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Physics30 questions

Q1JEE Main 2024MCQ4MUnits and Measurements

Given below are two statements : Statement (I) : Dimensions of specific heat is $[L^2 T^{-2} K^{-1}]$ . Statement (II) : Dimensions of gas constant is $[ML^2 T^{-1} K^{-1}]$ . In the light of the above statements, choose the most appropriate answer from the options given below.

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📖 Explanation

We need to evaluate two statements about dimensions. Statement I: Dimensions of specific heat is $[L^2 T^{-2} K^{-1}]$ . Specific heat = Energy / (mass \times temperature). Dimensions: $\frac{[ML^2T^{-2}]\$}{[M][K]} = [L^2T^{-2}K^{-1}]$ Statement I is correct . ✓ Statement II: Dimensions of gas constant is $[ML^2T^{-1}K^{-1}]$ . Gas constant R = Energy / (mol \times temperature) (from PV = nRT). Dimensions: $\frac{[ML^2T^{-2}]\$}{[\text{mol}][K]} = [ML^2T^{-2}K^{-1}\text{mol}^{-1}]$ The given dimension has $T^{-1}$ instead of $T^{-2}$ . Statement II is incorrect . ✗ The correct answer is Option 2 : Statement (I) is correct but Statement (II) is incorrect.

Q2JEE Main 2024MCQ4MMotion in a Straight Line

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in $t_1$ . If it is projected vertically downwards from the same point with the same speed, it reaches the ground \in $t_2$ . Time required to reach the ground, if it is dropped from the top of the tower, is :

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📖 Explanation

Let the tower height be $h$ , initial speed be $u$ , and $g$ be acceleration due to gravity. Projected upward: Taking downward as positive: $h = -ut_1 + \frac{1}{2}gt_1^2$ ... (1) Projected downward: $h = ut_2 + \frac{1}{2}gt_2^2$ ... (2) Dropped (u = 0): $h = \frac{1}{2}gt^2$ ... (3) From (1): $h = \frac{1}{2}gt_1^2 - ut_1$ From (2): $h = \frac{1}{2}gt_2^2 + ut_2$ Adding (1) and (2): $2h = \frac{1}{2}g(t_1^2 + t_2^2) + u(t_2 - t_1)$ Subtracting (1) from (2): $0 = \frac{1}{2}g(t_2^2 - t_1^2) + u(t_1 + t_2)$ $u = \frac{g(t_1^2 - t_2^2)}{2(t_1 + t_2)} = \frac{g(t_1 - t_2)}{2}$ Substituting back into (2): $h = \frac{g(t_1-t_2)}{2} \cdot t_2 + \frac{1}{2}gt_2^2 = \frac{g}{2}(t_1 t_2 - t_2^2 + t_2^2) = \frac{g t_1 t_2}{2}$ From (3): $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot \frac{g t_1 t_2}{2}}{g}} = \sqrt{t_1 t_2}$ . The correct answer is Option A: $\sqrt{t_1 t_2}$ .

Q3JEE Main 2024MCQ4MLaws of Motion

A body of weight $200 \text{ N}$ is suspended from a tree branch through a chain of mass $10 \text{ kg}$ . The branch pulls the chain by a force equal to (if $g = 10 \text{ m/s}^2$ ) :

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📖 Explanation

We need to find the force with which the branch pulls the chain. Weight of body = 200 N, mass of chain = 10 kg, $g = 10$ m/s $^2$ . $W_{chain} = mg = 10 \times 10 = 100 \text{ N}$ The branch must support both the chain and the body. The total downward force at the point where the chain meets the branch is: $F_{total} = W_{body} + W_{chain} = 200 + 100 = 300 \text{ N}$ The chain pulls the branch downward with 300 N. By Newton's third law, the branch pulls the chain upward with an equal and opposite force of 300 N . The correct answer is Option 3 : 300 N.

Q4JEE Main 2024MCQ4MCircular Motion

A car of $800 \text{ kg}$ is taking turn on a banked road of radius $300 \text{ m}$ and angle of banking $30°$ . If coefficient of static friction is $0.2$ then the maximum speed with which car can negotiate the turn safely: $(g = 10 \text{ m/s}^2, \sqrt{3} = 1.73)$

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📖 Explanation

Start from force resolution on a banked road at maximum speed. At this condition, friction acts down the plane (towards the center). Forces: Normal reaction N (perpendicular to surface) Friction f = μN (down the slope) Weight mg (vertical) Resolve along horizontal (towards center) and vertical. Horizontal: $N\sin\theta +\mu N\cos\theta =mv^2/r$ Vertical: $N\cos\theta −\mu N\sin\theta =mg$ Now divide the horizontal equation by the vertical equation: $(N\sin\theta +\mu N\cos\theta )/(N\cos\theta −\mu N\sin\theta )=(mv^2/r)/(mg)$ Cancel N and m: $(\sin\theta +\mu \cos\theta )/(\cos\theta −\mu \sin\theta )=v^2/(rg)$ Now divide numerator and denominator by cos\theta : $(\tan\theta +\mu )/(1−\mu \tan\theta )=v^2/(rg)$ So, $\frac{v^2}{r}=g\cdot\frac{\tan\theta+\mu}{1-\mu\tan\theta}$ Now substitute the given values $v^2=300\times \ 10\times \ \frac{\frac{1}{\sqrt{\ 3}}+0.2}{1-\frac{1}{\sqrt{\ 3}}\times \ 0.2}$ $v^2=300\times \ 10\times \ \frac{1.3464}{1.532}$ $v^2=3000\times0.8788$ $v^2=2636.4$ $v\approx\ 51.4\ \frac{m}{s}$

Q5JEE Main 2024MCQ4MWork, Power and Energy

When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be:

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📖 Explanation

We are given that the kinetic energy of a body becomes 36 times its original value and asked to determine the percentage increase in momentum. The relationship between kinetic energy ( $KE$ ) and momentum ( $p$ ) is $KE = \frac{p^2}{2m}$ , where $m$ is the mass of the body, and rearranging gives $p = \sqrt{2m \cdot KE}$ . Letting the initial kinetic energy be $KE$ with corresponding momentum $p$ leads to $p = \sqrt{2m \cdot KE}$ . When the kinetic energy increases to $KE' = 36 \, KE$ , the new momentum becomes $p' = \sqrt{2m \cdot KE'} = \sqrt{2m \cdot 36 \, KE} = \sqrt{36} \cdot \sqrt{2m \cdot KE} = 6p$ , since $\sqrt{36} = 6$ and $\sqrt{2m \cdot KE} = p$ . The increase \in momentum is then $\Delta p = p' - p = 6p - p = 5p$ . Therefore, the percentage increase \in momentum is $\frac{\Delta p}{p} \times 100 = \frac{5p}{p} \times 100 = 500\%$ . The correct answer is Option (4): 500%.

Q6JEE Main 2024MCQ4MGravitation

Assuming the earth to be a sphere of uniform mass density, a body weighed $300 \text{ N}$ on the surface of earth. How much it would weigh at $R/4$ depth under surface of earth?

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📖 Explanation

We need to find the weight of a body at a depth of $R/4$ below the Earth's surface, given that it weighs 300 N on the surface. We know that the acceleration due to gravity at a depth $d$ below the Earth's surface (assuming uniform density) is $g' = g\left(1 - \frac{d}{R}\right)$ , where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the Earth. This formula arises because at depth $d$ , only the spherical shell of radius $(R - d)$ contributes to the gravitational field (by the shell theorem, the outer shell exerts no net force). Substituting $d = R/4$ into this expression gives $g' = g\left(1 - \frac{R/4}{R}\right) = g\left(1 - \frac{1}{4}\right) = g \times \frac{3}{4}$ . Therefore, the weight of the body at this depth is $W' = m \times g' = m \times \frac{3g}{4}$ . Since the surface weight is $W = mg = 300 \,\text{N}$ , it follows that $W' = \frac{3}{4} \times W = \frac{3}{4} \times 300 = 225 \,\text{N}$ . The correct answer is Option (4): 225 N.

Q7JEE Main 2024MCQ4MProperties of Matter

Pressure inside a soap bubble is greater than the pressure outside by an amount : (given : $R$ = Radius of bubble, $S$ = Surface tension of bubble)

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📖 Explanation

We need to find the excess pressure inside a soap bubble over the outside pressure. Key Concept: Excess pressure in a bubble with surface tension. For a liquid drop (with one surface), the excess pressure inside due to surface tension is given by the Young-Laplace equation: $\Delta P_{drop} = \frac{2S}{R}$ where $S$ is the surface tension and $R$ is the radius. For a soap bubble: A soap bubble has two surfaces -- an inner surface and an outer surface. Both surfaces contribute to the excess pressure. Each surface contributes $\frac{2S}{R}$ . Therefore, the total excess pressure inside a soap bubble is: $\Delta P_{bubble} = \frac{2S}{R} + \frac{2S}{R} = \frac{4S}{R}$ This is double the excess pressure of a single-surface liquid drop because the soap film has two air-liquid interfaces. The correct answer is Option 4 : $\frac{4S}{R}$ .

Q8JEE Main 2024MCQ4MThermodynamics

A total of $48 \text{ J}$ heat is given to one mole of helium kept \in a cylinder. The temperature of helium increases by $2°C$ . The work done by the gas is: Given, $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$ .

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📖 Explanation

We need to find the work done by one mole of helium gas when 48 J of heat is supplied and the temperature increases by 2°C. We know that according to the First Law of Thermodynamics $Q = \Delta U + W$ , where $Q$ is heat supplied, $\Delta U$ is the change \in internal energy, and $W$ is the work done by the gas. For an ideal gas, the change \in internal energy depends only on temperature change, giving $\Delta U = n C_v \Delta T$ . Since helium is a monoatomic ideal gas with three translational degrees of freedom, its molar heat capacity at constant volume is $C_v = \frac{3}{2}R$ . Substituting $n = 1$ mol, $\Delta T = 2$ K (because 2°C corresponds to 2 K), and $R = 8.3 \, \text{J K}^{-1}\text{mol}^{-1}$ into the expression for the change \in internal energy yields $\Delta U = n C_v \Delta T = 1 \times \frac{3}{2} \times 8.3 \times 2$ . This gives $\Delta U = 1 \times 1.5 \times 8.3 \times 2 = 1 \times 12.45 \times 2 = 24.9 \, \text{J}$ . Applying the First Law then results \in the work done by the gas as $W = Q - \Delta U = 48 - 24.9 = 23.1 \, \text{J}$ . The correct answer is Option (4): 23.1 J.

Q9JEE Main 2024MCQ4MThermodynamics

Energy of 10 non rigid diatomic molecules at temperature $T$ is :

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📖 Explanation

We need to find the total energy of 10 non-rigid diatomic molecules at temperature $T$ by applying the equipartition theorem. According to the equipartition theorem, each degree of freedom contributes $\frac{1}{2}k_B T$ of energy per molecule, and the total energy per molecule is therefore $E = \frac{f}{2} k_B T$ , where $f$ is the number of degrees of freedom and $k_B$ is Boltzmann's constant. In a non-rigid diatomic molecule, there are three translational degrees of freedom (motion along the x, y, and z axes), two rotational degrees of freedom (rotation about the two axes perpendicular to the bond axis, since rotation about the bond axis is negligible), and two vibrational degrees of freedom (one for kinetic energy and one for potential energy). Hence, the total number of degrees of freedom is $f = 3 + 2 + 2 = 7$ . Substituting $f = 7$ into the expression for the energy per molecule gives $E_{\text{per molecule}} = \frac{7}{2} k_B T$ . For 10 molecules, this becomes $E_{\text{total}} = 10 \times \frac{7}{2} k_B T = \frac{70}{2} k_B T = 35 \, k_B T$ . The correct answer is Option (2): $35 K_B T$ .

Q10JEE Main 2024MCQ4MElectrostatics

Two identical conducting spheres $P$ and $S$ with charge $Q$ on each, repel each other with a force $16 \text{ N}$ . A third identical uncharged conducting sphere $R$ is successively brought \in contact with the two spheres. The new force of repulsion between $P$ and $S$ is :

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📖 Explanation

Initial: P and S each have Q. Force = $kQ^2/d^2 = 16$ N. R (uncharged) touches P: charge on each = Q/2. Then R (Q/2) touches S (Q): charge on each = 3Q/4. New charges: P = Q/2, S = 3Q/4. Force = $k(Q/2)(3Q/4)/d^2 = \frac{3}{8}kQ^2/d^2 = \frac{3}{8}\times 16 = 6$ N. The correct answer is Option (2): 6 N.

Q11JEE Main 2024MCQ4MCurrent Electricity

The number of electrons flowing per second in the filament of a $110 \text{ W}$ bulb operating at $220 \text{ V}$ is : (Given $e = 1.6 \times 10^{-19} \text{ C}$ )

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Q12JEE Main 2024MCQ4MMagnetic Effects of Current and Magnetism

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

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📖 Explanation

The dimensional formula for each quantity is derived from its definition.
(A) Latent heat is energy per unit mass, $L = Q/m$ . Its dimension is $\frac{[ML^2T^{-2}]\$}{[M]} = [M^0L^2T^{-2}]$ .
(B) Specific heat is energy per unit mass per unit temperature, $S = Q/(m\Delta T)$ . Its dimension is $\frac{[ML^2T^{-2}]\$}{[M][K]} = [M^0L^2T^{-2}K^{-1}]$ .
(C) Gravitational potential energy has the dimension of energy, which is $[M^1L^2T^{-2}]$ . Note that gravitational potential (energy per mass) would be $[L^2T^{-2}]$ .
(D) Molar heat capacity is energy per mole per unit temperature, $C = Q/(n\Delta T)$ . Its dimension is $\frac{[ML^2T^{-2}]\$}{[mol][K]} = [M^1L^2T^{-2}mol^{-1}K^{-1}]$ .

Q13JEE Main 2024MCQ4MElectromagnetic Induction

In a coil, the current changes from $-2 \text{ A}$ to $+2 \text{ A}$ \in $0.2 \text{ s}$ and induces an emf of $0.1 \text{ V}$ . The self inductance of the coil is :

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📖 Explanation

$|\varepsilon| = L|dI/dt|$ . $dI = 2-(-2) = 4$ A, $dt = 0.2$ s. $0.1 = L\times 4/0.2 = 20L$ . $L = 0.005$ H = 5 mH. The correct answer is Option (3): 5 mH.

Q14JEE Main 2024MCQ4MElectromagnetic Waves

In the given electromagnetic wave $E_y = 600 \sin(\omega t - kx) \text{ Vm}^{-1}$ , intensity of the associated light beam is (\in $\text{W/m}^2$ ) : (Given $\epsilon_0 = 9 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}$ )

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📖 Explanation

We are asked to find the intensity of an electromagnetic wave described by $E_y = 600 \sin(\omega t - kx) \, \text{Vm}^{-1}$ . The average intensity of an electromagnetic wave is related to the amplitude of the electric field by the formula $I = \frac{1}{2} \epsilon_0 c E_0^2$ Here, $\epsilon_0$ is the permittivity of free space, $c$ is the speed of light, and $E_0$ is the peak electric field amplitude. From the given wave equation, the amplitude is $E_0 = 600 \, \text{V/m}$ . The standard values are $\epsilon_0 = 9 \times 10^{-12} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}$ and $c = 3 \times 10^8 \, \text{m/s}$ . Substituting these into the intensity formula gives $I = \frac{1}{2} \times 9 \times 10^{-12} \times 3 \times 10^8 \times (600)^2$ First, compute $\epsilon_0 c$ : $\epsilon_0 \times c = 9 \times 10^{-12} \times 3 \times 10^8 = 27 \times 10^{-4} = 2.7 \times 10^{-3}$ Next, evaluate $E_0^2$ : $E_0^2 = 600^2 = 3.6 \times 10^5$ Therefore, the intensity is $I = \frac{1}{2} \times 2.7 \times 10^{-3} \times 3.6 \times 10^5 = \frac{1}{2} \times 972 = 486 \, \text{W/m}^2$ The correct answer is Option D: 486 W/m^2.

Q15JEE Main 2024MCQ4MRay Optics and Optical Instruments

In finding out refractive index of glass slab the following observations were made through travelling microscope 50 vernier scale division $= 49$ MSD; 20 divisions on main scale \in each cm. For mark on paper MSR $= 8.45 \text{ cm}$ , VC $= 26$ . For mark on paper seen through slab MSR $= 7.12 \text{ cm}$ , VC $= 41$ . For powder particle on the top surface of the glass slab MSR $= 4.05 \text{ cm}$ , VC $= 1$ . (MSR = Main Scale Reading, VC = Vernier Coincidence) Refractive index of the glass slab is :

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📖 Explanation

The microscope has a main scale and a vernier scale. Length of $1$ main scale division (MSD): 20 divisions occupy $1 \text{ cm}$ , hence $1 \text{ MSD} = \frac{1}{20} \text{ cm} = 0.05 \text{ cm}$ $-(1)$ Given: $50$ vernier scale divisions (VSD) $= 49$ MSD. Total length of $50$ VSD $= 49 \times 0.05 = 2.45 \text{ cm}$ Therefore length of $1$ VSD $= \frac{2.45}{50} = 0.049 \text{ cm}$ $-(2)$ Least count (LC) of the microscope is the difference of one MSD and one VSD: $\text{LC} = 1 \text{ MSD} - 1 \text{ VSD}$ Substituting from $(1)$ and $(2)$ : $\text{LC} = 0.05 - 0.049 = 0.001 \text{ cm}$ $-(3)$ The reading for any setting of the microscope is $\text{Reading} = \text{MSR} + (\text{VC} \times \text{LC})$ $-(4)$ Case 1: Mark on paper (no slab) $\text{MSR} = 8.45 \text{ cm}, \; \text{VC} = 26$ Using $(4)$ : $R_1 = 8.45 + 26 \times 0.001 = 8.476 \text{ cm}$ $-(5)$ Case 2: Same mark seen through the slab (apparent depth) $\text{MSR} = 7.12 \text{ cm}, \; \text{VC} = 41$ $R_2 = 7.12 + 41 \times 0.001 = 7.161 \text{ cm}$ $-(6)$ Case 3: Powder particle on the top surface of the slab $\text{MSR} = 4.05 \text{ cm}, \; \text{VC} = 1$ $R_3 = 4.05 + 1 \times 0.001 = 4.051 \text{ cm}$ $-(7)$ The true thickness $t$ of the slab is the distance between the top surface and the bottom surface (which is \in contact with the paper mark). Using $(5)$ and $(7)$ : $t = |R_3 - R_1| = |4.051 - 8.476| = 4.425 \text{ cm}$ $-(8)$ The apparent thickness $t'$ is the distance between the top surface and the apparent position of the mark as seen through the slab. Using $(6)$ and $(7)$ : $t' = |R_3 - R_2| = |4.051 - 7.161| = 3.110 \text{ cm}$ $-(9)$ The refractive index $\mu$ of the glass slab (normal incidence) is given by $\mu = \frac{t}{t'}$ $-(10)$ Substituting from $(8)$ and $(9)$ : $\mu = \frac{4.425}{3.110} \approx 1.42$ $-(11)$ Hence the refractive index of the glass slab is $1.42$ . Option C is correct.

Q16JEE Main 2024MCQ4MRay Optics and Optical Instruments

For the thin convex lens, the radii of curvature are at $15 \text{ cm}$ and $30 \text{ cm}$ respectively. The focal length the lens is $20 \text{ cm}$ . The refractive index of the material is :

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📖 Explanation

We need to find the refractive index of the lens material given the radii of curvature and focal length. The Lensmaker's equation relates the focal length of a thin lens to the radii of curvature and the refractive index: $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ For a convex lens, $R_1 = +15$ cm (first surface is convex, center of curvature on the right), $R_2 = -30$ cm (second surface is convex, center of curvature on the left, so negative by sign convention), and $f = 20$ cm. Substituting these values into the Lensmaker's equation gives: $\frac{1}{20} = (\mu - 1)\left(\frac{1}{15} - \frac{1}{-30}\right) = (\mu - 1)\left(\frac{1}{15} + \frac{1}{30}\right)$ The term \in parentheses simplifies as follows: $\frac{1}{15} + \frac{1}{30} = \frac{2}{30} + \frac{1}{30} = \frac{3}{30} = \frac{1}{10}$ Substituting this back yields: $\frac{1}{20} = (\mu - 1)\times \frac{1}{10}$ Solving for the refractive index: $\mu - 1 = \frac{10}{20} = \frac{1}{2} = 0.5$ $\mu = 1.5$ The correct answer is Option C: 1.5.

Q17JEE Main 2024MCQ4MDual Nature of Matter and Radiation

When UV light of wavelength $300 \text{ nm}$ is incident on the metal surface having work function $2.13 \text{ eV}$ , electron emission takes place. The stopping potential is: (Given $hc = 1240 \text{ eVnm}$ )

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📖 Explanation

We need to find the stopping potential when UV light of wavelength 300 nm falls on a metal with work function 2.13 eV. Key Formula: The photoelectric equation (Einstein's equation) is: $E = W + eV_0$ Here $E = \frac{hc}{\lambda}$ is the energy of the incident photon, $W$ is the work function of the metal, and $V_0$ is the stopping potential. Rearranging this equation yields: $eV_0 = \frac{hc}{\lambda} - W$ The energy of the incident photon is calculated as: $E = \frac{hc}{\lambda} = \frac{1240 \, \text{eV}\cdot\text{nm}}{300 \, \text{nm}} = 4.133 \, \text{eV}$ Substituting into the photoelectric equation to determine the stopping potential gives: $eV_0 = E - W = 4.133 - 2.13 = 2.003 \, \text{eV}$ Since $eV_0$ is expressed \in electronvolts, the stopping potential $V_0$ is approximately 2 V. The correct answer is Option C: 2 V.

Q18JEE Main 2024MCQ4MAtoms and Nuclei

The longest wavelength associated with Paschen series is : (Given $R_H = 1.097 \times 10^7 \text{ SI unit}$ )

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📖 Explanation

We need to find the longest wavelength associated with the Paschen series of hydrogen. The wavelengths of spectral lines in the hydrogen spectrum are given by the Rydberg formula: $\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ where $R_H = 1.097 \times 10^7 \,\text{m}^{-1}$ is the Rydberg constant, $n_1$ is the lower energy level, and $n_2$ is the upper energy level. For the Paschen series, transitions end at $n_1 = 3$ , and the lines correspond to $n_2 = 4, 5, 6, \ldots$ The longest wavelength corresponds to the smallest energy transition, which is the transition from $n_2 = 4$ to $n_1 = 3$ . Thus, $\frac{1}{\lambda} = R_H\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R_H\left(\frac{1}{9} - \frac{1}{16}\right)$ Computing the difference of fractions gives $\frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144}$ Substituting back into the Rydberg formula, $\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{7}{144} = \frac{7.679 \times 10^7}{144} = \frac{7.679}{144} \times 10^7$ $\frac{1}{\lambda} = 0.05333 \times 10^7 = 5.333 \times 10^5 \,\text{m}^{-1}$ Solving for $\lambda$ yields $\lambda = \frac{144}{7 \times 1.097 \times 10^7} = \frac{144}{7.679 \times 10^7} = 18.752 \times 10^{-7} \,\text{m} = 1.875 \times 10^{-6} \,\text{m}$ Rounding gives $1.876 \times 10^{-6}$ m, so the correct answer is Option B: $1.876 \times 10^{-6}$ m.

Q19JEE Main 2024MCQ4MSemiconductor Electronics

The acceptor level of a p-type semiconductor is $6 \text{ eV}$ . The maximum wavelength of light which can create a hole would be : Given $hc = 1242 \text{ eVnm}$ .

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📖 Explanation

To create a hole, an electron must be excited from the valence band to the acceptor level. The energy required for this transition is equal to the energy of the acceptor level above the valence band, which is given as $E = 6 \text{ eV}$ . The energy of a photon is related to its wavelength by $E = \frac{hc}{\lambda}$ . The maximum wavelength corresponds to the minimum energy required. Therefore, $\lambda_{max} = \frac{hc}{E} = \frac{1242 \text{ eVnm}}{6 \text{ eV}} = 207 \text{ nm}$ .

Q20JEE Main 2024MCQ4MUnits and Measurements

In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its $4^{th}$ division coincides exactly with a certain division on main scale. If 50 vernier scale divisions equal to 49 main scale divisions and zero error \in the instrument is $0.04 \text{ mm}$ then how many main scale divisions are there \in $1 \text{ cm}$ ?

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📖 Explanation

We need to find how many main scale divisions are in 1 cm for a vernier caliper. First, the least count LC is defined as one main scale division (MSD) minus one vernier scale division (VSD), and since 50 VSD equal 49 MSD, we have 1 VSD = 49/50 MSD. Moreover, the zero error equals the number of coinciding VSD multiplied by the least count. Expressing the least count in terms of MSD gives $\text{LC} = 1\,\text{MSD} - 1\,\text{VSD} = 1\,\text{MSD} - \frac{49}{50}\,\text{MSD} = \frac{1}{50}\,\text{MSD}$ Since the zero of the vernier has shifted left (indicating a positive zero error) and the fourth division on the vernier coincides with a main scale division, the zero error is four \times the least count: $\text{Zero error} = 4 \times \text{LC} = 4 \times \frac{1}{50}\,\text{MSD} = \frac{4}{50}\,\text{MSD}$ We are told this zero error measures 0.04 mm, so $\frac{4}{50}\,\text{MSD} = 0.04\,\text{mm} \implies \text{MSD} = 0.04 \times \frac{50}{4} = 0.04 \times 12.5 = 0.5\,\text{mm}$ Hence, the number of main scale divisions \in 1 cm (10 mm) is $\frac{10\,\text{mm}}{0.5\,\text{mm}} = 20$ The correct answer is Option (3): 20.

Q21JEE Main 2024NAT4MMotion in a Straight Line

A particle moves in a straight line so that its displacement $x$ at any time $t$ is given by $x^2 = 1 + t^2$ . Its acceleration at any time $t$ is $x^{-n}$ where $n =$ ___________

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📖 Explanation

$x^2 = 1+t^2$ . $2x\dot{x} = 2t \Rightarrow \dot{x} = t/x$ . $\ddot{x} = (x-t\dot{x})/x^2 = (x-t^2/x)/x^2 = (x^{2-}t^2)/x^3 = 1/x^3$ . So $n = 3$ . The answer is 3.

Q22JEE Main 2024NAT4MRotational Motion

Three balls of masses $2 \text{ kg}$ , $4 \text{ kg}$ and $6 \text{ kg}$ respectively are arranged at centre of the edges of an equilateral triangle of side $2 \text{ m}$ . The moment of inertia of the system about an axis through the centroid and perpendicular to the plane of triangle, will be ___________ $\text{kgm}^2$ .

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📖 Explanation

We need to find the moment of inertia of three balls placed at the midpoints of the edges of an equilateral triangle, about an axis through the centroid and perpendicular to the plane. First, recall that the moment of inertia of a system of point masses about an axis is given by $I = \sum m_i r_i^2$ , where $r_i$ is the perpendicular distance of each mass from the axis of rotation. To determine that distance, note that \in an equilateral triangle of side $a$ , the centroid divides each median \in a 2:1 ratio measured from the vertex. Since the median has length $\frac{\sqrt{3}}{2}a$ , the distance from the centroid to a vertex is $\frac{2}{3}\times \frac{\sqrt{3}}{2}a = \frac{a}{\sqrt{3}}$ , while the distance from the centroid to the midpoint of a side is $\frac{1}{3}\times \frac{\sqrt{3}}{2}a = \frac{a}{2\sqrt{3}} = \frac{a\sqrt{3}}{6}$ . Substituting $a = 2\text{ m}$ into the latter expression gives $r = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}\text{ m}$ . Because the centroid is equidistant from all three edges, each mass lies at the same distance $r = \frac{1}{\sqrt{3}}$ from the axis. Squaring this distance yields $r^2 = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}\,\text{m}^2.$ Therefore, the total moment of inertia can be written as $I = m_1r^2 + m_2r^2 + m_3r^2 = (m_1 + m_2 + m_3)\,r^2$ . Substituting the given masses and $r^2$ results \in $I = (2 + 4 + 6)\times \frac{1}{3} = 12\times \frac{1}{3} = 4\,\text{kg m}^2.$ The correct answer is 4 kg m $^2$ .

Q23JEE Main 2024NAT4MProperties of Matter

A wire of cross sectional area A, modulus of elasticity $2 \times 10^{11} \text{ Nm}^{-2}$ and length $2 \text{ m}$ is stretched between two vertical rigid supports. When a mass of $2 \text{ kg}$ is suspended at the middle, it sags lower from its original position making an angle $\theta = \frac{1}{100}$ radian on the points of support. The value of A is ___________ $\times 10^{-4} \text{ m}^2$ (consider $x << L$ ). (given : $g = 10 \text{ m/s}^2$ )

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📖 Explanation

When the mass $m$ is suspended, the vertical components of the tension ( $T$ ) \in the two halves of the wire balance the weight: $2T \sin \theta = mg$ For small angles ( $\theta = 1/100$ rad), we use the approximation $\sin \theta \approx \theta$ $2T\theta = mg \implies T = \frac{mg}{2\theta}$ The original length of half the wire is $L$ . The new length $L'$ is $L' = \frac{L}{\cos \theta}$ $\text{Strain} = \frac{L' - L}{L} = \frac{1}{\cos \theta} - 1$ Using the small angle approximation $\cos \theta \approx 1 - \frac{\theta^2}{2}$ : $\text{Strain} \approx \left(1 + \frac{\theta^2}{2}\right) - 1 = \frac{\theta^2}{2}$ $Y = \frac{T / A}{\theta^2 / 2} = \frac{2T}{A\theta^2}$ $Y = \frac{2(mg / 2\theta)}{A\theta^2} = \frac{mg}{A\theta^3}$ $A = \frac{mg}{Y\theta^3}$ $A = \frac{2 \times 10}{(2 \times 10^{11}) \times (10^{-2})^3}$ $A = \frac{20}{2 \times 10^{11} \times 10^{-6}} = \frac{10}{10^5} = 1 \times 10^{-4} \text{ m}^2$ The answer is 1.

Q24JEE Main 2024NAT4MWaves

Two open organ pipes of lengths $60 \text{ cm}$ and $90 \text{ cm}$ resonate at $6^{th}$ and $5^{th}$ harmonics respectively. The difference of frequencies for the given modes is ___________Hz. (Velocity of sound \in air $= 333 \text{ m/s}$ )

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📖 Explanation

We need to find the difference in frequencies between the 6th harmonic of a 60 cm open pipe and the 5th harmonic of a 90 cm open pipe. Recall that for an open organ pipe, the frequency of the $n$ th harmonic is given by $f_n = \frac{nv}{2L}$ where $v$ is the speed of sound, $L$ is the length of the pipe, and $n = 1, 2, 3, \ldots$ . First, applying this formula to the 6th harmonic of the first pipe, with $L_1$ = 60 cm = 0.6 m, we obtain $f_1 = \frac{6 \times 333}{2 \times 0.6} = \frac{1998}{1.2} = 1665 \, \text{Hz}$ . Next, for the 5th harmonic of the second pipe, with $L_2$ = 90 cm = 0.9 m, we find $f_2 = \frac{5 \times 333}{2 \times 0.9} = \frac{1665}{1.8} = 925 \, \text{Hz}$ . Subtracting these values gives $\Delta f = f_1 - f_2 = 1665 - 925 = 740 \, \text{Hz}$ . The correct answer is 740 Hz.

Q25JEE Main 2024NAT4MElectrostatic Potential and Capacitance

A capacitor of capacitance $10 \mu\text{F}$ whose plates are separated by $10 \text{ mm}$ through air and each plate has area $4 \text{ cm}^2$ is now filled equally with two dielectric media of $K_1 = 2$ , $K_2 = 3$ respectively as shown \in figure. If new force between the plates is $8 \text{ N}$ . The supply voltage is ___________ $\times 10^{-4} \text{ V}$ .

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📖 Explanation

The capacitor with two dielectrics is equivalent to two capacitors, $C_1$ and $C_2$ , connected in parallel.
The equivalent capacitance $C_{eq}$ is given by:
$C_{eq} = C_1 + C_2 = \frac{K_1 \epsilon_0 (A/2)}{d} + \frac{K_2 \epsilon_0 (A/2)}{d} = \frac{\epsilon_0 A}{2d}(K_1+K_2)$
The electrostatic force between the plates of a capacitor connected to a constant voltage supply $V$ is given by $F = \frac{1}{2} V^2 \left| \frac{dC_{eq}}{dd} \right|$ .
Since $C_{eq} \propto 1/d$ , we have $\left| \frac{dC_{eq}}{dd} \right| = \frac{C_{eq}}{d}$ . Thus, $F = \frac{C_{eq} V^2}{2d}$ .
Substituting the expression for $C_{eq}$ : $F = \left( \frac{\epsilon_0 A (K_1+K_2)}{2d} \right) \frac{V^2}{2d} = \frac{\epsilon_0 A (K_1+K_2) V^2}{4d^2}$ .
Solving for the supply voltage $V$ : $V^2 = \frac{4Fd^2}{\epsilon_0 A(K_1+K_2)}$ .
Plugging in the given values: $V^2 = \frac{4 \times 8 \text{ N} \times (10 \times 10^{-3} \text{ m})^2}{8.85 \times 10^{-12} \text{ F/m} \times 4 \times 10^{-4} \text{ m}^2 \times (2+3)} \approx 8649$ .
Therefore, $V = \sqrt{8649} = 93 \text{ V}$ .
The problem states the initial capacitance is $10 \mu\text{F}$ , which is inconsistent with the provided geometry, indicating that the geometric values should be used for the calculation. The question asks for the answer \in units of $\times 10^{-4} \text{ V}$ , which seems to be a typo \in the question's format, as the calculated voltage is $93 \text{ V}$ . The numerical answer is 93.

Q26JEE Main 2024NAT4MCurrent Electricity

In the given figure an ammeter A consists of a $240 \Omega$ coil connected \in parallel to a $10 \Omega$ shunt. The reading of the ammeter is ___________ mA.

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📖 Explanation

ammeter = coil (240 Ω) in parallel with shunt (10 Ω) step 1: equivalent resistance of ammeter $RA=\frac{\left(240\times10\right)}{240+10}=\frac{2400}{250}=9.6\Omega$ $R_{total}=140.4+9.6=150\Omega$ step 3: total current $I=\frac{24}{150}=0.16A$

Q27JEE Main 2024NAT4MMagnetic Effects of Current and Magnetism

A coil having 100 turns, area of $5 \times 10^{-3} \text{ m}^2$ , carrying current of $1 \text{ mA}$ is placed \in uniform magnetic field of $0.20 \text{ T}$ such a way that plane of coil is perpendicular to the magnetic field. The work done \in turning the coil through $90°$ is ___________ $\mu\text{J}$ .

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📖 Explanation

We need to find the work done in turning a current-carrying coil through 90° in a uniform magnetic field. First, recall that the magnetic moment of a coil is $M = NIA$ , where $N$ is the number of turns, $I$ is the current, and $A$ is the area. The potential energy of a magnetic dipole \in a field is $U = -MB\cos\theta$ , and the work done is the change \in potential energy, given by $W = U_f - U_i$ . Using the specified values, the magnetic moment is $M = NIA = 100 \times 1 \times 10^{-3} \times 5 \times 10^{-3}$ , which simplifies to $M = 100 \times 5 \times 10^{-6} = 5 \times 10^{-4}\,\text{A·m}^2$ . The plane of the coil is initially perpendicular to the magnetic field, so the magnetic moment, which is normal to the coil's plane, is parallel to the field. Therefore, $\theta_i = 0°$ . After rotating through 90°, the angle becomes $\theta_f = 90°$ . Applying the expression for work, we have $W = U_f - U_i = (-MB\cos 90°) - (-MB\cos 0°) = 0 - (-MB) = MB$ . Substituting $M = 5 \times 10^{-4}\,\text{A·m}^2$ and $B = 0.20\,\text{T}$ yields $W = 5 \times 10^{-4} \times 0.20 = 1 \times 10^{-4}\,\text{J} = 100\,\mu\text{J}$ . The correct answer is 100 $\mu$ J.

Q28JEE Main 2024NAT4MAlternating Current

For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is $2.5 \text{ nF}$ . If resistance of $200 \Omega$ and $100 \text{ mH}$ inductor is being used \in the given circuit. The frequency of ac source is ___________ $\times 10^3 \text{ Hz}$ . (given $\pi^2 = 10$ )

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📖 Explanation

At resonance in series LCR, $f=\frac{1}{2\pi\sqrt{LC}}$ Given, $L=100mH=0.1H$ $C=2.5\text{ nF}=2.5\times10^{-9}F$ So, $f=\frac{1}{2\pi\sqrt{(0.1)(2.5\times10^{-9})}}$ $=\frac{1}{2\pi\sqrt{2.5\times10^{-10}}}$ $\sqrt{2.5\times10^{-10}}=\frac{5\times10^{-5}}{\sqrt{10}}$ Thus, $f=\frac{1}{2\pi\cdot\frac{5\times10^{-5}}{\sqrt{10}}}$ Given $\pi^2=10\Rightarrow\sqrt{10}=\pi$ Hence, $f=\frac{1}{2\pi\cdot\frac{5\times10^{-5}}{\pi}}$ $=\frac{1}{10\times10^{-5}}$ $=10^4\text{ Hz}$ $=10\times10^3\text{ Hz}$

Q29JEE Main 2024NAT4MWave Optics

Two coherent monochromatic light beams of intensities I and 4I are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is $xI$ . The value of $x$ is ___________.

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📖 Explanation

We need to find the difference between maximum and minimum intensities when two coherent beams of intensities $I$ and $4I$ are superimposed. For two coherent light beams with intensities $I_1$ and $I_2$ , the maximum intensity for constructive interference is given by $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and the minimum intensity for destructive interference is given by $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$ . Substituting $I_1 = I$ and $I_2 = 4I$ into the expression for the maximum intensity gives $I_{\max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$ , while substituting into the minimum intensity yields $I_{\min} = (\sqrt{4I} - \sqrt{I})^2 = (2\sqrt{I} - \sqrt{I})^2 = (\sqrt{I})^2 = I$ . The difference between these extremes is $I_{\max} - I_{\min} = 9I - I = 8I$ , which shows that $x = 8$ . The correct answer is 8.

Q30JEE Main 2024NAT4MAtoms and Nuclei

In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at 10.2 V. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is ___________ nm. (Given $hc = 1245 \text{ eVnm}$ , $e = 1.6 \times 10^{-19} \text{ C}$ ).

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📖 Explanation

We need to find the wavelength of light emitted when hydrogen is excited to the first excitation level (10.2 eV). In the Franck-Hertz experiment, the first dip at 10.2 V corresponds to the energy required to excite hydrogen from the ground state $n = 1$ to the first excited state $n = 2$ , and when the atom de-excites back to $n = 1$ , it emits a photon with energy equal to 10.2 eV. The energy-wavelength relationship for a photon is given by $E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E}$ First, the energy of the emitted photon is $E = 10.2 \, \text{eV}$ . Next, using the value $hc = 1245 \, \text{eV·nm}$ , we find $\lambda = \frac{1245}{10.2} = 122.06 \approx 122 \, \text{nm}$ This corresponds to the Lyman-\alpha line of hydrogen (the $n = 2 \to n = 1$ transition), which lies in the ultraviolet region. The correct answer is 122 nm.

Chemistry30 questions

Q31JEE Main 2024MCQ4MSolutions

Molality ( $m$ ) of $3M$ aqueous solution of NaCl is : (Given : Density of solution $= 1.25 \text{ g mL}^{-1}$ , Molar mass \in $\text{gmol}^{-1}$ : Na $- 23$ , Cl $- 35.5$ )

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Q32JEE Main 2024MCQ4MChemical Equilibrium

The ratio $\frac{K_P}{K_C}$ for the reaction : $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons CO_{2(g)}$ is :

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📖 Explanation

We need to find the ratio $K_P/K_C$ for the reaction $CO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons CO_{2(g)}$ . The relationship between $K_P$ and $K_C$ is given by $K_P = K_C(RT)^{\Delta n_g}$ where $\Delta n_g$ is the change \in the number of moles of gaseous species (moles of gaseous products minus moles of gaseous reactants). Moles of gaseous products = 1 (CO $_2$ ) and moles of gaseous reactants = 1 (CO) + 1/2 (O $_2$ ) = 3/2, so $\Delta n_g = 1 - \frac{3}{2} = -\frac{1}{2}$ . Substituting into the formula gives $\frac{K_P}{K_C} = (RT)^{\Delta n_g} = (RT)^{-1/2} = \frac{1}{\sqrt{RT}}$ , therefore the correct answer is Option (1): $\frac{1}{\sqrt{RT}}$ .

Q33JEE Main 2024MCQ4MRedox Reactions

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

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📖 Explanation

To correctly match the reactions with their types, we analyze each one:

(A) $N_2(g) + O_2(g) \rightarrow 2NO(g)$ : Two elements combine to form a single compound. This is a combination reaction. Thus, (A) matches with (IV).

(B) $2Pb(NO_3)_2(s) \rightarrow 2PbO(s) + 4NO_2(g) + O_2(g)$ : A single compound breaks down into multiple simpler substances. This is a decomposition reaction. Thus, (B) matches with (I).

(C) $2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)$ : Sodium (Na), a more reactive element, displaces hydrogen (H) from water. This is a displacement reaction. Thus, (C) matches with (II).

(D) $2NO_2(g) + 2OH^-(aq) \rightarrow NO_2^-(aq) + NO_3^-(aq) + H_2O(l)$ : Nitrogen in $NO_2$ (oxidation state +4) is simultaneously reduced to +3 in $NO_2^-$ and oxidized to +5 in $NO_3^-$ . This is a disproportionation reaction. Thus, (D) matches with (III).

The correct matching is (A)-(IV), (B)-(I), (C)-(II), and (D)-(III).

Q34JEE Main 2024MCQ4MRedox Reactions

The number of ions from the following that are expected to behave as oxidising agent is : $Sn^{4+}, Sn^{2+}, Pb^{2+}, Tl^{3+}, Pb^{4+}, Tl^{+}$

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📖 Explanation

We need to identify how many ions from $Sn^{4+}, Sn^{2+}, Pb^{2+}, Tl^{3+}, Pb^{4+}, Tl^{+}$ behave as oxidising agents. Key Concept: Inert Pair Effect and Oxidising Behaviour An oxidising agent is a species that readily accepts electrons (gets reduced). Due to the inert pair effect, the higher oxidation states of heavier p-block elements are less stable, making them strong oxidising agents as they tend to get reduced to the lower oxidation state. Analysis of each ion: - $Sn^{4+}$ : Tin \in +4 state. While $Sn^{2+}$ is more stable for heavier members, $Sn^{4+}$ is actually quite stable for tin (Period 5), so it is not a particularly strong oxidising agent. - $Sn^{2+}$ : This is a reducing agent (gets oxidised to $Sn^{4+}$ ), not an oxidising agent. - $Pb^{2+}$ : Lead \in +2 state is the more stable oxidation state of lead. It is not an oxidising agent \in this context. - $Tl^{3+}$ : Thallium \in +3 state is unstable due to the strong inert pair effect \in Period 6. $Tl^{3+}$ readily gets reduced to $Tl^{+}$ , making it a strong oxidising agent . - $Pb^{4+}$ : Lead \in +4 state is unstable due to the inert pair effect. $Pb^{4+}$ readily gets reduced to $Pb^{2+}$ (e.g., $PbO_2$ is a strong oxidising agent). Strong oxidising agent. - $Tl^{+}$ : This is the more stable state of thallium. It is not an oxidising agent. The ions that act as oxidising agents are $Tl^{3+}$ and $Pb^{4+}$ -- a total of 2 ions. The correct answer is Option 2 : 2.

Q35JEE Main 2024MCQ4Mp Block Elements

Evaluate the following statements related to group 14 elements for their correctness. (A) Covalent radius decreases down the group from C to Pb in a regular manner. (B) Electronegativity decreases from C to Pb down the group gradually. (C) Maximum covalance of C is 4 whereas other elements can expand their covalance due to presence of d orbitals. (D) Heavier elements do not form $p\pi - p\pi$ bonds. (E) Carbon can exhibit negative oxidation states. Choose the correct answer from the options given below :

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📖 Explanation

We evaluate statements about Group 14 elements (C, Si, Ge, Sn, Pb): (A) Covalent radius decreases down the group from C to Pb in a regular manner. This is incorrect . The covalent radius generally increases down the group, not decreases. Also, the increase from Si to Ge is less pronounced due to the d-orbital contraction (scandide contraction). (B) Electronegativity decreases from C to Pb down the group gradually. This is incorrect . While the general trend is a decrease in electronegativity going down the group, it does not decrease gradually/uniformly. There are irregularities due to d- and f-orbital effects. (C) Maximum covalence of C is 4 whereas other elements can expand their covalence due to presence of d-orbitals. This is correct . Carbon has no d-orbitals in its valence shell, so its maximum covalence is 4. Si, Ge, Sn, and Pb have vacant d-orbitals and can expand their covalence beyond 4 (e.g., $SiF_6^{2-}$ ). (D) Heavier elements do not form $p\pi - p\pi$ bonds. This is correct . Due to their large atomic sizes, Si, Ge, Sn, and Pb cannot form effective $p\pi - p\pi$ multiple bonds, unlike carbon which readily forms $C=C$ and $C \equiv C$ bonds. (E) Carbon can exhibit negative oxidation states. This is correct . Carbon can show negative oxidation states, such as $-4$ \in $CH_4$ , $-1$ \in $C_2H_6$ , etc. The correct statements are (C), (D), and (E) only . The answer is Option B: (C), (D) and (E) Only.

Q36JEE Main 2024MCQ4MPractical Organic Chemistry

The correct statement among the following, for a "chromatography" purification method is :

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Q37JEE Main 2024MCQ4MHydrocarbons

The incorrect statement regarding the geometrical isomers of 2-butene is :

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📖 Explanation

We need to identify the incorrect statement about geometrical isomers of 2-butene. 2-Butene (CH $_3$ CH=CHCH $_3$ ) exists as two geometrical isomers due to restricted rotation around the C=C bond: (i) cis-2-butene: Both methyl groups are on the same side of the double bond. (ii) trans-2-butene: The methyl groups are on opposite sides of the double bond. Evaluate each statement: (1) "cis- and trans-2-butene are not interconvertible at room temperature": This is CORRECT . The energy barrier for rotation around a C=C bond is very high (~250 kJ/mol), preventing interconversion at room temperature. (2) "cis- and trans-2-butene are stereoisomers": This is CORRECT . They have the same molecular formula and connectivity but differ in the spatial arrangement of atoms. Geometrical isomers are a type of stereoisomers. (3) "cis-2-butene has less dipole moment than trans-2-butene": This is INCORRECT . In cis-2-butene, the two methyl groups are on the same side, so their bond dipoles add up, resulting in a net dipole moment. In trans-2-butene, the methyl groups are on opposite sides, and their dipoles cancel out, giving a dipole moment of approximately zero. Therefore, cis-2-butene has a HIGHER dipole moment than trans-2-butene. (4) "trans-2-butene is more stable than cis-2-butene": This is CORRECT . trans-2-butene is more stable because the methyl groups are farther apart, reducing steric strain. The correct answer is Option (3).

Q38JEE Main 2024MCQ4MElectrochemistry

How can an electrochemical cell be converted into an electrolytic cell?

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📖 Explanation

We need to determine how an electrochemical (galvanic) cell can be converted into an electrolytic cell. Key Concepts: A galvanic cell converts chemical energy to electrical energy through spontaneous redox reactions, producing an EMF equal to $E^\circ_{cell}$ . An electrolytic cell uses external electrical energy to drive non-spontaneous reactions. Analysis of each option: Option 1: Applying an external opposite potential lower than $E^\circ_{cell}$ -- This would reduce the current from the galvanic cell but not reverse the reaction. The cell would still function as a galvanic cell (albeit with reduced output). Incorrect. Option 2: Reversing the flow of ions \in the salt bridge -- The salt bridge maintains electrical neutrality. You cannot simply reverse ion flow without an external driving force. Incorrect. Option 3: Applying an external opposite potential greater than $E^\circ_{cell}$ -- When the external potential exceeds the cell's EMF and is applied \in the opposite direction, it overcomes the natural tendency of the cell and forces the reaction to proceed \in the reverse (non-spontaneous) direction. This is exactly the principle of electrolysis and charging of batteries. Correct. Option 4: Exchanging the electrodes -- Simply swapping electrode positions does not change the underlying chemistry or provide the energy needed to drive a non-spontaneous reaction. Incorrect. The correct answer is Option 3 : Applying an external opposite potential greater than $E^\circ_{cell}$ .

Q39JEE Main 2024MCQ4Md and f Block Elements

Arrange the following elements in the increasing order of number of unpaired electrons in it. (A) Sc (B) Cr (C) V (D) Ti (E) Mn. Choose the correct answer from the options given below :

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📖 Explanation

First note the atomic numbers: $Sc = 21$ , $Ti = 22$ , $V = 23$ , $Cr = 24$ , $Mn = 25$ . Use the Aufbau order $\text{4s} \rightarrow \text{3d}$ to write the ground-state electronic configurations: $Sc\;(21):\;[Ar]\,3d^{1}\,4s^{2}$ $Ti\;(22):\;[Ar]\,3d^{2}\,4s^{2}$ $V\;(23):\;[Ar]\,3d^{3}\,4s^{2}$ $Cr\;(24):\;[Ar]\,3d^{5}\,4s^{1}$ (half-filled 3d subshell gives extra stability) $Mn\;(25):\;[Ar]\,3d^{5}\,4s^{2}$ Count the unpaired electrons \in each case (Hund's rule: every degenerate orbital gets one electron before any pairing starts): Sc : one unpaired electron \in $3d$ \Rightarrow $1$ unpaired Ti : two unpaired electrons \in $3d$ \Rightarrow $2$ unpaired V : three unpaired electrons \in $3d$ \Rightarrow $3$ unpaired Cr : five unpaired \in $3d$ + one \in $4s$ \Rightarrow $6$ unpaired Mn : five unpaired \in $3d$ (the $4s^{2}$ pair is paired) \Rightarrow $5$ unpaired Arrange \in increasing order of the number of unpaired electrons: $1 \lt 2 \lt 3 \lt 5 \lt 6$ corresponds to $Sc \lt Ti \lt V \lt Mn \lt Cr$ . Using the given symbols: $(A) \lt (D) \lt (C) \lt (E) \lt (B)$ . Therefore the correct option is Option C .

Q40JEE Main 2024MCQ4MCoordination Compounds

The correct IUPAC name of $[PtBr_2(PMe_3)_2]$ is :

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📖 Explanation

We need to find the correct IUPAC name of $[PtBr_2(PMe_3)_2]$ . To determine this, we recall that ligands are named \in alphabetical order ignoring multiplicative prefixes, with simple ligands taking prefixes like di- and tri-, while more complex ligands use bis- or tris- within parentheses; anionic ligands end \in "-o" (for example, bromo for $Br^-$ ); and the metal name is followed by its oxidation state \in Roman numerals. First, the bromide anion $Br^-$ serves as an anionic ligand named "bromo," and since there are two bromides, we use "dibromo." The neutral ligand $PMe_3$ (trimethylphosphine) is already composite (it contains the prefix tri-), so we apply "bis" and enclose the ligand name \in parentheses to obtain "bis(trimethylphosphine)." Balancing the charges via $x + 2(-1) + 2(0) = 0$ shows that $x = +2$ , corresponding to platinum(II). Next, arranging the ligand names alphabetically places "bromo" before "trimethylphosphine," giving the name $\text{dibromobis(trimethylphosphine)platinum(II)}$ . The correct answer is Option (3): dibromobis(trimethylphosphine)platinum(II).

Q41JEE Main 2024MCQ4MChemical Bonding and Molecular Structure

Given below are two statements : Statement I : $PF_5$ and $BrF_5$ both exhibit $sp^3d$ hybridisation. Statement II : Both $SF_6$ and $[Co(NH_3)_6]^{3+}$ exhibit $sp^3d^2$ hybridisation. In the light of the above statements, choose the correct answer from the options given below :

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📖 Explanation

We need to evaluate two statements about hybridisation of certain molecules and complex ions. Analysis of Statement I: "PF $_5$ and BrF $_5$ both exhibit sp $^3$ d hybridisation." PF $_5$ : P has 5 valence electrons and forms 5 bonds with F atoms. There are 5 bond pairs and 0 lone pairs, giving sp $^3$ d hybridisation and a trigonal bipyramidal geometry. ✓ BrF $_5$ : Br has 7 valence electrons. With 5 F atoms, there are 5 bond pairs and 1 lone pair = 6 electron pairs total. This requires sp $^3$ d $^2$ hybridisation (not sp $^3$ d), giving a square pyramidal geometry. Since BrF $_5$ is sp $^3$ d $^2$ , not sp $^3$ d, Statement I is FALSE . Analysis of Statement II: "Both SF $_6$ and [Co(NH $_3$ ) $_6$ ] $^{3+}$ exhibit sp $^3$ d $^2$ hybridisation." SF $_6$ : S has 6 valence electrons and forms 6 bonds with F atoms. There are 6 bond pairs and 0 lone pairs. This gives sp $^3$ d $^2$ hybridisation and an octahedral geometry. ✓ [Co(NH $_3$ ) $_6$ ] $^{3+}$ : Co $^{3+}$ has the configuration [Ar]3d $^6$ . With NH $_3$ being a strong field ligand, the d-electrons pair up to occupy only 3 of the 5 d-orbitals, leaving 2 inner d-orbitals available for bonding. The hybridisation is d $^2$ sp $^3$ (inner orbital complex), not sp $^3$ d $^2$ (which would use outer d-orbitals). Since [Co(NH $_3$ ) $_6$ ] $^{3+}$ uses d $^2$ sp $^3$ (inner orbital) hybridisation, not sp $^3$ d $^2$ , Statement II is FALSE . The correct answer is Option (4): Both Statement I and Statement II are false.

Q42JEE Main 2024MCQ4MCoordination Compounds

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

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📖 Explanation

This question requires matching tetrahedral metal complexes with their d-orbital electronic configurations. For tetrahedral complexes, the d-orbitals split into a lower energy $e$ set and a higher energy $t_2$ set. The energy gap is small, so they are always high-spin.

First, determine the oxidation state and the number of d-electrons for the central metal atom in each complex.

(A) In $TiCl_4$ , Ti is in the +4 oxidation state ( $d^0$ ). The configuration is $e^0 t_2^0$ (III).
(B) In $[FeO_4]^{2-}$ , Fe is in the +6 oxidation state ( $d^2$ ). The configuration is $e^2 t_2^0$ (I).
(C) In $[FeCl_4]^{-}$ , Fe is in the +3 oxidation state ( $d^5$ ). The high-spin configuration is $e^2 t_2^3$ (IV).
(D) In $[CoCl_4]^{2-}$ , Co is in the +2 oxidation state ( $d^7$ ). The high-spin configuration is $e^4 t_2^3$ (II).

Matching these results gives the combination: (A)-(III), (B)-(I), (C)-(IV), and (D)-(II).

Q43JEE Main 2024MCQ4MHydrocarbons

The correct arrangement for decreasing order of electrophilic substitution for above compounds is :

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📖 Explanation

The rate of electrophilic aromatic substitution depends on the electron density of the benzene ring, which is influenced by the attached substituent.

The $-OCH_3$ group is strongly activating due to its strong +M (mesomeric) effect, which increases electron density significantly.
The $-CH_3$ group is a weakly activating group due to hyperconjugation and a weak +I (inductive) effect.
Benzene serves as the reference compound.
The $-NO_2$ group is strongly deactivating due to its strong -M and -I effects, which withdraw electron density from the ring.
Therefore, the decreasing order of reactivity is: Anisole > Toluene > Benzene > Nitrobenzene.

Q44JEE Main 2024MCQ4MAldehydes, Ketones and Carboxylic Acids

Consider the given reaction, identify the major product "P". $CH_3 - COOH \xrightarrow{(i) LiAlH_4 \; (ii) PCC \; (iii) HCN/OH^{-} \; (iv) H_2O/OH^{-}, \Delta}$ "P"

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📖 Explanation

This is a multi-step synthesis.

Step (i) $LiAlH_4$ : Lithium aluminium hydride is a strong reducing agent that reduces the carboxylic acid ( $CH_3COOH$ ) to a primary alcohol ( $CH_3CH_2OH$ ).
Step (ii) PCC: Pyridinium chlorochromate is a mild oxidizing agent that oxidizes the primary alcohol to an aldehyde ( $CH_3CHO$ ).
Step (iii) $HCN/OH^-$ : The aldehyde undergoes nucleophilic addition with cyanide ion to form a cyanohydrin, $CH_3CH(OH)CN$ .
Step (iv) $H_2O/OH^-, \Delta$ : The nitrile group of the cyanohydrin is hydrolyzed to a carboxylic acid group, yielding an $\alpha$ -hydroxy acid, $CH_3CH(OH)COOH$ (lactic acid). Heating this product can lead to further reactions like dehydration.

Q45JEE Main 2024MCQ4MHaloalkanes and Haloarenes

Consider the above chemical reaction. Product "A " is :

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📖 Explanation

The reaction is the nitration of anisole (methoxybenzene). The methoxy group ( $-OCH_3$ ) is an electron-donating group due to its +M (mesomeric) effect, which is stronger than its -I (inductive) effect. This makes the benzene ring highly activated towards electrophilic substitution. As an activating group, it directs the incoming electrophile ( $NO_2^+$ ) to the ortho and para positions. Due to steric hindrance from the bulky $-OCH_3$ group at the ortho positions, the para-substituted product is formed as the major product. Thus, product "A" is p-nitroanisole.

Q46JEE Main 2024MCQ4MAlcohols, Phenols and Ethers

The major products formed A and B respectively are:

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📖 Explanation

This is a two-step electrophilic aromatic substitution reaction.

Step 1 (Formation of A): Anisole reacts with a nitrating mixture (Conc. $HNO_3$ + Conc. $H_2SO_4$ ). The $-OCH_3$ group is ortho, para-directing and activating. The major product, A, is the para-isomer, p-nitroanisole, due to less steric hindrance.
Step 2 (Formation of B): Product A (p-nitroanisole) is treated with excess $Br_2$ in the presence of a Lewis acid (Fe, which forms $FeBr_3$ ). The position of the incoming electrophile ( $Br^+$ ) is directed by both substituents. The activating $-OCH_3$ group directs to its ortho positions (C2 and C6), and the deactivating $-NO_2$ group directs to its meta positions (also C2 and C6). Since excess bromine is used, both available positions are substituted, yielding 2,6-dibromo-4-nitroanisole as product B.

Q47JEE Main 2024MCQ4MAmines

Identify the product (A) in the following reaction.

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📖 Explanation

The reaction sequence transforms aniline into phenol.

Step (i) $NaNO_2 + HCl$ : This is the diazotization reaction, which converts the primary aromatic amine (aniline) into a benzenediazonium salt ( $C_6H_5N_2^+Cl^-$ ). This reaction is typically carried out at low temperatures (0-5 °C).
Step (ii) $Cu_2Cl_2 + HCl$ : This is a Sandmeyer reaction. The diazonium group is replaced by a chlorine atom, forming chlorobenzene.
Step (iii) $NaOH/\Delta$ : Chlorobenzene undergoes nucleophilic aromatic substitution with sodium hydroxide under high temperature and pressure (Dow's process) to form sodium phenoxide.
Step (iv) $H^+$ : Acidification of the sodium phenoxide solution protonates the phenoxide ion to yield the final product, phenol ( $C_6H_5OH$ ).

Q48JEE Main 2024MCQ4MSalt Analysis

During the detection of acidic radical present in a salt, a student gets a pale yellow precipitate soluble with difficulty in $NH_4OH$ solution when sodium carbonate extract was first acidified with dil. $HNO_3$ and then $AgNO_3$ solution was added. This indicates presence of :

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📖 Explanation

We need to identify the acidic radical that gives a pale yellow precipitate soluble with difficulty in NH $_4$ OH when tested with AgNO $_3$ . When the sodium carbonate extract of a salt is acidified with dilute HNO $_3$ and treated with AgNO $_3$ , the halide ions form silver halide precipitates with characteristic colours and solubilities: AgCl (from Cl $^-$ ) gives a white curdy precipitate easily soluble \in dilute NH $_4$ OH (forms soluble [Ag(NH $_3$ ) $_2$ ]Cl), AgBr (from Br $^-$ ) gives a pale yellow precipitate soluble with difficulty \in NH $_4$ OH (partially soluble, requires concentrated NH $_4$ OH), and AgI (from I $^-$ ) gives a yellow precipitate insoluble \in NH $_4$ OH. Since the observed precipitate is pale yellow, Cl $^-$ is ruled out and Br $^-$ is consistent; moreover, its solubility with difficulty \in NH $_4$ OH matches AgBr, whereas AgI would be completely insoluble and AgCl would dissolve easily. Furthermore, CO $_3^{2-}$ would not give a precipitate with AgNO $_3$ \in acidic medium, as it reacts with acid to form CO $_2$ , eliminating Option (4). The correct answer is Option (1) : Br $^-$ .

Q49JEE Main 2024MCQ4MClassification of Elements

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

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📖 Explanation

The problem requires matching reaction types with given chemical equations.
(A) In $Na^+ + H_2O \rightarrow NaOH + H_2$ , sodium displaces hydrogen from water. This is a displacement reaction.
(B) In $N_2 + O_2 \rightarrow 2NO$ , nitrogen and oxygen combine to form a single product. This is a combination reaction.
(C) In $NO_2 \rightarrow NO_2^- + NO_3^-$ , the oxidation state of nitrogen (+4) changes to +3 and +5. This is a disproportionation reaction.
(D) In $PbCO_3 \rightarrow PbO + CO_2$ , a single compound breaks down into simpler substances. This is a decomposition reaction.
The correct matching is (A)-Displacement, (B)-Combination, (C)-Disproportionation, (D)-Decomposition.

Q50JEE Main 2024MCQ4MBiomolecules

The incorrect statements regarding enzymes are : (A) Enzymes are biocatalysts. (B) Enzymes are non-specific and can catalyse different kinds of reactions. (C) Most Enzymes are globular proteins. (D) Enzyme - oxidase catalyses the hydrolysis of maltose into glucose. Choose the correct answer from the option given below :

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Q51JEE Main 2024NAT4MCoordination Compounds

Consider the following reactions $NiS + HNO_3 + HCl \rightarrow A + NO + S + H_2O$ . $A + NH_4OH + H_3C - C(=N-OH) - C(=N-OH) - H_3C \rightarrow B + NH_4Cl + H_2O$ . The number of protons that do not involve \in hydrogen bonding \in the product $B$ is ___________.

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📖 Explanation

Step 1: Identify the product "A" in the first reaction. In the reaction $NiS + HNO_3 + HCl \;\longrightarrow\; A + NO + S + H_2O$ nitrate ion acts as an oxidising agent, converting sulfide to elemental sulfur (S) and itself reduced to NO. The nickel remains \in the +2 oxidation state and combines with two chloride ions. Hence $A \;=\; NiCl_2.$ Step 2: Write the second reaction with the structure of the diacetyl dioxime ligand. Diacetyl dioxime has the formula $CH_3C(=N-OH)C(=N-OH)CH_3,$ and reacts with $NiCl_2$ \in the presence of ammonium hydroxide as follows: $NiCl_2 + 2\;NH_4OH + CH_3C(=N-OH)C(=N-OH)CH_3 \;\longrightarrow\; B + NH_4Cl + H_2O.$ In fact, one molecule of diacetyl dioxime provides two oxime groups that chelate to Ni $^{2+}$ , giving the neutral complex $B = Ni[CH_3C(=NOH)C(=NOH)CH_3]_2.$ Step 3: Draw or visualise the structure of product $B$ . It is a square-planar complex \in which each diacetyl dioxime ligand is bidentate, coordinating through the two nitrogen atoms. Each ligand retains two -OH groups, which form intramolecular hydrogen bonds to the adjacent C=N nitrogens: Intramolecular H-bonds: each $-OH$ proton is hydrogen-bonded to the adjacent $=N$ atom. Step 4: Count all the protons \in $B$ . Each ligand $CH_3C(=NOH)C(=NOH)CH_3$ has: \bullet Two methyl groups, each with 3 protons \to total 6 methyl protons per ligand. \bullet Two oxime -OH protons. Since there are two ligands, total protons \in $B$ = $2\times (6\;{\rm methyl}) + 2\times (2\;{\rm OH}) = 12\;{\rm CH_3} + 4\;{\rm OH} = 16\;{\rm H}.$ Step 5: Determine which protons are involved \in hydrogen bonding. The four $-OH$ protons are tied up \in intramolecular hydrogen bonds to the adjacent $=N$ groups. Therefore these 4 protons are involved \in hydrogen bonding. Step 6: Calculate the number of protons not involved \in hydrogen bonding. Subtracting the 4 hydrogen-bonded protons from the total 16 gives: $16 - 4 \;=\; 12.$ Final Answer: The number of protons \in product $B$ that do not participate in hydrogen bonding is 12.

Q52JEE Main 2024NAT4MStructure of Atom

For hydrogen atom, energy of an electron in first excited state is $-3.4 \text{ eV}$ , K.E. of the same electron of hydrogen atom is $x \text{ eV}$ . Value of $x$ is ___________ $\times 10^{-1} \text{ eV}$ . (Nearest integer)

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📖 Explanation

We need to find the kinetic energy of the electron in the first excited state of hydrogen, expressed as $x \times 10^{-1}$ eV. For a hydrogen atom, the total energy, kinetic energy, and potential energy are related by $E_n = -\frac{13.6}{n^2} \, \text{eV}$ , $KE_n = -E_n = \frac{13.6}{n^2} \, \text{eV}$ , and $PE_n = 2E_n = -\frac{27.2}{n^2} \, \text{eV}$ . The kinetic energy is always positive and equals the magnitude of the total energy, which follows from the virial theorem for a Coulomb potential: $KE = -E$ and $PE = 2E$ . Since the first excited state corresponds to $n = 2$ , we have $E_2 = -3.4 \, \text{eV}$ . Substituting into the expression for kinetic energy gives $KE = -E_2 = -(-3.4) = 3.4 \, \text{eV}$ . Expressing this value \in the form $x \times 10^{-1}\, \text{eV}$ yields $3.4 \, \text{eV} = 34 \times 10^{-1} \, \text{eV}$ , so $x = 34$ . Option 34

Q53JEE Main 2024NAT4MAmines

An amine (X) is prepared by ammonolysis of benzyl chloride. On adding p-toluenesulphonyl chloride to it the solution remains clear. Molar mass of the amine (X) formed is ___________ $\text{gmol}^{-1}$ . (Given molar mass \in $\text{gmol}^{-1}$ : C : 12, H : 1, O : 16, N : 14)

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📖 Explanation

We need to find the molar mass of amine X, prepared by ammonolysis of benzyl chloride, that gives a clear solution with the Hinsberg reagent (p-toluenesulphonyl chloride). The Hinsberg test distinguishes between primary, secondary, and tertiary amines: (i) Primary amines react with p-toluenesulphonyl chloride to form a sulphonamide that is soluble in NaOH (acidic NH, dissolves in alkali). (ii) Secondary amines form a sulphonamide that is insoluble in NaOH (precipitate). (iii) Tertiary amines do not react - the solution remains clear. Since the solution remains clear with p-toluenesulphonyl chloride, the amine X is a tertiary amine. In ammonolysis of benzyl chloride (C $_6$ H $_5$ CH $_2$ Cl) with excess benzyl chloride, the following sequence occurs: $C_6H_5CH_2Cl + NH_3 \rightarrow C_6H_5CH_2NH_2$ (primary), then $C_6H_5CH_2Cl + C_6H_5CH_2NH_2 \rightarrow (C_6H_5CH_2)_2NH$ (secondary), and finally $C_6H_5CH_2Cl + (C_6H_5CH_2)_2NH \rightarrow (C_6H_5CH_2)_3N$ (tertiary). Therefore, the tertiary amine is tribenzylamine: $(C_6H_5CH_2)_3N$ . The molecular formula of tribenzylamine is $C_{21}H_{21}N$ , and substituting into the molar mass calculation gives $M = 21(12) + 21(1) + 14 = 252 + 21 + 14 = 287 \, \text{g/mol}$ . Option A: 287 g mol $^{-1}$ .

Q54JEE Main 2024NAT4MChemical Thermodynamics

For the reaction at $298 \text{ K}$ , $2A + B \rightarrow C$ . $\Delta H = 400 \text{ kJ mol}^{-1}$ and $\Delta S = 0.2 \text{ kJ mol}^{-1} \text{K}^{-1}$ . The reaction will become spontaneous above ___________ K.

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📖 Explanation

A reaction is spontaneous when the Gibbs free energy change ( $\Delta G$ ) is negative. The relationship between $\Delta G$ , enthalpy change ( $\Delta H$ ), and entropy change ( $\Delta S$ ) is given by the equation $\Delta G = \Delta H - T\Delta S$ .
At equilibrium, $\Delta G = 0$ , so the equilibrium temperature ( $T_{eq}$ ) can be calculated as $T_{eq} = \frac{\Delta H}{\Delta S}$ .
Given $\Delta H = 400 \text{ kJ mol}^{-1}$ and $\Delta S = 0.2 \text{ kJ mol}^{-1} \text{K}^{-1}$ .
$T_{eq} = \frac{400}{0.2} = 2000 \text{ K}$
For the reaction to be spontaneous ( $\Delta G < 0$ ), we need $\Delta H - T\Delta S < 0$ , which implies $T > \frac{\Delta H}{\Delta S}$ . Therefore, the reaction becomes spontaneous at temperatures above 2000 K.

Q55JEE Main 2024NAT4MChemical Kinetics

Consider the two different first order reactions given below $A + B \rightarrow C$ (Reaction 1) $P \rightarrow Q$ (Reaction 2). The ratio of the half life of Reaction 1 : Reaction 2 is $5 : 2$ . If $t_1$ and $t_2$ represent the time taken to complete $\frac{2}{3}^{rd}$ and $\frac{4}{5}^{th}$ of Reaction 1 and Reaction 2, respectively, then the value of the ratio $t_1 : t_2$ is ___________ $\times 10^{-1}$ (nearest integer). [Given : $\log_{10}(3) = 0.477$ and $\log_{10}(5) = 0.699$ ]

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📖 Explanation

For a first-order reaction, the integrated rate law gives time $t = \frac{2.303}{k} \log\left(\frac{a}{a-x}\right)$ . The half-life is $t_{1/2} = \frac{0.693}{k}$ , so the rate constant $k$ is inversely proportional to $t_{1/2}$ .
The PDF solution is based on a half-life ratio of $(t_{1/2})_1 / (t_{1/2})_2 = 1/2$ , which means $k_2 / k_1 = 1/2$ .
For Reaction 1, time to complete 2/3rd ( $t_1$ ) is $t_1 = \frac{2.303}{k_1} \log(3)$ .
For Reaction 2, time to complete 4/5th ( $t_2$ ) is $t_2 = \frac{2.303}{k_2} \log(5)$ .
The ratio is $\frac{t_1}{t_2} = \frac{k_2}{k_1} \times \frac{\log(3)}{\log(5)} = \frac{1}{2} \times \frac{0.477}{0.699} \approx 0.5 \times 0.6824 \approx 0.34$ .
The value is 0.34, which is $3.4 \times 10^{-1}$ . The nearest integer is 3.

Q56JEE Main 2024NAT4Md and f Block Elements

Among $VO_2^{+}$ , $MnO_4^{-}$ and $Cr_2O_7^{2-}$ , the spin-only magnetic moment value of the species with least oxidising ability is ___________BM (Nearest integer). (Given atomic number $V = 23, Mn = 25, Cr = 24$ )

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📖 Explanation

We need to find the spin-only magnetic moment of the species with the least oxidising ability among $VO_2^+$ , $MnO_4^-$ , and $Cr_2O_7^{2-}$ . Oxidising ability is related to the ease with which the species can be reduced. The higher the reduction potential, the stronger the oxidising agent. Among common oxidising agents \in acidic medium: $MnO_4^- > Cr_2O_7^{2-} > VO_2^+$ \in terms of oxidising strength. The species with the least oxidising ability is $VO_2^+$ . Find the oxidation state and d-electron configuration of V \in $VO_2^+$ : Let the oxidation state of V be $x$ . Then $x + 2(-2) = +1$ , so $x = +5$ . Vanadium (Z = 23) has the configuration [Ar]3d $^3$ 4s $^2$ . V $^{5+}$ loses all 5 valence electrons, giving [Ar] = 3d $^0$ . Calculate the spin-only magnetic moment using the formula: $\mu = \sqrt{n(n+2)} \, \text{BM}$ where $n$ is the number of unpaired electrons. Since V $^{5+}$ has d $^0$ configuration, $n = 0$ : $\mu = \sqrt{0(0+2)} = 0 \, \text{BM}$ The answer is 0.

Q57JEE Main 2024NAT4MOrganic Chemistry - Some Basic Principles

Number of carbocations from the following that are not stabilized by hyperconjugation is __________ _

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📖 Explanation

Hyperconjugation is a stabilizing interaction that involves the delocalization of $\sigma$ -electrons of a C-H bond of an alkyl group directly attached to an atom with an empty p-orbital (like a carbocation). This requires the presence of at least one $\alpha$ -hydrogen.

The cyclopentadienyl cation (B) is destabilized by anti-aromaticity (4 $\pi$ electrons), a much stronger effect than any stabilization from its $\alpha$ -hydrogens.
The methoxymethyl cation ( $CH_2^+-O-CH_3$ ) (D) has no $\alpha$ -C-H bonds. It is stabilized by resonance with the oxygen lone pair.
The bridgehead carbocation (E) cannot be stabilized by hyperconjugation as it would require forming a double bond at the bridgehead, violating Bredt's rule.
The vinyl cation (F) has the positive charge on an sp-hybridized carbon and lacks an adjacent sp³ carbon with hydrogens.
Therefore, four carbocations (B, D, E, F) are not stabilized by hyperconjugation.

Q58JEE Main 2024NAT4MSolutions

When ' $x$ ' $\times 10^{-2}$ mL methanol (molar mass $= 32 \text{ g}$ ; density $= 0.792 \text{ g/cm}^3$ ) is added to $100 \text{ mL}$ water (density $= 1 \text{ g/cm}^3$ ), the following diagram is obtained. $x =$ ___________ (nearest integer). $[Given : Molal freezing point depression constant of water at $273.15 \text{ K}$ is $1.86 \text{ K kg mol}^{-1}$ ]$

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📖 Explanation

From the provided phase diagram, the freezing point of pure water is $273.15 \text{ K}$ and the freezing point of the methanolic solution is $270.65 \text{ K}$ . The depression \in freezing point ( $\Delta T_f$ ) is $273.15 - 270.65 = 2.5 \text{ K}$ .

Using the colligative property formula, $\Delta T_f = K_f \times m$ , we can find the molality ( $m$ ) of the solution:
$m = \frac{\Delta T_f}{K_f} = \frac{2.5 \text{ K}}{1.86 \text{ K kg mol}^{-1}} \approx 1.344 \text{ mol/kg}$

The mass of the solvent (water) is its volume multiplied by its density: $100 \text{ mL} \times 1 \text{ g/mL} = 100 \text{ g} = 0.1 \text{ kg}$ .
Moles of solute (methanol) = molality $\times$ mass of solvent (kg) $= 1.344 \times 0.1 = 0.1344 \text{ mol}$ .
Mass of methanol = moles $\times$ molar mass $= 0.1344 \text{ mol} \times 32 \text{ g/mol} \approx 4.301 \text{ g}$ .

The volume of methanol is $\frac{\text{Mass}}{\text{Density}} = \frac{4.301 \text{ g}}{0.792 \text{ g/mL}} \approx 5.43 \text{ mL}$ .
Given that the volume of methanol is $x \times 10^{-2} \text{ mL}$ , we have $x \times 10^{-2} = 5.43$ . Solving for $x$ , we get $x = 543$ .

Q59JEE Main 2024NAT4MChemical Bonding and Molecular Structure

Total number of species from the following with central atom utilising $sp^2$ hybrid orbitals for bonding is ___________. $NH_3$ , $SO_2$ , $SiO_2$ , $BeCl_2$ , $C_2H_2$ , $C_2H_4$ , $BCl_3$ , $HCHO$ , $C_6H_6$ , $BF_3$ , $C_2H_4Cl_2$

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📖 Explanation

We need to count the species with the central atom using sp $^2$ hybrid orbitals from the given list. sp $^2$ hybridisation occurs when the central atom has 3 electron domains (regions of electron density), giving a trigonal planar arrangement (or bent if lone pairs are present). Analyze each species: (i) $NH_3$ : N has 3 bond pairs + 1 lone pair = 4 electron domains \to sp $^3$ . ✗ (ii) $SO_2$ : S has 2 bond pairs + 1 lone pair = 3 electron domains \to sp $^2$ . ✓ (iii) $SiO_2$ : Each Si is bonded to 4 O atoms \in a tetrahedral arrangement \to sp $^3$ . ✗ (iv) $BeCl_2$ : Be has 2 bond pairs + 0 lone pairs = 2 electron domains \to sp. ✗ (v) $C_2H_4$ : Each C has 3 electron domains (2 C-H + 1 C=C) \to sp $^2$ . ✓ (vi) $C_2H_2$ : Each C has 2 electron domains (1 C-H + 1 C\equiv C) \to sp. ✗ (vii) $BCl_3$ : B has 3 bond pairs + 0 lone pairs = 3 electron domains \to sp $^2$ . ✓ (viii) $HCHO$ : C has 3 electron domains (2 C-H + 1 C=O) \to sp $^2$ . ✓ (ix) $C_6H_6$ : Each C has 3 electron domains \to sp $^2$ . ✓ (x) $BF_3$ : B has 3 bond pairs + 0 lone pairs \to sp $^2$ . ✓ (\xi) $C_2H_4Cl_2$ : Each C has 4 electron domains \to sp $^3$ . ✗ Count: SO $_2$ , C $_2$ H $_4$ , BCl $_3$ , HCHO, C $_6$ H $_6$ , BF $_3$ = 6 species . The answer is 6.

Q60JEE Main 2024NAT4MAlcohols, Phenols and Ethers

The ratio of number of oxygen atoms to bromine atoms in the product Q is ___________ $\times 10^{-1}$ .

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📖 Explanation

The reaction involves a two-step electrophilic aromatic substitution on ethoxybenzene.

Nitration: The starting material is treated with a nitrating mixture (Conc. H₂SO₄ + Conc. HNO₃). The ethoxy group (-OC₂H₅) is an ortho, para-directing group. Due to steric hindrance, the major product (X) is p-nitroethoxybenzene.
Bromination: Product (X) is then treated with two equivalents of Br₂ in the presence of FeBr₃. The ethoxy group is strongly activating and directs the incoming electrophiles (Br⁺) to the ortho positions relative to it. The nitro group is deactivating. Thus, two bromine atoms substitute at positions 2 and 6.

The final product (Y or Q) is 1-ethoxy-2,6-dibromo-4-nitrobenzene.

Number of oxygen atoms = 1 (from ethoxy) + 2 (from nitro) = 3.
Number of bromine atoms = 2.
The ratio of oxygen to bromine atoms is 3/2 = 1.5, which is equal to $15 \times 10^{-1}$ .

Mathematics30 questions

Q61JEE Main 2024MCQ4MComplex Numbers

If $z_1, z_2$ are two distinct complex numbers such that $\left|\frac{z_1 - 2z_2}{\frac{1}{2} - z_1\bar{z}_2}\right| = 2$ , then

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📖 Explanation

Given $\left|\frac{z_1 - 2z_2}{\frac{1}{2} - z_1\bar{z}_2}\right| = 2$ . Multiply numerator and denominator by 2: $\left|\frac{2(z_1 - 2z_2)}{1 - 2z_1\bar{z}_2}\right| = 2$ $\left|\frac{z_1 - 2z_2}{\frac{1}{2} - z_1\bar{z}_2}\right| = 2$ Squaring both sides: $|z_1 - 2z_2|^2 = 4\left|\frac{1}{2} - z_1\bar{z}_2\right|^2$ Expanding the left side: $|z_1|^2 - 2z_1\overline{2z_2} - 2z_2\bar{z}_1 + 4|z_2|^2 = |z_1|^2 - 2(z_1\bar{z}_2 + \bar{z}_1 z_2) + 4|z_2|^2$ Wait, let me be more careful. $|z_1 - 2z_2|^2 = (z_1 - 2z_2)(\bar{z}_1 - 2\bar{z}_2) = |z_1|^2 - 2z_1\bar{z}_2 - 2\bar{z}_1 z_2 + 4|z_2|^2$ . Right side: $4\left|\frac{1}{2} - z_1\bar{z}_2\right|^2 = 4\left(\frac{1}{2} - z_1\bar{z}_2\right)\left(\frac{1}{2} - \bar{z}_1 z_2\right)$ $= 4\left(\frac{1}{4} - \frac{1}{2}\bar{z}_1 z_2 - \frac{1}{2}z_1\bar{z}_2 + |z_1|^2|z_2|^2\right)$ $= 1 - 2\bar{z}_1 z_2 - 2z_1\bar{z}_2 + 4|z_1|^2|z_2|^2$ Setting LHS = RHS: $|z_1|^2 - 2z_1\bar{z}_2 - 2\bar{z}_1 z_2 + 4|z_2|^2 = 1 - 2\bar{z}_1 z_2 - 2z_1\bar{z}_2 + 4|z_1|^2|z_2|^2$ The $-2z_1\bar{z}_2 - 2\bar{z}_1 z_2$ terms cancel from both sides: $|z_1|^2 + 4|z_2|^2 = 1 + 4|z_1|^2|z_2|^2$ Rearranging: $|z_1|^2 - 1 + 4|z_2|^2 - 4|z_1|^2|z_2|^2 = 0$ $(|z_1|^2 - 1) - 4|z_2|^2(|z_1|^2 - 1) = 0$ $(|z_1|^2 - 1)(1 - 4|z_2|^2) = 0$ So either $|z_1|^2 = 1$ or $|z_2|^2 = \frac{1}{4}$ . This means either $|z_1| = 1$ ( $z_1$ lies on a circle of radius 1) or $|z_2| = \frac{1}{2}$ ( $z_2$ lies on a circle of radius $\frac{1}{2}$ ). The answer is Option D.

Q62JEE Main 2024MCQ4MPermutations and Combinations

Let $0 \leq r \leq n$ . If $^{n+1}C_{r+1} : ^{n}C_{r} : ^{n-1}C_{r-1} = 55 : 35 : 21$ , then $2n + 5r$ is equal to:

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📖 Explanation

$\frac{C(n+1,r+1)}{C(n,r)} = \frac{55}{35} = \frac{11}{7}$ . $\frac{n+1}{r+1} = 11/7 \Rightarrow 7n+7 = 11r+11 \Rightarrow 7n = 11r+4$ . $\frac{C(n,r)}{C(n-1,r-1)} = \frac{35}{21} = \frac{5}{3}$ . $\frac{n}{r} = 5/3 \Rightarrow n = 5r/3$ . From second: $r = 3n/5$ . Sub \in first: $7n = 11(3n/5)+4 = 33n/5+4$ . $35n-33n = 20$ . $2n = 20$ , $n = 10$ . $r = 6$ . $2n+5r = 20+30 = 50$ . The correct answer is Option (1): 50.

Q63JEE Main 2024MCQ4MPermutations and Combinations

If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at $315^{th}$ position in this arrangement is :

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Q64JEE Main 2024MCQ4MSequences and Series

Let $ABC$ be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle $ABC$ and the same process is repeated infinitely many \times . If $P$ is the \sum of perimeters and $Q$ is be the sum of areas of all the triangles formed in this process, then :

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📖 Explanation

Let the side length of the initial equilateral triangle $ABC$ be $a$ . Its perimeter is $P_1 = 3a$ and its area is $A_1 = \frac{\sqrt{3}}{4}a^2$ . The next triangle formed by joining the midpoints has a side length of $a/2$ , so its perimeter is $P_2 = \frac{1}{2}P_1$ and its area is $A_2 = \frac{1}{4}A_1$ . This process creates two infinite geometric series. The \sum of perimeters is $P = \sum_{n=0}^{\infty } P_1(\frac{1}{2})^n = \frac{P_1}{1 - 1/2} = 2P_1 = 6a$ . The \sum of areas is $Q = \sum_{n=0}^{\infty } A_1(\frac{1}{4})^n = \frac{A_1}{1 - 1/4} = \frac{4}{3}A_1 = \frac{4}{3}\left(\frac{\sqrt{3}}{4}a^2\right) = \frac{\sqrt{3}}{3}a^2$ . From $P = 6a$ , we get $a^2 = (\frac{P}{6})^2 = \frac{P^2}{36}$ . Substituting this into the equation for $Q$ , we get $Q = \frac{\sqrt{3}}{3}\left(\frac{P^2}{36}\right) = \frac{\sqrt{3}P^2}{108}$ . Rearranging the terms to solve for $P^2$ gives $P^2 = \frac{108Q}{\sqrt{3}} = \frac{108\sqrt{3}Q}{3} = 36\sqrt{3}Q$ .

Q65JEE Main 2024MCQ4MSequences and Series

A software company sets up $m$ number of computer systems to finish an assignment \in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of $m$ is equal to:

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📖 Explanation

We need to find the value of $m$ (number of computer systems) so that the assignment is completed \in 25 days after crashes. Total work = $17m$ system-days since $m$ systems would complete it \in 17 days. With 4 systems crashing at the start of each subsequent day: Day 1: $m$ systems working Day 2: $m - 4$ systems Day 3: $m - 8$ systems $\vdots$ Day $k$ : $m - 4(k-1)$ systems The total work done \in 25 days is $\sum_{k=1}^{25}[m - 4(k-1)] = 25m - 4\sum_{k=0}^{24}k = 25m - 4 \cdot \frac{24 \times 25}{2} = 25m - 1200$ . Setting this equal to the total work gives $25m - 1200 = 17m$ , so $8m = 1200$ and thus $m = 150$ . The correct answer is Option (1): 150.

Q66JEE Main 2024MCQ4MCircles

If $P(6, 1)$ be the orthocentre of the triangle whose vertices are $A(5, -2)$ , $B(8, 3)$ and $C(h, k)$ , then the point $C$ lies on the circle:

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Q67JEE Main 2024MCQ4MCircles

If the locus of the point, whose distances from the point $(2, 1)$ and $(1, 3)$ are \in the ratio $5 : 4$ , is $ax^2 + by^2 + cxy + dx + ey + 170 = 0$ , then the value of $a^2 + 2b + 3c + 4d + e$ is equal to :

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📖 Explanation

$\frac{(x-2)^{2+}(y-1)^2}{(x-1)^{2+}(y-3)^2} = 25/16$ . $16\$[(x-2)^{2+}(y-1)^2]\$ = 25[(x-1)^{2+}(y-3)^2]$ . $16(x²-4x+4+y²-2y+1) = 25(x²-2x+1+y²-6y+9)$ . $16x²-64x+80+16y²-32y = 25x²-50x+250+25y²-150y$ . $9x²+9y²+14x-118y+170 = 0$ . So a=9, b=9, c=0, d=14, e=-118. a²+2b+3c+4d+e = 81+18+0+56-118 = 37. The correct answer is Option (1): 37.

Q68JEE Main 2024MCQ4MLimits, Continuity and Differentiability

$\lim_{n \to \infty } \frac{(1^2 - 1)(n-1) + (2^2 - 2)(n-2) + \cdots + ((n-1)^2 - (n-1)) \cdot 1}{(1^3 + 2^3 + \cdots + n^3) - (1^2 + 2^2 + \cdots + n^2)}$ is equal to :

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📖 Explanation

General term: $(k^{2-}k)(n-k)=k(k-1)(n-k)$ So numerator: $\sum_{k=1}^nk(k-1)(n-k)$ Expand: $=\sum_{ }^{ }k(k-1)n-\sum_{ }^{ }k(k-1)k$ $=n\sum_{ }^{ }(k^{2-}k)-\sum_{ }^{ }(k^{3-}k^2)$ $=n\left(\sum_{ }^{ }k^{2-}\sum_{ }^{ }k\right)-\left(\sum_{ }^{ }k^{3-}\sum_{ }^{ }k^2\right)$ = $n\sum_{ }^{ }k^{2-}n\sum_{ }^{ }k-\sum_{ }^{ }k^{3+}\sum_{ }^{ }k^2$ $=(n+1)\sum_{ }^{ }k^{2-}n\sum_{ }^{ }k-\sum_{ }^{ }k^3$ Denominator $\sum_{ }^{ }k^{3-}\sum_{ }^{ }k^2$ $\sum_{ }^{ }k=\frac{n(n+1)}{2},\quad$ $\sum_{ }^{ }k^2=\frac{n(n+1)(2n+1)}{6},\quad$ $\sum_{ }^{ }k^3=\left(\frac{n(n+1)}{2}\right)^2$ $(\sum_{ }^{ }k\sim\frac{n^2}{2})$ $(\sum_{ }^{ }k^2\sim\frac{n^3}{3})$ $(\sum_{ }^{ }k^3\sim\frac{n^4}{4})$ Numerator leading term $(n+1)\frac{n^3}{3}-n\frac{n^2}{2}-\frac{n^4}{4}$ $\approx\frac{n^4}{3}-\frac{n^3}{2}-\frac{n^4}{4}$ $=\left(\frac{1}{3}-\frac{1}{4}\right)n^4=\frac{1}{12}n^4$ Denominator: $\frac{n^4}{4}-\frac{n^3}{3}\sim\frac{n^4}{4}$ $\lim_{n\to \infty }\frac{\frac{1}{12}n^4}{\frac{1}{4}n^4}$ $=\frac{1/12}{1/4}$ $=\frac{1}{3}$

Q69JEE Main 2024MCQ4MSets and Relations

Let $A = \{1, 2, 3, 4, 5\}$ . Let $R$ be a relation on $A$ defined by $xRy$ if and only if $4x \leq 5y$ . Let $m$ be the number of elements \in $R$ and $n$ be the minimum number of elements from $A \times A$ that are required to be added to $R$ to make it a symmetric relation. Then $m + n$ is equal to :

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📖 Explanation

The relation $R$ on $A = \{1,2,3,4,5\}$ is defined by $xRy$ iff $4x \leq 5y$ , i.e., $y \geq 4x/5$ . Count elements m: For $x=1$ : $y \geq 0.8$ , so $y \in \{1,2,3,4,5\}$ (5 pairs). $x=2$ : $y \geq 1.6$ (4). $x=3$ : $y \geq 2.4$ (3). $x=4$ : $y \geq 3.2$ (2). $x=5$ : $y \geq 4$ (2). Total $m = 16$ . All 5 diagonal elements $(i,i)$ are \in $R$ . Off-diagonal pairs \in R: 11. Check symmetry: For each off-diagonal $(x,y) \in R$ , check if $(y,x) \in R$ . The only symmetric off-diagonal pair is $\{4,5\}$ (both $(4,5)$ and $(5,4)$ are \in $R$ ). The remaining 9 off-diagonal pairs lack their reverse. So $n = 9$ elements must be added. $m + n = 16 + 9 = 25$ . The correct answer is Option 1 : $25$ .

Q70JEE Main 2024MCQ4MMatrices and Determinants

If $A$ is a square matrix of order 3 such that $\det(A) = 3$ and $\det(\text{adj}(-4 \text{adj}(-3 \text{adj}(3 \text{adj}((2A)^{-1}))))) = 2^m 3^n$ , then $m + 2n$ is equal to :

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📖 Explanation

To solve this problem, we use the properties of determinants for a square matrix $M$ of order $n$ (here, $n=3$ ): $\det(kM) = k^n \det(M)$ , $\det(\text{adj}(M)) = (\det(M))^{n-1}$ , and $\det(M^{-1}) = 1/\det(M)$ . Let's evaluate the determinant from the inside out.
First, $\det((2A)^{-1}) = \frac{1}{\det(2A)} = \frac{1}{2^3 \det(A)} = \frac{1}{8 \times 3} = \frac{1}{24}$ .
Let $D_1 = \det(3 \text{adj}((2A)^{-1})) = 3^3 (\det((2A)^{-1}))^2 = 27(\frac{1}{24})^2 = \frac{3}{64} = 3 \cdot 2^{-6}$ .
Let $D_2 = \det(-3 \text{adj}(\text{...})) = (-3)^3 (D_1)^2 = -27(3 \cdot 2^{-6})^2 = -3^5 \cdot 2^{-12}$ .
Let $D_3 = \det(-4 \text{adj}(\text{...})) = (-4)^3 (D_2)^2 = -64(-3^5 \cdot 2^{-12})^2 = -2^6 \cdot 3^{10} \cdot 2^{-24} = -3^{10} \cdot 2^{-18}$ .
The final expression is $\det(\text{adj}(\text{...})) = (D_3)^2 = (-3^{10} \cdot 2^{-18})^2 = 3^{20} \cdot 2^{-36}$ .
Comparing $2^{-36} 3^{20}$ with $2^m 3^n$ , we get $m = -36$ and $n = 20$ . Therefore, $m + 2n = -36 + 2(20) = -36 + 40 = 4$ .

Q71JEE Main 2024MCQ4MFunctions

Let $f(x) = \frac{1}{7 - \sin 5x}$ be a function defined on $\mathbb{R}$ . Then the range of the function $f(x)$ is equal to :

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📖 Explanation

To begin, we find the range of the function $f(x) = \frac{1}{7 - \sin 5x}$ . The sine function satisfies $-1 \leq \sin\theta \leq 1$ for all real $\theta$ , and since $5x$ spans all real values as $x$ ranges over $\mathbb{R}$ , it follows that $-1 \leq \sin 5x \leq 1$ The extreme values occur when $\sin 5x = 1$ (i.e.\ $5x = \frac{\pi}{2} + 2k\pi$ ) and $\sin 5x = -1$ (i.e.\ $5x = -\frac{\pi}{2} + 2k\pi$ ). Next, we consider the denominator $7 - \sin 5x$ . Multiplying the inequality $-1 \leq \sin 5x \leq 1$ by $-1$ reverses it, giving $-1 \leq -\sin 5x \leq 1$ and adding 7 yields $6 \leq 7 - \sin 5x \leq 8$ Since $7 - \sin 5x$ lies \in $[6,8]$ and is always positive, we can take reciprocals. Because the function $g(t) = \frac{1}{t}$ is strictly decreasing for $t > 0$ , reversing the inequality gives $\frac{1}{8} \leq \frac{1}{7 - \sin 5x} \leq \frac{1}{6}$ We verify that these bounds are attained: when $\sin 5x = -1$ , one finds $f(x) = \frac{1}{7 - (-1)} = \frac{1}{8}$ , and when $\sin 5x = 1$ , $f(x) = \frac{1}{7 - 1} = \frac{1}{6}$ . Therefore, the range of $f(x)$ is $\left[\frac{1}{8}, \frac{1}{6}\right]$ . The correct answer is Option D: $\left[\frac{1}{8}, \frac{1}{6}\right]$ .

Q72JEE Main 2024MCQ4MDifferentiation

Suppose for a differentiable function $h$ , $h(0) = 0$ , $h(1) = 1$ and $h'(0) = h'(1) = 2$ . If $g(x) = h(e^x)e^{h(x)}$ , then $g'(0)$ is equal to:

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Q73JEE Main 2024MCQ4MApplication of Derivatives

If the function $f(x) = \left(\frac{1}{x}\right)^{2x}; x \gt 0$ attains the maximum value at $x = \frac{1}{e}$ then :

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📖 Explanation

The given information states that the function $f(x) = \left(\frac{1}{x}\right)^{2x}$ attains its maximum value at $x = \frac{1}{e}$ . Let's analyze this function by making a substitution $u = \frac{1}{x}$ . The function can be rewritten \in terms of $u$ as $g(u) = u^{2/u} = (u^{1/u})^2$ . The maximum of $f(x)$ at $x = \frac{1}{e}$ corresponds to a maximum of $g(u)$ at $u = e$ .

This implies that the simpler related function $h(u) = u^{1/u}$ also has its maximum value at $u=e$ . This means that for any other positive value $a \neq e$ , we have $h(e) > h(a)$ . Let's choose $a=\pi$ , since $\pi \neq e$ .

We can therefore state the inequality $h(e) > h(\pi)$ , which translates to $e^{1/e} > \pi^{1/\pi}$ . To compare $e^\pi$ and $\pi^e$ , we can raise both sides of this inequality to the power of $e\pi$ (a positive number), which preserves the inequality direction.

This gives us $(e^{1/e})^{e\pi} > (\pi^{1/\pi})^{e\pi}$ , which simplifies to $e^{\frac{e\pi}{e}} > \pi^{\frac{e\pi}{\pi}}$ , resulting \in the final inequality $e^\pi > \pi^e$ .

Q74JEE Main 2024MCQ4MIndefinite Integrals

If $\int \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} dx = \frac{1}{12} \tan^{-1}(3 \tan x) +$ constant, then the maximum value of $a \sin x + b \cos x$ , is :

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📖 Explanation

$\int \frac{dx}{a²sin²x+b²cos²x} = \frac{1}{ab}tan⁻¹(\frac{a}{b}tanx) + C$ . Comparing: 1/(ab) = 1/12 and a/b = 3. So ab = 12, a = 3b. 3b² = 12, b² = 4, b = 2, a = 6. Max of a sin x + b cos x = √(a²+b²) = √(36+4) = √40. The correct answer is Option (1): √40.

Q75JEE Main 2024MCQ4MArea Under The Curves

If the area of the region $\{(x, y) : \frac{a}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1\}$ is $(\log_e 2) - \frac{1}{7}$ then the value of $7a - 3$ is equal to:

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📖 Explanation

We need to find the value of $7a - 3$ given that the area of the region $\{(x,y): \frac{a}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1\}$ equals $\ln 2 - \frac{1}{7}$ . To determine this value, we first set up the area integral by noting that the area between the curves $y_1 = \frac{a}{x^2}$ (lower curve) and $y_2 = \frac{1}{x}$ (upper curve) over the interval $x \in [1,2]$ is given by $A = \int_1^2 \left(\frac{1}{x} - \frac{a}{x^2}\right) dx$ . We integrate the difference $y_2 - y_1$ because $y_2 \geq y_1$ \in this region (for $0 < a < 1$ ). We now evaluate each integral separately. First, $\int_1^2 \frac{1}{x}\,dx = [\ln x]_1^2 = \ln 2 - \ln 1 = \ln 2$ . For the second integral, recall that $\int x^{-2}\,dx = \frac{x^{-1}}{-1} = -\frac{1}{x}$ : $\int_1^2 \frac{a}{x^2}\,dx = a\left[-\frac{1}{x}\right]_1^2 = a\left(-\frac{1}{2} - \left(-\frac{1}{1}\right)\right) = a\left(-\frac{1}{2} + 1\right) = \frac{a}{2}$ . Combining these results, the total area satisfies $A = \ln 2 - \frac{a}{2} = \ln 2 - \frac{1}{7}$ . Solving for $a$ by subtracting $\ln 2$ from both sides gives $-\frac{a}{2} = -\frac{1}{7}$ . Hence, $\frac{a}{2} = \frac{1}{7}$ and therefore $a = \frac{2}{7}$ . Finally, $7a - 3 = 7 \times \frac{2}{7} - 3 = 2 - 3 = -1$ . The correct answer is Option C: $-1$ .

Q76JEE Main 2024MCQ4MDifferential Equations

Suppose the solution of the differential equation $\frac{dy}{dx} = \frac{(2+\alpha)x - \beta y + 2}{\beta x - 2\alpha y - (\beta\gamma - 4\alpha)}$ represents a circle passing through origin. Then the radius of this circle is :

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📖 Explanation

We rewrite the given differential equation in differential form as $\frac{dy}{dx} \;=\;\frac{(2+\alpha)x \;-\;\beta y \;+\;2}{\beta x \;-\;2\alpha y \;-\;(\beta\gamma -4\alpha)}.$ Putting all terms on one side gives $\bigl[-\bigl((2+\alpha)x -\beta y +2\bigr)\bigr]\,dx \;+\;\bigl[\beta x -2\alpha y -(\beta\gamma -4\alpha)\bigr]\,dy \;=\;0.$ Thus we identify $M(x,y) \;=\;-\,\bigl((2+\alpha)x -\beta y +2\bigr),\quad N(x,y) \;=\;\beta x -2\alpha y -(\beta\gamma -4\alpha).$ We check for exactness. A differential form $M\,dx + N\,dy=0$ is exact if $\frac{\partial M}{\partial y} \;=\;\frac{\partial N}{\partial x}.$ Here $\frac{\partial M}{\partial y} = -(-\beta)=\beta,\qquad \frac{\partial N}{\partial x} = \beta.$ Since these are equal, the equation is exact. Because it is exact, there exists a potential function $\phi(x,y)$ such that $\frac{\partial \phi}{\partial x}=M,\quad \frac{\partial \phi}{\partial y}=N.$ We find $\phi$ by integrating $M$ with respect to $x$ : $\phi(x,y)=\int M\,dx =\int -\bigl((2+\alpha)x -\beta y +2\bigr)\,dx =-\frac{(2+\alpha)x^2}{2}+\beta x y -2x +h(y),$ where $h(y)$ is an unknown function of $y$ . Next we impose $\displaystyle \frac{\partial \phi}{\partial y}=N$ . From our $\phi$ we get $\frac{\partial \phi}{\partial y} =\beta x +h'(y) \quad\stackrel{!}{=} \; \beta x -2\alpha y -(\beta\gamma -4\alpha).$ Hence $h'(y) = -2\alpha y -(\beta\gamma -4\alpha).$ Integrating with respect to $y$ gives $h(y) = -\alpha y^2 -(\beta\gamma -4\alpha)\,y +C_1.$ Thus the general solution $\phi(x,y)=C$ is $-\frac{(2+\alpha)x^2}{2}+\beta x y -2x \;-\;\alpha y^2 \;-\;(\beta\gamma -4\alpha)y +C_1 =0.$ Multiplying by $-2$ and renaming the constant yields the family of curves $(2+\alpha)x^2 \;-\;2\beta x y \;+\;2\alpha y^2 \;+\;4x \;+\;2\beta\gamma\,y \;-\;8\alpha\,y +C_2 \;=\;0.$ For this family to represent a circle, two conditions must hold: Case 1: the coefficient of $xy$ must be zero, yielding $-2\beta =0\;\Longrightarrow\;\beta=0.$ Case 2: the coefficients of $x^2$ and $y^2$ must be equal, giving $2+\alpha =2\alpha\;\Longrightarrow\;\alpha=2.$ Substituting $\alpha=2$ and $\beta=0$ into the equation, we get $4x^2 +4y^2 +4x -16y +C_2 =0.$ Dividing by 4 simplifies to $x^2 +y^2 +x -4y +C_3 =0.$ We now use the condition that the circle passes through the origin $(0,0)$ . Substituting $(x,y)=(0,0)$ gives $0^2 +0^2 +0 -0 +C_3 =0\;\Longrightarrow\;C_3=0.$ Hence the specific circle is $x^2 +y^2 +x -4y =0.$ We complete the square to find its centre and radius: $x^2 +x +y^2 -4y =0 \;\Longrightarrow\;(x+\tfrac12)^2 -\tfrac14 + (y-2)^2 -4 =0$ $\Longrightarrow (x+\tfrac12)^2 + (y-2)^2 = \tfrac14 +4 = \tfrac{17}{4}.$ Therefore the radius is $\boxed{\frac{\sqrt{17}}{2}}.$

Q77JEE Main 2024MCQ4MVector Algebra

Let $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ , $\vec{b} = ((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}) \times \hat{i}$ . Then the square of the projection of $\vec{a}$ on $\vec{b}$ is :

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📖 Explanation

Given $\vec{a}=2\,\hat{i}+\,\hat{j}-\,\hat{k}$ and $\vec{b}=((\vec{a}\times (\hat{i}+\hat{j}))\times \hat{i})\times \hat{i}$ . We first evaluate the cross products one by one. Step 1: Compute $\vec{c}=\vec{a}\times (\hat{i}+\hat{j})$ . Write $\hat{i}+\hat{j}=(1,1,0)$ and $\vec{a}=(2,1,-1)$ . Using the determinant formula for $\vec{u}\times \vec{v}$ : $\vec{c}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 1 & -1\\ 1 & 1 & 0 \end{vmatrix} =\hat{i}(1\cdot0-(-1)\cdot1)-\hat{j}(2\cdot0-(-1)\cdot1)+\hat{k}(2\cdot1-1\cdot1)$ $\Rightarrow\;\vec{c}=1\,\hat{i}-1\,\hat{j}+1\,\hat{k}=(1,-1,1)$ . Step 2: Compute $\vec{d}=\vec{c}\times \hat{i}$ . Take $\hat{i}=(1,0,0)$ : $\vec{d}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & -1 & 1\\ 1 & 0 & 0 \end{vmatrix} =\hat{i}((-1)\cdot0-1\cdot0)-\hat{j}(1\cdot0-1\cdot1)+\hat{k}(1\cdot0-(-1)\cdot1)$ $\Rightarrow\;\vec{d}=0\,\hat{i}+1\,\hat{j}+1\,\hat{k}=(0,1,1)$ . Step 3: Compute $\vec{b}=\vec{d}\times \hat{i}$ . Again cross with $\hat{i}=(1,0,0)$ : $\vec{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 0 & 1 & 1\\ 1 & 0 & 0 \end{vmatrix} =\hat{i}(1\cdot0-1\cdot0)-\hat{j}(0\cdot0-1\cdot1)+\hat{k}(0\cdot0-1\cdot1)$ $\Rightarrow\;\vec{b}=0\,\hat{i}+1\,\hat{j}-1\,\hat{k}=(0,1,-1)$ . Step 4: Square of the projection of $\vec{a}$ on $\vec{b}$ . Projection length formula: the component of $\vec{a}$ along $\vec{b}$ is $\dfrac{\vec{a}\cdot\vec{b}}{\lvert\vec{b}\rvert}$ . Hence $\text{(projection)}^{2}=\dfrac{(\vec{a}\cdot\vec{b})^{2}}{\lvert\vec{b}\rvert^{2}}$ . Dot product: $\vec{a}\cdot\vec{b}=(2,1,-1)\cdot(0,1,-1)=0+1+1=2$ . Magnitude squared of $\vec{b}$ : $\lvert\vec{b}\rvert^{2}=0^{2}+1^{2}+(-1)^{2}=2$ . Therefore $\text{(projection)}^{2}=\dfrac{2^{2}}{2}=2$ . Hence the square of the projection of $\vec{a}$ on $\vec{b}$ is $2$ , which matches Option C.

Q78JEE Main 2024MCQ4MVector Algebra

Let $\vec{a} = 6\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$ . If $\vec{c}$ is a vector such that $|\vec{c}| \geq 6$ , $\vec{a} \cdot \vec{c} = 6|\vec{c}|$ , $|\vec{c} - \vec{a}| = 2\sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $60°$ , then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is equal to:

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Q79JEE Main 2024MCQ4MThree Dimensional Geometry

Let $P(\alpha, \beta, \gamma)$ be the image of the point $Q(3, -3, 1)$ \in the line $\frac{x-0}{1} = \frac{y-3}{1} = \frac{z-1}{-1}$ and $R$ be the point $(2, 5, -1)$ . If the area of the triangle $PQR$ is $\lambda$ and $\lambda^2 = 14K$ , then $K$ is equal to :

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📖 Explanation

First, write the parametric form of the given line. We have $\frac{x-0}{1} = \frac{y-3}{1} = \frac{z-1}{-1} = t$ . So a point on the line is $A(0,3,1)$ and direction vector is $\mathbf d = (1,1,-1)$ . To find the reflection point $P$ of $Q(3,-3,1)$ across the line, we first find the foot of the perpendicular from $Q$ to the line. The formula for the foot $H$ from a point $Q$ to a line passing through $A$ with direction $\mathbf d$ is: $H = A + \frac{\mathbf d \cdot (Q - A)}{\mathbf d \cdot \mathbf d}\,\mathbf d\quad-(1)$ . Compute $Q - A = (3,-3,1)-(0,3,1) = (3,-6,0)$ . Then $\mathbf d \cdot (Q-A) = (1,1,-1)\cdot(3,-6,0) = -3$ and $\mathbf d\cdot\mathbf d = 1^{2+}1^{2+}(-1)^2 = 3$ . Substituting into $(1)$ gives: $H = (0,3,1) + \frac{-3}{3}(1,1,-1) = (0,3,1) - (1,1,-1) = (-1,2,2)\,$ . The reflection point $P$ is related to the foot $H$ by $P = 2H - Q$ . Hence: $P=2(-1,2,2)-(3,-3,1)=(-2,4,4)-(3,-3,1)=(-5,7,3)\,.$ Next, let $R=(2,5,-1)$ . To find the area of triangle $PQR$ , we use the formula: $\text{Area}=\tfrac12\bigl\lVert\overrightarrow{PQ}\times \overrightarrow{RQ}\bigr\rVert\quad-(2)$ . Compute the vectors: $\overrightarrow{PQ}=P-Q=(-5,7,3)-(3,-3,1)=(-8,10,2)\,,\quad \overrightarrow{RQ}=R-Q=(2,5,-1)-(3,-3,1)=(-1,8,-2)\,.$ Compute the cross product using $\overrightarrow{a}\times \overrightarrow{b}=(a_2b_3-a_3b_2,\;a_3b_1-a_1b_3,\;a_1b_2-a_2b_1)\,.$ We get: $\overrightarrow{PQ}\times \overrightarrow{RQ}=(-36,-18,-54)\,.$ Its magnitude is $\bigl\lVert(-36,-18,-54)\bigr\rVert=\sqrt{(-36)^{2+}(-18)^{2+}(-54)^2}=\sqrt{4536}=18\sqrt{14}\,.$ Substituting into $(2)$ yields the area: $\lambda=\tfrac12\cdot18\sqrt{14}=9\sqrt{14}\,.$ Hence $\lambda^2=(9\sqrt{14})^2=81\cdot14=14\cdot81\,.$ Comparing with $\lambda^2=14K$ gives $K=81$ . Therefore the correct option is Option B.

Q80JEE Main 2024MCQ4MProbability

If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is:

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📖 Explanation

We need to find the probability that three letters posted to 5 addresses go to exactly two addresses. We start by counting the total outcomes. Each letter can go to any of 5 addresses independently, so total ways = $5^3 = 125$ . Next, we count the favorable outcomes where exactly two addresses are used. We first choose which 2 addresses: $\binom{5}{2} = 10$ ways. Then we count surjective maps from 3 letters to 2 addresses. Total ways to post 3 letters to 2 addresses = $2^3 = 8$ . From these, we subtract the cases where all letters go to the same address, of which there are 2, giving $8 - 2 = 6$ . Therefore, the total favorable outcomes = $10 \times 6 = 60$ . Substituting these values into the probability formula gives $P = \frac{60}{125} = \frac{12}{25}$ . The correct answer is Option B: $\frac{12}{25}$ .

Q81JEE Main 2024NAT4MQuadratic Equation and Inequalities

Let $\alpha, \beta$ be roots of $x^2 + \sqrt{2}x - 8 = 0$ . If $U_n = \alpha^n + \beta^n$ , then $\frac{U_{10} + \sqrt{2}U_9}{2U_8}$ is equal to ___________

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📖 Explanation

We are given that $\alpha, \beta$ are roots of $x^2 + \sqrt{2}x - 8 = 0$ and $U_n = \alpha^n + \beta^n$ , and we wish to find $\frac{U_{10} + \sqrt{2}U_9}{2U_8}$ . Since $\alpha$ satisfies $\alpha^2 + \sqrt{2}\alpha - 8 = 0$ , it follows that $\alpha^2 = -\sqrt{2}\alpha + 8$ , and multiplying both sides by $\alpha^{n-2}$ gives $\alpha^n = -\sqrt{2}\alpha^{n-1} + 8\alpha^{n-2}$ . A similar argument for $\beta$ yields $\beta^n = -\sqrt{2}\beta^{n-1} + 8\beta^{n-2}$ , and adding these two results establishes the recurrence $U_n = -\sqrt{2}U_{n-1} + 8U_{n-2}$ . Applying this recurrence for $n = 10$ gives $U_{10} = -\sqrt{2}U_9 + 8U_8$ . Therefore, $U_{10} + \sqrt{2}U_9 = (-\sqrt{2}U_9 + 8U_8) + \sqrt{2}U_9 = 8U_8$ . It follows that $\frac{U_{10} + \sqrt{2}U_9}{2U_8} = \frac{8U_8}{2U_8} = 4$ . The answer is 4.

Q82JEE Main 2024NAT4MSequences and Series

If $S(x) = (1+x) + 2(1+x)^2 + 3(1+x)^3 + \cdots + 60(1+x)^{60}$ , $x \neq 0$ , and $(60)^2 S(60) = a(b)^b + b$ , where $a, b \in N$ , then $(a + b)$ equal to ___________

🚀 Solve in Practice Mode

📖 Explanation

We have $S(x) = \sum_{k=1}^{60} k(1+x)^k$ and need to find $(a+b)$ where $(60)^2 S(60) = a(b)^b + b$ . The standard result for the \sum is $\sum_{k=1}^{n} k r^k = \frac{r(1 - (n+1)r^n + nr^{n+1})}{(1-r)^2}$ . In our case $r = (1+x)$ and $n = 60$ , so with $x = 60$ we have $r = 61$ . Substituting into the formula gives $S(60) = \frac{61(1 - 61\cdot61^{60} + 60\cdot61^{61})}{(1-61)^2} = \frac{61(1 - 61^{61} + 60\cdot61^{61})}{3600}$ . This simplifies to $S(60) = \frac{61(1 + 59\cdot61^{61})}{3600}$ . Multiplying by $(60)^2 = 3600$ yields $(60)^2 S(60) = 3600\cdot S(60) = 61(1 + 59\cdot61^{61}) = 61 + 59\cdot61^{62}$ . Comparing with the form $a(b)^b + b$ shows that $59\cdot61^{62} + 61 = a\cdot b^b + b$ . Choosing $b = 61$ gives $a\cdot61^{61} + 61 = 59\cdot61^{62} + 61$ , so $a\cdot61^{61} = 59\cdot61^{62}$ and hence $a = 59\cdot61 = 3599$ . It follows that $a+b = 3599 + 61 = 3660$ . The answer is 3660.

Q83JEE Main 2024NAT4MHyperbola

The length of the latus rectum and directrices of a hyperbola with eccentricity $e$ are 9 and $x = \pm \frac{4}{\sqrt{13}}$ , respectively. Let the line $y - \sqrt{3}x + \sqrt{3} = 0$ touch this hyperbola at $(x_0, y_0)$ . If $m$ is the product of the focal distances of the point $(x_0, y_0)$ , then $4e^2 + m$ is equal to ___________

🚀 Solve in Practice Mode

📖 Explanation

We are given a hyperbola with eccentricity $e$ , latus rectum length 9, and directrices $x = \pm \dfrac{4}{\sqrt{13}}$ . For a standard hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ the directrices are $x = \pm \dfrac{a}{e}$ and the length of the latus rectum is $\dfrac{2b^2}{a}$ . From the directrices $\dfrac{a}{e} = \dfrac{4}{\sqrt{13}}$ so $a = \dfrac{4e}{\sqrt{13}}$ , and from the latus rectum $\dfrac{2b^2}{a} = 9$ so $b^2 = \dfrac{9a}{2}$ . Using $b^2 = a^2(e^2 - 1)$ we obtain $\dfrac{9a}{2} = a^2(e^2 - 1)$ $\dfrac{9}{2} = a(e^2 - 1)$ Substituting $a = \dfrac{4e}{\sqrt{13}}$ gives $\dfrac{9}{2} = \dfrac{4e}{\sqrt{13}}(e^2 - 1)$ $\dfrac{9\sqrt{13}}{8} = e(e^2 - 1) = e^3 - e$ If we try $e = \dfrac{\sqrt{13}}{2}$ then $e^3 - e = \dfrac{13\sqrt{13}}{8} - \dfrac{\sqrt{13}}{2} = \dfrac{13\sqrt{13} - 4\sqrt{13}}{8} = \dfrac{9\sqrt{13}}{8}$ This works, so $e = \dfrac{\sqrt{13}}{2}$ . Hence $a = \dfrac{4 \cdot \frac{\sqrt{13}}{2}}{\sqrt{13}} = 2$ and $b^2 = a^2(e^2 - 1) = 4\left(\dfrac{13}{4} - 1\right) = 9$ . Thus the equation of the hyperbola is $\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$ . The tangent line is given by $y - \sqrt{3}x + \sqrt{3} = 0$ , or $y = \sqrt{3}(x - 1)$ . Its slope is $m_t = \sqrt{3}$ . The tangent to $\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$ at a point $(x_0,y_0)$ has equation $\dfrac{x\,x_0}{4} - \dfrac{y\,y_0}{9} = 1$ or equivalently $y = \dfrac{9x_0}{4y_0}x - \dfrac{9}{y_0}$ . Comparing with $y = \sqrt{3}x - \sqrt{3}$ gives $\dfrac{9x_0}{4y_0} = \sqrt{3}$ and $-\dfrac{9}{y_0} = -\sqrt{3}$ , so $y_0 = 3\sqrt{3}$ . Then $x_0 = \dfrac{4\sqrt{3}\,y_0}{9} = 4$ . Hence the point of tangency is $(4,3\sqrt{3})$ , which indeed lies on the hyperbola since $\dfrac{16}{4} - \dfrac{27}{9} = 1$ . The foci are at $(\pm ae,0) = (\pm\sqrt{13},0)$ . For the point $(x_0,y_0)$ on the right branch the focal distances are $d_1 = ex_0 - a = 2\sqrt{13}-2$ and $d_2 = ex_0 + a = 2\sqrt{13}+2$ . Their product is $m = d_1\,d_2 = (2\sqrt{13})^2 - 4 = 52 - 4 = 48$ . Finally, $4e^2 + m = 4 \cdot \dfrac{13}{4} + 48 = 13 + 48 = 61$ Therefore the required value is $\boxed{61}$ .

Q84JEE Main 2024NAT4MTrigonometric Ratios and Identities

In a triangle $ABC$ , $BC = 7$ , $AC = 8$ , $AB = \alpha \in \mathbb{N}$ and $\cos A = \frac{2}{3}$ . If $49\cos(3C) + 42 = \frac{m}{n}$ , where $\gcd(m, n) = 1$ , then $m + n$ is equal to ___________

🚀 Solve in Practice Mode

📖 Explanation

In triangle $ABC$ , $BC = a = 7$ , $AC = b = 8$ , $AB = c = \alpha \in \mathbb{N}$ , and $\cos A = 2/3$ . Find $\alpha$ using the cosine rule. $a^2 = b^2 + c^2 - 2bc\cos A$ $49 = 64 + \alpha^2 - 2(8)(\alpha)(2/3) = 64 + \alpha^2 - \frac{32\alpha}{3}$ $\alpha^2 - \frac{32\alpha}{3} + 15 = 0$ $3\alpha^2 - 32\alpha + 45 = 0$ $\alpha = \frac{32 \pm \sqrt{1024 - 540}}{6} = \frac{32 \pm \sqrt{484}}{6} = \frac{32 \pm 22}{6}$ $\alpha = 9$ or $\alpha = 5/3$ (not a natural number). So $\alpha = 9$ . Find $\cos C$ using the cosine rule. $c^2 = a^2 + b^2 - 2ab\cos C$ $81 = 49 + 64 - 2(7)(8)\cos C$ $81 = 113 - 112\cos C$ $\cos C = \frac{32}{112} = \frac{2}{7}$ Compute $\cos 3C$ . $\cos 3C = 4\cos^3 C - 3\cos C = 4\left(\frac{2}{7}\right)^3 - 3\left(\frac{2}{7}\right) = \frac{32}{343} - \frac{6}{7} = \frac{32 - 294}{343} = \frac{-262}{343}$ Compute the expression. $49\cos 3C + 42 = 49 \cdot \frac{-262}{343} + 42 = \frac{-262}{7} + 42 = \frac{-262 + 294}{7} = \frac{32}{7}$ So $\frac{m}{n} = \frac{32}{7}$ with $\gcd(32, 7) = 1$ . Therefore $m + n = 32 + 7 = 39$ . The answer is 39 .

Q85JEE Main 2024NAT4MMatrices and Determinants

If the system of equations $2x + 7y + \lambda z = 3$ , $3x + 2y + 5z = 4$ , $x + \mu y + 32z = -1$ has infinitely many solutions, then $(\lambda - \mu)$ is equal to ___________

🚀 Solve in Practice Mode

📖 Explanation

To have infinitely many solutions, the coefficient matrix must be singular (det = 0) and the augmented matrix must have the same rank as the coefficient matrix. The coefficient matrix is $A = \begin{pmatrix}2 & 7 & \lambda \\ 3 & 2 & 5 \\ 1 & \mu & 32\end{pmatrix}$ and its determinant $\Delta = \det(A)$ can be found by expansion along the first row: $\Delta = 2\bigl(2\cdot 32 - 5\mu\bigr)\;-\;7\bigl(3\cdot 32 - 5\cdot 1\bigr)\;+\;\lambda\bigl(3\mu - 2\cdot 1\bigr).$ Computing each term yields $2(64 - 5\mu) = 128 - 10\mu$ , $7(96 - 5) = 7\cdot 91 = 637$ , and $\lambda(3\mu - 2)$ . Thus $\Delta = 128 - 10\mu \;-\; 637\;+\;\lambda(3\mu - 2) = -509 - 10\mu + \lambda(3\mu - 2). \quad-(1)$ For infinitely many solutions, $\Delta = 0$ , so $\lambda(3\mu - 2) = 509 + 10\mu. \quad-(2)$ The augmented matrix determinant must also be zero. $\det\begin{pmatrix}2 & 7 & 3\\3 & 2 & 4\\1 & \mu & -1\end{pmatrix} = 0.$ Expansion along the first row gives $2\bigl(2\cdot(-1) - 4\mu\bigr)\;-\;7\bigl(3\cdot(-1)-4\cdot1\bigr)\;+\;3\bigl(3\mu - 2\cdot1\bigr).$ Evaluating each term: $2(-2 - 4\mu) = -4 - 8\mu$ , $-7(-3 - 4) = 49$ , and $3(3\mu - 2) = 9\mu - 6$ . Summing gives $-4 - 8\mu + 49 + 9\mu - 6 = 39 + \mu.$ Setting this to zero yields $39 + \mu = 0 \quad\Longrightarrow\quad \mu = -39. \quad-(3)$ Substituting $\mu = -39$ into equation $(2)$ gives $\lambda\bigl(3(-39) - 2\bigr) = 509 + 10(-39).$ Since $3(-39) - 2 = -119$ and $509 - 390 = 119$ , it follows that $-119\,\lambda = 119\quad\Longrightarrow\quad \lambda = -1. \quad-(4)$ Finally, $\lambda - \mu = (-1) - (-39) = 38.$ Final Answer: $\lambda - \mu = 38.$

Q86JEE Main 2024NAT4MLimits, Continuity and Differentiability

Let $[t]$ denote the greatest integer less than or equal to $t$ . Let $f : [0, \infty ) \rightarrow \mathbb{R}$ be a function defined by $f(x) = \left[\frac{x}{2} + 3\right] - [\sqrt{x}]$ . Let $S$ be the set of all points \in the interval $[0, 8]$ at which $f$ is not continuous. Then $\sum_{a \in S} a$ is equal to ___________

🚀 Solve in Practice Mode

📖 Explanation

We need to find the sum of all points in $[0, 8]$ where $f(x) = [x/2 + 3] - [\sqrt{x}]$ is discontinuous, where $[t]$ is the greatest integer function. The function $[x/2 + 3]$ is discontinuous when $x/2 + 3$ is an integer, i.e., $x/2$ is an integer, which gives $x = 0, 2, 4, 6, 8$ . The function $[\sqrt{x}]$ is discontinuous when $\sqrt{x}$ is an integer, i.e., $x = 0, 1, 4$ (\in $[0,8]$ , we have $\sqrt{x} = 0, 1, 2$ at $x = 0, 1, 4$ ; at $9$ , $\sqrt{x} = 3$ but $9 > 8$ ). Potential discontinuity points of $f = [x/2 + 3] - [\sqrt{x}]$ are the union $\{0, 1, 2, 4, 6, 8\}$ , though some discontinuities may cancel. At $x = 0$ (the left endpoint), we check right continuity. We have $f(0) = [3] - [0] = 3$ , and as $x \to 0^+$ , $f = [x/2 + 3] - [\sqrt{x}] = 3 - 0 = 3$ , so $f$ is continuous from the right at 0. At $x = 1$ , near the left we get $f(1^-) = [3 + \epsilon] - [1 - \epsilon] = 3 - 0 = 3$ . At $x = 1$ , $f(1) = [3.5] - [1] = 3 - 1 = 2$ , so a jump occurs and $f$ is discontinuous at 1. At $x = 2$ , $f(2^-) = [4 - \epsilon] - [\sqrt{2 - \epsilon}] = 3 - 1 = 2$ , while $f(2) = [4] - [\sqrt{2}] = 4 - 1 = 3$ , so there is a jump and $f$ is discontinuous at 2. At $x = 4$ , we calculate carefully: $f(4) = [4/2 + 3] - [\sqrt{4}] = [5] - [2] = 5 - 2 = 3$ . The left-hand limit is $f(4^-) = [2 - \epsilon + 3] - [\sqrt{4 - \epsilon}] = [5 - \epsilon] - [2 - \delta] = 4 - 1 = 3$ , and the right-hand limit is $f(4^+) = [2 + \epsilon + 3] - [\sqrt{4 + \epsilon}] = [5 + \epsilon] - [2 + \delta] = 5 - 2 = 3$ . Since both one-sided limits equal $f(4)$ , the discontinuities cancel and $f$ is continuous at 4. At $x = 6$ , $f(6^-) = [6 - \epsilon] - [\sqrt{6 - \epsilon}] = 5 - 2 = 3$ , while $f(6) = [6] - [\sqrt{6}] = 6 - 2 = 4$ , so a jump occurs and $f$ is discontinuous at 6. At $x = 8$ , $f(8^-) = [7 - \epsilon] - [\sqrt{8 - \epsilon}] = 6 - 2 = 4$ and $f(8) = [7] - [2\sqrt{2}] = 7 - 2 = 5$ , so $f$ is discontinuous at 8. Thus the discontinuity points \in $[0,8]$ are $\{1,2,6,8\}$ , and their \sum is $1 + 2 + 6 + 8 = 17$ , which is the final answer.

Q87JEE Main 2024NAT4MDefinite Integrals

Let $[t]$ denote the largest integer less than or equal to $t$ . If $\int_0^3 \left([x^2]\$ + \left[\frac{x^2}{2}\right]\$\right) dx = a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7}$ , where $a, b, c \in \mathbb{Z}$ , then $a + b + c$ is equal to ___________

🚀 Solve in Practice Mode

📖 Explanation

The integral is evaluated by splitting it into two parts: $I = \int_0^3 \$[x^2]\$ dx + \int_0^3 \left[\frac{x^2}{2}\right]\$ dx$ . The domain of integration $[0, 3]$ is broken into subintervals at points where the value of the greatest integer function changes.

For $\int_0^3 \$[x^2]\$ dx$ , the break points are $x = \sqrt{n}$ for integers $n \in [1, 8]$ .
For $\int_0^3 \left[\frac{x^2}{2}\right]\$ dx$ , the break points are $x = \sqrt{2n}$ for integers $n \in [1, 4]$ .

Summing the integrals of the resulting piecewise constant functions gives:
$I = (21 - 3\sqrt{2} - \sqrt{3} - \sqrt{5} - \sqrt{6} - \sqrt{7}) + (10 - 3\sqrt{2} - \sqrt{6})$
$I = 31 - 6\sqrt{2} - \sqrt{3} - \sqrt{5} - 2\sqrt{6} - \sqrt{7}$
Comparing this with the given form $a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7}$ , we find $a=31, b=-6, c=-2$ .
Therefore, $a+b+c = 31 - 6 - 2 = 23$ .

Q88JEE Main 2024NAT4MDifferential Equations

If the solution $y(x)$ of the given differential equation $(e^y + 1)\cos x \, dx + e^y \sin x \, dy = 0$ passes through the point $\left(\frac{\pi}{2}, 0\right)$ , then the value of $e^{y\left(\frac{\pi}{6}\right)}$ is equal to ___________

🚀 Solve in Practice Mode

📖 Explanation

We need to solve $(e^y + 1)\cos x\, dx + e^y \sin x\, dy = 0$ with the condition that it passes through $\left(\frac{\pi}{2}, 0\right)$ , and find $e^{y(\pi/6)}$ . Consider the function $F(x, y) = (e^y + 1)\sin x$ whose total differential is $dF = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy = (e^y + 1)\cos x\, dx + e^y \sin x\, dy$ , exactly matching the given equation. Hence $d\left[(e^y + 1)\sin x\right]\$ = 0\,.$ Integration yields $(e^y + 1)\sin x = C\,.$ Applying the condition at $\left(\frac{\pi}{2}, 0\right)$ gives $(e^0 + 1)\sin\frac{\pi}{2} = C \implies (1 + 1)\cdot 1 = C \implies C = 2\,.$ At $x = \frac{\pi}{6}$ , the equation becomes $(e^y + 1)\sin\frac{\pi}{6} = 2 \implies (e^y + 1)\cdot\frac{1}{2} = 2 \implies e^y + 1 = 4 \implies e^y = 3\,.$ Therefore, $e^{y(\pi/6)} = 3\,.$

Q89JEE Main 2024NAT4MThree Dimensional Geometry

If the shortest distance between the lines $\frac{x - \lambda}{3} = \frac{y - 2}{-1} = \frac{z - 1}{1}$ and $\frac{x + 2}{-3} = \frac{y + 5}{2} = \frac{z - 4}{4}$ is $\frac{44}{\sqrt{30}}$ , then the largest possible value of $|\lambda|$ is equal to ___________

🚀 Solve in Practice Mode

📖 Explanation

We need to find the largest $|\lambda|$ if the shortest distance between two lines equals $\frac{44}{\sqrt{30}}$ . The first line passes through $(\lambda, 2, 1)$ with direction vector $\vec{b_1} = (3, -1, 1)$ and the second line passes through $(-2, -5, 4)$ with direction vector $\vec{b_2} = (-3, 2, 4)$ . The cross product of the direction vectors is $\vec{b_1}\times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} = (-4 - 2)\hat{i} - (12 + 3)\hat{j} + (6 - 3)\hat{k} = (-6, -15, 3)$ and its magnitude is $|\vec{b_1}\times \vec{b_2}| = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$ . The vector from the point on the first line to the point on the second line is $\vec{a_2}-\vec{a_1} = (-2-\lambda, -7, 3)$ , and its dot product with the cross product is $(-2-\lambda)(-6) + (-7)(-15) + (3)(3) = 6(2+\lambda) + 105 + 9 = 6\lambda + 126\,.$ Using the formula for the shortest distance between skew lines, we have $d = \frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times \vec{b_2})|}{|\vec{b_1}\times \vec{b_2}|} = \frac{|6\lambda + 126|}{3\sqrt{30}} = \frac{44}{\sqrt{30}}\,,$ which implies $|6\lambda + 126| = 132\,.$ Hence, $6\lambda + 126 = 132 \implies \lambda = 1$ or $6\lambda + 126 = -132 \implies \lambda = -43\,.$ The larger value of $|\lambda|$ is $\max(|1|, |-43|) = 43$ , so the correct answer is 43.

Q90JEE Main 2024NAT4MProbability

From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items \in the sample. Let items \in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$ , where $\gcd(m, n) = 1$ , then $n - m$ is equal to ___________

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📖 Explanation

We need the variance of $X$ (number of defectives \in a sample of 5 from 12 items with 3 defectives, drawn without replacement). Here, $X$ follows a Hypergeometric distribution with parameters $N=12$ , $K=3$ (defectives), and $n=5$ (sample size). The variance for a Hypergeometric distribution is given by $\text{Var}(X) = n \cdot \frac{K}{N} \cdot \frac{N-K}{N} \cdot \frac{N-n}{N-1}$ . Substituting the values yields $= 5 \cdot \frac{3}{12} \cdot \frac{9}{12} \cdot \frac{7}{11} = \frac{5 \times 3 \times 9 \times 7}{12 \times 12 \times 11} = \frac{945}{1584}.$ Since $\gcd(945, 1584) = 9$ , we have $\frac{945}{1584} = \frac{105}{176}.$ Verifying the factors, $105 = 3\times5\times7$ and $176 = 2^4\times11$ , confirms that $\gcd(105,176)=1$ . Setting $m=105$ and $n=176$ gives $n - m = 176 - 105 = 71$ . The correct answer is 71 .

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