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Physics30 questions

Q1JEE Main 2024MCQ4MUnits and Measurements

The resistance $R = \frac{V}{I}$ , where $V = (200 \pm 5)$ V and $I = (20 \pm 0.2)$ A, the percentage error \in the measurement of $R$ is :

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📖 Explanation

$R = V/I$ . Percentage error \in R: $\frac{\Delta R}{R} \times 100 = \frac{\Delta V}{V} \times 100 + \frac{\Delta I}{I} \times 100$ $= \frac{5}{200} \times 100 + \frac{0.2}{20} \times 100 = 2.5\% + 1\% = 3.5\%$ The answer is $3.5\%$ , which corresponds to Option (1).

Q2JEE Main 2024MCQ4MMotion in a Straight Line

A body starts moving from rest with constant acceleration covers displacement $S_1$ \in first $(p - 1)$ seconds and $S_2$ \in first $p$ seconds. The displacement $S_1 + S_2$ will be made in time :

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📖 Explanation

$S_1 = \frac{1}{2}a(p-1)^2$ , $S_2 = \frac{1}{2}ap^2$ . $S_1 + S_2 = \frac{a}{2}\$[(p-1)^2 + p^2]\$ = \frac{a}{2}(2p^2 - 2p + 1)$ If this equals $\frac{1}{2}at^2$ : $t^2 = 2p^2 - 2p + 1$ , $t = \sqrt{2p^{2-}2p+1}$ . The answer corresponds to Option (2).

Q3JEE Main 2024MCQ4MCircular Motion

If the radius of curvature of the path of two particles of same mass are in the ratio $3 : 4$ , then in order to have constant centripetal force, their velocities will be in the ratio of:

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📖 Explanation

We need to find the ratio of velocities of two particles with the same mass, given that their radii of curvature are in the ratio 3:4 and the centripetal force is constant. The centripetal force on a particle of mass $m$ moving with speed $v$ along a circular path of radius $r$ is: $F = \frac{mv^2}{r}$ Since both particles have the same mass and the same centripetal force: $\frac{mv_1^2}{r_1} = \frac{mv_2^2}{r_2}$ The mass cancels: $\frac{v_1^2}{r_1} = \frac{v_2^2}{r_2}$ $\frac{v_1^2}{v_2^2} = \frac{r_1}{r_2}$ Taking the square root: $\frac{v_1}{v_2} = \sqrt{\frac{r_1}{r_2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$ Therefore, $v_1 : v_2 = \sqrt{3} : 2$ . The correct answer is Option (1): $\sqrt{3} : 2$ .

Q4JEE Main 2024MCQ4MWork, Power and Energy

A block of mass $100$ kg slides over a distance of $10$ m on a horizontal surface. If the co-efficient of friction between the surfaces is $0.4$ , then the work done against friction (in J) is:

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📖 Explanation

We need to find the work done against friction when a 100 kg block slides 10 m on a horizontal surface with a coefficient of friction of 0.4. The kinetic friction force on a horizontal surface is: $f = \mu m g$ where $\mu$ is the coefficient of friction, $m$ is the mass, and $g$ is the acceleration due to gravity. $f = 0.4 \times 100 \times 10 = 400 \text{ N}$ (using $g = 10$ m/s $^2$ ) Work done against friction is: $W = f \times d = 400 \times 10 = 4000 \text{ J}$ where $d = 10$ m is the distance. Note: The work done against friction is positive because the friction force opposes the direction of motion. The energy is converted into heat. The correct answer is Option (3): 4000 J .

Q5JEE Main 2024MCQ4MWork, Power and Energy

The potential energy function (in J) of a particle in a region of space is given as $U = (2x^2 + 3y^3 + 2z)$ . Here $x, y$ and $z$ are \in meter. The magnitude of $x$ -component of force (\in N) acting on the particle at point $P(1, 2, 3)$ m is:

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📖 Explanation

We need to find the magnitude of the x-component of force at point P(1, 2, 3) given the potential energy function $U = 2x^2 + 3y^3 + 2z$ . The force is the negative gradient of the potential energy: $\vec{F} = -\nabla U = -\left(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k}\right)$ Therefore, the x-component of force is: $F_x = -\frac{\partial U}{\partial x}$ $\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2 + 3y^3 + 2z) = 4x$ (The terms $3y^3$ and $2z$ are treated as constants when differentiating with respect to $x$ .) $F_x = -4x$ At $x = 1$ : $F_x = -4(1) = -4 \text{ N}$ The magnitude of the x-component of force is $|F_x| = 4$ N. The correct answer is Option (3): 4 N .

Q6JEE Main 2024MCQ4MGravitation

At what distance above and below the surface of the earth a body will have same weight? (Take radius of earth as $R$ )

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📖 Explanation

At height $h$ above surface: $g_h = g\frac{R^2}{(R+h)^2}$ . At depth $d$ below surface: $g_d = g\left(1 - \frac{d}{R}\right)$ . Setting equal and using $h = d$ : $\frac{R^2}{(R+h)^2} = 1 - \frac{h}{R} = \frac{R-h}{R}$ $R^3 = (R-h)(R+h)^2$ Let $x = h/R$ : $1 = (1-x)(1+x)^2 = (1-x^2)(1+x) = 1+x-x^{2-}x^3$ $x + x^2(-1) - x^3 = 0$ ... Actually: $x - x^2 - x^3 = 0$ , $x(1-x-x^2) = 0$ . $x^2 + x - 1 = 0$ , $x = \frac{-1+\sqrt{5}}{2}$ . $h = \frac{(\sqrt{5}-1)R}{2} = \frac{\sqrt{5}R - R}{2}$ . The answer corresponds to Option (4).

Q7JEE Main 2024MCQ4MProperties of Matter

Given below are two statements: Statement I : If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in hot water. Statement II : If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in cold water. In the light of the above statements, choose the most appropriate from the options given below

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📖 Explanation

We need to evaluate two statements about capillary rise in cold and hot water. The height of liquid rise in a capillary tube is given by Jurin's law: $h = \frac{2T\cos\theta}{\rho g r}$ where $T$ is the surface tension of the liquid, $\theta$ is the contact angle, $\rho$ is the liquid density, $g$ is acceleration due to gravity, and $r$ is the radius of the capillary tube. Surface tension of a liquid decreases with increasing temperature . This is because as temperature increases, the kinetic energy of molecules increases, weakening the intermolecular attractive forces at the surface. Therefore: $T_{hot} < T_{cold}$ Since $h \propto T$ (surface tension), and the surface tension decreases with temperature: $h_{hot} < h_{cold}$ This means the height of capillary rise is smaller in hot water than in cold water. Statement I: "The height of capillary rise will be smaller in hot water." This is TRUE , as shown above. Statement II: "The height of capillary rise will be smaller in cold water." This is FALSE ; it is the opposite of what actually happens. The correct answer is Option (3): Statement I is true but Statement II is false .

Q8JEE Main 2024MCQ4MThermodynamics

A thermodynamic system is taken from an original state $A$ to an intermediate state $B$ by a linear process as shown \in the figure. Its volume is then reduced to the original value from $B$ to $C$ by an isobaric process. The total work done by the gas from $A$ to $B$ and $B$ to $C$ would be :

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📖 Explanation

Work done is the area under the P-V curve. From A→BA→ BA→B, pressure changes linearly, so work done is area of trapezium: $W_{AB}=\frac{P_A+P_B}{2}(V_B-V_A)$ Given $P_A=8000\ \text{dyne/cm}^2,\quad P_B=4000\ \text{dyne/cm}^2$ $V_A=3m^3\ and\ V_B=7m^3$ So $W_{AB}=\frac{8000+4000}{2}(7-3)$ $=6000\times4=24000$ From B\to C, process is isobaric compression, so $W_{BC}=P(V_C-V_B)$ $=4000(3−7)$ =−16000 Therefore total work done, $W=W_{AB}+W_{BC}$ $=24000−16000=8000$ Now converting units: $1\ \text{dyne/cm}^2=0.1\ \text{Pa}$ So $8000\ \text{dyne/cm}^2=800\ \text{Pa}$ Thus, $W=800(m^3)=800J$

Q9JEE Main 2024MCQ4MThermodynamics

Two vessels $A$ and $B$ are of the same size and are at same temperature. $A$ contains $1$ g of hydrogen and $B$ contains $1$ g of oxygen. $P_A$ and $P_B$ are the pressures of the gases \in $A$ and $B$ respectively, then $\frac{P_A}{P_B}$ is :

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📖 Explanation

We need to find the ratio of pressures $P_A/P_B$ where vessel A contains 1 g of hydrogen and vessel B contains 1 g of oxygen, with both vessels being the same size and at the same temperature. $PV = nRT$ Since both vessels have the same volume ( $V$ ) and temperature ( $T$ ), and $R$ is a constant: $P \propto n$ Therefore: $\frac{P_A}{P_B} = \frac{n_A}{n_B}$ The number of moles is given by $n = \frac{\text{mass}}{\text{molar mass}}$ . For hydrogen ( $H_2$ ), molar mass = 2 g/mol: $n_A = \frac{1}{2} = 0.5 \text{ mol}$ For oxygen ( $O_2$ ), molar mass = 32 g/mol: $n_B = \frac{1}{32} \text{ mol}$ $\frac{P_A}{P_B} = \frac{n_A}{n_B} = \frac{0.5}{1/32} = 0.5 \times 32 = 16$ The correct answer is Option (1): 16 .

Q10JEE Main 2024MCQ4MElectrostatics

Two charges of $5Q$ and $-2Q$ are situated at the points $(3a, 0)$ and $(-5a, 0)$ respectively. The electric flux through a sphere of radius $4a$ having centre at origin is:

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📖 Explanation

We need to find the electric flux through a sphere of radius $4a$ centered at origin with charges $5Q$ at $(3a, 0)$ and $-2Q$ at $(-5a, 0)$ . The sphere has radius $4a$ and center at origin. Charge $5Q$ at $(3a, 0)$ : distance from origin = $3a < 4a$ . Inside the sphere. Charge $-2Q$ at $(-5a, 0)$ : distance from origin = $5a > 4a$ . Outside the sphere. $\Phi = \frac{Q_{enclosed}}{\varepsilon_0} = \frac{5Q}{\varepsilon_0}$ The electric flux is $\frac{5Q}{\varepsilon_0}$ , which matches Option B. Therefore, the answer is Option B.

Q11JEE Main 2024MCQ4MElectromagnetic Waves

Match List I with List II Choose the correct answer from the options given below

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📖 Explanation

This problem requires matching Maxwell's equations in their integral form (List I) with their corresponding names (List II).

A. The equation $\oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$ is the Ampere-Maxwell law. It relates the magnetic field around a closed loop to the conduction current ( $i_c$ ) and the displacement current (due to a changing electric flux, $\frac{d\phi_E}{dt}$ ). So, A matches with IV.
B. The equation $\oint \vec{E} \cdot d\vec{l} = \frac{d\phi_B}{dt}$ (properly written as $-\frac{d\phi_B}{dt}$ ) is Faraday's law of induction. It states that a changing magnetic flux through a surface induces an electromotive force (the line integral of $\vec{E}$ ) around the boundary of that surface. So, B matches with III.
C. The equation $\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$ is Gauss's law for electricity. It states that the net electric flux through a closed surface is proportional to the total electric charge $Q$ enclosed by the surface. So, C matches with I.
D. The equation $\oint \vec{B} \cdot d\vec{A} = 0$ is Gauss's law for magnetism. It states that the net magnetic flux through any closed surface is always zero, implying the non-existence of magnetic monopoles. So, D matches with II.

Based on this analysis, the correct matching is A - IV, B - III, C - I, and D - II.

Q12JEE Main 2024MCQ4MAlternating Current

A capacitor of capacitance $100 \mu F$ is charged to a potential of $12$ V and connected to a $6.4$ mH inductor to produce oscillations. The maximum current in the circuit would be :

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📖 Explanation

We need to find the maximum current in an LC oscillation circuit where a capacitor of $100 \, \mu F$ charged to $12$ V is connected to a $6.4$ mH inductor. In an LC circuit, energy oscillates between the electric field of the capacitor and the magnetic field of the inductor. The total energy is conserved: $\frac{1}{2}CV^2 = \frac{1}{2}LI_{max}^2$ The left side is the maximum energy stored \in the capacitor (when current is zero), and the right side is the maximum energy stored \in the inductor (when the capacitor is fully discharged). From the energy conservation equation: $I_{max} = V\sqrt{\frac{C}{L}}$ $V = 12$ V, $C = 100 \, \mu F = 100 \times 10^{-6}$ F, $L = 6.4$ mH $= 6.4 \times 10^{-3}$ H: $I_{max} = 12\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}}$ $= 12\sqrt{\frac{10^{-4}}{6.4 \times 10^{-3}}} = 12\sqrt{\frac{1}{64}}$ $= 12 \times \frac{1}{8} = 1.5 \text{ A}$ The correct answer is Option (2): 1.5 A .

Q13JEE Main 2024MCQ4MCurrent Electricity

The electric current through a wire varies with time as $I = I_0 + \beta t$ , where $I_0 = 20$ A and $\beta = 3$ A s $^{-1}$ . The amount of electric charge crossed through a section of the wire \in $20$ s is:

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📖 Explanation

We need to find the total electric charge that crosses a section of wire in 20 seconds, given that the current varies as $I = I_0 + \beta t$ . Current is the rate of flow of charge: $I = \frac{dQ}{dt}$ . Therefore, the total charge is the integral of current over time: $Q = \int_0^{t} I \, dt$ Given $I = I_0 + \beta t$ with $I_0 = 20$ A and $\beta = 3$ A/s: $Q = \int_0^{20} (I_0 + \beta t) \, dt = \int_0^{20} (20 + 3t) \, dt$ $Q = \left[20t + \frac{3t^2}{2}\right]_0^{20}$ $= \left(20 \times 20 + \frac{3 \times 20^2}{2}\right) - (0)$ $= 400 + \frac{3 \times 400}{2}$ $= 400 + 600 = 1000 \text{ C}$ The correct answer is Option (2): 1000 C .

Q14JEE Main 2024MCQ4MCurrent Electricity

A galvanometer having coil resistance $10 \Omega$ shows a full scale deflection for a current of $3$ mA. For it to measure a current of $8$ A, the value of the shunt should be:

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📖 Explanation

We need to find the shunt resistance to convert a galvanometer (coil resistance $G = 10 \, \Omega$ , full-scale deflection current $I_g = 3$ mA) into an ammeter reading up to $I = 8$ A. A shunt is a low resistance connected \in parallel with the galvanometer. The voltage across both must be equal: $I_g \cdot G = (I - I_g) \cdot S$ Solving for $S$ : $S = \frac{I_g \cdot G}{I - I_g}$ $I_g = 3 \times 10^{-3}$ A, $G = 10 \, \Omega$ , $I = 8$ A: $S = \frac{3 \times 10^{-3} \times 10}{8 - 3 \times 10^{-3}} = \frac{0.03}{7.997}$ $S = \frac{0.03}{7.997} \approx 3.75 \times 10^{-3} \, \Omega$ Note: Since $I_g \ll I$ , we can approximate $I - I_g \approx I = 8$ , giving $S \approx 0.03/8 = 3.75 \times 10^{-3} \, \Omega$ , which confirms our answer. The correct answer is Option (3): $3.75 \times 10^{-3} \, \Omega$ .

Q15JEE Main 2024MCQ4MCurrent Electricity

The deflection in moving coil galvanometer falls from $25$ divisions to $5$ division when a shunt of $24 \Omega$ is applied. The resistance of galvanometer coil will be:

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📖 Explanation

We need to find the resistance of a galvanometer coil, given that the deflection falls from 25 divisions to 5 divisions when a shunt of $24 \, \Omega$ is applied. In a galvanometer, the deflection is directly proportional to the current passing through it: $\theta \propto I_g$ So if the deflection drops from 25 to 5 divisions, the current through the galvanometer has reduced by a factor of 5: $\frac{I_g'}{I_g} = \frac{5}{25} = \frac{1}{5}$ Without shunt: All the current flows through the galvanometer: $I = I_g$ . With shunt: The current divides between the galvanometer (resistance $G$ ) and the shunt (resistance $S = 24 \, \Omega$ ) \in parallel. The total current $I$ remains the same (assuming the external circuit is unchanged), but now: $I_g' = I \times \frac{S}{G + S}$ This is because \in a parallel combination, the current divides inversely as the resistance. $\frac{I_g'}{I} = \frac{S}{G + S} = \frac{1}{5}$ (Since $I_g' = I_g/5$ and $I_g = I$ without the shunt) $\frac{24}{G + 24} = \frac{1}{5}$ $5 \times 24 = G + 24$ $120 = G + 24$ $G = 96 \, \Omega$ The correct answer is Option (2): $96 \, \Omega$ .

Q16JEE Main 2024MCQ4MRay Optics and Optical Instruments

A convex mirror of radius of curvature $30$ cm forms an image that is half the size of the object. The object distance is :

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📖 Explanation

For a convex mirror, the radius of curvature $R$ and focal length $f$ are positive. The focal length is $f = \frac{R}{2} = \frac{+30}{2} = +15$ cm.

The image is half the size of the object, and for a convex mirror, the image is always virtual and erect. Thus, the magnification is $m = +\frac{1}{2}$ .

The magnification is also given by the formula $m = -\frac{v}{u}$ .
$\frac{1}{2} = -\frac{v}{u} \implies v = -\frac{u}{2}$

Using the mirror formula, $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ , we substitute the values of $f$ and $v$ :
$\frac{1}{15} = \frac{1}{-u/2} + \frac{1}{u} = -\frac{2}{u} + \frac{1}{u}$

Solving the equation gives $\frac{1}{15} = -\frac{1}{u}$ , which means the object distance is $u = -15$ cm.

Q17JEE Main 2024MCQ4MRay Optics and Optical Instruments

A biconvex lens of refractive index $1.5$ has a focal length of $20$ cm \in air. Its focal length when immersed \in a liquid of refractive index $1.6$ will be:

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📖 Explanation

We need to find the focal length of a biconvex lens when immersed in a liquid, given its focal length in air. For a biconvex lens with equal radii of curvature $R$ : $\frac{1}{f} = (\mu_{rel} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (\mu_{rel} - 1)\frac{2}{R}$ where $\mu_{rel}$ is the refractive index of the lens relative to the surrounding medium. In air ( $\mu_{medium} = 1$ ), $\mu_{rel} = 1.5/1 = 1.5$ : $\frac{1}{20} = (1.5 - 1)\frac{2}{R} = 0.5 \times \frac{2}{R} = \frac{1}{R}$ $R = 20 \text{ cm}$ In liquid ( $\mu_{liquid} = 1.6$ ), $\mu_{rel} = 1.5/1.6 = 15/16$ : $\frac{1}{f'} = \left(\frac{15}{16} - 1\right)\frac{2}{20} = \left(-\frac{1}{16}\right)\frac{1}{10} = -\frac{1}{160}$ $f' = -160 \text{ cm}$ The negative sign indicates that the lens behaves as a diverging lens \in this liquid. This happens because the refractive index of the liquid (1.6) is greater than that of the lens (1.5), so the lens bends light \in the opposite direction compared to air. The correct answer is Option (2): $-160$ cm .

Q18JEE Main 2024MCQ4MDual Nature of Matter and Radiation

The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is $25\%$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

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📖 Explanation

We need to find the ratio of kinetic energies of an electron and a photon that have the same de Broglie wavelength, given that the electron's velocity is 25% of the speed of light. For the electron (using classical kinetic energy since $v = 0.25c$ ): $KE_e = \frac{1}{2}m_e v_e^2$ We can also write this as: $KE_e = \frac{p_e v_e}{2}$ (since $p_e = m_e v_e$ ). For the photon: $KE_{ph} = E_{ph} = pc = \frac{h}{\lambda} \cdot c = \frac{hc}{\lambda}$ Since both have the same de Broglie wavelength $\lambda$ : For the electron: $\lambda = \frac{h}{p_e}$ , so $p_e = \frac{h}{\lambda}$ For the photon: $\lambda = \frac{h}{p_{ph}}$ , so $p_{ph} = \frac{h}{\lambda}$ Therefore, $p_e = p_{ph}$ (equal momenta). $\frac{KE_e}{KE_{ph}} = \frac{p_e v_e / 2}{p_{ph} \cdot c}$ Since $p_e = p_{ph}$ , these cancel: $\frac{KE_e}{KE_{ph}} = \frac{v_e}{2c} = \frac{0.25c}{2c} = \frac{1}{8}$ The correct answer is Option (2): $\frac{1}{8}$ .

Q19JEE Main 2024MCQ4MAtoms and Nuclei

The explosive in a Hydrogen bomb is a mixture of $_1H^2$ , $_1H^3$ and $_3Li^6$ \in some condensed form. The chain reaction is given by $_3Li^6 + _0n^1 \rightarrow _2He^4 + _1H^3$ ; $_1H^2 + _1H^3 \rightarrow _2He^4 + _0n^1$ During the explosion the energy released is approximately $[Given: $M(Li) = 6.01690$ amu, $M(_1H^2) = 2.01471$ amu, $M(_2He^4) = 4.00388$ amu and $1$ amu $= 931.5$ MeV]$

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📖 Explanation

We need to find the total energy released in the hydrogen bomb chain reaction. Reaction 1: $_3Li^6 + _0n^1 \rightarrow \, _2He^4 + \, _1H^3$ Reaction 2: $_1H^2 + \, _1H^3 \rightarrow \, _2He^4 + \, _0n^1$ Adding both reactions (the $_1H^3$ and $_0n^1$ cancel as they appear on opposite sides): $_3Li^6 + \, _1H^2 \rightarrow 2 \, _2He^4$ Mass of reactants: $m_{reactants} = M(Li^6) + M(H^2) = 6.01690 + 2.01471 = 8.03161 \text{ amu}$ Mass of products: $m_{products} = 2 \times M(He^4) = 2 \times 4.00388 = 8.00776 \text{ amu}$ Mass defect: $\Delta m = m_{reactants} - m_{products} = 8.03161 - 8.00776 = 0.02385 \text{ amu}$ Using the conversion factor $1 \text{ amu} = 931.5$ MeV: $E = \Delta m \times 931.5 = 0.02385 \times 931.5 = 22.22 \text{ MeV}$ The correct answer is Option (4): 22.22 MeV .

Q20JEE Main 2024MCQ4MSemiconductor Electronics

In the given circuit, the breakdown voltage of the Zener diode is $3.0$ V. What is the value of $I_z$ ?

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📖 Explanation

This is a Zener regulator circuit. step 1: check if Zener conducts Zener breakdown voltage = 3 V Supply = 10 V So Zener will be in breakdown (active). Thus voltage across A-B = 3 V step 2: current through series resistor (1 kΩ) Voltage drop across 1 kΩ: $10-3=7\text{ V}$ $I_{total}=\frac{7}{1000}=7\text{ mA}$ step 3: current through load (2 k\Omega ) Load is across A-B, so: $I_{load}=\frac{3}{2000}=1.5\text{ mA}$ step 4: Zener current $I_z=I_{total}-I_{load}=7-1.5=5.5\text{ mA}$

Q21JEE Main 2024NAT4MMotion in a Plane

A ball rolls off the top of a stairway with horizontal velocity $u$ . The steps are $0.1$ m high and $0.1$ m wide. The minimum velocity $u$ with which that ball just hits the step 5 of the stairway will be $\sqrt{x}$ m s $^{-1}$ , where $x =$ _______ $[use $g = 10$ m s $^{-2}$ ]$.

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📖 Explanation

Let the ball leave the edge of the topmost step with only a horizontal velocity $u$ . Choose origin at this edge: horizontal forward direction is $+x$ , vertical downward is $+y$ . The riser (height) and tread (width) of every step are each $h = 0.1\text{ m}$ . Hence the corner (outer edge) of step $n$ is located at $x_n = n\,h$ and $y_n = n\,h$ for $n = 1,2,3,\dots$ For horizontal projection, the co-ordinates of the projectile after time $t$ are $x = u\,t$ , $y = \tfrac12 g t^{2}$ . Eliminating $t$ gives the trajectory equation $y = \frac{g\,x^{2}}{2u^{2}} \qquad -(1)$ The problem asks for the minimum $u$ such that the ball clears steps 1 to 4 and first lands on step 5. Step 5 has vertical drop $y_5 = 5h = 0.5\text{ m}$ , horizontal range $0.4\text{ m} \le x \le 0.5\text{ m}$ . Condition at landing (step 5) Let the ball meet step 5 at some $x = a$ , where $0.4 \le a \le 0.5$ . Substitute $y = 0.5$ and $x = a$ into (1): $0.5 = \frac{g\,a^{2}}{2u^{2}}$ $\Rightarrow \; u^{2} = g\,a^{2} \qquad -(2)$ $\Rightarrow \; u = a\,\sqrt{g}$ Clearance over the earlier steps At the corner of step $n$ (for $n = 1,2,3,4$ ) the co-ordinates are $(x_n , y_n) = (nh , nh).$ The ball must be above each such corner, i.e. $y(x_n) &lt y_n$ $\Longrightarrow \frac{g\,x_n^{2}}{2u^{2}} &lt nh$ Using $(2)$ (replace $u^{2}$ by $g\,a^{2}$ ) and $x_n = nh$ : $\frac{g\,(nh)^{2}}{2g\,a^{2}} &lt nh$ $\Rightarrow \frac{n^{2}h^{2}}{2a^{2}} &lt nh$ $\Rightarrow \frac{nh}{2a^{2}} &lt 1$ $\Rightarrow a^{2} &gt \frac{nh}{2} \qquad -(3)$ Compute the right-hand side of (3) for $h = 0.1\text{ m}$ : n = 1: $a^{2} &gt 0.05$ \to $a &gt 0.223$ n = 2: $a^{2} &gt 0.10$ \to $a &gt 0.316$ n = 3: $a^{2} &gt 0.15$ \to $a &gt 0.387$ n = 4: $a^{2} &gt 0.20$ \to $a &gt 0.447$ The most stringent requirement comes from step 4: $a \ge \sqrt{0.20} = 0.447\text{ m}$ . Choosing the landing point for minimum $u$ Equation (2) shows $u = a\sqrt{g}$ , so smaller $a$ means smaller $u$ . Subject to both conditions $0.4 \le a \le 0.5$ (must be on step 5) and $a \ge 0.447\text{ m}$ (must clear step 4), the minimum admissible value is $a_{\min} = \sqrt{0.20}\text{ m} = 0.447\text{ m}$ . Insert this $a_{\min}$ into (2): $u_{\min} = a_{\min}\sqrt{g} = \sqrt{0.20\,g}$ With $g = 10\text{ m s}^{-2}$ , $u_{\min} = \sqrt{0.20 \times 10} = \sqrt{2}\text{ m s}^{-1}$ . Thus the minimum horizontal velocity is $\sqrt{2}\text{ m s}^{-1}$ , which corresponds to $x = 2$ \in the given form $\sqrt{x}$ . Answer: $x = 2$

Q22JEE Main 2024NAT4MRotational Motion

A cylinder is rolling down on an inclined plane of inclination $60°$ . Its acceleration during rolling down will be $\frac{x}{\sqrt{3}}$ m s $^{-2}$ , where $x =$ _______ (use $g = 10$ m s $^{-2}$ ).

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📖 Explanation

For a body rolling down an inclined plane without slipping, the linear acceleration ( $a$ ) of the COM is given by: $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$ For a solid cylinder, the moment of inertia about its central axis is $I = \frac{1}{2}MR^2$ . $a = \frac{g \sin 60^\circ}{1 + \frac{1}{2}} = \frac{g \sin 60^\circ}{3/2} = \frac{2}{3} g \sin 60^\circ$ $a = \frac{2}{3} \times 10 \times \frac{\sqrt{3}}{2} = \frac{10\sqrt{3}}{3} = \frac{10}{\sqrt{3}} \text{ m s}^{-2}$ $x = 10$

Q23JEE Main 2024NAT4MProperties of Matter

In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are $70$ m s $^{-1}$ and $65$ m s $^{-1}$ respectively. If the wing area is $2$ m $^2$ , the lift of the wing is _______N. (Given density of air $= 1.2$ kg m $^{-3}$ )

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📖 Explanation

Lift = $\frac{1}{2}\rho(v_1^2 - v_2^2) \times A = \frac{1}{2}(1.2)(70^2 - 65^2)(2) = 1.2(4900-4225) = 1.2 \times 675 = 810$ N. The answer is $\boxed{810}$ .

Q24JEE Main 2024NAT4MOscillations

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $\frac{x}{8}$ , where $x =$ _______.

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📖 Explanation

Total energy = $\frac{1}{2}kA^2$ . KE = $\frac{1}{2}k(A^2 - x^2)$ . $\frac{E}{KE} = \frac{A^2}{A^{2-}x^2} = \frac{A^2}{A^2 - A^2/9} = \frac{1}{8/9} = \frac{9}{8}$ . $x = 9$ . The answer is $\boxed{9}$ .

Q25JEE Main 2024NAT4MElectrostatics

An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet $S$ having surface charge density $+\sigma$ . The electron at $t = 0$ is at a distance of $1$ m from $S$ and has a speed of $1$ m s $^{-1}$ . The maximum value of $\sigma$ , if the electron strikes $S$ at $t = 1$ s is $\alpha \left[\frac{m\varepsilon_0}{e}\right] \frac{C}{m^2}$ . The value of $\alpha$ is _______.

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📖 Explanation

Take away from sheet as positive. Then electron starts moving away: u=1 Displacement to reach sheet: s=−1 Acceleration is toward sheet, hence negative: $a=-\frac{e\sigma}{2m\epsilon_0}$ Using $s=ut+\frac{1}{2}at^2$ with t=1, $-1=1+\frac{1}{2}a$ $\frac{a}{2}=-2$ a=−4 Now $-\frac{e\sigma}{2m\epsilon_0}=-4$ $\frac{e\sigma}{2m\epsilon_0}=4$ $\sigma=8\frac{m\epsilon_0}{e}$

Q26JEE Main 2024NAT4MElectrostatic Potential and Capacitance

A $16 \Omega$ wire is bend to form a square loop. A $9$ V battery with internal resistance $1 \Omega$ is connected across one of its sides. If a $4 \mu F$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $\frac{x}{2} \mu J$ , where $x =$ _______.

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📖 Explanation

Let the square be $ABCD$ , taken \in the clockwise sense. Since the total length of the wire has resistance $16\ \Omega$ , each side of the square has resistance $R = \frac{16\ \Omega}{4} = 4\ \Omega$ The battery (emf $E = 9\ \text{V}$ , internal resistance $r = 1\ \Omega$ ) is connected across side $AB$ . Thus the terminals of the battery are joined to the same two nodes $A$ and $B$ that are already connected by the 4 $\Omega$ side. For easier calculation, replace the battery by its Norton equivalent across the nodes $A$ and $B$ . Step 1: Norton conversion An emf $E$ \in series with resistance $r$ is equivalent to a current source $I_N$ \in parallel with the same resistance $r$ , where $I_N = \frac{E}{r} = \frac{9}{1} = 9\ \text{A}$ Therefore, across the nodes $A$ (positive) and $B$ (negative) we now have \bullet a current source of $9\ \text{A}$ from $A \rightarrow B$ , and \bullet a resistor $1\ \Omega$ \in parallel with that source. Step 2: Find the equivalent resistance between $A$ and $B$ of the rest of the square. Besides the internal $1\ \Omega$ resistor, two purely resistive paths join $A$ and $B$ : \bullet Direct side $AB$ : $4\ \Omega$ \bullet The remaining three sides $A \rightarrow D \rightarrow C \rightarrow B$ : $4 + 4 + 4 = 12\ \Omega$ Hence the three resistances $1\ \Omega, 4\ \Omega, 12\ \Omega$ are all \in parallel between the same nodes. Their combined resistance is $\frac{1}{R_{\text{eq}}} \;=\; \frac{1}{1} + \frac{1}{4} + \frac{1}{12} \;=\; 1 + 0.25 + 0.0833 \;=\; 1.3333$ $R_{\text{eq}} = \frac{1}{1.3333} = 0.75\ \Omega$ Step 3: Node voltage between $A$ and $B$ By KCL at the parallel combination, the current source of $9\ \text{A}$ drives the currents through the parallel resistors, producing a node voltage $V_{AB} = I_N \, R_{\text{eq}} = 9\ \text{A} \times 0.75\ \Omega = 6.75\ \text{V}$ Take $V_B = 0$ ; hence $V_A = +6.75\ \text{V}$ Step 4: Currents and potentials around the other three sides The only path from $A$ to $B$ that does not go through the direct side is the series string $A \rightarrow D \rightarrow C \rightarrow B$ of total resistance $12\ \Omega$ . The current through this path is $I = \frac{V_{AB}}{12\ \Omega} = \frac{6.75}{12} = 0.5625\ \text{A}$ Voltage drop on each of the three equal resistors (4 $\Omega$ ) is $\Delta V = I \times 4 = 0.5625 \times 4 = 2.25\ \text{V}$ Therefore, moving clockwise: $V_D = V_A - 2.25 = 6.75 - 2.25 = 4.50\ \text{V}$ $V_C = V_D - 2.25 = 4.50 - 2.25 = 2.25\ \text{V}$ Step 5: Potential difference across a diagonal Place the $4\ \mu\text{F}$ capacitor across any diagonal; choose diagonal $AC$ . The voltage across this diagonal is $V_{AC} = V_A - V_C = 6.75 - 2.25 = 4.50\ \text{V}$ (If the capacitor had been connected across the other diagonal $BD$ , the magnitude is the same, $4.50\ \text{V}$ .) Step 6: Energy stored \in the capacitor The energy \in a capacitor is $U = \tfrac{1}{2} C V^2$ . Here $C = 4\ \mu\text{F} = 4 \times 10^{-6}\ \text{F}$ , $V = 4.50\ \text{V}$ . Hence $U = \tfrac{1}{2} \times 4 \times 10^{-6} \times (4.5)^2$ $\qquad= 2 \times 10^{-6} \times 20.25$ $\qquad= 40.5 \times 10^{-6}\ \text{J}$ Expressing \in microjoules: $U = 40.5\ \mu \text{J}$ . The question states this energy equals $\dfrac{x}{2}\ \mu \text{J}$ , so $\frac{x}{2} = 40.5 \quad\Longrightarrow\quad x = 81$ Therefore, $x = 81$ .

Q27JEE Main 2024NAT4MMagnetism and Matter

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of $20$ cm from its center is $1.5 \times 10^{-5}$ T m. The magnetic moment of the dipole is _______A m $^2$ . (Given: $\frac{\mu_0}{4\pi} = 10^{-7}$ T m A $^{-1}$ )

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📖 Explanation

Magnetic potential on axis: $V = \frac{\mu_0}{4\pi}\frac{M\cos\theta}{r^2}$ . On axis, $\theta = 0$ : $V = \frac{\mu_0 M}{4\pi r^2} = 10^{-7} \times \frac{M}{(0.2)^2} = 10^{-7} \times 25M$ $1.5 \times 10^{-5} = 25 \times 10^{-7} M$ . $M = \frac{1.5 \times 10^{-5}}{25 \times 10^{-7}} = 6$ Am². The answer is $\boxed{6}$ .

Q28JEE Main 2024NAT4MElectromagnetic Induction

A square loop of side $10$ cm and resistance $0.7 \Omega$ is placed vertically \in the east-west plane. A uniform magnetic field of $0.20$ T is set up across the plane \in the north-east direction. The magnetic field is decreased to zero \in $1$ s at a steady rate. Then, the magnitude of induced emf is $\sqrt{x} \times 10^{-3}$ V. The value of $x$ is _______.

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📖 Explanation

Square loop in east-west plane, B in NE direction. B makes 45° with the plane of the loop. $\Phi = BA\cos 45° = 0.20 \times 0.01 \times \frac{1}{\sqrt{2}} = \frac{0.002}{\sqrt{2}}$ $\varepsilon = \frac{\Delta\Phi}{\Delta t} = \frac{0.002/\sqrt{2}}{1} = \frac{0.002}{\sqrt{2}} = \sqrt{2} \times 10^{-3}$ V. $\sqrt{x} \times 10^{-3} = \sqrt{2} \times 10^{-3}$ . $x = 2$ . The answer is $\boxed{2}$ .

Q29JEE Main 2024NAT4MWave Optics

In a double slit experiment shown in figure, when light of wavelength $400$ nm is used, dark fringe is observed at $P$ . If $D = 0.2$ m, the minimum distance between the slits $S_1$ and $S_2$ is $\alpha$ mm. Write the value of $10\alpha$ to the nearest integer.

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📖 Explanation

The path difference between the two rays reaching point P is Δx = (Source→S₁→P) - (Source→S₂→P). The path lengths are L₁ = 2\sqrt{D^2 + d^2} and L₂ = 2D.
So, the path difference is Δx = 2\sqrt{D^2 + d^2} - 2D = 2D\left(1 + \frac{d^2}{D^2}\right)^{1/2} - 2D.
Using the binomial approximation (1+x)^n \approx 1+nx for d \ll D, the path difference simplifies to:
$\Delta x \approx 2D\left(1 + \frac{1}{2}\frac{d^2}{D^2}\right) - 2D = \frac{d^2}{D}$
For a dark fringe to be observed at P, the path difference must be an odd multiple of λ/2. For the minimum distance d, we consider the first dark fringe:
$\Delta x = \frac{\lambda}{2}$
Equating the expressions for the path difference, we get d^2/D = λ/2, which gives d = \sqrt{Dλ/2}.
Substituting the given values D = 0.2 m and λ = 400 nm = 4 × 10^{-7} m:
$d = \sqrt{\frac{0.2 \times 4 \times 10^{-7}}{2}} = \sqrt{4 \times 10^{-8}} = 2 \times 10^{-4} \text{ m} = 0.2 \text{ mm}$
The problem states this minimum distance is α mm, so α = 0.2. The value of 10α is 10 × 0.2 = 2.

Q30JEE Main 2024NAT4MAtoms and Nuclei

When a hydrogen atom going from $n = 2$ to $n = 1$ emits a photon, its recoil speed is $\frac{x}{5}$ m s $^{-1}$ . Where $x =$ _______. (Use: mass of hydrogen atom $= 1.6 \times 10^{-27}$ kg, charge of electron $e = 1.6 \times 10^{-19}$ C)

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📖 Explanation

Energy of photon (n=2→n=1): $E = 13.6(1-1/4) = 10.2$ eV $= 10.2 \times 1.6 \times 10^{-19}$ J. Momentum: $p = E/c = 10.2 \times 1.6 \times 10^{-19}/(3 \times 10^8) = 5.44 \times 10^{-27}$ kg\cdot m/s. Recoil speed = $p/m = 5.44 \times 10^{-27}/(1.6 \times 10^{-27}) = 3.4$ m/s = $17/5$ . $x = 17$ . The answer is $\boxed{17}$ .

Chemistry30 questions

Q31JEE Main 2024MCQ4MStructure of Atom

The correct set of four quantum numbers for the valence electron of rubidium atom $(Z = 37)$ is:

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📖 Explanation

We need to find the four quantum numbers for the valence electron of rubidium ( $Z = 37$ ). Determine the electronic configuration of Rb. Rubidium has 37 electrons. Its electronic configuration is: $Rb: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^6 \, 5s^1$ Or equivalently: $[Kr] \, 5s^1$ The valence (outermost) electron is \in the $5s$ orbital. Assign the quantum numbers. For an electron \in the $5s$ orbital: - Principal quantum number ( $n$ ) : $n = 5$ (the electron is \in the 5th shell) - Azimuthal quantum number ( $l$ ) : $l = 0$ (for an s-orbital, $l$ is always 0) - Magnetic quantum number ( $m_l$ ) : $m_l = 0$ (for $l = 0$ , the only possible value is 0) - Spin quantum number ( $m_s$ ) : $m_s = +\frac{1}{2}$ (for the first electron \in an orbital, by convention) The set of quantum numbers is: $5, 0, 0, +\frac{1}{2}$ . The correct answer is Option (1): $5, 0, 0, +\frac{1}{2}$ .

Q32JEE Main 2024MCQ4MClassification of Elements

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: The first ionisation enthalpy decreases across a period. Reason R: The increasing nuclear charge outweighs the shielding across the period. In the light of the above statements, choose the most appropriate from the options given below:

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Q33JEE Main 2024MCQ4MChemical Thermodynamics

Which of the following is not correct?

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📖 Explanation

We need to identify which statement about Gibbs free energy ( $\Delta G$ ) is NOT correct. Recall the significance of $\Delta G$ . The Gibbs free energy change ( $\Delta G$ ) is the criterion for spontaneity of a process at constant temperature and pressure: $\Delta G = \Delta H - T\Delta S$ Evaluate each statement. Option (1): " $\Delta G$ is negative for a spontaneous reaction." This is CORRECT . A negative $\Delta G$ means the reaction decreases the free energy of the system, which is the thermodynamic condition for spontaneity. Option (2): " $\Delta G$ is positive for a spontaneous reaction." This is INCORRECT . A positive $\Delta G$ means the reaction is non-spontaneous - it requires external energy input to proceed. This directly contradicts the fundamental criterion for spontaneity. Option (3): " $\Delta G$ is zero for a reversible reaction." This is CORRECT . At equilibrium (which is the condition for a reversible process), $\Delta G = 0$ . The system has reached the minimum of free energy and has no driving force \in either direction. Option (4): " $\Delta G$ is positive for a non-spontaneous reaction." This is CORRECT . A positive $\Delta G$ indicates that the reaction requires energy input and will not proceed on its own. The incorrect statement is Option (2): $\Delta G$ is positive for a spontaneous reaction .

Q34JEE Main 2024MCQ4MRedox Reactions

Chlorine undergoes disproportionation in alkaline medium as shown below : $aCl_2(g) + bOH^-(aq) \rightarrow cClO^-(aq) + dCl^-(aq) + eH_2O(l)$ The values of $a, b, c$ and $d$ in a balanced redox reaction are respectively :

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📖 Explanation

This is a disproportionation reaction where chlorine is both oxidized and reduced. We balance it using the half-reaction method in an alkaline medium.

Half-Reactions:
- Oxidation: $Cl_2 \rightarrow ClO^-$ (Oxidation state of $Cl$ changes from $0$ to $+1$ )
- Reduction: $Cl_2 \rightarrow Cl^-$ (Oxidation state of $Cl$ changes from $0$ to $-1$ )
Balance the Reduction Half-Reaction:
$Cl_2 + 2e^- \rightarrow 2Cl^-$
Balance the Oxidation Half-Reaction (in alkaline medium):
First, balance atoms other than O and H: $Cl_2 \rightarrow 2ClO^-$
Balance O by adding $OH^-$ and H with $H_2O$ : $Cl_2 + 4OH^- \rightarrow 2ClO^- + 2H_2O$
Balance charge by adding electrons: $Cl_2 + 4OH^- \rightarrow 2ClO^- + 2H_2O + 2e^-$
Combine Half-Reactions:
The number of electrons lost (2) equals the number of electrons gained (2). Add the two half-reactions:
$(Cl_2 + 2e^-) + (Cl_2 + 4OH^-) \rightarrow (2Cl^-) + (2ClO^- + 2H_2O + 2e^-)$
$2Cl_2 + 4OH^- \rightarrow 2ClO^- + 2Cl^- + 2H_2O$
Simplify and Find Coefficients:
Divide the entire equation by 2 to get the simplest integer coefficients:
$Cl_2 + 2OH^- \rightarrow ClO^- + Cl^- + H_2O$
Comparing this to $aCl_2 + bOH^- \rightarrow cClO^- + dCl^- + eH_2O$ , we get $a=1, b=2, c=1, d=1$ .

Q35JEE Main 2024MCQ4Md and f Block Elements

$KMnO_4$ decomposes on heating at $513$ K to form $O_2$ along with

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📖 Explanation

We need to identify the products formed when $KMnO_4$ is heated at 513 K (along with $O_2$ ). Write the thermal decomposition reaction. When potassium permanganate is heated, it undergoes the following decomposition: $2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2$ Verify the balancing. Left side: K: 2, Mn: 2, O: 8 Right side: K: 2, Mn: 1+1=2, O: 4+2+2=8. Balanced. Understand the chemistry. In $KMnO_4$ , manganese is \in the +7 oxidation state. During thermal decomposition: - One Mn goes from +7 to +6 \in $K_2MnO_4$ (potassium manganate) - this is a reduction - Another Mn goes from +7 to +4 \in $MnO_2$ (manganese dioxide) - this is also a reduction - Oxygen goes from -2 to 0 \in $O_2$ - this is an oxidation This is a disproportionation-type reaction where the oxygen is oxidised while manganese is reduced to two different lower oxidation states. The products (along with $O_2$ ) are $K_2MnO_4$ and $MnO_2$ . The correct answer is Option (4): $K_2MnO_4$ and $MnO_2$ .

Q36JEE Main 2024MCQ4Mp Block Elements

Given below are two statements : Statement I : The electronegativity of group 14 elements from Si to Pb gradually decreases. Statement II : Group 14 contains non-metallic, metallic, as well as metalloid elements. In the light of the above statements, choose the most appropriate from the options given below :

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Q37JEE Main 2024MCQ4MOrganic Chemistry - Some Basic Principles

The interaction between $\pi$ bond and lone pair of electrons present on an adjacent atom is responsible for

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📖 Explanation

Question: The interaction between a $\pi$ bond and a lone pair of electrons present on an adjacent atom is responsible for which effect? Review the electronic effects \in organic chemistry. Hyperconjugation (Option 1): This involves the interaction between $\sigma$ bond electrons (particularly C-H bonds) and an adjacent empty or partially filled p-orbital or $\pi$ system. It does not involve lone pairs directly. Inductive effect (Option 2): This is the transmission of charge through a chain of atoms via $\sigma$ bonds. It involves $\sigma$ bonds only, not $\pi$ bonds or lone pairs interacting with each other. Electromeric effect (Option 3): This is a temporary effect that occurs when a reagent approaches a $\pi$ bond, causing complete transfer of the $\pi$ electron pair. It does not specifically involve lone pairs on adjacent atoms. Resonance effect (Option 4): Resonance (also called the mesomeric effect) involves the delocalisation of electrons through $\pi$ bonds and lone pairs on adjacent atoms. When a lone pair on an atom adjacent to a $\pi$ bond participates \in conjugation, it is the resonance effect . For example, \in aniline ( $C_6H_5NH_2$ ), the lone pair on N interacts with the $\pi$ system of the benzene ring. The correct answer is Option 4 : Resonance effect.

Q38JEE Main 2024MCQ4MOrganic Chemistry - Some Basic Principles

The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is:

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📖 Explanation

Understanding resonance energy: Many organic compounds can be represented by two or more Lewis structures (called resonance structures or canonical structures). The actual molecule is a resonance hybrid of all these structures. The lowest energy resonance structure (also called the most stable contributing structure) is a hypothetical reference. The actual molecule, being a hybrid of all resonance structures, has a lower energy than any single resonance structure - including the most stable one. Definition of resonance energy: Resonance energy is defined as the difference in energy between the actual structure (resonance hybrid) and the most stable (lowest energy) resonance structure: $\text{Resonance Energy} = E_{\text{lowest energy resonance structure}} - E_{\text{actual structure (hybrid)}}$ Since the actual structure is always more stable (lower energy) than any individual resonance structure, the resonance energy is always a positive quantity. For example, benzene has a resonance energy of approximately 150 kJ/mol, meaning the actual benzene molecule is 150 kJ/mol more stable than the hypothetical cyclohexatriene (Kekulé) structure. Among the given options: Option A (electromeric energy): This refers to the temporary shift of electrons \in a multiple bond under the influence of an attacking reagent. It is not related to the stability difference between resonance structures. Option B (resonance energy): This is exactly the energy difference between the actual structure and the lowest energy resonance structure. This is correct. Option C (ionization energy): This is the energy required to remove an electron from an isolated gaseous atom. Unrelated to resonance. Option D (hyperconjugation energy): This refers to the stabilization due to the interaction of electrons \in a $\sigma$ bond with an adjacent empty or partially filled p-orbital. While it contributes to stability, it is not defined as the difference between the actual and resonance structures. The correct answer is Option B: resonance energy.

Q39JEE Main 2024MCQ4MPractical Organic Chemistry

Appearance of blood red colour, on treatment of the sodium fusion extract of an organic compound with $FeSO_4$ \in presence of concentrated $H_2SO_4$ indicates the presence of element/s

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📖 Explanation

We need to determine which element(s) are indicated by a blood red colour when the sodium fusion extract is treated with $FeSO_4$ and concentrated $H_2SO_4$ . Understand the Lassaigne (sodium fusion) test. In the sodium fusion test, the organic compound is fused with sodium metal. During fusion: - Nitrogen \in the compound forms $NaCN$ (sodium cyanide) - Sulphur forms $Na_2S$ (sodium sulphide) - If both N and S are present, $NaSCN$ (sodium thiocyanate) is formed Describe the reaction that produces blood red colour. When $FeSO_4$ is added to the sodium fusion extract containing $NaSCN$ : $FeSO_4 + 2NaSCN \rightarrow Fe(SCN)_2 + Na_2SO_4$ In the presence of concentrated $H_2SO_4$ (which oxidises $Fe^{2+}$ to $Fe^{3+}$ ): $Fe^{3+} + SCN^- \rightarrow [Fe(SCN)]^{2+}$ The ferric thiocyanate complex $[Fe(SCN)]^{2+}$ has a characteristic blood red colour . Identify the elements. The formation of thiocyanate ( $SCN^-$ ) requires BOTH nitrogen (for the CN part) and sulphur (for the S part). Therefore, the blood red colour indicates the presence of both nitrogen and sulphur \in the organic compound. Note: If only nitrogen were present (without sulphur), the test with $FeSO_4$ would give Prussian blue colour (from ferric ferrocyanide), not blood red. The correct answer is Option (3): N and S .

Q40JEE Main 2024MCQ4MHydrocarbons

Identify product A and product B

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📖 Explanation

The reaction of cyclohexene with $Cl_2$ under UV light (hν) proceeds via a free-radical substitution mechanism. The light initiates the formation of chlorine radicals ( $Cl\bullet$ ), which preferentially abstract hydrogens from the allylic positions (C3 and C6) of cyclohexene. This is favored because the resulting allylic radical is stabilized by resonance. This pathway leads to Product A, 3,6-dichlorocyclohexene.

In contrast, the reaction with $Cl_2$ in an inert solvent like $CCl_4$ is an electrophilic addition reaction. The electron-rich double bond of cyclohexene attacks the $Cl_2$ molecule, forming a cyclic chloronium ion intermediate. Subsequent nucleophilic attack by the chloride ion ( $Cl^−$ ) from the opposite face (anti-addition) breaks the double bond and adds two chlorine atoms to the adjacent carbons. This yields Product B, 1,2-dichlorocyclohexane.
$\text{Product A: 3,6-Dichlorocyclohexene}$
$\text{Product B: 1,2-Dichlorocyclohexane}$
Therefore, the correct pair of products is given in option D.

Q41JEE Main 2024MCQ4MAlcohols, Phenols and Ethers

The major product (P) in the following reaction is

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Q42JEE Main 2024MCQ4MAmines

The arenium ion which is not involved in the bromination of Aniline is

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📖 Explanation

The amino group ( $-NH_2$ ) in aniline is a strong activating group and is ortho- and para-directing for electrophilic aromatic substitution. Bromination, therefore, occurs at the ortho and para positions. Let's analyze the arenium ion (sigma complex) intermediates formed.

Ortho Attack: When the electrophile ( $Br^+$ ) attacks the ortho position (e.g., C2), the positive charge is delocalized by resonance onto carbons 3, 5, and 1. Delocalization to C1 allows the nitrogen's lone pair to participate, creating a highly stable resonance structure. Structures A and D are valid resonance contributors for the ortho-attack intermediate.
Para Attack: When $Br^+$ attacks the para position (C4), the positive charge is delocalized onto carbons 3, 5, and 1. Again, delocalization to C1 allows for stabilization by the nitrogen lone pair. Structure B is the most stable resonance contributor for the para-attack intermediate.
Analysis of Structure C: Structure C shows the bromine at an ortho position (e.g., C6) and the positive charge at C4. For an attack at an ortho position, the positive charge is delocalized to the other ortho (C2) and para (C4) positions relative to the point of attack, and the carbon bearing the substituent (C1). So for attack at C6, the charge would be at C1, C3, and C5. It is never located at C4. Therefore, structure C is not a valid resonance structure for the arenium ion formed during the bromination of aniline.

Q43JEE Main 2024MCQ4MAldehydes, Ketones and Carboxylic Acids

The final product A formed in the following multistep reaction sequence is:

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📖 Explanation

The reaction begins with the acid-catalyzed hydration of styrene ( $C_6H_5CH=CH_2$ ). This follows Markovnikov's rule, forming the more stable benzylic carbocation, which leads to 1-phenylethanol ( $C_6H_5CH(OH)CH_3$ ). The second step is the oxidation of this secondary alcohol with $CrO_3$ , a strong oxidizing agent, to form the ketone acetophenone ( $C_6H_5C(=O)CH_3$ ). The final step is a Wolff-Kishner reduction. Hydrazine ( $H_2N-NH_2$ ) in the presence of a strong base ( $KOH$ ) and heat reduces the carbonyl group completely to a methylene group ( $CH_2$ ), yielding the final product, ethylbenzene ( $C_6H_5CH_2CH_3$ ).

Q44JEE Main 2024MCQ4Mp Block Elements

Identify the incorrect pair from the following :

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📖 Explanation

We need to identify the incorrect pair of mineral name and chemical formula. Check each pair. Option (1): Fluorspar - $BF_3$ This is INCORRECT . Fluorspar (also called fluorite) is $CaF_2$ (calcium fluoride), NOT $BF_3$ (boron trifluoride). $BF_3$ is a gaseous compound, not a mineral. Fluorspar is a naturally occurring mineral and is the principal source of fluorine and hydrofluoric acid. Option (2): Cryolite - $Na_3AlF_6$ This is correct. Cryolite is sodium hexafluoroaluminate, used as a flux \in the electrolytic extraction of aluminium (Hall-Heroult process). Option (3): Fluoroapatite - $3Ca_3(PO_4)_2 \cdot CaF_2$ This is correct. Fluoroapatite is a phosphate mineral containing fluoride, found \in teeth and bones. Option (4): Carnallite - $KCl \cdot MgCl_2 \cdot 6H_2O$ This is correct. Carnallite is a hydrated potassium magnesium chloride mineral, an important source of potassium. The incorrect pair is Option (1): Fluorspar - $BF_3$ .

Q45JEE Main 2024MCQ4Md and f Block Elements

In chromyl chloride test for confirmation of $Cl^-$ ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and $10\% H_2O_2$ turns organic layer blue indicating formation of chromium pentoxide. The oxidation state of chromium in that is

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📖 Explanation

We need to find the oxidation state of chromium in chromium pentoxide ( $CrO_5$ ). Identify the structure of $CrO_5$ . Chromium pentoxide ( $CrO_5$ ) has the structure with one oxide ion ( $O^{2-}$ ) and two peroxide ions ( $O_2^{2-}$ ). $CrO_5 = CrO(O_2)_2$ Calculate oxidation state. Let the oxidation state of Cr be $x$ . $x + (-2) + 2(-2) = 0$ $x - 2 - 4 = 0$ $x = +6$ Alternatively: One $O^{2-}$ contributes $-2$ , two $O_2^{2-}$ contribute $2 \times (-2) = -4$ . Total oxygen charge = $-6$ . So $x = +6$ . Conclusion. The oxidation state of chromium \in $CrO_5$ is $+6$ , which matches Option A. Therefore, the answer is Option A.

Q46JEE Main 2024MCQ4MRedox Reactions

In alkaline medium, $MnO_4^-$ oxidises $I^-$ to

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📖 Explanation

We need to determine the product when $MnO_4^-$ oxidises $I^-$ \in alkaline medium. Understand the oxidising power of permanganate \in alkaline medium. In alkaline medium, permanganate ion ( $MnO_4^-$ ) is a strong oxidising agent. It is reduced from Mn(+7) to Mn(+6) (as $MnO_4^{2-}$ ) or to MnO $_2$ depending on conditions. Determine the oxidation product of iodide. In alkaline medium, $MnO_4^-$ being a strong enough oxidiser can take iodide ( $I^-$ , oxidation state -1) all the way up to iodate ( $IO_3^-$ , oxidation state +5). The reaction is: $2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-$ Verify the oxidation states. - Iodine goes from -1 \in $I^-$ to +5 \in $IO_3^-$ (loss of 6 electrons) - Manganese goes from +7 \in $MnO_4^-$ to +4 \in $MnO_2$ (gain of 3 electrons each, 6 total for 2 atoms) Electrons are balanced. The correct answer is Option (4): $IO_3^-$ .

Q47JEE Main 2024MCQ4MCoordination Compounds

In which one of the following metal carbonyls, CO forms a bridge between metal atoms?

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📖 Explanation

Question: In which metal carbonyl does CO form a bridge between metal atoms? Analyse each metal carbonyl. $[Co_2(CO)_8]$ (Option 1): Dicobalt octacarbonyl exists \in two isomeric forms. One isomer has two bridging CO groups connecting the two Co atoms (along with six terminal CO groups), and the other has a direct Co-Co bond with all terminal COs. The bridged form is well-known and commonly depicted \in textbooks. This compound does have bridging CO . $[Mn_2(CO)_{10}]$ (Option 2): This has a direct Mn-Mn bond with all ten CO groups \in terminal positions. No bridging CO . $[Os_3(CO)_{12}]$ (Option 3): This has a triangular Os₃ framework with all twelve CO groups \in terminal positions (four per Os atom). No bridging CO . $[Ru_3(CO)_{12}]$ (Option 4): Similar to the osmium analogue, this has all terminal CO groups. No bridging CO . The correct answer is Option 1 : $[Co_2(CO)_8]$ .

Q48JEE Main 2024MCQ4MAlcohols, Phenols and Ethers

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : Aryl halides cannot be prepared by replacement of hydroxyl group of phenol by halogen atom. Reason R : Phenols react with halogen acids violently. In the light of the above statements, choose the most appropriate from the options given below:

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Q49JEE Main 2024MCQ4MBiomolecules

Type of amino acids obtained by hydrolysis of proteins is :

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📖 Explanation

We need to identify the type of amino acids obtained by hydrolysis of proteins. Understand the structure of proteins. Proteins are polymers of amino acids linked by peptide bonds ( $-CO-NH-$ ). The peptide bond is formed between the $\alpha$ -carboxyl group ( $-COOH$ ) of one amino acid and the $\alpha$ -amino group ( $-NH_2$ ) of another. Identify the type of amino acids. In $\alpha$ -amino acids, the amino group ( $-NH_2$ ) is attached to the $\alpha$ -carbon (the carbon atom directly adjacent to the carboxyl group). The general structure is: $H_2N - \underset{|}{C}H(R) - COOH$ where $R$ is the side chain that varies between different amino acids. Why $\alpha$ -amino acids specifically? All naturally occurring amino acids that make up proteins are $\alpha$ -amino acids (with the exception of proline, which is technically an $\alpha$ -imino acid). While $\beta$ , $\gamma$ , and $\delta$ -amino acids exist (where the amino group is on the 2nd, 3rd, or 4th carbon from the carboxyl group respectively), they are not found \in natural proteins. The correct answer is Option (2): $\alpha$ .

Q50JEE Main 2024MCQ4MCoordination Compounds

Match List I with List II Choose the correct answer from the options given below:

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📖 Explanation

To solve this, we must identify the key metallic element present in each substance.

Ziegler catalyst is used in polymerization and is based on a titanium compound, typically titanium tetrachloride ( $TiCl_4$ ). Thus, A matches with IV (Titanium).
Blood Pigment, specifically hemoglobin, is responsible for oxygen transport and contains a central iron ( $Fe$ ) atom in its heme group. So, B matches with III (Iron).
Wilkinson catalyst is a homogeneous catalyst for hydrogenation, with the chemical formula $RhCl(PPh_3)_3$ . It contains a central rhodium ( $Rh$ ) atom. Hence, C matches with I (Rhodium).
Vitamin B12, also called cobalamin, is a complex coordination compound with a cobalt ( $Co$ ) atom at its core. This means D matches with II (Cobalt).

Combining these matches, we get A-IV, B-III, C-I, and D-II.

Q51JEE Main 2024NAT4MChemical Bonding and Molecular Structure

Number of compounds with one lone pair of electrons on central atom amongst following is _______ $O_3, H_2O, SF_4, ClF_3, NH_3, BrF_5, XeF_4$

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📖 Explanation

To determine the number of lone pairs on the central atom, we use the formula: Lone Pairs $= \frac{1}{2}(V - B)$ , where $V$ is the number of valence electrons of the central atom and $B$ is the number of electrons used for bonding.

$O_3$ : The central O atom forms 3 effective bonds (in resonance) and has 1 lone pair.
$H_2O$ : Oxygen has $\frac{1}{2}(6 - 2) = 2$ lone pairs.
$SF_4$ : Sulfur has $\frac{1}{2}(6 - 4) = 1$ lone pair.
$NH_3$ : Nitrogen has $\frac{1}{2}(5 - 3) = 1$ lone pair.
$BrF_5$ : Bromine has $\frac{1}{2}(7 - 5) = 1$ lone pair.
$ClF_3$ has $\frac{1}{2}(7 - 3) = 2$ lone pairs, and $XeF_4$ has $\frac{1}{2}(8 - 4) = 2$ lone pairs.

The compounds with exactly one lone pair on the central atom are $O_3, SF_4, NH_3$ , and $BrF_5$ . Thus, the total count is 4.

Q52JEE Main 2024NAT4MChemical Bonding and Molecular Structure

The number of species from the following which are paramagnetic and with bond order equal to one is _______ $H_2, He_2^+, O_2^+, N_2^{2-}, O_2^{2-}, F_2, Ne_2^+, B_2$

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📖 Explanation

We need to find the number of species that are paramagnetic AND have bond order = 1 from: $H_2, He_2^+, O_2^+, N_2^{2-}, O_2^{2-}, F_2, Ne_2^+, B_2$ . Analyze each species. $H_2$ : $\sigma_{1s}^2$ . Bond order = 1. Diamagnetic (all paired). No. $He_2^+$ : $\sigma_{1s}^2 {\sigma^*_{1s}}^{1}$ . Bond order = 0.5. No (BO \ne 1). $O_2^+$ : Bond order = 2.5. No. $N_2^{2-}$ : 16 electrons. Same as $O_2$ . Bond order = 2. No. $O_2^{2-}$ : 18 electrons. $\sigma_{1s}^2{\sigma^*_{1s}}^{2}\sigma_{2s}^2{\sigma^*_{2s}}^{2}\sigma_{2p}^2\pi_{2p}^4{\pi^*_{2p}}^{4}$ . Bond order = (10-8)/2 = 1. All electrons paired. Diamagnetic. No. $F_2$ : 18 electrons. Bond order = 1. Diamagnetic. No. $Ne_2^+$ : 19 electrons. Bond order = (10-9)/2 = 0.5. No. $B_2$ : 10 electrons. $\sigma_{1s}^2{\sigma^*_{1s}}^{2}\sigma_{2s}^2{\sigma^*_{2s}}^{2}\pi_{2p}^1\pi_{2p}^1$ . Bond order = (6-4)/2 = 1. Two unpaired electrons \in $\pi_{2p}$ . Paramagnetic with bond order 1. Yes! Count. Only $B_2$ satisfies both conditions = 1 species. Conclusion. The number is 1. Therefore, the answer is 1.

Q53JEE Main 2024NAT4MChemical Equilibrium

For the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ , $K_p = 0.492$ atm at $300$ K. $K_c$ for the reaction at same temperature is _______ $\times 10^{-2}$ . (Given: $R = 0.082$ L atm mol $^{-1}$ K $^{-1}$ )

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📖 Explanation

$K_p = K_c(RT)^{\Delta n}$ . $\Delta n = 2-1 = 1$ . $K_c = K_p/(RT) = 0.492/(0.082 \times 300) = 0.492/24.6 = 0.02 = 2 \times 10^{-2}$ . The answer is $\boxed{2}$ .

Q54JEE Main 2024NAT4MOrganic Chemistry - Some Basic Principles

Number of compounds among the following which contain sulphur as heteroatom is _______. Furan, Thiophene, Pyridine, Pyrrole, Cysteine, Tyrosine

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📖 Explanation

We need to count the number of compounds containing sulphur as a heteroatom from: Furan, Thiophene, Pyridine, Pyrrole, Cysteine, Tyrosine. Define heteroatom. A heteroatom is any atom in an organic compound that is not carbon or hydrogen. In heterocyclic chemistry, it specifically refers to atoms like N, O, S that replace carbon in a ring. Analyse each compound. Furan: A five-membered heterocyclic ring containing oxygen as the heteroatom. No sulphur. Thiophene: A five-membered heterocyclic ring containing sulphur as the heteroatom. Counted. Pyridine: A six-membered heterocyclic ring containing nitrogen as the heteroatom. No sulphur. Pyrrole: A five-membered heterocyclic ring containing nitrogen as the heteroatom. No sulphur. Cysteine: An amino acid with the structure $HSCH_2CH(NH_2)COOH$ . It contains a thiol group ( $-SH$ ), so sulphur is present as a heteroatom. Counted. Tyrosine: An amino acid with a phenol group ( $HO-C_6H_4-CH_2CH(NH_2)COOH$ ). It contains oxygen and nitrogen but no sulphur . Count. Compounds containing sulphur as heteroatom: Thiophene and Cysteine = 2 . The answer is 2 .

Q55JEE Main 2024NAT4MHydrocarbons

Consider the given reaction. Trans-2-butene $\xrightarrow{(i) O_3}$ $\xrightarrow{(ii) Zn/H_2O}$ (P) The total number of oxygen atoms present per molecule of the product (P) is _______.

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Q56JEE Main 2024NAT4MSolutions

A solution of $H_2SO_4$ is $31.4\%$ $H_2SO_4$ by mass and has a density of $1.25$ g/mL. The molarity of the $H_2SO_4$ solution is M (nearest integer) $[Given molar mass of $H_2SO_4 = 98$ g mol $^{-1}$ ]$

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📖 Explanation

31.4% H₂SO₄ by mass, density 1.25 g/mL. In 1 L: mass = 1250 g. Mass of H₂SO₄ = 0.314 × 1250 = 392.5 g. Moles = 392.5/98 = 4.005 ≈ 4 M. The answer is $\boxed{4}$ .

Q57JEE Main 2024NAT4MSolutions

The osmotic pressure of a dilute solution is $7 \times 10^5$ Pa at $273$ K. Osmotic pressure of the same solution at $283$ K is _______ $\times 10^4$ Nm $^{-2}$ . (Nearest integer)

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📖 Explanation

We need to find the osmotic pressure of a dilute solution at 283 K, given its osmotic pressure at 273 K. Recall the relationship between osmotic pressure and temperature. For a dilute solution, osmotic pressure is given by the van't Hoff equation: $\pi = CRT = \frac{n}{V}RT$ where $C$ is the molar concentration, $R$ is the gas constant, and $T$ is the absolute temperature. For the same solution (same $C$ ): $\pi \propto T$ Set up the proportion. $\frac{\pi_2}{\pi_1} = \frac{T_2}{T_1}$ $\pi_2 = \pi_1 \times \frac{T_2}{T_1} = 7 \times 10^5 \times \frac{283}{273}$ Calculate. $\pi_2 = 7 \times 10^5 \times \frac{283}{273} = \frac{7 \times 283}{273} \times 10^5$ $= \frac{1981}{273} \times 10^5 = 7.2564 \times 10^5 \text{ Pa}$ $= 72.564 \times 10^4 \text{ Nm}^{-2} \approx 73 \times 10^4 \text{ Nm}^{-2}$ The answer is 73 .

Q58JEE Main 2024NAT4MElectrochemistry

The mass of zinc produced by the electrolysis of zinc sulphate solution with a steady current of $0.015$ A for $15$ minutes is _______ $\times 10^{-4}$ g. (Atomic mass of zinc $= 65.4$ amu)

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📖 Explanation

$Q = It = 0.015 \times 15 \times 60 = 13.5$ C. Moles of electrons = $13.5/96500$ . Zn²⁺ + 2e⁻ \to Zn. Moles of Zn = $13.5/(2 \times 96500) = 6.994 \times 10^{-5}$ . Mass = $6.994 \times 10^{-5} \times 65.4 = 45.74 \times 10^{-4} \approx 46 \times 10^{-4}$ g. The answer is $\boxed{46}$ .

Q59JEE Main 2024NAT4MChemical Kinetics

For a reaction taking place in three steps at same temperature, overall rate constant $K = \frac{K_1 K_2}{K_3}$ . If $Ea_1, Ea_2$ and $Ea_3$ are $40, 50$ and $60$ kJ/mol respectively, the overall $Ea$ is _______ kJ/mol.

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📖 Explanation

We need to find the overall activation energy when the overall rate constant $K = \frac{K_1 K_2}{K_3}$ . Recall the Arrhenius equation. The Arrhenius equation relates the rate constant to activation energy: $K_i = A_i \, e^{-Ea_i/RT}$ Express the overall rate constant. $K = \frac{K_1 K_2}{K_3} = \frac{A_1 e^{-Ea_1/RT} \cdot A_2 e^{-Ea_2/RT}}{A_3 e^{-Ea_3/RT}}$ $= \frac{A_1 A_2}{A_3} \cdot e^{-(Ea_1 + Ea_2 - Ea_3)/RT}$ Identify the overall activation energy. Comparing with $K = A \, e^{-Ea/RT}$ , the overall activation energy is: $Ea = Ea_1 + Ea_2 - Ea_3$ Substitute the values. $Ea = 40 + 50 - 60 = 30 \text{ kJ/mol}$ The answer is 30 kJ/mol.

Q60JEE Main 2024NAT4MAldehydes, Ketones and Carboxylic Acids

From the compounds given below, number of compounds which give positive Fehling's test is _______ Benzaldehyde, Acetaldehyde, Acetone, Acetophenone, Methanal, 4-nitrobenzaldehyde, cyclohexane carbaldehyde.

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📖 Explanation

We need to count the number of compounds that give a positive Fehling's test from: Benzaldehyde, Acetaldehyde, Acetone, Acetophenone, Methanal, 4-nitrobenzaldehyde, Cyclohexane carbaldehyde. Recall what Fehling's test detects. Fehling's test uses Fehling's reagent (an alkaline solution of cupric tartrate, deep blue in colour) to detect aliphatic aldehydes . Aldehydes reduce $Cu^{2+}$ (blue) to $Cu_2O$ (red-brown precipitate). Important: aromatic aldehydes do NOT give a positive Fehling's test because the benzene ring reduces the reactivity of the aldehyde group. Ketones also do not give a positive test. Analyse each compound. - Benzaldehyde ( $C_6H_5CHO$ ): Aromatic aldehyde. Does NOT give positive Fehling's test. - Acetaldehyde ( $CH_3CHO$ ): Aliphatic aldehyde. Positive. - Acetone ( $CH_3COCH_3$ ): Ketone. Does NOT give positive test. - Acetophenone ( $C_6H_5COCH_3$ ): Aromatic ketone. Does NOT give positive test. - Methanal ( $HCHO$ ): Aliphatic aldehyde (simplest one). Positive. - 4-Nitrobenzaldehyde ( $O_2N-C_6H_4-CHO$ ): Aromatic aldehyde. Does NOT give positive Fehling's test. - Cyclohexane carbaldehyde ( $C_6H_{11}CHO$ ): Aliphatic aldehyde (cyclohexane ring is not aromatic). Positive. Count. Positive Fehling's test: Acetaldehyde, Methanal, Cyclohexane carbaldehyde = 3 . The answer is 3 .

Mathematics30 questions

Q61JEE Main 2024MCQ4MComplex Numbers

If $z = \frac{1}{2} - 2i$ , is such that $|z + 1| = \alpha z + \beta(1 + i)$ , $i = \sqrt{-1}$ and $\alpha, \beta \in R$ , then $\alpha + \beta$ is equal to

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📖 Explanation

$z = 1/2 - 2i$ . $|z+1| = |3/2 - 2i| = \sqrt{9/4+4} = \sqrt{25/4} = 5/2$ . $\alpha z + \beta(1+i) = \alpha/2 - 2\alpha i + \beta + \beta i = (\alpha/2+\beta) + (-2\alpha+\beta)i$ . This must equal 5/2 (real): $-2\alpha+\beta = 0$ and $\alpha/2+\beta = 5/2$ . $\beta = 2\alpha$ . $\alpha/2 + 2\alpha = 5/2$ , $5\alpha/2 = 5/2$ , $\alpha = 1, \beta = 2$ . $\alpha + \beta = 3$ . Option (2).

Q62JEE Main 2024MCQ4MSequences and Series

In an A.P., the sixth term $a_6 = 2$ . If the $a_1 a_4 a_5$ is the greatest, then the common difference of the A.P., is equal to

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📖 Explanation

Let the first term of the A.P. be $a_1$ and the common difference be $d$ .
We are given that the sixth term $a_6 = a_1 + 5d = 2$ , which implies $a_1 = 2 - 5d$ .
The product to be maximized is $P = a_1 a_4 a_5$ . Expressing the terms using $d$ :
$a_4 = a_1 + 3d = (2 - 5d) + 3d = 2 - 2d$
$a_5 = a_1 + 4d = (2 - 5d) + 4d = 2 - d$
So, $P(d) = (2 - 5d)(2 - 2d)(2 - d)$ . To find the value of $d$ that makes this product the greatest, we find the derivative of $P(d)$ with respect to $d$ and set it to zero.
The expanded form is $P(d) = -10d^3 + 34d^2 - 32d + 8$ . The derivative is $P'(d) = -30d^2 + 68d - 32$ .
Setting $P'(d) = 0$ gives $15d^2 - 34d + 16 = 0$ , which factors as $(5d-8)(3d-2)=0$ . The solutions are $d=\frac{8}{5}$ and $d=\frac{2}{3}$ .
The second derivative, $P''(d) = -60d + 68$ , is negative for $d=\frac{8}{5}$ (indicating a maximum) and positive for $d=\frac{2}{3}$ (a minimum). Thus, the product is greatest when $d = \frac{8}{5}$ .

Q63JEE Main 2024MCQ4MSequences and Series

If in a G.P. of $64$ terms, the \sum of all the terms is $7$ times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

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📖 Explanation

Sum of all 64 terms = $S_{64} = a\frac{r^{64}-1}{r-1}$ . Sum of odd terms (32 terms, GP with first term a, ratio r²): $S_{odd} = a\frac{r^{64}-1}{r^{2-}1}$ . $S_{64}/S_{odd} = 7$ . $\frac{r^{2-}1}{r-1} \cdot \frac{r-1}{1}$ ... Actually $S_{64}/S_{odd} = \frac{(r^{64}-1)/(r-1)}{(r^{64}-1)/(r^{2-}1)} = \frac{r^{2-}1}{r-1} = r+1 = 7$ . So $r = 6$ . Option (4).

Q64JEE Main 2024MCQ4MTrigonometric Ratios and Identities

If $\alpha$ , $-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$ is the solution of $4\cos\theta + 5\sin\theta = 1$ , then the value of $\tan\alpha$ is

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📖 Explanation

Begin with the equation $4\cos\alpha + 5\sin\alpha = 1$ . Divide by $\cos\alpha$ (which is non-zero for any solution) to get $4 + 5\tan\alpha = \sec\alpha$ . Squaring both sides gives $(4 + 5\tan\alpha)^2 = \sec^2\alpha$ . Using the identity $\sec^2\alpha = 1 + \tan^2\alpha$ , we can form a quadratic equation in terms of $\tan\alpha$ : $16 + 40\tan\alpha + 25\tan^2\alpha = 1 + \tan^2\alpha$ . This simplifies to $24\tan^2\alpha + 40\tan\alpha + 15 = 0$ . Using the quadratic formula, we find $\tan\alpha = \frac{-40 \pm \sqrt{40^2 - 4(24)(15)}}{2(24)} = \frac{-10 \pm \sqrt{10}}{12}$ . Since $-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$ , $\sec\alpha > 0$ , which implies from our first step that $4 + 5\tan\alpha > 0$ . Only the solution $\tan\alpha = \frac{-10 + \sqrt{10}}{12}$ satisfies this condition.

Q65JEE Main 2024MCQ4MStraight Lines and Pair of Straight Lines

Let $(5, \frac{a}{4})$ , be the circumcenter of a triangle with vertices $A(a, -2)$ , $B(a, 6)$ and $C(\frac{a}{4}, -2)$ . Let $\alpha$ denote the circumradius, $\beta$ denote the area and $\gamma$ denote the perimeter of the triangle. Then $\alpha + \beta + \gamma$ is

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📖 Explanation

We are given the circumcenter $(5, a/4)$ of a triangle with vertices $A(a, -2)$ , $B(a, 6)$ , and $C(a/4, -2)$ . We need to find $\alpha + \beta + \gamma$ where $\alpha$ is the circumradius, $\beta$ is the area, and $\gamma$ is the perimeter. Find the value of $a$ . Since $A(a, -2)$ and $B(a, 6)$ have the same x-coordinate, side $AB$ is vertical (along $x = a$ ). The perpendicular bisector of $AB$ is horizontal, passing through the midpoint $(a, 2)$ , so it is the line $y = 2$ . The circumcenter lies on this perpendicular bisector, so: $\frac{a}{4} = 2 \implies a = 8$ Find the vertices and sides. $A(8, -2)$ , $B(8, 6)$ , $C(2, -2)$ . Circumcenter: $(5, 2)$ . $AB = |6 - (-2)| = 8$ $AC = |8 - 2| = 6$ $BC = \sqrt{(8-2)^2 + (6-(-2))^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ Note: $6^2 + 8^2 = 100 = 10^2$ , so this is a right triangle with the right angle at $A$ . Calculate circumradius $\alpha$ . For a right triangle, the circumradius equals half the hypotenuse: $\alpha = \frac{BC}{2} = \frac{10}{2} = 5$ Verification: Distance from circumcenter $(5,2)$ to $A(8,-2)$ : $\sqrt{9+16} = 5$ . Correct. Calculate area $\beta$ . $\beta = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 8 \times 6 = 24$ Calculate perimeter $\gamma$ . $\gamma = AB + AC + BC = 8 + 6 + 10 = 24$ Final answer. $\alpha + \beta + \gamma = 5 + 24 + 24 = 53$ The correct answer is Option (2): 53 .

Q66JEE Main 2024MCQ4MStraight Lines and Pair of Straight Lines

In a $\Delta ABC$ , suppose $y = x$ is the equation of the bisector of the angle $B$ and the equation of the side $AC$ is $2x - y = 2$ . If $2AB = BC$ and the point $A$ and $B$ are respectively $(4, 6)$ and $(\alpha, \beta)$ , then $\alpha + 2\beta$ is equal to

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📖 Explanation

The angle bisector from vertex $B$ is given by the line $y = x$ , and it intersects side $AC$ at a point $D$ . The equation of side $AC$ is $2x - y = 2$ , and point $A$ is $(4, 6)$ . Since $D$ lies on both the angle bisector $y = x$ and side $AC$ , substitute $y = x$ into $2x - y = 2$ : $2x - x = 2 \implies x = 2, \quad y = 2$ Thus, $D = (2, 2)$ . Given $2AB = BC$ , it follows that $AB/BC = 1/2$ . By the angle bisector theorem, $AD/DC = AB/BC = 1/2$ . Therefore, $D$ divides $AC$ \in the ratio $AD:DC = 1:2$ . Let $C = (x, y)$ . Using the section formula, the coordinates of $D$ are: $D_x = \frac{2 \cdot x_A + 1 \cdot x_C}{1 + 2} = \frac{2 \cdot 4 + x_C}{3} = \frac{8 + x_C}{3}$ $D_y = \frac{2 \cdot y_A + 1 \cdot y_C}{1 + 2} = \frac{2 \cdot 6 + y_C}{3} = \frac{12 + y_C}{3}$ Substitute $D = (2, 2)$ : $\frac{8 + x_C}{3} = 2 \implies 8 + x_C = 6 \implies x_C = -2$ $\frac{12 + y_C}{3} = 2 \implies 12 + y_C = 6 \implies y_C = -6$ Thus, $C = (-2, -6)$ . Since the angle bisector $y = x$ passes through $B$ , and $B = (\alpha, \beta)$ , it follows that $\beta = \alpha$ , so $B = (\alpha, \alpha)$ . Given $2AB = BC$ , compute the distances: $AB = \sqrt{(\alpha - 4)^2 + (\alpha - 6)^2}$ $BC = \sqrt{(\alpha - (-2))^2 + (\alpha - (-6))^2} = \sqrt{(\alpha + 2)^2 + (\alpha + 6)^2}$ So: $2 \sqrt{(\alpha - 4)^2 + (\alpha - 6)^2} = \sqrt{(\alpha + 2)^2 + (\alpha + 6)^2}$ Square both sides: $4 \left[ (\alpha - 4)^2 + (\alpha - 6)^2 \right]\$ = (\alpha + 2)^2 + (\alpha + 6)^2$ Expand: Left side: $4 \left[ (\alpha^2 - 8\alpha + 16) + (\alpha^2 - 12\alpha + 36) \right]\$ = 4 \left[ 2\alpha^2 - 20\alpha + 52 \right]\$ = 8\alpha^2 - 80\alpha + 208$ Right side: $(\alpha^2 + 4\alpha + 4) + (\alpha^2 + 12\alpha + 36) = 2\alpha^2 + 16\alpha + 40$ Set equal: $8\alpha^2 - 80\alpha + 208 = 2\alpha^2 + 16\alpha + 40$ Bring all terms to the left: $6\alpha^2 - 96\alpha + 168 = 0$ Divide by 6: $\alpha^2 - 16\alpha + 28 = 0$ Solve using the quadratic formula: $\alpha = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 28}}{2} = \frac{16 \pm \sqrt{256 - 112}}{2} = \frac{16 \pm \sqrt{144}}{2} = \frac{16 \pm 12}{2}$ Thus: $\alpha = \frac{16 + 12}{2} = 14, \quad \alpha = \frac{16 - 12}{2} = 2$ So $B = (14, 14)$ or $B = (2, 2)$ . If $B = (2, 2)$ , it lies on $AC$ (since $2(2) - 2 = 2$ ), making the triangle degenerate. Thus, discard $B = (2, 2)$ . Therefore, $B = (14, 14)$ , so $\alpha = 14$ , $\beta = 14$ , and: $\alpha + 2\beta = 14 + 2 \cdot 14 = 14 + 28 = 42$ Verification: $AB = \sqrt{(14-4)^2 + (14-6)^2} = \sqrt{100 + 64} = \sqrt{164} = 2\sqrt{41}$ $BC = \sqrt{(14+2)^2 + (14+6)^2} = \sqrt{256 + 400} = \sqrt{656} = 4\sqrt{41}$ $2AB = 2 \cdot 2\sqrt{41} = 4\sqrt{41} = BC$ , satisfying the condition. The angle bisector $y = x$ passes through $B(14, 14)$ and $D(2, 2)$ , and the angle bisector theorem holds as $AD/DC = 1/2$ and $AB/BC = 1/2$ . Thus, $\alpha + 2\beta = 42$ .

Q67JEE Main 2024MCQ4MLimits, Continuity and Differentiability

$\lim_{x \rightarrow \frac{\pi}{2}} \left(\frac{1}{(x - \frac{\pi}{2})^2} \int_{x^3}^{(\frac{\pi}{2})^3} \cos\left(\frac{1}{t^3}\right) dt\right)$ is equal to

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📖 Explanation

This limit is an indeterminate form of $\frac{0}{0}$ , so we can use L'Hôpital's Rule. The integrand is likely a typo and should be $\cos(t^{1/3})$ for the limit to result in one of the given options. Let $N(x) = \int_{x^3}^{(\frac{\pi}{2})^3} \cos(t^{1/3}) dt$ and $D(x) = (x - \frac{\pi}{2})^2$ . By the Fundamental Theorem of Calculus and the chain rule, $N'(x) = -\cos((x^3)^{1/3}) \cdot \frac{d}{dx}(x^3) = -3x^2 \cos(x)$ . The derivative of the denominator is $D'(x) = 2(x - \frac{\pi}{2})$ .

Applying L'Hôpital's Rule once gives the limit $L = \lim_{x \rightarrow \frac{\pi}{2}} \frac{-3x^2 \cos(x)}{2(x - \frac{\pi}{2})}$ . This is still a $\frac{0}{0}$ form. Applying L'Hôpital's Rule a second time yields $L = \lim_{x \rightarrow \frac{\pi}{2}} \frac{-6x\cos(x) + 3x^2\sin(x)}{2}$ . Evaluating this limit by substituting $x = \frac{\pi}{2}$ gives $\frac{-6(\frac{\pi}{2})\cos(\frac{\pi}{2}) + 3(\frac{\pi}{2})^2\sin(\frac{\pi}{2})}{2}$ . This simplifies to $\frac{0 + 3(\frac{\pi^2}{4})(1)}{2} = \frac{3\pi^2}{8}$ .

Q68JEE Main 2024MCQ4MSets and Relations

Let $R$ be a relation on $Z \times Z$ defined by $(a, b)R(c, d)$ if and only if $ad - bc$ is divisible by $5$ . Then $R$ is

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📖 Explanation

The relation $R$ on $\mathbb{Z} \times \mathbb{Z}$ is defined by $(a, b)R(c, d)$ if and only if $ad - bc$ is divisible by 5. Check Reflexivity. $(a, b)R(a, b) \iff ab - ba = 0$ is divisible by 5. $\checkmark$ $R$ is reflexive. Check Symmetry. If $(a, b)R(c, d)$ , then $5 \mid (ad - bc)$ . We need to check: is $5 \mid (cb - da)$ ? $cb - da = -(ad - bc)$ . Since $5 \mid (ad - bc)$ , we have $5 \mid (-(ad-bc)) = cb - da$ . $\checkmark$ $R$ is symmetric. Check Transitivity. We need: if $(a,b)R(c,d)$ and $(c,d)R(e,f)$ , then $(a,b)R(e,f)$ . Given: $5 \mid (ad - bc)$ and $5 \mid (cf - de)$ . Need to show: $5 \mid (af - be)$ . Counterexample: Take $(a, b) = (1, 1)$ , $(c, d) = (5, 5)$ , $(e, f) = (1, 2)$ . Check $(1,1)R(5,5)$ : $ad - bc = 1 \cdot 5 - 1 \cdot 5 = 0$ . Divisible by 5. $\checkmark$ Check $(5,5)R(1,2)$ : $cf - de = 5 \cdot 2 - 5 \cdot 1 = 5$ . Divisible by 5. $\checkmark$ Check $(1,1)R(1,2)$ : $af - be = 1 \cdot 2 - 1 \cdot 1 = 1$ . NOT divisible by 5. $\times$ So transitivity fails. $R$ is reflexive and symmetric but not transitive . The correct answer is Option (1): $\boxed{\text{Reflexive and symmetric but not transitive}}$ .

Q69JEE Main 2024MCQ4MMatrices and Determinants

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix}$ and $|2A|^3 = 2^{21}$ where $\alpha, \beta \in Z$ , Then a value of $\alpha$ is

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📖 Explanation

Given matrix $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix}$ and the condition $|2A|^3 = 2^{21}$ , with $\alpha, \beta \in \mathbb{Z}$ . First, recall that for an $n \times n$ matrix, $|kA| = k^n |A|$ . Here, $A$ is a 3x3 matrix, so $n=3$ . Thus, $|2A| = 2^3 |A| = 8 |A|$ . Substitute into the given condition: $(8 |A|)^3 = 2^{21}$ . Since $8 = 2^3$ , this becomes: $(2^3 |A|)^3 = 2^{21}$ . Simplifying the exponents: $2^{9} |A|^3 = 2^{21}$ . Divide both sides by $2^9$ : $|A|^3 = 2^{12}$ . Taking the cube root: $|A| = 2^{4} = 16$ . Therefore, the determinant of $A$ is 16. Now, compute the determinant of $A$ . Expanding along the first row: $|A| = 1 \cdot \begin{vmatrix} \alpha & \beta \\ \beta & \alpha \end{vmatrix} - 0 \cdot (\ldots) + 0 \cdot (\ldots) = \alpha^2 - \beta^2$ . So, $\alpha^2 - \beta^2 = 16$ . Factorizing: $(\alpha - \beta)(\alpha + \beta) = 16$ . Since $\alpha$ and $\beta$ are integers, both $\alpha - \beta$ and $\alpha + \beta$ are integers and must be even (as their product is even and both must have the same parity). Set $x = \alpha - \beta$ and $y = \alpha + \beta$ , so $x \cdot y = 16$ , and both $x$ and $y$ are even integers. The even factor pairs of 16 are: $(2,8)$ , $(4,4)$ , $(8,2)$ , $(-2,-8)$ , $(-4,-4)$ , $(-8,-2)$ . Solve for $\alpha$ and $\beta$ \in each pair using $\alpha = \frac{x + y}{2}$ and $\beta = \frac{y - x}{2}$ : For $(2,8)$ : $\alpha = \frac{2+8}{2} = 5$ , $\beta = \frac{8-2}{2} = 3$ For $(4,4)$ : $\alpha = \frac{4+4}{2} = 4$ , $\beta = \frac{4-4}{2} = 0$ For $(8,2)$ : $\alpha = \frac{8+2}{2} = 5$ , $\beta = \frac{2-8}{2} = -3$ For $(-2,-8)$ : $\alpha = \frac{-2-8}{2} = -5$ , $\beta = \frac{-8-(-2)}{2} = -3$ For $(-4,-4)$ : $\alpha = \frac{-4-4}{2} = -4$ , $\beta = \frac{-4-(-4)}{2} = 0$ For $(-8,-2)$ : $\alpha = \frac{-8-2}{2} = -5$ , $\beta = \frac{-2-(-8)}{2} = 3$ The possible values of $\alpha$ are $5, 4, -5, -4$ . Comparing with the options: A. $3$ , B. $5$ , C. $17$ , D. $9$ , the value $\alpha = 5$ is present (option B). Verification for $\alpha = 5$ , $\beta = 3$ : $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 5 & 3 \\ 0 & 3 & 5 \end{bmatrix}$ , $|A| = 5^2 - 3^2 = 25 - 9 = 16$ . Then $|2A| = 8 \times 16 = 128$ , and $|2A|^3 = 128^3 = (2^7)^3 = 2^{21}$ , which satisfies the condition. Checking other options: For $\alpha = 3$ : $\alpha^2 - \beta^2 = 9 - \beta^2 = 16 \implies \beta^2 = -7$ , not real. For $\alpha = 17$ : $289 - \beta^2 = 16 \implies \beta^2 = 273$ , not a perfect square. For $\alpha = 9$ : $81 - \beta^2 = 16 \implies \beta^2 = 65$ , not a perfect square. Thus, the only valid value from the options is $\alpha = 5$ .

Q70JEE Main 2024MCQ4MMatrices and Determinants

Let A be a square matrix such that $AA^T = I$ . Then $\frac{1}{2}A\left[(A + A^T)^2 + (A - A^T)^2\right]$ is equal to

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📖 Explanation

$A$ is a square matrix with $AA^T = I$ , i.e., $A$ is orthogonal, so $A^T = A^{-1}$ . Find $\frac{1}{2}A\left[(A+A^T)^2 + (A-A^T)^2\right]$ . Expand the squares. $(A+A^T)^2 = A^2 + AA^T + A^TA + (A^T)^2$ $(A-A^T)^2 = A^2 - AA^T - A^TA + (A^T)^2$ Add the two expressions. $(A+A^T)^2 + (A-A^T)^2 = 2A^2 + 2(A^T)^2$ Multiply by $\frac{1}{2}A$ . $\frac{1}{2}A \cdot \$[2A^2 + 2(A^T)^2]\$ = A \cdot \$[A^2 + (A^T)^2]\$ = A^3 + A(A^T)^2$ Simplify $A(A^T)^2$ . $A(A^T)^2 = (AA^T) \cdot A^T = I \cdot A^T = A^T$ Final result. $\frac{1}{2}A\$[(A+A^T)^2 + (A-A^T)^2]\$ = A^3 + A^T$ The correct answer is Option 4 : $A^3 + A^T$ .

Q71JEE Main 2024MCQ4MFunctions

If $

f(x) = \begin{cases} 2 + 2x, & -1 \leq x < 0 \\ 1 - \frac{x}{3}, & 0 \leq x \leq 3 \end{cases}

$;$

g(x) = \begin{cases} -x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{cases}

$, then range of $(f \circ g(x))$ is

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📖 Explanation

We need to find the range of the composite function $f \circ g(x) = f(g(x))$ . Determine the domain and range of $g(x)$ . $

g(x) = \begin{cases} -x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{cases}

$ For $x \in [-3, 0]$ : $g(x) = -x \in [0, 3]$ For $x \in (0, 1]$ : $g(x) = x \in (0, 1]$ So the range of $g$ is $[0, 3]$ . Evaluate $f(g(x))$ where $g(x) \in [0, 3]$ . Since $g(x) \in [0, 3]$ , we use the second piece of $f$ : $f(t) = 1 - \frac{t}{3}$ for $0 \leq t \leq 3$ When $t = 0$ : $f(0) = 1$ When $t = 3$ : $f(3) = 1 - 1 = 0$ Since $f(t) = 1 - t/3$ is continuous and decreasing on $[0, 3]$ , its range on this interval is $[0, 1]$ . Verify that all values \in $[0, 1]$ are achieved. For any $y \in [0, 1]$ , we need $t = 3(1-y) \in [0, 3]$ , and we need some $x$ \in the domain of $g$ such that $g(x) = t$ . Since $g$ maps $[-3, 0]$ onto $[0, 3]$ (via $g(x) = -x$ ), every value $t \in [0, 3]$ is achieved. Therefore, $f(g(x))$ achieves all values \in $[0, 1]$ , including both endpoints. The range of $f \circ g(x)$ is $[0, 1]$ . The correct answer is Option (3): $[0, 1]$ .

Q72JEE Main 2024MCQ4MFunctions

Consider the function $f : [\frac{1}{2}, 1] \to R$ defined by $f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1$ . Consider the statements (I) The curve $y = f(x)$ intersects the $x$ -axis exactly at one point (II) The curve $y = f(x)$ intersects the $x$ -axis at $x = \cos\frac{\pi}{12}$ Then

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📖 Explanation

$f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1$ on $[\frac{1}{2}, 1]$ . Statement (I): The curve intersects the x-axis exactly at one point. Statement (II): The curve intersects the x-axis at $x = \cos\frac{\pi}{12}$ . Check Statement (II) by recognising the triple angle formula. Recall that $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ . If $x = \cos\theta$ , then $4x^3 - 3x = \cos 3\theta$ . $f(x) = \sqrt{2}(4x^3 - 3x) - 1 = \sqrt{2}\cos 3\theta - 1$ Setting $f(x) = 0$ : $\sqrt{2}\cos 3\theta = 1$ , so $\cos 3\theta = \frac{1}{\sqrt{2}}$ . $3\theta = \frac{\pi}{4} \implies \theta = \frac{\pi}{12}$ $x = \cos\frac{\pi}{12} \approx 0.9659$ , which lies \in $[\frac{1}{2}, 1]$ . Statement (II) is correct . Check Statement (I) - uniqueness. $f'(x) = 12\sqrt{2}x^2 - 3\sqrt{2} = 3\sqrt{2}(4x^2 - 1)$ $f'(x) = 0 \implies x = \frac{1}{2}$ (the only critical point \in the domain). $f(\frac{1}{2}) = 4\sqrt{2} \cdot \frac{1}{8} - 3\sqrt{2} \cdot \frac{1}{2} - 1 = \frac{\sqrt{2}}{2} - \frac{3\sqrt{2}}{2} - 1 = -\sqrt{2} - 1 < 0$ $f(1) = 4\sqrt{2} - 3\sqrt{2} - 1 = \sqrt{2} - 1 > 0$ Since $f(\frac{1}{2}) < 0$ and $f(1) > 0$ , and $f'(x) > 0$ for $x > \frac{1}{2}$ , the function is strictly increasing on $(\frac{1}{2}, 1]$ . There is exactly one root. Statement (I) is correct . Both statements are correct. The answer is Option 4 .

Q73JEE Main 2024MCQ4MDifferentiation

Suppose $f(x) = \frac{(2^x + 2^{-x})\tan x \sqrt{\tan^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^3}$ . Then the value of $f'(0)$ is equal to

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📖 Explanation

To find $f'(0)$ for the function $f(x) = \frac{(2^x + 2^{-x})\tan x \sqrt{\tan^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^3},$ we use the definition of the derivative at $x = 0$ : $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}.$ First, evaluate $f(0)$ : Numerator: $(2^0 + 2^{-0}) \tan 0 \sqrt{\tan^{-1}(0^2 - 0 + 1)} = (1 + 1) \cdot 0 \cdot \sqrt{\tan^{-1}(1)} = 2 \cdot 0 \cdot \sqrt{\pi/4} = 0$ , since $\tan^{-1}(1) = \pi/4$ . Denominator: $(7(0)^2 + 3(0) + 1)^3 = 1^3 = 1$ . Thus, $f(0) = 0/1 = 0$ . So, $f'(0) = \lim_{h \to 0} \frac{f(h)}{h} = \lim_{h \to 0} \frac{1}{h} \cdot \frac{(2^h + 2^{-h}) \tan h \sqrt{\tan^{-1}(h^2 - h + 1)}}{(7h^2 + 3h + 1)^3}.$ Rewrite the expression as: $\frac{f(h)}{h} = (2^h + 2^{-h}) \cdot \left( \frac{\tan h}{h} \right) \cdot \sqrt{\tan^{-1}(h^2 - h + 1)} \cdot \frac{1}{(7h^2 + 3h + 1)^3}.$ Now, evaluate the limit as $h \to 0$ for each part: $\lim_{h \to 0} (2^h + 2^{-h}) = 2^0 + 2^0 = 1 + 1 = 2$ . $\lim_{h \to 0} \frac{\tan h}{h} = 1$ , since $\tan h \sim h$ near 0. $\lim_{h \to 0} \sqrt{\tan^{-1}(h^2 - h + 1)} = \sqrt{\tan^{-1}(1)} = \sqrt{\pi/4} = \sqrt{\pi}/2$ , as $\tan^{-1}(1) = \pi/4$ . $\lim_{h \to 0} \frac{1}{(7h^2 + 3h + 1)^3} = \frac{1}{(1)^3} = 1$ , since the denominator approaches 1. Therefore, the limit is: $f'(0) = 2 \cdot 1 \cdot \frac{\sqrt{\pi}}{2} \cdot 1 = \sqrt{\pi}.$ The function and its components are well-behaved at $x = 0$ : $2^x + 2^{-x}$ is continuous and differentiable. $\tan x$ is continuous and differentiable near 0 (as $\cos x \neq 0$ ). $\tan^{-1}(x^2 - x + 1)$ is continuous and differentiable since $x^2 - x + 1 \geq 3/4 > 0$ for all $x$ , and its square root is defined and differentiable. The denominator $(7x^2 + 3x + 1)^3$ is never zero (minimum value at $x = -3/14$ is positive). Thus, the limit exists and equals $\sqrt{\pi}$ . The value of $f'(0)$ is $\sqrt{\pi}$ , which corresponds to option C.

Q74JEE Main 2024MCQ4MDefinite Integrals

If the value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{x^2 \cos x}{1 + \pi^x} + \frac{1 + \sin^2 x}{1 + e^{(\sin x)^{2023}}}\right) dx = \frac{\pi}{4}(\pi + a) - 2$ , then the value of $a$ is

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📖 Explanation

We need to evaluate $\int_{-\pi/2}^{\pi/2}\left(\frac{x^2\cos x}{1+\pi^x} + \frac{1+\sin^2 x}{1+e^{(\sin x)^{2023}}}\right)dx$ . Evaluate $I_1 = \int_{-\pi/2}^{\pi/2}\frac{x^2\cos x}{1+\pi^x}\,dx$ . We use the property: for even function $f(x)$ and constant $b > 0$ , $\int_{-a}^{a}\frac{f(x)}{1+b^x}\,dx = \int_0^a f(x)\,dx$ Here $f(x) = x^2\cos x$ is even (product of two even functions) and $b = \pi$ . $I_1 = \int_0^{\pi/2} x^2\cos x\,dx$ Using integration by parts twice: $\int x^2\cos x\,dx = x^2\sin x - 2\int x\sin x\,dx = x^2\sin x - 2\left[-x\cos x + \sin x\right]$ $= x^2\sin x + 2x\cos x - 2\sin x$ Evaluating from $0$ to $\frac{\pi}{2}$ : $I_1 = \left[\frac{\pi^2}{4} \cdot 1 + 2 \cdot \frac{\pi}{2} \cdot 0 - 2 \cdot 1\right]\$ - [0] = \frac{\pi^2}{4} - 2$ Evaluate $I_2 = \int_{-\pi/2}^{\pi/2}\frac{1+\sin^2 x}{1+e^{(\sin x)^{2023}}}\,dx$ . Let $g(x) = (\sin x)^{2023}$ . Since 2023 is odd, $g(-x) = (-\sin x)^{2023} = -g(x)$ , so $g$ is an odd function. The function $h(x) = 1 + \sin^2 x$ is even. Using the same property (with $e^{g(x)}$ playing the role of $b^x$ where $g$ is odd): $I_2 = \int_0^{\pi/2}(1+\sin^2 x)\,dx$ $= \int_0^{\pi/2} 1\,dx + \int_0^{\pi/2}\sin^2 x\,dx = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$ (Using $\int_0^{\pi/2}\sin^2 x\,dx = \frac{\pi}{4}$ .) Combine and compare. $I_1 + I_2 = \frac{\pi^2}{4} - 2 + \frac{3\pi}{4} = \frac{\pi}{4}(\pi + 3) - 2$ Comparing with $\frac{\pi}{4}(\pi + a) - 2$ , we get: $a = 3$ The correct answer is Option (1): $\boxed{3}$ .

Q75JEE Main 2024MCQ4MIndefinite Integrals

For $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$ , if $y(x) = \int \frac{\csc x + \sin x}{\csc x \sec x + \tan x \sin^2 x} dx$ and $\lim_{x \to (\frac{\pi}{2})^-} y(x) = 0$ then $y(\frac{\pi}{4})$ is equal to

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📖 Explanation

We need to evaluate $y(x) = \int\frac{\csc x + \sin x}{\csc x\sec x + \tan x\sin^2 x}\,dx$ with $\lim_{x \to (\pi/2)^-} y(x) = 0$ . Simplify the integrand. Multiply numerator and denominator by $\sin x\cos x$ : Numerator: $(\csc x + \sin x) \cdot \sin x\cos x = \cos x + \sin^2 x\cos x = \cos x(1 + \sin^2 x)$ Denominator: $\left(\frac{1}{\sin x\cos x} + \frac{\sin x}{\cos x} \cdot \sin^2 x\right) \cdot \sin x\cos x = 1 + \sin^4 x$ So the integrand simplifies to $\frac{\cos x(1 + \sin^2 x)}{1 + \sin^4 x}$ . Substitute $t = \sin x$ , $dt = \cos x\,dx$ . $y = \int\frac{1 + t^2}{1 + t^4}\,dt$ Evaluate the integral. Divide numerator and denominator by $t^2$ : $\frac{1 + 1/t^2}{t^2 + 1/t^2} = \frac{1 + 1/t^2}{(t - 1/t)^2 + 2}$ Let $u = t - \frac{1}{t}$ , so $du = \left(1 + \frac{1}{t^2}\right)dt$ . $y = \int\frac{du}{u^2 + 2} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x - \csc x}{\sqrt{2}}\right) + C$ Apply the boundary condition. As $x \to \left(\frac{\pi}{2}\right)^-$ : $\sin x \to 1$ , so $\sin x - \csc x \to 0$ . $\lim_{x \to (\pi/2)^-} y(x) = \frac{1}{\sqrt{2}}\tan^{-1}(0) + C = C = 0$ So $C = 0$ . Evaluate $y(\pi/4)$ . At $x = \frac{\pi}{4}$ : $\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ , $\csc\frac{\pi}{4} = \sqrt{2}$ . $\sin x - \csc x = \frac{1}{\sqrt{2}} - \sqrt{2} = \frac{1 - 2}{\sqrt{2}} = \frac{-1}{\sqrt{2}}$ $y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{-1/\sqrt{2}}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(-\frac{1}{2}\right)$ The correct answer is Option (4): $\boxed{\frac{1}{\sqrt{2}}\tan^{-1}\left(-\frac{1}{2}\right)}$ .

Q76JEE Main 2024MCQ4MDifferential Equations

A function $y = f(x)$ satisfies $f(x)\sin 2x + \sin x - (1 + \cos^2 x)f'(x) = 0$ with condition $f(0) = 0$ . Then $f(\frac{\pi}{2})$ is equal to

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📖 Explanation

Given the ODE: $f(x)\sin 2x + \sin x - (1 + \cos^2 x)f'(x) = 0$ with $f(0) = 0$ . Rearrange the equation. $(1 + \cos^2 x)f'(x) - \sin 2x \cdot f(x) = \sin x$ Recognize the exact derivative. Note that $\frac{d}{dx}\$[\cos^2 x]\$ = -2\sin x\cos x = -\sin 2x$ . Therefore: $\frac{d}{dx}\left[(1 + \cos^2 x)f(x)\right]\$ = (1 + \cos^2 x)f'(x) + (-\sin 2x)f(x)$ This is exactly the left-hand side! So the equation becomes: $\frac{d}{dx}\left[(1 + \cos^2 x)f(x)\right]\$ = \sin x$ Integrate both sides. $(1 + \cos^2 x)f(x) = \int \sin x\,dx = -\cos x + C$ Apply the initial condition $f(0) = 0$ . $(1 + \cos^2 0) \cdot 0 = -\cos 0 + C$ $0 = -1 + C \implies C = 1$ So the solution is: $(1 + \cos^2 x)f(x) = 1 - \cos x$ Find $f(\pi/2)$ . $(1 + \cos^2(\pi/2)) \cdot f(\pi/2) = 1 - \cos(\pi/2)$ $(1 + 0) \cdot f(\pi/2) = 1 - 0$ $f(\pi/2) = 1$ The correct answer is Option (1): $\boxed{1}$ .

Q77JEE Main 2024MCQ4MVector Algebra

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b}$ and $\vec{c}$ are non-collinear. If $\vec{a} + 5\vec{b}$ is collinear with $\vec{c}$ , $\vec{b} + 6\vec{c}$ is collinear with $\vec{a}$ and $\vec{a} + \alpha\vec{b} + \beta\vec{c} = \vec{0}$ , then $\alpha + \beta$ is equal to

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📖 Explanation

$\vec{a} + 5\vec{b}$ is collinear with $\vec{c}$ , and $\vec{b} + 6\vec{c}$ is collinear with $\vec{a}$ . Also $\vec{a} + \alpha\vec{b} + \beta\vec{c} = \vec{0}$ . Express the collinearity conditions. Since $\vec{a} + 5\vec{b}$ is collinear with $\vec{c}$ : $\vec{a} + 5\vec{b} = \lambda\vec{c}$ for some scalar $\lambda$ ... (i) Since $\vec{b} + 6\vec{c}$ is collinear with $\vec{a}$ : $\vec{b} + 6\vec{c} = \mu\vec{a}$ for some scalar $\mu$ ... (ii) Express $\vec{a}$ from equation (ii). $\vec{a} = \frac{1}{\mu}\vec{b} + \frac{6}{\mu}\vec{c}$ Substitute into equation (i). $\frac{1}{\mu}\vec{b} + \frac{6}{\mu}\vec{c} + 5\vec{b} = \lambda\vec{c}$ $\left(\frac{1}{\mu} + 5\right)\vec{b} + \left(\frac{6}{\mu} - \lambda\right)\vec{c} = \vec{0}$ Since $\vec{b}$ and $\vec{c}$ are non-collinear, both coefficients must be zero: $\frac{1}{\mu} + 5 = 0 \implies \mu = -\frac{1}{5}$ $\frac{6}{\mu} = \lambda \implies \lambda = 6 \times (-5) = -30$ Find $\alpha$ and $\beta$ . From (i): $\vec{a} = -5\vec{b} + \lambda\vec{c} = -5\vec{b} - 30\vec{c}$ So $\vec{a} + 5\vec{b} + 30\vec{c} = \vec{0}$ Therefore $\alpha = 5$ and $\beta = 30$ . $\alpha + \beta = 5 + 30 = 35$ The answer is Option 1 : 35.

Q78JEE Main 2024MCQ4MVector Algebra

Let $O$ be the origin and the position vector of $A$ and $B$ be $2\hat{i} + 2\hat{j} + \hat{k}$ and $2\hat{i} + 4\hat{j} + 4\hat{k}$ respectively. If the internal bisector of $\angle AOB$ meets the line $AB$ at $C$ , then the length of $OC$ is

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📖 Explanation

We need to find the length of $OC$ , where $C$ is the point where the internal bisector of $\angle AOB$ meets line $AB$ . Find the position vectors and distances. $\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$ , so $|\vec{OA}| = \sqrt{4+4+1} = 3$ $\vec{OB} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ , so $|\vec{OB}| = \sqrt{4+16+16} = 6$ Apply the angle bisector theorem. By the angle bisector theorem, the internal bisector of $\angle AOB$ divides $AB$ \in the ratio $OA:OB = 3:6 = 1:2$ . Find the position vector of $C$ . Using the section formula (internal division \in ratio $1:2$ ): $\vec{OC} = \frac{1 \cdot \vec{OB} + 2 \cdot \vec{OA}}{1 + 2} = \frac{(2\hat{i}+4\hat{j}+4\hat{k}) + 2(2\hat{i}+2\hat{j}+\hat{k})}{3}$ $= \frac{(2+4)\hat{i} + (4+4)\hat{j} + (4+2)\hat{k}}{3} = \frac{6\hat{i} + 8\hat{j} + 6\hat{k}}{3} = 2\hat{i} + \frac{8}{3}\hat{j} + 2\hat{k}$ Calculate $|OC|$ . $|OC| = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}} = \sqrt{\frac{72 + 64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{136}}{3} = \frac{2\sqrt{34}}{3}$ The correct answer is Option (2): $\frac{2}{3}\sqrt{34}$ .

Q79JEE Main 2024MCQ4MThree Dimensional Geometry

Let $PQR$ be a triangle with $R(-1, 4, 2)$ . Suppose $M(2, 1, 2)$ is the mid point of $PQ$ . The distance of the centroid of $\Delta PQR$ from the point of intersection of the line $\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$ and $\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$ is

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📖 Explanation

Given triangle PQR with vertex R(-1, 4, 2) and M(2, 1, 2) as the midpoint of PQ. The centroid of triangle PQR needs to be found, and then its distance from the intersection point of the lines $\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$ and $\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$ . Let the coordinates of P and Q be $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ . Since M(2, 1, 2) is the midpoint of PQ: $\frac{x_1 + x_2}{2} = 2 \implies x_1 + x_2 = 4$ $\frac{y_1 + y_2}{2} = 1 \implies y_1 + y_2 = 2$ $\frac{z_1 + z_2}{2} = 2 \implies z_1 + z_2 = 4$ The centroid G of triangle PQR is the average of the coordinates of P, Q, and R(-1, 4, 2): $G_x = \frac{x_1 + x_2 + (-1)}{3} = \frac{4 + (-1)}{3} = \frac{3}{3} = 1$ $G_y = \frac{y_1 + y_2 + 4}{3} = \frac{2 + 4}{3} = \frac{6}{3} = 2$ $G_z = \frac{z_1 + z_2 + 2}{3} = \frac{4 + 2}{3} = \frac{6}{3} = 2$ Thus, the centroid G is at (1, 2, 2). Next, find the intersection point of the two lines. The first line is $\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$ . Since the denominator of x is 0, x is constant: x = 2. Let the parameter be \lambda : $\frac{y}{2} = \lambda \implies y = 2\lambda$ $\frac{z+3}{-1} = \lambda \implies z = - \lambda - 3$ So parametric equations are: x = 2, y = 2\lambda , z = -\lambda - 3. The second line is $\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$ . Let the parameter be \mu : $\frac{x-1}{1} = \mu \implies x = 1 + \mu$ $\frac{y+3}{-3} = \mu \implies y = -3 - 3\mu$ $\frac{z+1}{1} = \mu \implies z = -1 + \mu$ So parametric equations are: x = 1 + \mu , y = -3 - 3\mu , z = -1 + \mu . Set the coordinates equal to find intersection: From x-coordinates: 2 = 1 + \mu ⟹ \mu = 1 Substitute \mu = 1 into z-coordinate: z = -1 + 1 = 0 Set z from first line: -\lambda - 3 = 0 ⟹ \lambda = -3 Substitute \lambda = -3 into y-coordinate: y = 2(-3) = -6 Verify with second line y-coordinate: y = -3 - 3(1) = -6, which matches. Thus, the intersection point S is (2, -6, 0). The distance between G(1, 2, 2) and S(2, -6, 0) is: $d = \sqrt{(2-1)^2 + (-6-2)^2 + (0-2)^2} = \sqrt{(1)^2 + (-8)^2 + (-2)^2} = \sqrt{1 + 64 + 4} = \sqrt{69}$ Therefore, the distance is $\sqrt{69}$ .

Q80JEE Main 2024MCQ4MProbability

A fair die is thrown until $2$ appears. Then the probability, that $2$ appears in even number of throws, is

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📖 Explanation

Let $P$ be the probability that a 2 appears on an even-numbered throw. For this to occur, the first throw must not be a 2, which has a probability of $\frac{5}{6}$ . After the first throw fails, the sequence of throws effectively resets. For the total number of throws to be even, the subsequent appearance of a 2 must happen in an odd number of throws from this new starting point. The probability for this is $(1-P)$ .

We can thus set up the recurrence relation: $P = (\text{Prob. of not 2 on 1st throw}) \times (\text{Prob. of 2 \in odd throws after 1st})$ .
This gives the equation $P = \frac{5}{6}(1-P)$ .
Solving for $P$ , we get $P = \frac{5}{6} - \frac{5}{6}P$ .
This simplifies to $P + \frac{5}{6}P = \frac{5}{6}$ , or $\frac{11}{6}P = \frac{5}{6}$ .
Therefore, the probability is $P = \frac{5}{11}$ .

Q81JEE Main 2024NAT4MQuadratic Equation and Inequalities

Let $\alpha, \beta$ be the roots of the equation $x^2 - x + 2 = 0$ with $\text{Im}(\alpha) > \text{Im}(\beta)$ . Then $\alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$ is equal to _______

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📖 Explanation

$x^2 - x + 2 = 0$ : $\alpha + \beta = 1, \alpha\beta = 2$ . $\alpha^2 + \beta^2 = 1 - 4 = -3$ . $\alpha^4 + \beta^4 = (\alpha^{2+}\beta^2)^2 - 2(\alpha\beta)^2 = 9-8=1$ . $\alpha^6 = (\alpha^2)^3$ . $\alpha^2 = \alpha - 2$ . $\alpha^3 = \alpha^2 - 2\alpha = -\alpha-2$ . $\alpha^6 = (\alpha^3)^2 = (-\alpha-2)^2 = \alpha^{2+}4\alpha+4 = \alpha-2+4\alpha+4 = 5\alpha+2$ . $\alpha^{6+}\alpha^{4+}\beta^{4-}5\alpha^2 = 5\alpha+2+1-5(\alpha-2) = 5\alpha+3-5\alpha+10 = 13$ . The answer is $\boxed{13}$ .

Q82JEE Main 2024NAT4MPermutations and Combinations

All the letters of the word $GTWENTY$ are written \in all possible ways with or without meaning and these words are written as \in a dictionary. The serial number of the word $GTWENTY$ is _______

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📖 Explanation

To find the serial number of the word $GTWENTY$ , we count the number of words that come before it in a dictionary listing. The letters of the word, in alphabetical order, are E, G, N, T, T, W, Y. The total number of letters is 7, with T repeated twice.

Words starting with a letter before G (i.e., E): With E fixed in the first position, the remaining 6 letters (G, N, T, T, W, Y) can be arranged in $\frac{6!}{2!} = 360$ ways.
Words starting with G, but the second letter comes before T (i.e., E, N):
- Words starting with GE: The remaining 5 letters (N, T, T, W, Y) can be arranged in $\frac{5!}{2!} = 60$ ways.
- Words starting with GN: The remaining 5 letters (E, T, T, W, Y) can be arranged in $\frac{5!}{2!} = 60$ ways.
Words starting with GT, but the third letter comes before W (i.e., E, N, T):
- Words starting with GTE, GTN, or GTT: For each prefix, the remaining 4 distinct letters can be arranged in $4! = 24$ ways. This gives $3 \times 24 = 72$ words.

After GTW, the remaining letters in $GTWENTY$ are E, N, T, Y, which are already in alphabetical order, so no more preceding words are formed.
The total number of words before $GTWENTY$ is $360 + 60 + 60 + 72 = 552$ . Therefore, the serial number of the word $GTWENTY$ is $552 + 1 = 553$ .

Q83JEE Main 2024NAT4MBinomial Theorem

If $\frac{^{11}C_1}{2} + \frac{^{11}C_2}{3} + \ldots + \frac{^{11}C_9}{10} = \frac{n}{m}$ with $\gcd(n, m) = 1$ , then $n + m$ is equal to _______

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📖 Explanation

We need to evaluate $\sum_{r=1}^{9}\frac{^{11}C_r}{r+1}$ and express it as $\frac{n}{m}$ with $\gcd(n,m) = 1$ , then find $n + m$ . Use the identity $\frac{\binom{n}{r}}{r+1} = \frac{1}{n+1}\binom{n+1}{r+1}$ . This identity comes from: $\frac{\binom{n}{r}}{r+1} = \frac{n!}{r!(n-r)!(r+1)} = \frac{1}{n+1} \cdot \frac{(n+1)!}{(r+1)!(n-r)!} = \frac{1}{n+1}\binom{n+1}{r+1}$ With $n = 11$ : $\frac{\binom{11}{r}}{r+1} = \frac{1}{12}\binom{12}{r+1}$ Substitute and evaluate the \sum. $\sum_{r=1}^{9}\frac{\binom{11}{r}}{r+1} = \frac{1}{12}\sum_{r=1}^{9}\binom{12}{r+1} = \frac{1}{12}\sum_{k=2}^{10}\binom{12}{k}$ where $k = r + 1$ . Use the binomial \sum. We know: $\sum_{k=0}^{12}\binom{12}{k} = 2^{12} = 4096$ $\sum_{k=2}^{10}\binom{12}{k} = 4096 - \binom{12}{0} - \binom{12}{1} - \binom{12}{11} - \binom{12}{12}$ $= 4096 - 1 - 12 - 12 - 1 = 4070$ Final calculation. $\frac{n}{m} = \frac{4070}{12} = \frac{2035}{6}$ Check: $\gcd(2035, 6)$ . $2035 = 5 \times 407 = 5 \times 11 \times 37$ . $6 = 2 \times 3$ . No common factors, so $\gcd = 1$ . $n + m = 2035 + 6 = 2041$ The answer is 2041 .

Q84JEE Main 2024NAT4MCircles

Equations of two diameters of a circle are $2x - 3y = 5$ and $3x - 4y = 7$ . The line joining the points $(-\frac{22}{7}, -4)$ and $(-\frac{1}{7}, 3)$ intersects the circle at only one point $P(\alpha, \beta)$ . Then $17\beta - \alpha$ is equal to _______

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📖 Explanation

We have two diameters of a circle: $2x - 3y = 5$ and $3x - 4y = 7$ . Find the center of the circle (intersection of diameters). From $2x - 3y = 5$ and $3x - 4y = 7$ : Multiply first by 4: $8x - 12y = 20$ Multiply second by 3: $9x - 12y = 21$ Subtracting: $x = 1$ , then $2(1) - 3y = 5 \implies y = -1$ . Center: $(1, -1)$ . Find the equation of the line joining $\left(-\frac{22}{7}, -4\right)$ and $\left(-\frac{1}{7}, 3\right)$ . Slope: $\frac{3 - (-4)}{-\frac{1}{7} - (-\frac{22}{7})} = \frac{7}{\frac{21}{7}} = \frac{7}{3}$ Line equation: $y - 3 = \frac{7}{3}\left(x + \frac{1}{7}\right) \implies y = \frac{7x}{3} + \frac{1}{3} + 3 = \frac{7x + 10}{3}$ Or: $7x - 3y + 10 = 0$ Since the line intersects the circle at only one point, it is tangent to the circle. The point $P(\alpha, \beta)$ is the foot of perpendicular from the center to the line. $\alpha = 1 - \frac{7(7 \cdot 1 - 3(-1) + 10)}{49 + 9} = 1 - \frac{7 \times 20}{58} = 1 - \frac{140}{58} = 1 - \frac{70}{29} = -\frac{41}{29}$ $\beta = -1 + \frac{3(7 \cdot 1 - 3(-1) + 10)}{58} = -1 + \frac{3 \times 20}{58} = -1 + \frac{60}{58} = -1 + \frac{30}{29} = \frac{1}{29}$ Compute $17\beta - \alpha$ . $17\beta - \alpha = \frac{17}{29} - \left(-\frac{41}{29}\right) = \frac{17 + 41}{29} = \frac{58}{29} = 2$ The answer is $\boxed{2}$ .

Q85JEE Main 2024NAT4MEllipse

If the points of intersection of two distinct conics $x^2 + y^2 = 4b$ and $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ lie on the curve $y^2 = 3x^2$ , then $3\sqrt{3}$ times the area of the rectangle formed by the intersection points is _______.

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📖 Explanation

The points common to the circle $x^{2}+y^{2}=4b\quad -(1)$ and the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1\quad -(2)$ are stated to lie on the pair of straight lines $y^{2}=3x^{2}\; \Longrightarrow\; y=\pm\sqrt{3}\,x\quad -(3)$ . Substitute $y^{2}=3x^{2}$ from $(3)$ \in the circle $(1)$ : $x^{2}+3x^{2}=4b\;\Longrightarrow\;4x^{2}=4b\;\Longrightarrow\;x^{2}=b\;\Longrightarrow\;|x|=\sqrt{b}$ . Using $(3)$ again, $y^{2}=3x^{2}=3b\;\Longrightarrow\;|y|=\sqrt{3b}$ . Hence the four intersection points are $\bigl(\pm\sqrt{b},\,\pm\sqrt{3b}\bigr)$ , one \in each quadrant. These points must also satisfy the ellipse $(2)$ . Substitute $x^{2}=b,\;y^{2}=3b$ \in $(2)$ : $\frac{b}{16}+\frac{3b}{b^{2}}=1 \;\Longrightarrow\;\frac{b}{16}+\frac{3}{b}=1 \;\Longrightarrow\;b^{2}-16b+48=0 \;\Longrightarrow\;(b-12)(b-4)=0$ . This gives $b=12$ or $b=4$ . When $b=4$ the ellipse becomes $\dfrac{x^{2}}{16}+\dfrac{y^{2}}{16}=1 \;\Longrightarrow\;x^{2}+y^{2}=16$ , identical to the circle. The problem states the conics are distinct, so we reject $b=4$ and take $b=12$ . Width of the rectangle formed by the four points: $2|x|=2\sqrt{b}=2\sqrt{12}=4\sqrt{3}$ . Height of the rectangle: $2|y|=2\sqrt{3b}=2\sqrt{36}=12$ . Area of the rectangle $=4\sqrt{3}\times12=48\sqrt{3}$ . Required expression: $3\sqrt{3}\times (\text{area})=3\sqrt{3}\times48\sqrt{3}=3\sqrt{3}\times48\sqrt{3}=3\times48\times3=432$ . Hence, $3\sqrt{3}$ \times the required area equals $432$ .

Q86JEE Main 2024NAT4MStatistics

If the mean and variance of the data $65, 68, 58, 44, 48, 45, 60, \alpha, \beta, 60$ where $\alpha > \beta$ are $56$ and $66.2$ respectively, then $\alpha^2 + \beta^2$ is equal to _______

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📖 Explanation

Sum $= 65+68+58+44+48+45+60+\alpha+\beta+60 = 448+\alpha+\beta$ . Mean=56: $448+\alpha+\beta=560$ , $\alpha+\beta=112$ . Σx²=65²+68²+...+60²+\alpha ²+\beta ²=\sum of known squares+\alpha ²+\beta ². Known: $4225+4624+3364+1936+2304+2025+3600+3600=25678$ . Variance=66.2: Σx²/10 - 56² = 66.2. Σx²=10(3136+66.2)=32022. \alpha ²+\beta ²=32022-25678=6344. The answer is $\boxed{6344}$ .

Q87JEE Main 2024NAT4MApplication of Derivatives

Let $f(x) = 2^x - x^2$ , $x \in R$ . If $m$ and $n$ are respectively the number of points at which the curves $y = f(x)$ and $y = f'(x)$ intersects the $x$ -axis, then the value of $m + n$ is _______

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📖 Explanation

We need the number of real roots of the equations $f(x)=0 \quad\Longleftrightarrow\quad 2^{x}=x^{2}$ $f'(x)=0 \quad\Longleftrightarrow\quad 2^{x}\,\ln 2=2x$ Case 1: Points where $y=f(x)$ cuts the $x$ -axis (roots of $f(x)$ ) Define $g(x)=2^{x}-x^{2}\;.$ We inspect its sign \in different regions. \bullet As $x\rightarrow -\infty$ , $2^{x}\rightarrow 0$ while $x^{2}\rightarrow \infty$ , so $g(x)\rightarrow-\infty \;(\text{negative}).$ \bullet $g(0)=2^{0}-0^{2}=1 \gt 0$ (positive). \bullet $g(-1)=2^{-1}-(-1)^{2}=0.5-1=-0.5 \lt 0$ (negative). Because $g(-1)\lt 0$ and $g(0)\gt 0$ , by the Intermediate Value Theorem there is one root \in $(-1,0)$ . \bullet $g(2)=2^{2}-2^{2}=0$ and $g(4)=2^{4}-4^{2}=0$ . A quick check of the sign between these points: $g(3)=2^{3}-3^{2}=8-9=-1\lt 0,$ so the function changes sign at both $x=2$ and $x=4$ but not again between them. \bullet For $x\gt 4$ , the exponential term dominates: $g(6)=2^{6}-6^{2}=64-36=28\gt 0$ and the value keeps increasing, so no more zeros occur. Hence the three distinct real roots are one \in $(-1,0)$ , and the obvious ones $x=2$ and $x=4$ . Therefore $m=3$ . Case 2: Points where $y=f'(x)$ cuts the $x$ -axis (roots of $f'(x)$ ) Differentiate: $f'(x)=2^{x}\ln 2-2x.$ Let $h(x)=2^{x}\ln 2-2x.$ \bullet $h(0)=\ln 2\;(\approx0.693)\gt0.$ \bullet As $x\rightarrow -\infty$ : $2^{x}\ln 2\rightarrow 0$ and $-2x\rightarrow+\infty ,$ hence $h(x)\rightarrow+\infty \;(\text{positive}).$ \bullet Evaluate at $x=1$ : $h(1)=2\ln 2-2\approx1.386-2=-0.614\lt0.$ So a sign change occurs between $x=0$ and $x=1$ \Rightarrow one root \in $(0,1).$ \bullet To check for more roots, find extrema of $h(x)$ : $h'(x)=2^{x}(\ln 2)^{2}-2.$ Set $h'(x)=0$ : $2^{x}=\dfrac{2}{(\ln 2)^{2}}\approx4.166\;,$ giving $x_{0}=\log_{2}(4.166)\approx2.06.$ Because $h''(x)=2^{x}(\ln 2)^{3}\gt0,$ this point is the unique minimum of $h(x).$ \bullet Value at the minimum: $h(2.06)=2^{2.06}\ln 2-2(2.06)\approx2.888-4.12\approx-1.23\lt0.$ Thus $h(x)$ is negative \in a neighborhood of this minimum. \bullet Check farther right: $h(4)=2^{4}\ln 2-8=16\ln 2-8\approx11.09-8=3.09\gt0.$ So $h(x)$ changes from negative to positive between $x\approx3$ and $x=4$ , giving a second root \in $(3,4).$ \bullet Since $h(x)$ is positive for $x\lt0$ and again for sufficiently large $x\gt4$ , and has only one minimum, no additional sign changes are possible. Hence $h(x)=0$ has exactly two real solutions, so $n=2$ . Finally, $m+n=3+2=5$ . Answer : 5

Q88JEE Main 2024NAT4MArea Under The Curves

The area (in sq. units) of the part of circle $x^2 + y^2 = 169$ which is below the line $5x - y = 13$ is $\frac{\pi\alpha}{2\beta} - \frac{65}{2} + \frac{\alpha}{\beta}\sin^{-1}(\frac{12}{13})$ where $\alpha, \beta$ are coprime numbers. Then $\alpha + \beta$ is equal to _______

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📖 Explanation

The circle is given by $x^2 + y^2 = 169$ , which has center $(0,0)$ and radius $13$ . The line is $5x - y = 13$ . The area below this line and inside the circle needs to be found. First, find the points of intersection between the circle and the line by substituting $y = 5x - 13$ into the circle equation: $x^2 + (5x - 13)^2 = 169$ $x^2 + 25x^2 - 130x + 169 = 169$ $26x^2 - 130x = 0$ $2x(13x - 65) = 0$ So, $x = 0$ or $x = 5$ . When $x = 0$ , $y = 5(0) - 13 = -13$ . When $x = 5$ , $y = 5(5) - 13 = 12$ . The points of intersection are $(0, -13)$ and $(5, 12)$ . The chord joining these points subtends an angle $\theta$ at the center. The vectors are $\overrightarrow{OA} = (0, -13)$ and $\overrightarrow{OB} = (5, 12)$ . The dot product is: $\overrightarrow{OA} \cdot \overrightarrow{OB} = (0)(5) + (-13)(12) = -156$ $|\overrightarrow{OA}| = 13, \quad |\overrightarrow{OB}| = 13$ $\cos \theta = \frac{-156}{13 \times 13} = -\frac{12}{13}$ Thus, $\theta = \cos^{-1}\left(-\frac{12}{13}\right)$ . Since $\cos \theta = -\frac{12}{13}$ , $\theta$ is obtuse, and $\sin \theta = \sqrt{1 - \left(-\frac{12}{13}\right)^2} = \sqrt{\frac{25}{169}} = \frac{5}{13}$ (positive as $\theta$ is \in the second quadrant). Let $\gamma = \cos^{-1}\left(\frac{12}{13}\right)$ , so $\theta = \pi - \gamma$ and $\sin \gamma = \frac{5}{13}$ . The area of the minor segment (below the chord) is given by the formula $\frac{r^2}{2} (\theta - \sin \theta)$ : $\text{Area} = \frac{169}{2} \left( \theta - \frac{5}{13} \right)$ Substitute $\theta = \pi - \gamma$ : $\text{Area} = \frac{169}{2} \left( (\pi - \gamma) - \frac{5}{13} \right) = \frac{169}{2} (\pi - \gamma) - \frac{169}{2} \cdot \frac{5}{13}$ $\frac{169}{2} \cdot \frac{5}{13} = \frac{169 \times 5}{2 \times 13} = \frac{13 \times 13 \times 5}{2 \times 13} = \frac{65}{2}$ So, $\text{Area} = \frac{169}{2} (\pi - \gamma) - \frac{65}{2}$ Now, $\gamma = \cos^{-1}\left(\frac{12}{13}\right) = \sin^{-1}\left(\frac{5}{13}\right)$ . Let $\beta = \sin^{-1}\left(\frac{12}{13}\right)$ , so $\sin \beta = \frac{12}{13}$ and $\cos \beta = \frac{5}{13}$ . Then $\gamma = \frac{\pi}{2} - \beta$ , since $\sin(\gamma + \beta) = \sin \gamma \cos \beta + \cos \gamma \sin \beta = \left(\frac{5}{13}\right)\left(\frac{5}{13}\right) + \left(\frac{12}{13}\right)\left(\frac{12}{13}\right) = 1$ , and $\gamma + \beta = \frac{\pi}{2}$ . Substitute $\gamma = \frac{\pi}{2} - \beta$ : $\text{Area} = \frac{169}{2} \left( \pi - \left(\frac{\pi}{2} - \beta\right) \right) - \frac{65}{2} = \frac{169}{2} \left( \pi - \frac{\pi}{2} + \beta \right) - \frac{65}{2}$ $= \frac{169}{2} \left( \frac{\pi}{2} + \beta \right) - \frac{65}{2} = \frac{169}{2} \cdot \frac{\pi}{2} + \frac{169}{2} \beta - \frac{65}{2} = \frac{169\pi}{4} + \frac{169}{2} \beta - \frac{65}{2}$ Since $\beta = \sin^{-1}\left(\frac{12}{13}\right)$ , $\text{Area} = \frac{169\pi}{4} + \frac{169}{2} \sin^{-1}\left(\frac{12}{13}\right) - \frac{65}{2}$ The given expression is $\frac{\pi \alpha}{2\beta} - \frac{65}{2} + \frac{\alpha}{\beta} \sin^{-1}\left(\frac{12}{13}\right)$ , where $\alpha$ and $\beta$ are coprime integers. Comparing coefficients: Coefficient of $\pi$ : $\frac{\alpha}{2\beta} = \frac{169}{4}$ Coefficient of $\sin^{-1}\left(\frac{12}{13}\right)$ : $\frac{\alpha}{\beta} = \frac{169}{2}$ Constant term: $-\frac{65}{2}$ matches. From $\frac{\alpha}{\beta} = \frac{169}{2}$ , and $\frac{\alpha}{2\beta} = \frac{1}{2} \cdot \frac{\alpha}{\beta} = \frac{1}{2} \cdot \frac{169}{2} = \frac{169}{4}$ , which matches the coefficient of $\pi$ . Thus, $\frac{\alpha}{\beta} = \frac{169}{2}$ , so $\alpha = 169$ , $\beta = 2$ . These are coprime since $169 = 13^2$ and $2$ have no common factors. Therefore, $\alpha + \beta = 169 + 2 = 171$ .

Q89JEE Main 2024NAT4MDifferential Equations

If the solution curve $y = y(x)$ of the differential equation $(1 + y^2)(1 + \log_e x)dx + xdy = 0$ , $x > 0$ passes through the point $(1, 1)$ and $y(e) = \frac{\alpha - \tan(\frac{3}{2})}{\beta + \tan(\frac{3}{2})}$ , then $\alpha + 2\beta$ is _______

🚀 Solve in Practice Mode

📖 Explanation

We need to solve the differential equation $(1+y^2)(1+\ln x)dx + xdy = 0$ with $y(1) = 1$ , then find $\alpha + 2\beta$ . Separate the variables. Rearranging: $xdy = -(1+y^2)(1+\ln x)dx$ $\frac{dy}{1+y^2} = -\frac{(1+\ln x)}{x}dx$ Integrate both sides. Left side: $\int \frac{dy}{1+y^2} = \tan^{-1}y$ Right side: $-\int \frac{1+\ln x}{x}dx$ . Let $u = \ln x$ , $du = dx/x$ : $-\int (1+u)du = -(u + u^2/2) = -\ln x - \frac{(\ln x)^2}{2}$ So: $\tan^{-1}y = -\ln x - \frac{(\ln x)^2}{2} + C$ Apply initial condition $y(1) = 1$ . At $x = 1$ : $\tan^{-1}(1) = 0 - 0 + C$ , so $C = \pi/4$ . Find $y(e)$ . At $x = e$ : $\ln e = 1$ , so: $\tan^{-1}y = -1 - \frac{1}{2} + \frac{\pi}{4} = \frac{\pi}{4} - \frac{3}{2}$ $y = \tan\left(\frac{\pi}{4} - \frac{3}{2}\right) = \frac{\tan(\pi/4) - \tan(3/2)}{1 + \tan(\pi/4)\tan(3/2)} = \frac{1 - \tan(3/2)}{1 + \tan(3/2)}$ Comparing with $y(e) = \frac{\alpha - \tan(3/2)}{\beta + \tan(3/2)}$ : $\alpha = 1$ , $\beta = 1$ $\alpha + 2\beta = 1 + 2(1) = 3$ The answer is 3 .

Q90JEE Main 2024NAT4MThree Dimensional Geometry

A line with direction ratio $2, 1, 2$ meets the lines $x = y + 2 = z$ and $x + 2 = 2y = 2z$ respectively at the point $P$ and $Q$ . If the length of the perpendicular from the point $(1, 2, 12)$ to the line $PQ$ is $l$ , then $l^2$ is _______

🚀 Solve in Practice Mode

📖 Explanation

We need to find $l^2$ , where $l$ is the perpendicular distance from $(1, 2, 12)$ to line $PQ$ . Parametrise the given lines. Line 1: $x = y + 2 = z$ . Let $x = t$ , then $y = t-2$ , $z = t$ . Direction: $(1, 1, 1)$ . Point: $(t, t-2, t)$ . Line 2: $x + 2 = 2y = 2z$ . Let $2y = s$ , then $y = s/2$ , $z = s/2$ , $x = s-2$ . Direction: $(2, 1, 1)$ . Point: $(s-2, s/2, s/2)$ . Find P and Q. The line through P and Q has direction ratios $2:1:2$ . P is on Line 1: $P = (t, t-2, t)$ . Q is on Line 2: $Q = (s-2, s/2, s/2)$ . Direction $PQ$ : $(s-2-t, s/2-t+2, s/2-t)$ must be proportional to $(2, 1, 2)$ . $\frac{s-2-t}{2} = \frac{s/2-t+2}{1} = \frac{s/2-t}{2} = k$ From equations 1 and 3: $s-2-t = s/2-t$ , so $s/2 = 2$ , $s = 4$ . From equation 3: $k = (2-t)/2$ . From equation 2: $2-t+2 = k = (2-t)/2$ , so $4-t = (2-t)/2$ , $8-2t = 2-t$ , $t = 6$ . So $P = (6, 4, 6)$ and $Q = (2, 2, 2)$ . Find the perpendicular distance from $(1, 2, 12)$ to line PQ. Direction of PQ: $(6-2, 4-2, 6-2) = (4, 2, 4)$ or simplified $(2, 1, 2)$ . $\vec{AQ} = (1-2, 2-2, 12-2) = (-1, 0, 10)$ where $A = (1, 2, 12)$ . $\vec{AQ} \times \vec{d} = (-1, 0, 10) \times (2, 1, 2)$ $= (0 \cdot 2 - 10 \cdot 1, 10 \cdot 2 - (-1) \cdot 2, (-1) \cdot 1 - 0 \cdot 2) = (-10, 22, -1)$ $|\vec{AQ} \times \vec{d}| = \sqrt{100 + 484 + 1} = \sqrt{585}$ $|\vec{d}| = \sqrt{4+1+4} = 3$ $l = \frac{\sqrt{585}}{3}$ $l^2 = \frac{585}{9} = 65$ The answer is 65 .

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