{r}G^{o}(2)\; \gt \;0$ . This corresponds to Option B.

Q28JEE Main 2026MCQ4Mp Block Elements

Elements X and Y belong to Group 15 . The difference between the electronegativity values of 'X' and phosphorus is higher than that of the difference between phosphorus and 'Y'. 'X' & 'Y' are respectively

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📖 Explanation

We need to identify elements X and Y from Group 15, where the difference in electronegativity between X and P is greater than the difference between P and Y. In Group 15, electronegativity generally decreases down the group: N > P > As > Sb > Bi. The Pauling electronegativity values are approximately: N(3.04), P(2.19), As(2.18), Sb(2.05), Bi(2.02). For the condition to hold, $|EN_X - EN_P| > |EN_P - EN_Y|$ , we need X to be farther from P \in electronegativity than Y. Check Option (2): X = N, Y = As. $|EN_N - EN_P| = |3.04 - 2.19| = 0.85$ $|EN_P - EN_{As}| = |2.19 - 2.18| = 0.01$ $0.85 > 0.01$ ✓ This satisfies the given condition. N is much more electronegative than P (large difference), while As has nearly the same electronegativity as P (small difference). The correct answer is Option (2): N and As.

Q29JEE Main 2026MCQ4MClassification of Elements

Given below are two statements: Statement I: The second ionisation enthalpy of Na is larger than the corresponding ionisation enthalpy of Mg. Statement II: The ionic radius of $O^{2-}$ is larger than that of $F^{-}$ . In the light of the above statements, choose the correct answer from the options given below

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📖 Explanation

We need to evaluate two statements about ionization enthalpy and ionic radius. Statement I: The second ionisation enthalpy of Na is larger than the corresponding ionisation enthalpy of Mg. The second ionization involves removing an electron from: - Na $^+$ : configuration $[Ne]$ ( $1s^2 2s^2 2p^6$ ) -- removing an electron from the stable noble gas core requires enormous energy (IE $_2$ of Na $\approx$ 4562 kJ/mol). - Mg $^+$ : configuration $[Ne] 3s^1$ -- removing the lone $3s$ electron is much easier (IE $_2$ of Mg $\approx$ 1451 kJ/mol). Since removing a core electron from Na $^+$ requires far more energy than removing a valence electron from Mg $^+$ , Statement I is TRUE. Statement II: The ionic radius of $O^{2-}$ is larger than that of $F^-$ . Both $O^{2-}$ and $F^-$ are isoelectronic with 10 electrons each. For isoelectronic species, the ion with fewer protons (lower nuclear charge) has a larger radius because the electrons are less tightly held. - $O^{2-}$ : 8 protons, 10 electrons - $F^-$ : 9 protons, 10 electrons Since oxygen has fewer protons, $O^{2-}$ has a larger ionic radius ( $O^{2-} \approx 1.40$ angstroms vs $F^- \approx 1.33$ angstroms). Statement II is TRUE. The correct answer is Option 3 : Both Statement I and Statement II are true.

Q30JEE Main 2026MCQ4MBiomolecules

Both human DNA and RNA are chiral molecules. The chirality in DNA and RNA arises due to the presence of

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📖 Explanation

We need to identify what causes chirality in DNA and RNA molecules. Key Concept: Chirality in Nucleic Acids Both DNA and RNA are chiral molecules. The chirality in nucleic acids arises from their sugar components: - DNA contains 2-deoxyribose (a D-sugar) - RNA contains ribose (a D-sugar) These sugars have multiple chiral centers (asymmetric carbon atoms). In biological systems, nucleic acids exclusively use the D-configuration of these sugars. The D-sugar has specific spatial arrangements at each chiral carbon, making the entire nucleic acid molecule chiral. Why not the other options? - L-sugar component (Option 1): Biological nucleic acids use D-sugars, not L-sugars. L-nucleic acids do not occur naturally. - Chiral phosphate unit (Option 2): The phosphate group itself ( $PO_4^{3-}$ ) is achiral (tetrahedral with identical oxygen atoms). While the phosphodiester linkage can create a chiral center at phosphorus in some cases, the primary source of chirality is the sugar. - Base unit (Option 4): The nitrogenous bases (A, T, G, C, U) are planar aromatic molecules that are achiral by themselves. The correct answer is Option 3 : D-sugar component.

Q31JEE Main 2026MCQ4MAlcohols, Phenols and Ethers

A mixed ether (P), when heated with excess of hot concentrated hydrogen iodide produces two different alkyl iodides which when treated with aq. NaOH give compounds (Q) and (R). Both (Q) and (R) give yellow precipitate with NaOI. Identify the mixed ether (P):

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📖 Explanation

The key concept is the cleavage of ethers by hot concentrated HI, followed by hydrolysis to alcohols, and the iodoform test. Ether (P) breaks into two different alkyl iodides, which convert to alcohols (Q) and (R). Both (Q) and (R) must give a positive iodoform test, indicating they contain the $CH_3CH(OH)R$ group.

Option A, ethyl sec-butyl ether ( $CH_3CH_2-O-CH(CH_3)CH_2CH_3$ ), yields ethyl iodide and sec-butyl iodide. These hydrolyze to ethanol ( $CH_3CH_2OH$ ) and butan-2-ol ( $CH_3CH(OH)CH_2CH_3$ ). Both ethanol and butan-2-ol possess the necessary $CH_3CH(OH)R$ structure and thus give a yellow precipitate with NaOI. The other options yield at least one alcohol that does not satisfy this condition.

Q32JEE Main 2026MCQ4MCoordination Compounds

Identify the CORRECT set of details from the following: A. $[Co(NH_3)_6]^{3+}$ : Inner orbital compex; $d^{2}sp^{3}$ hybridized B. $[MnCl_6]^{3-}$ : Outer orbital complex; $sp^{3}d^{2}$ hybridized C. $[CoF_6]^{3-}$ : Outer orbital complex; $d^{2}sp^{3}$ hybridized D. $[FeF_6]^{3-}$ : Outer orbital complex; $sp^{3}d^{2}$ hybridized E. $[Ni(CN)_4]^{2-}$ : Inner orbital complex; $sp^{3}d^{2}$ hybridized Choose the correct answer from the options given below:

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📖 Explanation

We need to identify which statements about coordination compounds are correct. Statement A: $[Co(NH_3)_6]^{3+}$ - Inner orbital complex, $d^2sp^3$ hybridized. Co \in $[Co(NH_3)_6]^{3+}$ : Co is $\text{Co}^{3+}$ with configuration $3d^6$ . $NH_3$ is a strong field ligand, causing electron pairing. All 6 electrons pair up \in three $3d$ orbitals, leaving two $3d$ orbitals empty. Hybridization: $d^2sp^3$ (using two inner 3d, one 4s, and three 4p orbitals). This is an inner orbital complex. Correct. Statement B: $[MnCl_6]^{3-}$ - Outer orbital complex, $sp^3d^2$ hybridized. Mn \in $[MnCl_6]^{3-}$ : Mn is $\text{Mn}^{3+}$ with configuration $3d^4$ . $Cl^-$ is a weak field ligand, so no pairing occurs. With 4 unpaired electrons \in the 3d orbitals, inner d orbitals are not available. Hybridization: $sp^3d^2$ (using 4s, 4p, and two 4d outer orbitals). This is an outer orbital complex. Correct. Statement C: $[CoF_6]^{3-}$ - Outer orbital complex, $d^2sp^3$ hybridized. Co \in $[CoF_6]^{3-}$ : $\text{Co}^{3+}$ with $3d^6$ . $F^-$ is a weak field ligand, so electrons remain unpaired (4 unpaired electrons). Since inner d-orbitals are occupied, hybridization should be $sp^3d^2$ (outer orbital), not $d^2sp^3$ . Incorrect - the hybridization is wrong. Statement D: $[FeF_6]^{3-}$ - Outer orbital complex, $sp^3d^2$ hybridized. Fe \in $[FeF_6]^{3-}$ : $\text{Fe}^{3+}$ with $3d^5$ . $F^-$ is a weak field ligand - all five 3d electrons remain unpaired. No inner d orbitals available. Hybridization: $sp^3d^2$ (outer orbital). Correct. Statement E: $[Ni(CN)_4]^{2-}$ - Inner orbital complex, $sp^3d^2$ hybridized. Ni \in $[Ni(CN)_4]^{2-}$ : $\text{Ni}^{2+}$ with $3d^8$ . $CN^-$ is a strong field ligand causing pairing: all 8 electrons pair \in four 3d orbitals, leaving one 3d orbital empty. For 4 ligands, hybridization is $dsp^2$ (square planar), not $sp^3d^2$ . Also $sp^3d^2$ implies 6 coordination, but there are only 4 ligands. Incorrect. Conclusion: Statements A, B, and D are correct. Hence the correct answer is Option 4: A, B & D Only .

Q33JEE Main 2026MCQ4MAmines

A student has been given a compound "x" of molecular formula- $C_{6}H_{7}N$ . 'x' is sparingly soluble \in water. However, on addition of dilute mineral acid, 'x' becomes soluble \in water. 'x' when treated with $CHCl_{3}$ and KOH(alc), 'Y' is produced. 'y' has a specific unpleasant smell. On treatment with benzenesulphonyl chloride, 'x' gives a compound 'z' which is soluble in alkali. The number of different "H" atoms present in 'z' is :-

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📖 Explanation

The compound 'x' is a primary amine because it has the formula $C_6H_7N$ , is sparingly soluble in water but soluble in acid, gives the carbylamine test (unpleasant smell 'Y'), and reacts with benzenesulphonyl chloride to form 'z' which is soluble in alkali (indicating an acidic N-H proton). Thus, 'x' is aniline ( $C_6H_5NH_2$ ). When aniline reacts with benzenesulphonyl chloride, 'z' is N-phenylbenzenesulphonamide ( $C_6H_5SO_2NHC_6H_5$ ). This molecule has one unique hydrogen on the nitrogen atom. Each of the two phenyl rings is monosubstituted, giving three different types of hydrogens on each ring (ortho, meta, para), which are distinct due to their different chemical environments. Therefore, 'z' has 1 (N-H) + 3 (from one phenyl ring) + 3 (from the other phenyl ring) = 7 different types of hydrogen atoms.

The final answer is $\boxed{D}$

Q34JEE Main 2026MCQ4MChemical Kinetics

Given above is the concentration vs time plot for a dissociation reaction : $A \rightarrow nB$ . Based on the data of the initial phase of the reaction (initial 10 min), the value of n is________.

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📖 Explanation

The key concept here is the stoichiometric relationship between the change in concentration of reactant and product. For the reaction $A \rightarrow nB$ , the amount of A consumed is related to the amount of B produced by the coefficient 'n'.

From the graph:

Initial concentration of A at $t=0$ is $[A]_0 = 0.15$ M.

Q35JEE Main 2026MCQ4MPractical Organic Chemistry

Iodoform test can differentiate between A. Methanol and Ethanol B. $CH_{3}COOH$ and $CH_{3}CH_{2}COOH$ C. Cyclohexene and cyclohexanone D. Diethyl ether and Pentan-3-one E. Anisole and acetone Choose the correct answer from the options given below:

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📖 Explanation

The iodoform test is used to detect the presence of a methyl group attached to a carbonyl group ( $CH_3CO-$ ) or a secondary alcohol group ( $CH_3CH(OH)-$ ), as well as ethanol and acetaldehyde. We need to identify which pairs can be differentiated by this test. Key Principle: The iodoform test is positive for compounds containing the $CH_3CO-$ group (or structures that can be oxidized to it). Specifically, it is positive for: - Acetaldehyde ( $CH_3CHO$ ) - Methyl ketones ( $CH_3COR$ ) - Ethanol ( $CH_3CH_2OH$ , oxidized to $CH_3CHO$ ) - Secondary alcohols of the form $CH_3CH(OH)R$ Pair A: Methanol and Ethanol - Methanol ( $CH_3OH$ ): Does NOT give iodoform test (no $CH_3CO$ or $CH_3CHOH$ group) - Ethanol ( $CH_3CH_2OH$ ): Gives POSITIVE iodoform test (oxidized to acetaldehyde) Can be differentiated. Pair B: $CH_3COOH$ and $CH_3CH_2COOH$ - Both are carboxylic acids. Carboxylic acids do not give the iodoform test (the $CH_3CO-$ \in acetic acid is $CH_3COOH$ , not a ketone or aldehyde). Both are negative. Cannot be differentiated. Pair C: Cyclohexene and Cyclohexanone - Cyclohexene: No $CH_3CO$ group. Negative. - Cyclohexanone: It is a cyclic ketone but does NOT have a $CH_3CO-$ group (the carbonyl is flanked by $CH_2$ groups). Negative. Cannot be differentiated. Pair D: Diethyl ether and Pentan-3-one - Diethyl ether ( $C_2H_5OC_2H_5$ ): Negative. - Pentan-3-one ( $CH_3CH_2COCH_2CH_3$ ): No $CH_3CO$ group (the carbonyl is flanked by $CH_2$ groups). Negative. Cannot be differentiated. Pair E: Anisole and Acetone - Anisole ( $C_6H_5OCH_3$ ): Negative. - Acetone ( $CH_3COCH_3$ ): Has $CH_3CO$ group. Positive. Can be differentiated. The pairs that can be differentiated are A and E. The correct answer is Option 3 : A & E Only.

Q36JEE Main 2026MCQ4MStructure of Atom

The work functions of two metals ( $M_{A}$ and $M_{B}$ ) are \in the 1 : 2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of $M_{A}$ : $M_{B}$ is \in the ratio of 2.642 : 1. The work functions (\in eV) of $M_{A}$ and $M_{A}$ are respectively.

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📖 Explanation

The photoelectric effect equation states that the kinetic energy $K$ of an emitted electron is given by $K = h\nu - \phi$ , where $h\nu$ is the photon energy and $\phi$ is the work function of the metal. Given that the work functions of metals $M_A$ and $M_B$ are \in the ratio 1:2, let $\phi_A$ be the work function of $M_A$ and $\phi_B$ be the work function of $M_B$ . Therefore, $\phi_B = 2\phi_A$ . The photon energy is 6 eV for both metals. Thus: For $M_A$ : $K_A = 6 - \phi_A$ For $M_B$ : $K_B = 6 - \phi_B = 6 - 2\phi_A$ The ratio of kinetic energies is given as $K_A : K_B = 2.642 : 1$ , so: $\frac{K_A}{K_B} = \frac{6 - \phi_A}{6 - 2\phi_A} = 2.642$ Solving for $\phi_A$ : $6 - \phi_A = 2.642 \times (6 - 2\phi_A)$ $6 - \phi_A = 2.642 \times 6 - 2.642 \times 2\phi_A$ $6 - \phi_A = 15.852 - 5.284\phi_A$ Bringing like terms together: $-\phi_A + 5.284\phi_A = 15.852 - 6$ $4.284\phi_A = 9.852$ $\phi_A = \frac{9.852}{4.284} \approx 2.3 \text{ eV}$ Then, $\phi_B = 2 \times \phi_A = 2 \times 2.3 = 4.6 \text{ eV}$ . Verifying the kinetic energies: $K_A = 6 - 2.3 = 3.7 \text{ eV}$ $K_B = 6 - 4.6 = 1.4 \text{ eV}$ Ratio: $\frac{K_A}{K_B} = \frac{3.7}{1.4} = 2.642857... \approx 2.642$ This matches the given ratio. Thus, the work functions of $M_A$ and $M_B$ are 2.3 eV and 4.6 eV, respectively. Comparing with the options: A. 3.1, 6.2 B. 1.4, 2.8 C. 1.5, 3.0 D. 2.3, 4.6 The correct answer is option D.

Q37JEE Main 2026MCQ4MStructure of Atom

Identify the INCORRECT statements from the following: A. Notation $_{12}^{24}Mg$ represents 24 protons and 12 neutrons. B. Wavelength of a radiation of frequency $4.5\times10^{15}s^{-1}$ is $6.7\times10^{-8}$ m. C. One radiation has wavelength = $\lambda_1(900nm)$ and energy= $E_{1}$ . Other radiation has wavelength = $\lambda_2(300nm)$ and energy= $E2 \cdot E1 : E_2= 3 : 1$ . D. Number of photons of light of wavelength 2000 pm that provides 1 J of energy is $1.006 x 10^{16}$ . Choose the correct answer from the options given below:

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📖 Explanation

We need to identify the INCORRECT statements from the given options about atomic structure. Notation $_{12}^{24}Mg$ represents 24 protons and 12 neutrons. This is INCORRECT. In the notation $_Z^A X$ , Z (subscript) = atomic number = number of protons = 12, and A (superscript) = mass number = 24. Number of neutrons = A - Z = 24 - 12 = 12. So it represents 12 protons and 12 neutrons, not 24 protons and 12 neutrons. ✗ Wavelength of radiation of frequency $4.5 \times 10^{15} \, \text{s}^{-1}$ is $6.7 \times 10^{-8}$ m. $\lambda = c/\nu = 3 \times 10^8 / 4.5 \times 10^{15} = 6.67 \times 10^{-8}$ m. This is CORRECT. ✓ $\lambda_1 = 900$ nm, $\lambda_2 = 300$ nm. $E_1 : E_2 = 3 : 1$ . Since $E = hc/\lambda$ , we get $E_1/E_2 = \lambda_2/\lambda_1 = 300/900 = 1/3$ , so $E_1 : E_2 = 1 : 3$ , not $3 : 1$ . This is INCORRECT. ✗ Number of photons of wavelength 2000 pm providing 1 J of energy is $1.006 \times 10^{16}$ . $E_{\text{photon}} = hc/\lambda = 6.63 \times 10^{-34} \times 3 \times 10^8 / (2000 \times 10^{-12}) = 9.945 \times 10^{-17}$ J. $n = 1 / 9.945 \times 10^{-17} = 1.006 \times 10^{16}$ . This is CORRECT. ✓ Incorrect statements are A and C. The correct answer is Option (2): A and C Only.

Q38JEE Main 2026MCQ4MOrganic Chemistry - Some Basic Principles

Given below are two statements: Statement I: $(CH_3)_3C^+$ is more stable than $CH_3^+$ as nine hyperconjugation interactions are possible \in $(CH_3)_3C^+$ . Statement II: $CH_3^+$ is less stable than $(CH_3)_3C^+$ as only three hyperconjugation interactions are possible \in $CH_3^+$ . In the light of the above statements, choose the correct answer from the options given below

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📖 Explanation

Hyperconjugation is a stabilizing effect in carbocations where electrons from the C-H bond adjacent to the carbocation carbon (alpha hydrogens) are donated into the empty p-orbital of the carbocation. The number of hyperconjugation interactions depends on the number of alpha hydrogens. Consider the methyl carbocation, $CH_3^+$ . The carbocation carbon has no alkyl groups attached, so there are no \alpha carbons and hence no \alpha hydrogens. Therefore, the number of hyperconjugation interactions \in $CH_3^+$ is zero. Now, consider the tert-butyl carbocation, $(CH_3)_3C^+$ . The carbocation carbon is attached to three methyl groups. Each methyl group has three hydrogens, so the total number of \alpha hydrogens is 3 \times 3 = 9. Therefore, nine hyperconjugation interactions are possible. Statement I claims that $(CH_3)_3C^+$ is more stable than $CH_3^+$ because it has nine hyperconjugation interactions. This is correct because more hyperconjugation interactions lead to greater stability. Thus, Statement I is true. Statement II claims that $CH_3^+$ is less stable than $(CH_3)_3C^+$ because it has only three hyperconjugation interactions. However, as established, $CH_3^+$ has zero hyperconjugation interactions, not three. Therefore, Statement II is false. In summary, Statement I is true and Statement II is false. The correct option is C: Statement I is true but Statement II is false.

Q39JEE Main 2026MCQ4MElectrochemistry

Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming $M^{+}(M\rightarrow M^{+}+e^{-})$ . The cation $M^{+}$ is present \in two different concentrations $c_{1}$ and $c_{2}$ as shown above. which of the following statement is correct for generating a positive cell potential?

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📖 Explanation

This is a concentration cell, where the standard cell potential $E^\circ_{cell}$ is zero. For a positive cell potential ( $E_{cell} > 0$ ), the Nernst equation simplifies to $E_{cell} = \frac{RT}{nF} \ln \frac{[M^+]_{cathode}}{[M^+]_{anode}}$ . To achieve $E_{cell} > 0$ , the ratio $\frac{[M^+]_{cathode}}{[M^+]_{anode}}$ must be greater than 1. This means the concentration of $M^+$ ions at the cathode must be higher than at the anode ([M^+]{cathode} > $[M^+]{anode} $). If$ c_1 $is at the cathode, then$ c_2 $is at the anode, so$ c_1 > c_2$ is required, making option B correct.

Q40JEE Main 2026MCQ4MAmines

In the light of the above statements, choose the correct answer from the options given below

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📖 Explanation

Key Concept:** This question evaluates knowledge of organic reaction mechanisms, specifically oxidation of alkylbenzenes, amide formation, Hofmann bromamide degradation, electrophilic aromatic substitution (bromination), diazotization, and reduction of diazonium salts.

Statement I is True:

1,2-dipropylbenzene undergoes oxidation with $\text{KMnO}_4$ to form phthal

Q41JEE Main 2026MCQ4MPractical Organic Chemistry

In Carius method 0.2425 g of an organic compound gave 0.5253 g silver chloride. The percentage of chlorine in the organic compound is

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📖 Explanation

The Carius method involves converting the chlorine in an organic compound to silver chloride (AgCl). The mass of chlorine in the AgCl precipitate comes from the organic compound. Given: - Mass of organic compound = 0.2425 g - Mass of AgCl obtained = 0.5253 g The molar mass of AgCl is calculated using atomic masses: Ag = 108 g/mol, Cl = 35.5 g/mol. Thus, molar mass of AgCl = 108 + 35.5 = 143.5 g/mol. In 143.5 g of AgCl, the mass of chlorine is 35.5 g. Therefore, the mass of chlorine in 0.5253 g of AgCl is found using proportion: Mass of chlorine = $\left( \frac{35.5}{143.5} \right) \times 0.5253$ First, compute the ratio: $\frac{35.5}{143.5} = \frac{355}{1435} = \frac{71}{287} \approx 0.2473867595818815$ Then, multiply by the mass of AgCl: $0.2473867595818815 \times 0.5253 = 0.1299522648083624 \text{ g}$ This mass of chlorine comes from 0.2425 g of the organic compound. The percentage of chlorine is: Percentage of chlorine = $\left( \frac{\text{mass of chlorine}}{\text{mass of compound}} \right) \times 100 = \left( \frac{0.1299522648083624}{0.2425} \right) \times 100$ Compute the division: $\frac{0.1299522648083624}{0.2425} = 0.5358762886597938$ Multiply by 100: $0.5358762886597938 \times 100 = 53.58762886597938\%$ Rounding to two decimal places gives 53.59%. However, comparing with the options, 53.58% is the closest match. The options are: - A. 37.57% - B. 53.58% - C. 87.65% - D. 34.79% Thus, the correct answer is option B, 53.58%.

Q42JEE Main 2026MCQ4MChemical Bonding and Molecular Structure

Which statements are NOT TRUE about $XeO_2 F_2$ ? A. It has a see-saw shape. B. Xe has 5 electron pairs \in its valence sheU \in XeO 2 F 2. C. The $O - Xe- O$ bond angle is close to $180^{o}$ . D. The $F- Xe -F$ bond angle is close to $180^{o}$ . E. $Xe$ has 16 valence electrons \in $XeO_2 F_2$ . Choose the correct answer from the options given below:

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📖 Explanation

To determine which statements are not true about $XeO_2F_2$ , we analyze its molecular structure using VSEPR theory and Lewis structure concepts. First, recall that xenon (Xe) has 8 valence electrons. In $XeO_2F_2$ , the central Xe atom is bonded to two oxygen (O) atoms and two fluorine (F) atoms. The Lewis structure shows: Two Xe-F single bonds (each bond is a single pair of electrons). Two Xe-O double bonds (each double bond consists of two pairs of electrons). One lone pair on Xe (since Xe contributes 6 electrons to bonding and has 2 electrons left). Total valence electrons \in $XeO_2F_2$ : Xe contributes 8, each O contributes 6 (total 12 for two O), each F contributes 7 (total 14 for two F), summing to $8 + 12 + 14 = 34$ electrons. After accounting for bonding and lone pairs on terminal atoms, Xe has one lone pair. Now, evaluate each statement: Statement A: It has a see-saw shape. Using VSEPR theory, the steric number (SN) is the number of electron domains around Xe. Each bond (single or double) counts as one domain, and the lone pair counts as one domain. Thus, SN = 4 bonding domains (two from Xe-F and two from Xe-O) + 1 lone pair domain = 5. The electron geometry is trigonal bipyramidal. With one lone pair \in an equatorial position, the molecular shape is see-saw. Thus, this statement is TRUE. Statement B: Xe has 5 electron pairs \in its valence shell \in $XeO_2F_2$ . "Electron pairs" refer to both bonding pairs and lone pairs. In the Lewis structure: - Lone pair on Xe: 1 pair. - Bonding pairs: Two Xe-F single bonds contribute 2 bonding pairs, and two Xe-O double bonds contribute 4 bonding pairs (since each double bond has two pairs). Total = $1 + 2 + 4 = 7$ electron pairs. The statement says 5, which is incorrect. Thus, this statement is NOT TRUE. Statement C: The $O - Xe - O$ bond angle is close to $180^\circ$ . In the see-saw shape (trigonal bipyramidal with one lone pair), the lone pair occupies an equatorial position. The two O atoms are typically \in equatorial positions (due to bond length and electronegativity considerations). The ideal equatorial bond angle is $120^\circ$ , but the lone pair repulsion reduces it to less than $120^\circ$ . It is not close to $180^\circ$ . Thus, this statement is NOT TRUE. Statement D: The $F - Xe - F$ bond angle is close to $180^\circ$ . In the trigonal bipyramidal arrangement, the two F atoms are \in axial positions. The axial bond angle is $180^\circ$ . Thus, this statement is TRUE. Statement E: Xe has 16 valence electrons \in $XeO_2F_2$ . The number of valence electrons assigned to Xe \in the Lewis structure is calculated as: - Lone pair electrons: 2 electrons. - Half the bonding electrons: Each Xe-F bond (2 electrons) contributes 1 electron to Xe, so two bonds give 2 electrons. Each Xe-O double bond (4 electrons) contributes 2 electrons to Xe, so two double bonds give 4 electrons. Total = $2 + 2 + 4 = 8$ valence electrons. The statement says 16, which is incorrect. Thus, this statement is NOT TRUE. The NOT TRUE statements are B, C, and E. The correct option is A. B, C and E Only.

Q43JEE Main 2026MCQ4MChemical Kinetics

Observe the following reactions at T(K). I. $A\rightarrow$ products II. $5Br^{-}(aq)+BrO_{3}\text{ } ^{-}(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$ Both the reactions are started at 10.00 am. The rates of these reactions at 10.10 am are same. The value of $-\frac{\triangle\$[Br^{-}]\$}{\triangle t}$ at 10.10am. is $2\times 10^{-4} mol \text{ }L^{-1}min^{-1}$ . The concentration of A at 10.10am is $10^{-1}mol \text{ } L^{-1}$ . What is the first order rate constant (in min^{-1}) of reaction I?

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📖 Explanation

At 10:10 am, the rates of reactions I and II are equal. Reaction I: $A \rightarrow$ products (first order) Reaction II: $5Br^-(aq) + BrO_3^-(aq) \rightarrow 3Br_2(aq) + 3H_2O(l)$ At 10:10 am, $-\frac{\Delta\$[Br^-]\$}{\Delta t} = 2 \times 10^{-4}$ mol L $^{-1}$ min $^{-1}$ and $[A] = 10^{-1}$ mol L $^{-1}$ . We first note that the rate of disappearance of $A$ \in reaction I equals the rate of disappearance of $Br^-$ \in reaction II: $-\frac{d[A]}{dt} = -\frac{d\$[Br^-]\$}{dt} = 2 \times 10^{-4} \text{ mol L}^{-1}\text{min}^{-1}$ To find the rate constant $k$ , we apply the first-order rate law: For a first-order reaction: $-\frac{d[A]}{dt} = k[A]$ $k \times 10^{-1} = 2 \times 10^{-4}$ $k = \frac{2 \times 10^{-4}}{10^{-1}} = 2 \times 10^{-3} \text{ min}^{-1}$ The correct answer is Option B: $2 \times 10^{-3}$ min $^{-1}$ .

Q44JEE Main 2026MCQ4MPractical Organic Chemistry

Which of the following statements are TRUE about Haloform reaction?: A. Sodium hypochlorite reacts with KI to give KOI. B. KOI is a reducing agent. C. $\alpha,\beta$ -unsaturated methylketone $\mathrm{CH_3 - CH = CH - C(=O) - CH_3}$ will give iodoform reaction. D. Isopropyl alcohol will not give iodoform test. E. Methanoic acid will give positive iodoform test. Choose the correct answer from the options given below:

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📖 Explanation

The haloform reaction involves compounds with a methyl ketone group (R-CO-CH3) or compounds that can be oxidized to methyl ketones (like secondary alcohols with the structure R-CH(OH)-CH3), reacting with halogen (I2, Br2, Cl2) in the presence of base to form a haloform (CHX3) and a carboxylate ion. Evaluating each statement: Statement A: "Sodium hypochlorite reacts with KI to give KOI." Sodium hypochlorite (NaOCl) reacts with potassium iodide (KI) in the reaction: $\text{NaOCl} + \text{KI} \rightarrow \text{NaCl} + \text{KOI}$ where KOI is potassium hypoiodite. This reaction occurs because hypochlorite oxidizes iodide to hypoiodite. Thus, statement A is true. Statement B: "KOI is a reducing agent." Potassium hypoiodite (KOI) contains the hypoiodite ion (OI-), which acts as an oxidizing agent \in reactions like the haloform reaction. It oxidizes methyl ketones to carboxylates and is not a reducing agent. Thus, statement B is false. Statement C: "\alpha ,\beta -unsaturated methylketone $\mathrm{CH_3 - CH = CH - C(=O) - CH_3}$ will give iodoform reaction." The compound $\mathrm{CH_3 - CH = CH - C(=O) - CH_3}$ is a methyl ketone because it has the -CO-CH3 group. The \alpha ,\beta -unsaturation does not prevent enolization of the methyl group, so it undergoes the haloform reaction. Thus, statement C is true. Statement D: "Isopropyl alcohol will not give iodoform test." Isopropyl alcohol $\mathrm{(CH_3)_2CHOH}$ is a secondary alcohol that can be oxidized to acetone $\mathrm{(CH_3)_2C=O}$ , a methyl ketone, which gives a positive iodoform test. Therefore, isopropyl alcohol gives a positive test, and statement D is false. Statement E: "Methanoic acid will give positive iodoform test." Methanoic acid (HCOOH) lacks a methyl group or a group oxidizable to a methyl ketone. It does not undergo the haloform reaction and gives a negative test. Thus, statement E is false. True statements are A and C only. The correct option is D.

Q45JEE Main 2026MCQ4MRedox Reactions

The oxidation state of chromium in the final product formed in the reaction between $KI$ and acidified $K_2 Cr_2 O_7$ solution is:

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📖 Explanation

The reaction between $KI$ and acidified $K_2Cr_2O_7$ yields a chromium-containing product whose oxidation state must be determined. In acidic medium, $K_2Cr_2O_7$ acts as a strong oxidizing agent, oxidizing $I^-$ to $I_2$ , while $Cr_2O_7^{2-}$ (with Cr \in the +6 oxidation state) undergoes reduction. The balanced redox equation under these conditions is: $Cr_2O_7^{2-} + 6I^- + 14H^+ \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O$ From this equation, chromium is reduced from +6 \in $Cr_2O_7^{2-}$ to +3 \in $Cr^{3+}$ , which subsequently forms $Cr_2(SO_4)_3$ or $CrCl_3$ depending on the acid employed. Each chromium atom thus gains 3 electrons. Hence, the oxidation state of chromium in the product is Option C) +3 .

Q46JEE Main 2026NAT4MChemical Equilibrium

$X_2(g ) + Y_2(g ) \rightleftharpoons 2Z(g)$ $X_2(g )$ and $Y_2(g )$ are added to a 1 L flask and it is found that the system attains the above equilibrium at T(K) with the number of moles of $X_2(g ),\text{ } Y_2(g )$ and $Z(g)$ being 3, 3 and 9 mol respectively (equilibrium moles). Under this condition of equilibrium, 10 mol of $Z(g$ ) is added to the flask and the temperature is maintained at $T(K)$ . Then the number of moles of $Z(g)$ in the flask when the new equilibrium is established is __ . (Nearest integer)

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📖 Explanation

We have the equilibrium $X_2 + Y_2 \rightleftharpoons 2Z$ \in a 1 L flask with equilibrium moles: $[X_2] = 3$ , $[Y_2] = 3$ , $[Z] = 9$ mol. Then 10 mol of Z is added and we need the new equilibrium moles of Z. To begin, we calculate the equilibrium constant $K_c$ . Since the volume is 1 L, concentrations equal the number of moles: $K_c = \frac{[Z]^2}{[X_2][Y_2]} = \frac{9^2}{3 \times 3} = \frac{81}{9} = 9$ Next, we set up the ICE table after adding 10 mol of Z. After adding 10 mol Z, the initial (non-equilibrium) concentrations are: $[X_2] = 3$ , $[Y_2] = 3$ , $[Z] = 9 + 10 = 19$ mol. The reaction quotient $Q = \frac{19^2}{3 \times 3} = \frac{361}{9} \approx 40.1 > K_c = 9$ . Since $Q > K_c$ , the reaction will shift backward (to the left), consuming Z and producing $X_2$ and $Y_2$ . $X_2$ $Y_2$ $Z$ Initial: 3 3 19 Change: $+x$ $+x$ $-2x$ Equilibrium: $3+x$ $3+x$ $19-2x$ We then apply the equilibrium condition. $K_c = \frac{(19 - 2x)^2}{(3 + x)(3 + x)} = \frac{(19 - 2x)^2}{(3 + x)^2} = 9$ We solve the resulting equation. Taking the square root of both sides (both numerator and denominator are positive): $\frac{19 - 2x}{3 + x} = 3$ $19 - 2x = 3(3 + x) = 9 + 3x$ $19 - 9 = 3x + 2x$ $10 = 5x$ $x = 2$ Finally, we find the new equilibrium moles of Z. $[Z] = 19 - 2x = 19 - 4 = 15$ The answer is 15.

Q47JEE Main 2026NAT4MAldehydes, Ketones and Carboxylic Acids

Consider the following reaction of benzene. In compound (Q), the percentage of oxygen is ___ %. (Nearest integer)

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Q48JEE Main 2026NAT4MRedox Reactions

200 cc of $x\times 10^{-3} M$ potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution \in acidic mediun. Here $x$ =___________

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📖 Explanation

This problem involves a redox titration where the number of equivalents of oxidant equals the number of equivalents of reductant. For dichromate ( $Cr_2O_7^{2-}$ ), the n-factor (change in oxidation state per mole) is 6 (from $Cr^{6+}$ to $Cr^{3+}$ ), and for ferrous ion ( $Fe^{2+}$ ) in Mohr's salt, the n-factor is 1 (from $Fe^{2+}$ to $Fe^{3+}$ ). Applying the equivalence principle, $M_1V_1n_1 = M_2V_2n_2$ : $(x \times 10^{-3} M) \times (200 \text{ cc}) \times 6 = (0.6 M) \times (750 \text{ cc}) \times 1$ . Solving this equation gives $1.2x = 450$ , therefore $x = 375$ .

Q49JEE Main 2026NAT4MSolutions

Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is farther added to this solution, vapour pressure of the solution increases by 20 mm Hg. Vapour pressure (in mm Hg) of B in pure state is ____ . (Nearest integer)

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📖 Explanation

To begin, we apply Raoult's law to find the vapour pressure of pure B. For an ideal solution, the total vapour pressure is $P_{\text{total}} = x_A P_A^* + x_B P_B^*$ where $x_A, x_B$ are mole fractions and $P_A^*, P_B^*$ are the vapour pressures of the pure components. In the first mixture, there are 3 mol of A and 1 mol of B, giving a total of 4 mol. Thus $x_A = 3/4$ and $x_B = 1/4$ . Substituting into Raoult's law gives $\frac{3}{4}P_A^* + \frac{1}{4}P_B^* = 500 \quad \cdots (1)$ After adding 1 mol of A, the amounts become 4 mol of A and 1 mol of B (total 5 mol), and the vapour pressure rises to 520 mm Hg. The new mole fractions are $x_A = 4/5$ and $x_B = 1/5$ . Substituting again yields $\frac{4}{5}P_A^* + \frac{1}{5}P_B^* = 520 \quad \cdots (2)$ We now solve these equations simultaneously. Rewriting equation (1) gives $3P_A^* + P_B^* = 2000 \quad \cdots (1')$ and equation (2) becomes $4P_A^* + P_B^* = 2600 \quad \cdots (2')$ Subtracting (1') from (2') yields $P_A^* = 2600 - 2000 = 600 \text{ mm Hg}$ Substituting this value back into (1') gives $3(600) + P_B^* = 2000$ $1800 + P_B^* = 2000$ $P_B^* = 200 \text{ mm Hg}$ The answer is 200 mm Hg.

Q50JEE Main 2026NAT4MCoordination Compounds

Total number of unpaired electrons present in the central met al atoms/ions of $[Ni(CO)_4],[NiCl_4]^{2-},[PtCl_2(NH_3)_2],[Ni(CN)_4]^{2-}$ and $[Pt(CN)_4]^{2-}$ is_____

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📖 Explanation

To find the total number of unpaired electrons in the central metal atoms/ions of the given complexes, we analyze each complex individually by determining the oxidation state of the central metal, its electron configuration, and the geometry and ligand field strength to assess the number of unpaired electrons. Complex 1: [Ni(CO)4] - Oxidation state of Ni: CO is neutral, so Ni has oxidation state 0. - Ni(0) has electron configuration [Ar] 3d^8 4s^2. In coordination, CO is a strong field ligand, and the complex is tetrahedral. - Strong field ligands cause pairing of electrons. The complex is diamagnetic, so unpaired electrons = 0. Complex 2: [NiCl_4]^{2-} - Oxidation state of Ni: Let oxidation state be x. Charge from Cl is -1 each, so x + 4(-1) = -2 → x = +2. - Ni^{2+} has electron configuration [Ar] 3d^8. - Cl^- is a weak field ligand, and the complex is tetrahedral. In a tetrahedral field with weak field ligands, the complex is high-spin. - For d^8 in tetrahedral splitting: the t2 orbitals (lower energy, 3 orbitals) hold 6 electrons (paired), and the remaining 2 electrons occupy the e orbitals (higher energy, 2 orbitals) with one electron each, resulting in 2 unpaired electrons. Complex 3: [PtCl_2(NH_3)2] - Oxidation state of Pt: Let oxidation state be x. Cl has -1 charge, NH_3 is neutral, so x + 2(-1) + 2(0) = 0 → x = +2. - Pt^{2+} has electron configuration [Xe] 4f^{14} 5d^8. - The complex is square planar (common for Pt^{2+}). Both Cl^- and NH_3 are ligands that lead to strong field splitting in square planar geometry. - For d^8 in square planar field: the d{xy}, d{xz}, d_{yz} orbitals hold 6 electrons (paired), and the d_{z^2} orbital holds the remaining 2 electrons (paired), with d_{x^{2-}y^2} empty. Thus, unpaired electrons = 0. Complex 4: [Ni(CN)_4]^{2-} - Oxidation state of Ni: Let oxidation state be x. CN has -1 charge, so x + 4(-1) = -2 → x = +2. - Ni^{2+} has electron configuration [Ar] 3d^8. - CN^- is a strong field ligand, and the complex is square planar. - For d^8 in square planar field with strong field ligands: electrons pair completely (similar to Pt^{2+} case), so unpaired electrons = 0. Complex 5: [Pt(CN)_4]^{2-} - Oxidation state of Pt: Let oxidation state be x. CN has -1 charge, so x + 4(-1) = -2 → x = +2. - Pt^{2+} has electron configuration [Xe] 4f^{14} 5d^8. - CN^- is a strong field ligand, and the complex is square planar. - For d^8 in square planar field: electrons pair completely, so unpaired electrons = 0. Summary of unpaired electrons: - [Ni(CO)_4]: 0 - [NiCl_4]^{2-}: 2 - [PtCl_2(NH_3)_2]: 0 - [Ni(CN)_4]^{2-}: 0 - [Pt(CN)_4]^{2-}: 0 Total unpaired electrons = 0 + 2 + 0 + 0 + 0 = 2. Final Answer: 2

Mathematics25 questions

Q51JEE Main 2026MCQ4MLogarithms

The sum of all the real solutions of the equation $\log_{(x+3)}{(6x^{2}+28x+30)}=5-2\log_{(6x+10)}{(x^{2}+6x+9)}$ is equal to

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📖 Explanation

Given: $\log_{(x+3)}{(6x^{2}+28x+30)}=5-2\log_{(6x+10)}{(x^{2}+6x+9)}$ $6x^2 + 28x + 30 = (x +3)(6x + 10)$ $x^{2}+6x+9 = (x + 3)^2$ $\Rightarrow$ $\log_{(x+3)}{(x+3)(6x+10)} = 5 - 2\log_{(6x+10)}{(x + 3)^2}$ $\Rightarrow$ $\log_{(x+3)}{(x+3)} + \log_{(x+3)}{(6x+10)} = 5 - 4\log_{(6x+10)}{(x + 3)}$ $\Rightarrow$ $1 + \log_{(x+3)}{(6x+10)} = 5 - 4\log_{(6x+10)}{(x + 3)}$ $\Rightarrow$ $\log_{(x+3)}{(6x+10)} + 4\log_{(6x+10)}{(x + 3)} = 4$ Let $\log_{(x+3)}{(6x+10)} = t$ $\Rightarrow$ $\log_{(6x+10)}{(x + 3)} = \dfrac{1}{t}$ $\Rightarrow$ $t + \dfrac{4}{t} = 4$ $\Rightarrow$ $t^2 - 4t + 4 = 0$ $\Rightarrow$ $(t - 2)^2 = 0$ $\Rightarrow$ $t = 2$ $\therefore$ $\log_{(x+3)}{(6x+10)} = 2$ $\Rightarrow$ $(6x + 10) = (x + 3)^2$ $\Rightarrow$ $x^2 + 6x + 9 = 6x + 10$ $\Rightarrow$ $x^2 = 1$ $\Rightarrow$ $x = \pm 1$ Both of these values of $x$ satisfy the given equation $\therefore$ Sum of real solutions $= (1) + (-1) = 0$ Hence, option B is the correct choice

Q52JEE Main 2026MCQ4MVector Algebra

Let $\overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k}, \overrightarrow{b}=2\widehat{i}+\widehat{j}-\widehat{k}, \overrightarrow{c}=\lambda \widehat{i}+\widehat{j}+\widehat{k}$ and $\overrightarrow{v}= \overrightarrow{a} \times \overrightarrow{b}$ . If $\overrightarrow{v}\cdot\overrightarrow{c}=11$ and the length of the projection of $\overrightarrow{b}$ on $\overrightarrow{c}$ is p, then $9p^{2}$ is equal to

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📖 Explanation

Given: $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ , $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$ , $\vec{c} = \lambda\hat{i} + \hat{j} + \hat{k}$ , and $\vec{v} = \vec{a} \times \vec{b}$ . We need to find $9p^2$ where $\vec{v} \cdot \vec{c} = 11$ and $p$ is the length of projection of $\vec{b}$ on $\vec{c}$ . Computing the cross product $\vec{v} = \vec{a} \times \vec{b}$ , we evaluate the determinant $\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{vmatrix}$ which expands as $= \hat{i}[(-2)(-1) - (3)(1)] - \hat{j}[(1)(-1) - (3)(2)] + \hat{k}[(1)(1) - (-2)(2)]$ then simplifies to $= \hat{i}[2 - 3] - \hat{j}[-1 - 6] + \hat{k}[1 + 4]$ and finally yields $= -\hat{i} + 7\hat{j} + 5\hat{k}.$ Using $\vec{v} \cdot \vec{c} = 11$ we write $\vec{v} \cdot \vec{c} = (-1)(\lambda) + (7)(1) + (5)(1) = -\lambda + 7 + 5 = 12 - \lambda$ so that $12 - \lambda = 11$ , which implies $\lambda = 1$ . Thus $\vec{c} = \hat{i} + \hat{j} + \hat{k}$ . The length of projection of $\vec{b}$ on $\vec{c}$ is given by $p = \frac{|\vec{b} \cdot \vec{c}|}{|\vec{c}|},$ with $\vec{b} \cdot \vec{c} = (2)(1) + (1)(1) + (-1)(1) = 2 + 1 - 1 = 2$ and $|\vec{c}| = \sqrt{1 + 1 + 1} = \sqrt{3},$ hence $p = \frac{|2|}{\sqrt{3}} = \frac{2}{\sqrt{3}}.$ Since $p^2 = \frac{4}{3},$ it follows that $9p^2 = 9 \times \frac{4}{3} = 12.$ The correct answer is Option (1): $\boxed{12}$ .

Q53JEE Main 2026MCQ4MLimits, Continuity and Differentiability

If $

f(x)= \begin{cases}\frac{a|x|+x^{2}-2(\sin|x|)(\cos|x|)}{x} & ,x \neq 0\\b & ,x = 0\end{cases}

$ is continuous at x = 0, then a + b is equal to

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📖 Explanation

For continuity at x = 0: $\lim_{x \to 0^+} \frac{ax + x^2 - 2\sin x\cos x}{x} = \lim_{x \to 0^+} \frac{ax + x^2 - \sin 2x}{x}$ $= a + 0 - 2 = a - 2$ (using $\sin 2x/x \to 2$ ) $\lim_{x \to 0^-} \frac{-ax + x^2 - 2\sin(-x)\cos(-x)}{x} = \frac{-ax + x^2 + \sin 2x}{x} = -a + 0 + 2 = 2 - a$ For continuity: $a - 2 = 2 - a \implies 2a = 4 \implies a = 2$ . Then $b = a - 2 = 0$ . $a + b = 2 + 0 = 2$ The answer is Option 3 : 2.

Q54JEE Main 2026MCQ4MProbability

Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A. Then a ball is randomly drawn from the bag A. If the probability, that the ball drawn is white, is $\dfrac{p}{q},gcd(p,q)=1,$ then $p+q$ is equal to

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📖 Explanation

Bag A contains 9 white and 8 black balls and Bag B contains 6 white and 4 black balls. It is given that one ball is picked from Bag B and mixed with the balls in Bag A. Then one ball is picked from Bag A and we have to find the probability that the ball picked is white. Two cases will be formed. Case 1: White ball is picked from Bag B and mixed with the other balls in Bag A. Probability of picking a white ball from Bag B = $\dfrac{6}{10}$ Now, it is mixed with the balls \in Bag A. Bag A: 10 white and 8 black balls. Probability of picking a white ball from Bag A = $\dfrac{10}{18}$ Probability of picking a white ball from Bag A given that a white ball is picked from Bag B = $\dfrac{6}{10}\times \dfrac{10}{18}=\dfrac{1}{3}$ Case 2: Black ball is picked from Bag B and mixed with the other balls \in Bag A. Probability of picking a black ball from Bag B = $\dfrac{4}{10}$ Now, it is mixed with the balls \in Bag A. Bag A: 9 white and 9 black balls. Probability of picking a white ball from Bag A = $\dfrac{9}{18}$ Probability of picking a white ball from Bag A given that a black ball is picked from Bag B = $\dfrac{4}{10}\times \dfrac{9}{18}=\dfrac{1}{5}$ So, the probability of picking a white ball from Bag A after picking a ball from Bag B = $\dfrac{1}{3}+\dfrac{1}{5}=\dfrac{8}{15}$ It is \in the form of $\dfrac{p}{q}$ where the greatest common divisor of $(p\ ,\ q)=1$ So, $\dfrac{p}{q}=\dfrac{8}{15}$ $p+q=8+15=23$ Hence, the \sum of $p$ and $q$ is 23. $\therefore\$ The required answer is A.

Q55JEE Main 2026MCQ4MComplex Numbers

If $z = \frac{\sqrt{3}}{2}+\frac{i}{2},i=\sqrt{-1},\text{ then }(z^{201}-i)^{8}\text{ is equal to }$

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📖 Explanation

$z = \frac{\sqrt{3}}{2} + \frac{i}{2} = \cos 30° + i\sin 30° = e^{i\pi/6}$ $z^{201} = e^{i \cdot 201\pi/6} = e^{i \cdot 33.5\pi} = e^{i(33\pi + \pi/2)} = e^{i\pi/2} \cdot (e^{i\pi})^{33} = i \cdot (-1)^{33} = -i$ $(z^{201} - i)^8 = (-i - i)^8 = (-2i)^8 = 2^8 \cdot i^8 = 256 \cdot 1 = 256$ The answer is Option 3 : 256.

Q56JEE Main 2026MCQ4MPermutations and Combinations

The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange , is

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📖 Explanation

Firstly, we can give one orange to each child. There are four children. So, the remaining oranges now are 16 - 4 = 12. Now we distribute 12 oranges among 4 children with no restriction. We can directly apply the formula here, i.e. n things can be distributed among r candidates in $^{n+r-1}C_{r-1}$ ways. Here, n = 12 and r = 4 Applying the formula, we get: $^{12+4-1}C_{4-1}=^{15}C_3=\frac{15\times14\times13}{3\times2\times1}$ = 455

Q57JEE Main 2026MCQ4MSets and Relations

Let A = {0 ,1,2,...,9}. Let R be a relation on A defined by (x,y) $\in$ R if and only if $\mid x - y \mid$ is a multiple of 3. Given below are two statements: Statement I: $n (R) = 36.$ Statement II: R is an equivalence relation. In the light of the above statements, choose the correct answer from the options given below

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📖 Explanation

Let $A = \{0, 1, 2, \ldots, 9\}$ . The relation $R$ on $A$ is defined by $(x, y) \in R$ if and only if $|x - y|$ is a multiple of 3. We verify Statement I ( $n(R) = 36$ ) and Statement II (R is an equivalence relation). Since two elements $x, y \in A$ are related precisely when $3 \mid (x - y)$ , they have the same remainder upon division by 3. Hence the equivalence class for remainder 0 is $\{0, 3, 6, 9\}$ , which has 4 elements; the class for remainder 1 is $\{1, 4, 7\}$ with 3 elements; and the class for remainder 2 is $\{2, 5, 8\}$ with 3 elements. Now, to show that $R$ is an equivalence relation, observe that $|x - x| = 0$ is a multiple of 3, so $R$ is reflexive. Moreover, if $|x - y|$ is a multiple of 3 then $|y - x| = |x - y|$ is also a multiple of 3, giving symmetry. Furthermore, if $3 \mid (x - y)$ and $3 \mid (y - z)$ , then $3 \mid \bigl((x - y) + (y - z)\bigr) = 3 \mid (x - z)$ , which establishes transitivity. Therefore, $R$ is an equivalence relation. Next, the total number of ordered pairs \in $R$ is the \sum of $k^2$ over the sizes of the equivalence classes, giving $n(R) = 4^2 + 3^2 + 3^2 = 16 + 9 + 9 = 34.$ Since $34 \neq 36$ , Statement I is incorrect. The correct answer is Option (2): Statement I is incorrect but Statement II is correct .

Q58JEE Main 2026MCQ4MFunctions

Consider two sets $A=\left\{x\in Z:|(|x-3|-3)\leq1\right\}$ and $B=\left\{x \in \mathbb R-\left\{1,2\right\}:\frac{(x-2)(x-4)}{x-1}\log_{e}(|x-2|)=0 \right\}$ . Then the number of onto functions $f:A\rightarrow B$ is equal to

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Q59JEE Main 2026MCQ4MEllipse

If the points of intersection of the ellipses $x^{2}+2y^{2}-6x-12y+23=0$ and $4x^{2}+2y^{2}-20x-12y+35=0$ lie on a circle of radius r and centre (a, b), then the value of $ab+18r^{2}$ is

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📖 Explanation

Ellipse 1: $x^2 + 2y^2 - 6x - 12y + 23 = 0$ Ellipse 2: $4x^2 + 2y^2 - 20x - 12y + 35 = 0$ To find the intersection points, subtract Ellipse 1 from Ellipse 2: $3x^2 - 14x + 12 = 0$ Using quadratic formula: $x = \frac{14 \pm \sqrt{196 - 144}}{6} = \frac{14 \pm \sqrt{52}}{6} = \frac{14 \pm 2\sqrt{13}}{6} = \frac{7 \pm \sqrt{13}}{3}$ To find the circle through the intersection points, we use the family $\lambda(\text{Ellipse 1}) + \mu(\text{Ellipse 2}) = 0$ and choose $\lambda, \mu$ so the coefficients of $x^2$ and $y^2$ are equal (circle condition). Ellipse 1: $x^2 + 2y^2 - 6x - 12y + 23 = 0$ Ellipse 2: $4x^2 + 2y^2 - 20x - 12y + 35 = 0$ Subtract: $3x^2 - 14x + 12 = 0$ . This is the radical axis (a pair of vertical lines). Adding $k$ \times this to Ellipse 1: $(1+3k)x^2 + 2y^2 + (-6-14k)x - 12y + (23+12k) = 0$ For a circle: $1 + 3k = 2$ , so $k = 1/3$ . $2x^2 + 2y^2 + (-6 - 14/3)x - 12y + (23 + 4) = 0$ $2x^2 + 2y^2 - \frac{32}{3}x - 12y + 27 = 0$ $x^2 + y^2 - \frac{16}{3}x - 6y + \frac{27}{2} = 0$ Center: $a = \frac{8}{3}$ , $b = 3$ . $r^2 = \left(\frac{8}{3}\right)^2 + 9 - \frac{27}{2} = \frac{64}{9} + 9 - \frac{27}{2} = \frac{128 + 162 - 243}{18} = \frac{47}{18}$ $ab + 18r^2 = \frac{8}{3} \times 3 + 18 \times \frac{47}{18} = 8 + 47 = 55$ The correct answer is Option C: 55.

Q60JEE Main 2026MCQ4MParabola

An equilateral triangle OAB is inscribed in the parabola $y^{2} = 4x$ with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having AB as a diameter from the origin is

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📖 Explanation

We need to find the minimum distance from the origin to the circle having AB as diameter, where equilateral triangle OAB is inscribed in the parabola $y^2 = 4x$ with O at the vertex. Since the parabola $y^2 = 4x$ has vertex at the origin O(0,0) and parameter $a = 1$ , any point on it can be written as $(t^2, 2t)$ . By symmetry about the x-axis, let $A = (t^2, 2t)$ and $B = (t^2, -2t)$ for some $t > 0$ . Next, since $AB = |2t - (-2t)| = 4t$ and $OA = \sqrt{t^4 + 4t^2} = t\sqrt{t^2 + 4}$ , imposing the equilateral condition $OA = AB$ gives $t\sqrt{t^2 + 4} = 4t$ which simplifies to $\sqrt{t^2 + 4} = 4 \implies t^2 + 4 = 16 \implies t^2 = 12 \implies t = 2\sqrt{3}$ . Using this value yields $A = (12, 4\sqrt{3})$ and $B = (12, -4\sqrt{3})$ . The midpoint of $AB$ , which is the center of the circle having $AB$ as diameter, is $M = (12, 0)$ , and its radius is $\frac{AB}{2} = \frac{4 \times 2\sqrt{3}}{2} = 4\sqrt{3}$ . Since the distance from the origin to the center is $OM = \sqrt{12^2 + 0^2} = 12$ , the minimum distance from the origin to the circle is $OM - r = 12 - 4\sqrt{3} = 4(3 - \sqrt{3})$ . The correct answer is Option 4: $4(3 - \sqrt{3})$ .

Q61JEE Main 2026MCQ4MStraight Lines and Pair of Straight Lines

Let A (1, 2) and C(- 3, -6) be two diagonally opposite vertices of a rhombus, whose sides AD and BC are parallel to the line $7x - y = 14$ . If B ( $\alpha, \beta$ ) and D ( $\gamma, \delta$ ) are the other two vertices, then $|\alpha+ \beta+\gamma+\delta |$ is equal to

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📖 Explanation

The key concept is that the diagonals of a rhombus bisect each other, meaning their midpoints coincide. Let $M$ be the midpoint of the diagonal $AC$ . Using the midpoint formula for $A(1,2)$ and $C(-3,-6)$ , we find $M = (\frac{1+(-3)}{2}, \frac{2+(-6)}{2}) = (-1, -2)$ .

Since $M$ is also the midpoint of the diagonal $BD$ , where $B(\alpha, \beta)$ and $D(\gamma, \delta)$ , we have $\frac{\alpha+\gamma}{2} = -1$ and $\frac{\beta+\delta}{2} = -2$ . This implies $\alpha+\gamma = -2$ and $\beta+\delta = -4$ . Therefore, the sum $\alpha+\beta+\gamma+\delta$ is $(\alpha+\gamma) + (\beta+\delta) = -2 + (-4) = -6$ . Finally, $|\alpha+\beta+\gamma+\delta| = |-6| = 6$ .

Q62JEE Main 2026MCQ4MTrigonometric Ratios and Identities

Let $\frac{\pi}{2} < \theta < \pi$ and $\cot\theta=-\frac{1}{2\sqrt{2}}.$ Then the value of $\sin\left( \frac{150}{2}\right)\left(\cos 80 + \sin 80\right)+\cos\left( \frac{150}{2}\right)\left(\cos 80 - \sin 80\right)$ is equal to

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Q63JEE Main 2026MCQ4MStatistics

If the mean and the variance of the data are $\mu$ and 19 respectively, then the value of $\lambda$ $+\mu$ is

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📖 Explanation

To find $\lambda + \mu$ , we first determine the class marks ( $x_i$ ) and then set up equations for the mean ( $\mu$ ) and variance.
The class marks are 6, 10, 14, 18. The total frequency is $\sum f_i = 3 + \lambda + 4 + 7 = 14 +

Q64JEE Main 2026MCQ4MIndefinite Integrals

Let $I(x)=\int\frac{3dx}{\left(4x+6\right)\left(\sqrt{4x^{2}}+8x+3\right)}$ and $I(0)=\frac{{\sqrt{3}}}{4}+20.$ If $I\left( \frac{1}{2} \right)=\frac{a\sqrt{2}}{b}+c, \text { Where a,b,c } \in N,gcd(a,b)=1, \text{ a+b+c is equal to}$

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📖 Explanation

We need to evaluate $I(x) = \int \frac{3\,dx}{(4x+6)\sqrt{4x^{2+}8x+3}}$ . First, simplify the expression under the square root: $4x^2 + 8x + 3 = 4(x^2 + 2x) + 3 = 4(x+1)^2 - 4 + 3 = 4(x+1)^2 - 1$ Also, $4x + 6 = 2(2x + 3)$ . So the integral becomes: $I(x) = \int \frac{3\,dx}{2(2x+3)\sqrt{4(x+1)^2 - 1}}$ Let us try the substitution $t = \frac{1}{2x+3}$ . Then $2x + 3 = \frac{1}{t}$ , so $x = \frac{1-3t}{2t}$ , and $dx = -\frac{1}{2t^2}dt$ . Now compute $4x^2 + 8x + 3$ : $4x^2 + 8x + 3 = (2x+3)(2x+1)$ We have $2x + 3 = \frac{1}{t}$ , and $2x + 1 = \frac{1}{t} - 2 = \frac{1-2t}{t}$ . So $4x^2 + 8x + 3 = \frac{1}{t} \cdot \frac{1-2t}{t} = \frac{1-2t}{t^2}$ . $\sqrt{4x^2 + 8x + 3} = \frac{\sqrt{1-2t}}{t}$ (taking $t > 0$ for now). Substituting into the integral: $I = \int \frac{3}{2 \cdot \frac{1}{t} \cdot \frac{\sqrt{1-2t}}{t}} \cdot \left(-\frac{1}{2t^2}\right) dt$ $= \int \frac{3t^2}{2\sqrt{1-2t}} \cdot \left(-\frac{1}{2t^2}\right) dt$ $= \int \frac{-3}{4\sqrt{1-2t}} dt$ $= \frac{-3}{4} \cdot \frac{\sqrt{1-2t}}{-1} + C = \frac{3}{4}\sqrt{1-2t} + C$ Substituting back $t = \frac{1}{2x+3}$ : $I(x) = \frac{3}{4}\sqrt{1 - \frac{2}{2x+3}} + C = \frac{3}{4}\sqrt{\frac{2x+3-2}{2x+3}} + C = \frac{3}{4}\sqrt{\frac{2x+1}{2x+3}} + C$ Now use the initial condition $I(0) = \frac{\sqrt{3}}{4} + 20$ : $I(0) = \frac{3}{4}\sqrt{\frac{1}{3}} + C = \frac{3}{4} \cdot \frac{1}{\sqrt{3}} + C = \frac{\sqrt{3}}{4} + C$ So $C = 20$ . Therefore $I(x) = \frac{3}{4}\sqrt{\frac{2x+1}{2x+3}} + 20$ . Now compute $I\left(\frac{1}{2}\right)$ : $I\left(\frac{1}{2}\right) = \frac{3}{4}\sqrt{\frac{2 \cdot \frac{1}{2} + 1}{2 \cdot \frac{1}{2} + 3}} + 20 = \frac{3}{4}\sqrt{\frac{2}{4}} + 20 = \frac{3}{4} \cdot \frac{\sqrt{2}}{2} + 20 = \frac{3\sqrt{2}}{8} + 20$ So $I\left(\frac{1}{2}\right) = \frac{3\sqrt{2}}{8} + 20$ . Comparing with $\frac{a\sqrt{2}}{b} + c$ : we get $a = 3$ , $b = 8$ , $c = 20$ . Check: $\gcd(3, 8) = 1$ ✓, and $a, b, c \in \mathbb{N}$ ✓. $a + b + c = 3 + 8 + 20 = 31$ The answer is Option B: $31$ .

Q65JEE Main 2026MCQ4MArea Under The Curves

The area of the region enclosed between the circles $x^{2}+y^{2}=4 \text{ and } x^{2}+(y-2)^{2}=4$ is

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📖 Explanation

The area of the region enclosed between the circles is found by summing the areas of two identical circular segments. The circles $x^{2+}y^2=4$ and $x^{2+}(y-2)^2=4$ both have a radius $r=2$ . Their centers are $(0,0)$ and $(0,2)$ respectively, with the distance between centers

Q66JEE Main 2026MCQ4MVector Algebra

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be three vectors such that $\overrightarrow{a}\times \overrightarrow{b}=2(\overrightarrow{a}\times \overrightarrow{c}).$ If $\mid \overrightarrow{a}\mid, \mid\overrightarrow{b}\mid = 4, \mid \overrightarrow{c}\mid = 2,$ and the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $60^{o}$ , then $\mid\overrightarrow{a}\cdot\overrightarrow{c}$ is

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📖 Explanation

We are given three vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ such that $\overrightarrow{a} \times \overrightarrow{b} = 2(\overrightarrow{a} \times \overrightarrow{c})$ , with $|\overrightarrow{a}| = |\overrightarrow{b}| = 4$ , $|\overrightarrow{c}| = 2$ , and the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $60°$ . Since $\overrightarrow{a} \times \overrightarrow{b} = 2(\overrightarrow{a} \times \overrightarrow{c})$ , it follows that $\overrightarrow{a} \times \overrightarrow{b} - 2(\overrightarrow{a} \times \overrightarrow{c}) = \overrightarrow{0}$ and hence $\overrightarrow{a} \times (\overrightarrow{b} - 2\overrightarrow{c}) = \overrightarrow{0}$ . This implies that $\overrightarrow{a}$ is parallel to $(\overrightarrow{b} - 2\overrightarrow{c})$ , so one can write $(\overrightarrow{b} - 2\overrightarrow{c}) = \lambda\,\overrightarrow{a}$ for some scalar $\lambda$ . Taking magnitudes squared of both sides gives $|\overrightarrow{b} - 2\overrightarrow{c}|^2 = \lambda^2 |\overrightarrow{a}|^2$ , which expands to $|\overrightarrow{b}|^2 - 4\,\overrightarrow{b}\cdot\overrightarrow{c} + 4|\overrightarrow{c}|^2 = 16\lambda^2$ . Now $\overrightarrow{b}\cdot\overrightarrow{c} = |\overrightarrow{b}|\,|\overrightarrow{c}|\cos 60° = 4 \times 2 \times \tfrac12 = 4$ , so this becomes $16 - 16 + 16 = 16\lambda^2 \implies \lambda^2 = 1 \implies \lambda = \pm 1$ . Next, taking the dot product with $\overrightarrow{c}$ \in $(\overrightarrow{b} - 2\overrightarrow{c}) = \lambda\,\overrightarrow{a}$ yields $\overrightarrow{a}\cdot\overrightarrow{c} = \frac{(\overrightarrow{b} - 2\overrightarrow{c})\cdot\overrightarrow{c}}{\lambda} = \frac{\overrightarrow{b}\cdot\overrightarrow{c} - 2|\overrightarrow{c}|^2}{\lambda} = \frac{4 - 8}{\lambda} = \frac{-4}{\lambda}$ . When $\lambda = 1$ , $\overrightarrow{a}\cdot\overrightarrow{c} = -4$ and thus $|\overrightarrow{a}\cdot\overrightarrow{c}| = 4$ ; similarly, when $\lambda = -1$ , $\overrightarrow{a}\cdot\overrightarrow{c} = 4$ so again $|\overrightarrow{a}\cdot\overrightarrow{c}| = 4$ . The correct answer is Option A: 4 .

Q67JEE Main 2026MCQ4MTrigonometric Ratios and Identities

The least value of $(\cos^{2} \theta- 6\sin \theta \cos \theta + 3\sin^{2} \theta +2)$ is

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📖 Explanation

We need to find the minimum value of $E = \cos^2\theta - 6\sin\theta\cos\theta + 3\sin^2\theta + 2$ . Since $\cos^2\theta = \frac{1+\cos 2\theta}{2}$ , $\sin^2\theta = \frac{1-\cos 2\theta}{2}$ and $\sin\theta\cos\theta = \frac{\sin 2\theta}{2}$ , substituting these into the expression gives $E = \frac{1+\cos 2\theta}{2} - 6 \cdot \frac{\sin 2\theta}{2} + 3 \cdot \frac{1-\cos 2\theta}{2} + 2$ which simplifies to $= \frac{1}{2} + \frac{\cos 2\theta}{2} - 3\sin 2\theta + \frac{3}{2} - \frac{3\cos 2\theta}{2} + 2$ and further to $= 4 - \cos 2\theta - 3\sin 2\theta$ . For an expression of the form $a\cos\phi + b\sin\phi$ , the range is $[-\sqrt{a^{2+}b^2}, \sqrt{a^{2+}b^2}]$ , so here with $a = -1$ and $b = -3$ we have $\sqrt{(-1)^2 + (-3)^2} = \sqrt{1+9} = \sqrt{10}$ and hence $-\cos 2\theta - 3\sin 2\theta$ varies between $-\sqrt{10}$ and $\sqrt{10}$ . Therefore the minimum value of $E$ is $E_{min} = 4 + (-\sqrt{10}) = 4 - \sqrt{10}$ The correct answer is Option 1 : $4 - \sqrt{10}$ .

Q68JEE Main 2026MCQ4MHyperbola

Let PQ be a chord of the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{b^{2}}=1$ , perpendicular to the x-axis such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If the eccentricity of the hyperbola is $\sqrt{3}.$ then the area of the triangle OPQ is

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📖 Explanation

For the hyperbola $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$ with eccentricity $e = \sqrt{1 + \frac{b^2}{4}} = \sqrt{3}$ , we find $b^2 = 4$ . The vertices along the $y$ -axis, where $x = 0$ , give points $P(0, h)$ and $Q(0, -h)$ for some height $h$ . Since OPQ forms an equilateral triangle, the height $h$ relates to the length of side which can be determined using the equilateral triangle area formula $A = \frac{\sqrt{3}}{4} s^2$ , leading to $A = \frac{8\sqrt{3}}{5}$ for the coordinates given. Thus, the area of triangle OPQ is $\frac{8\sqrt{3}}{5}$ .

Q69JEE Main 2026MCQ4MSequences and Series

$\text{Let }\sum_{k=1}^n a_k=\alpha n ^2 +\beta n.$ If $a_{10}=59$ and $a_6 = 7a_1,$ then $\alpha+\beta$ is equal to

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📖 Explanation

$S_n = \alpha n^2 + \beta n$ . $a_n = S_n - S_{n-1} = \alpha(2n-1) + \beta$ . $a_{10} = 19\alpha + \beta = 59$ . $a_1 = \alpha + \beta$ . $a_6 = 11\alpha + \beta = 7(\alpha + \beta) = 7\alpha + 7\beta$ . $4\alpha = 6\beta \implies \alpha = 3\beta/2$ . From $19(3\beta/2) + \beta = 59$ : $57\beta/2 + \beta = 59$ , $59\beta/2 = 59$ , $\beta = 2$ , $\alpha = 3$ . $\alpha + \beta = 5$ . The answer is Option 3 : 5.

Q70JEE Main 2026MCQ4MMatrices and Determinants

The system of linear equations $x + y + z = 6$ $2x + 5y + az =36$ $x + 2y + 3z = b$

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📖 Explanation

We are given the system of linear equations: $x + y + z = 6 \quad \cdots (1)$ $2x + 5y + az = 36 \quad \cdots (2)$ $x + 2y + 3z = b \quad \cdots (3)$ The coefficient matrix is: $A = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{vmatrix}$ Its determinant is $\det(A) = 1(15 - 2a) - 1(6 - a) + 1(4 - 5) = 15 - 2a - 6 + a - 1 = 8 - a.$ Since $\det(A) = 8 - a\,,$ for $a = 8$ we have $\det(A) = 0$ , so the system does not have a unique solution and Options 1 and 3 are eliminated. Substituting $a = 8$ into the augmented matrix gives

\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 2 & 5 & 8 & | & 36 \\ 1 & 2 & 3 & | & b \end{pmatrix}.$$ Applying $$R_2 \to R_2 - 2R_1$$ yields $$(0, 3, 6 \mid 24)$$ and then $$R_3 \to R_3 - R_1$$ yields $$(0, 1, 2 \mid b - 6).$$ Finally, $$R_3 \to R_3 - \tfrac{1}{3}R_2$$ gives $$(0, 0, 0 \mid b - 14).$$ For consistency (infinitely many solutions) we require $$b - 14 = 0\,, $$ so $$b = 14.$$ If instead $$b = 16\,, $$ then $$b - 14 = 2 \neq 0$$ and the system is inconsistent (no solution). Therefore the correct answer is Option 4 : infinitely many solutions for $$a = 8$$ and $$b = 14$$.

Q71JEE Main 2026NAT4MDifferential Equations

If the solution curve $y =f (x)$ of the differential equation $(x^{2}-4)y^{'}-2xy+2x(4-x^{2})^{2}=0,x>2,$ passes through the point (3, 15), then the local maximum value of $f$ is __________

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📖 Explanation

The given differential equation is $(x^{2}-4)\,y' \;-\;2x\,y \;+\;2x\,(4-x^{2})^{2}=0,\qquad x\gt 2$ Rewrite it \in the standard linear form $y'+P(x)\,y=Q(x)$ . Divide the whole equation by $x^{2}-4$ (which is positive for $x\gt 2$ ): $y' \;-\;\frac{2x}{x^{2}-4}\,y \;+\;\frac{2x(4-x^{2})^{2}}{x^{2}-4}=0$ Because $(4-x^{2})^{2}=(x^{2}-4)^{2}$ , the last term simplifies to $2x(x^{2}-4)$ . Thus $y' \;-\;\frac{2x}{x^{2}-4}\,y \;+\;2x(x^{2}-4)=0$ Move the non-homogeneous term to the right: $y' \;-\;\frac{2x}{x^{2}-4}\,y = -\,2x(x^{2}-4)\qquad -(1)$ Here $P(x)=-\dfrac{2x}{x^{2}-4}$ and $Q(x)=-\,2x(x^{2}-4)$ . Integrating factor The integrating factor (I.F.) is $\displaystyle \mu(x)=\exp\!\bigl(\int P(x)\,dx\bigr)$ . $\int P(x)\,dx=\int\!-\frac{2x}{x^{2}-4}\,dx$ Put $u=x^{2}-4\;\Longrightarrow\;du=2x\,dx$ , giving $\int\!-\frac{du}{u}=-\ln|u|=-\ln|x^{2}-4|$ Hence $\mu(x)=e^{-\ln|x^{2}-4|}=(x^{2}-4)^{-1},\qquad x\gt 2$ Using the integrating factor Multiplying equation $(1)$ by $(x^{2}-4)^{-1}$ gives $\frac{1}{x^{2}-4}\,y' -\frac{2x}{(x^{2}-4)^{2}}\,y = -\,2x$ The left side is the derivative of $y\,(x^{2}-4)^{-1}$ , so $\frac{d}{dx}\Bigl\$[\frac{y}{x^{2}-4}\Bigr]\$= -\,2x$ Integrate with respect to $x$ : $\frac{y}{x^{2}-4}= -\int 2x\,dx + C = -x^{2}+C$ Therefore $y=(x^{2}-4)\,(-x^{2}+C)= -x^{4}+4x^{2}+C\,x^{2}-4C$ Simplify: $y=-x^{4}+(4+C)\,x^{2}-4C\qquad -(2)$ Finding the constant using the point (3, 15) Substitute $x=3,\;y=15$ into $(2)$ : $15=-81+(4+C)\,9-4C=-81+36+9C-4C=-45+5C$ $5C=60\;\Longrightarrow\;C=12$ Hence the explicit solution is $y=-x^{4}+16x^{2}-48\qquad -(3)$ Locating critical points Differentiate $(3)$ : $y'=\frac{dy}{dx}=-4x^{3}+32x=-4x\,(x^{2}-8)$ Set $y'=0$ : $-4x\,(x^{2}-8)=0\;\;\Longrightarrow\;\;x=0,\;\;x=\pm\sqrt{8}= \pm2\sqrt{2}$ Because the domain is $x\gt 2$ , the only relevant critical point is $x=2\sqrt{2}\ (\approx2.828).$ Second-derivative test Compute $y''$ : $y''=\frac{d}{dx}(-4x^{3}+32x)=-12x^{2}+32$ At $x=2\sqrt{2}$ , $x^{2}=8$ , hence $y''=-12(8)+32=-96+32=-64\lt 0$ Since $y''\lt 0$ , $x=2\sqrt{2}$ gives a local maximum. Maximum value of $y$ Substitute $x=2\sqrt{2},\;x^{2}=8,\;x^{4}=64$ into $(3)$ : $y_{\text{max}}=-64+16\!\times \!8-48=-64+128-48=16$ Thus, the local maximum value of $f$ is $16$ .

Q72JEE Main 2026NAT4MThree Dimensional Geometry

If the image of the point $P(a, 2, a)$ \in the line $\frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}$ is Q and the image of Q \in the line $\frac{x-2b}{2}=\frac{y-a}{1}=\frac{z+2b}{-5}$ is P, then a + b is equal to _____.

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📖 Explanation

The key concept here involves reflecting a point across two lines in three-dimensional space. The coordinates of point $P(a, 2, a)$ can be expressed in terms of the lines provided to find point $Q$ . The relationships given by the equations allow us to set up systems of equations that ultimately simplify to reveal the values of $a$ and $b$ . Solving these equations, we find that $a + b = 3$ , which is consistent with the reflection properties of points across lines in the specified format.

Q73JEE Main 2026NAT4MPermutations and Combinations

Let S denote the set of 4-digit numbers $abcd$ such that $a > b > c > d$ and P denote the set of 5-digit numbers having product of its digits equal to 20. Then $n(S) + n(P)$ is equal to ______

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📖 Explanation

Calculating $n(S)$ : The number of 4-digit numbers $abcd$ such that $a > b > c > d$ can be calculated by selecting 4 digits out of the digits 0, 1, 2,..., 9 and arranging them \in descending order. The 4-digit numbers obtained \in this manner satisfy all the above conditions and include all possible 4-digit numbers that satisfy the above conditions. There are 10 digits from 0 to 9, and the selection of 4 digits out of them can be done \in $^{10}C_4$ ways. $n(S)=\ ^{10}C_4\ =\ \dfrac{10!}{6!\ \times \ 4!}\ =\ \dfrac{10\times \ 9\times \ 8\times \ 7}{1\times \ 2\times \ 3\times \ 4}\ =\ 210$ Calculating $n(P)$ : We need to find the sets of 5 single-digit numbers whose product equals 20. $20\ =\ 2^2\ \times \ 5\ =\ 5\ \times \ 2\ \times \ 2$ So, the only possible sets of 5 digits with a product equal to 20 are {5, 2, 2, 1, 1} and {5, 4, 1, 1, 1}. The number of 5-digit numbers possible using the 5 digits {5, 2, 2, 1, 1} can be calculated as $\dfrac{5!}{2!\ \times \ 2!}\ =\ 30$ as there are 2 twos and 2 ones. The number of 5-digit numbers possible using the 5 digits {5, 4, 1, 1, 1} can be calculated as $\dfrac{5!}{3!}\ =\ 20$ as there are 3 ones. $n\left(P\right)\ =\ 20\ +\ 30\ =\ 50$ $n\left(S\right)\ +\ n\left(P\right)\ =\ 210\ +\ 50\ =\ 260$ Hence, the correct answer is 260.

Q74JEE Main 2026NAT4MDefinite Integrals

The number of elements in the set $S=\left\{ x:x\in [0,100] \text{ and } \int_{0}^{x} t^{2} \sin(x-t)dt=x^{2}\right\}$ is _________

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📖 Explanation

We need to find the number of elements in $S = \{x : x \in [0, 100] \text{ and } \int_0^x t^2 \sin(x-t)\,dt = x^2\}$ . Using the identity $\sin(x-t) = \sin x \cos t - \cos x \sin t$ , we write the integral as $\int_0^x t^2 \sin(x-t)\,dt = \sin x \int_0^x t^2 \cos t\,dt - \cos x \int_0^x t^2 \sin t\,dt$ . To evaluate $\int_0^x t^2 \cos t\,dt$ we apply integration by parts twice, taking $u = t^2$ and $dv = \cos t\,dt$ to obtain $\int t^2 \cos t\,dt = t^2 \sin t - \int 2t \sin t\,dt$ . In the integral $\int 2t \sin t\,dt$ we set $u = 2t$ and $dv = \sin t\,dt$ , giving $\int 2t \sin t\,dt = -2t\cos t + \int 2\cos t\,dt = -2t\cos t + 2\sin t$ and hence $\int t^2 \cos t\,dt = t^2 \sin t + 2t\cos t - 2\sin t + C$ so that $\int_0^x t^2 \cos t\,dt = x^2 \sin x + 2x\cos x - 2\sin x$ . For $\int_0^x t^2 \sin t\,dt$ we use integration by parts with $u = t^2$ and $dv = \sin t\,dt$ , yielding $\int t^2 \sin t\,dt = -t^2 \cos t + \int 2t\cos t\,dt$ . Since $\int 2t\cos t\,dt = 2t\sin t - \int 2\sin t\,dt = 2t\sin t + 2\cos t$ we have $\int t^2 \sin t\,dt = -t^2\cos t + 2t\sin t + 2\cos t + C$ and thus $\int_0^x t^2 \sin t\,dt = -x^2\cos x + 2x\sin x + 2\cos x - 2$ . Substituting these results into the expression for the integral gives $\text{LHS} = \sin x(x^2\sin x + 2x\cos x - 2\sin x) - \cos x(-x^2\cos x + 2x\sin x + 2\cos x - 2)$ . Expanding and simplifying leads to $= x^2\sin^2 x + 2x\sin x\cos x - 2\sin^2 x + x^2\cos^2 x - 2x\sin x\cos x - 2\cos^2 x + 2\cos x = x^2(\sin^2 x + \cos^2 x) - 2(\sin^2 x + \cos^2 x) + 2\cos x = x^2 - 2 + 2\cos x$ . Setting this equal to $x^2$ gives $2\cos x = 2$ , so $\cos x = 1$ and hence $x = 2k\pi, \quad k = 0, 1, 2, \ldots$ . Requiring $2k\pi \leq 100 \implies k \leq \frac{50}{\pi} \approx 15.915$ , we find $k = 0, 1, 2, \ldots, 15$ , giving 16 elements. The answer is $\mathbf{16}$ .

Q75JEE Main 2026NAT4MMatrices and Determinants

Let $A = \begin{bmatrix}0 & 2 & -3 \\-2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix}$ and B be a matrix such that $B(I- A)=I+A.$ Then the sumof the diagonal elements of $B^{T}B$ is equal to _________

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📖 Explanation

We need to find the sum of diagonal elements of $B^TB$ . Given $A = \begin{bmatrix}0&2&-3\\-2&0&1\\3&-1&0\end{bmatrix}$ and $B(I-A) = I+A$ . Since $A^T = \begin{bmatrix}0&-2&3\\2&0&-1\\-3&1&0\end{bmatrix} = -A$ , it follows that A is skew-symmetric ( $A^T = -A$ ). Substituting this result into $B(I-A) = I+A$ gives $B = (I+A)(I-A)^{-1}$ . For a skew-symmetric matrix A, one computes $B^T = ((I-A)^{-1})^T(I+A)^T = ((I-A)^T)^{-1}(I+A^T) = (I-A^T)^{-1}(I-A) = (I+A)^{-1}(I-A)$ and therefore $B^TB = (I+A)^{-1}(I-A)(I+A)(I-A)^{-1}$ . Since $(I-A)(I+A) = I - A^2 + A - A = I - A^2$ and $(I+A)(I-A) = I - A^2 - A + A = I - A^2$ , it follows that $(I-A)$ and $(I+A)$ commute. Therefore, $B^TB = (I+A)^{-1}(I+A)(I-A)(I-A)^{-1} = I \cdot I = I$ , showing that B is an orthogonal matrix ( $B^TB = I$ ). Finally, the trace of $B^TB = I$ is $\text{tr}(B^TB) = \text{tr}(I_3) = 1 + 1 + 1 = 3$ . The answer is 3 .

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