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Physics30 questions

Q1JEE Main 2024MCQ4MUnits and Measurements

If mass is written as $m = kc^{P}G^{-1/2}h^{1/2}$ , then the value of $P$ will be : (Constants have their usual meaning with $k$ a dimensionless constant)

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📖 Explanation

Given $m = kc^P G^{-1/2} h^{1/2}$ . Using dimensional analysis: $[m] = M$ , $[c] = LT^{-1}$ , $[G] = M^{-1}L^3T^{-2}$ , $[h] = ML^2T^{-1}$ . $M = (LT^{-1})^P (M^{-1}L^3T^{-2})^{-1/2} (ML^2T^{-1})^{1/2}$ $= L^P T^{-P} \cdot M^{1/2}L^{-3/2}T^{1} \cdot M^{1/2}L^{1}T^{-1/2}$ $= M^{1} L^{P-3/2+1} T^{-P+1-1/2}$ Matching M: $1 = 1$ ✓ Matching L: $0 = P - 1/2 \Rightarrow P = 1/2$ Matching T: $0 = -P + 1/2 \Rightarrow P = 1/2$ ✓ The answer is Option (1): $\boxed{\frac{1}{2}}$ .

Q2JEE Main 2024MCQ4MMotion in a Plane

Projectiles $A$ and $B$ are thrown at angles of $45°$ and $60°$ with vertical respectively from top of a 400 m high tower. If their \times of flight are same, the ratio of their speeds of projection $v_A : v_B$ is:

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📖 Explanation

Given, Two Projectiles A and B are thrown with an angle of $45°$ and $60°$ with vertical In the question it is specified that the angle is with respect to the vertical For the Time of flight (T) to be same their vertical velocities must be equal So As it is with respect to vertical the vertical velocities are:- For A $v_A=v_a\cos\theta\ _A$ ( $\cos\theta\ \ as\ \ \theta\ \ is\ with\ respect\ to\ vertical$ ) and For B $v_B=v_b\cos\theta\ _A$ Now these both should be equal so , $v_a\cos\left(45^{\circ\ }\right)=v_b\cos\left(60^{\circ\ }\right)$ $v_a\times \ \frac{1}{\sqrt{\ 2}}=v_b\times \ \frac{1}{2}$ $\frac{v_a}{v_b}=\ \frac{\sqrt{\ 2}}{2}=\frac{1}{\sqrt{\ 2}}$

Q3JEE Main 2024MCQ4MLaws of Motion

Three blocks $A$ , $B$ and $C$ are pulled on a horizontal smooth surface by a force of 80 N as shown \in figure. The tensions $T_1$ and $T_2$ in the string are respectively:

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📖 Explanation

Let us write equations for all three blocks A,B and C For A:- $F-T_2=2a$ $80-T_2=2a$ For B:- $T_2-T_1=3a$ and For C:- $T_1=5a$ Now on adding all 3 equations We get $80\ =\ 10a$ $a = 8 \text{ m s}^{-2}$ now substitute value of a \in equation 1 $80-T_2=2a$ $80-T_2=2\times \ 8$ $80-16\ =T_2$ $T_2=64N$ Substitute the value of a \in equation 3 $T_1=5a$ $T_1=5\times \ 8$ $T_1=40N$

Q4JEE Main 2024MCQ4MCentre of Mass and Collision

A block of mass $m$ is placed on a surface having vertical cross section given by $y = \frac{x^2}{4}$ . If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

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📖 Explanation

Surface is given by $y=\frac{x^2}{4}$ At any point, slope: $\frac{dy}{dx}=\frac{x}{2}$ So, $\tan\theta =\frac{x}{2}$ For no slipping: $\tan\theta \le \mu$ Given $\mu =0.5$ So: $\frac{x}{2}\le0.5$ $x\le1$ Now find maximum height: $y=\frac{x^2}{4}$ $=\frac{(1)^2}{4}$ = $\frac{1}{4}$ Final answer: $\frac{1}{4}$

Q5JEE Main 2024MCQ4MWork, Power and Energy

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of $60°$ by a force of 10 N parallel to the inclined surface as shown \in figure. When the block is pushed up by 10 m along inclined surface, the work done against frictional force is : $g = 10$ m s $^{-2}$

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📖 Explanation

The work done against friction is calculated using the formula $W_{f} = f_k \times d$ , where $f_k$ is the kinetic friction force and $d$ is the distance moved. First, we determine the normal force ( $N$ ) acting on the block, which is the component of gravity perpendicular to the incline: $N = mg \cos(\theta)$ . Substituting the given values: $N = 1 \text{ kg} \times 10 \text{ m/s}^2 \times \cos(60°) = 10 \times 0.5 = 5$ N. The kinetic friction force is $f_k = \mu N$ . Using the given coefficient $\mu = 0.1$ , we get $f_k = 0.1 \times 5 \text{ N} = 0.5$ N. Therefore, the work done against this friction over a distance of 10 m is $W_{f} = 0.5 \text{ N} \times 10 \text{ m} = 5$ J.

Q6JEE Main 2024MCQ4MGravitation

Escape velocity of a body from earth is 11.2 km s $^{-1}$ . If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is:

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📖 Explanation

Escape velocity: $v_e = \sqrt{\frac{2GM}{R}}$ . Given: $R_p = \frac{R_e}{3}$ and $M_p = \frac{M_e}{6}$ . $\frac{v_p}{v_e} = \sqrt{\frac{M_p/R_p}{M_e/R_e}} = \sqrt{\frac{M_p \cdot R_e}{M_e \cdot R_p}} = \sqrt{\frac{(M_e/6) \cdot R_e}{M_e \cdot (R_e/3)}} = \sqrt{\frac{1/6}{1/3}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$ $v_p = \frac{11.2}{\sqrt{2}} = \frac{11.2 \times \sqrt{2}}{2} = 5.6\sqrt{2} \approx 7.92 \text{ km/s}$ The answer is Option (4): $\boxed{7.9 \text{ km s}^{-1}}$ .

Q7JEE Main 2024MCQ4MThermodynamics

A block of ice at $-10°C$ is slowly heated and converted to steam at $100°C$ . Which of the following curves represent the phenomenon qualitatively:

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Q8JEE Main 2024MCQ4MThermodynamics

Choose the correct statement for processes $A$ & $B$ shown in figure.

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📖 Explanation

Yes - this question was asking to identify the processes, not compare work/heat. From the graph, process A is the less steep hyperbola, corresponding to an isothermal process: $PV=k$ For process BBB, the curve is steeper than an isotherm, so it represents an adiabatic process: $PV^{\gamma}=k$ because adiabatic curves are steeper than isotherms on a P-V diagram.

Q9JEE Main 2024MCQ4MThermodynamics

If three moles of monoatomic gas $\left(\gamma = \frac{5}{3}\right)$ is mixed with two moles of a diatomic gas $\left(\gamma = \frac{7}{5}\right)$ , the value of adiabatic exponent $\gamma$ for the mixture is:

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📖 Explanation

For a mixture of gases, the effective $\gamma$ is found using: $\frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1} = \frac{n_1 + n_2}{\gamma_{mix} - 1}$ $\frac{3}{5/3 - 1} + \frac{2}{7/5 - 1} = \frac{5}{\gamma - 1}$ $\frac{3}{2/3} + \frac{2}{2/5} = \frac{5}{\gamma - 1}$ $\frac{9}{2} + 5 = \frac{5}{\gamma - 1}$ $\frac{19}{2} = \frac{5}{\gamma - 1}$ $\gamma - 1 = \frac{10}{19}$ $\gamma = 1 + \frac{10}{19} = \frac{29}{19} \approx 1.526$ The answer is Option (3): $\boxed{1.52}$ .

Q10JEE Main 2024MCQ4MElectrostatics

A particle of charge $-q$ and mass $m$ moves \in a circle of radius $r$ around an infinitely long line charge of linear density $+\lambda$ . Then time period will be given as: (Consider $k$ as Coulomb's constant)

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📖 Explanation

Electric field due to infinite line charge at distance $r$ : $E = \frac{2k\lambda}{r}$ . For circular motion: $qE = \frac{mv^2}{r}$ . $q \cdot \frac{2k\lambda}{r} = \frac{mv^2}{r} \Rightarrow v^2 = \frac{2k\lambda q}{m}$ $v = \sqrt{\frac{2k\lambda q}{m}}$ Time period: $T = \frac{2\pi r}{v} = 2\pi r\sqrt{\frac{m}{2k\lambda q}}$ The answer is Option (2): $\boxed{T = 2\pi r\sqrt{\frac{m}{2k\lambda q}}}$ .

Q11JEE Main 2024MCQ4MCurrent Electricity

When a potential difference $V$ is applied across a wire of resistance $R$ , it dissipates energy at a rate $W$ . If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the energy dissipation rate will become:

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📖 Explanation

Original: $W = \frac{V^2}{R}$ . Wire cut \in two halves: each has resistance $R/2$ . Two halves \in parallel: $R_{eq} = \frac{(R/2)(R/2)}{R/2 + R/2} = \frac{R/4}{R} \cdot R = \frac{R}{4}$ . New power: $W' = \frac{V^2}{R/4} = \frac{4V^2}{R} = 4W$ . The answer is Option (4): $\boxed{4W}$ .

Q12JEE Main 2024MCQ4MAlternating Current

An alternating voltage $V(t) = 220\sin 100\pi t$ volt is applied to a purely resistive load of $50 \Omega$ . The time taken for the current to rise from half of the peak value to the peak value is:

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📖 Explanation

$V(t) = 220\sin(100\pi t)$ , $R = 50\Omega$ . Peak current: $I_0 = V_0/R = 220/50 = 4.4$ A. Current: $I = I_0\sin(100\pi t)$ . Half peak: $I_0/2$ when $\sin(100\pi t_1) = 1/2 \Rightarrow 100\pi t_1 = \pi/6 \Rightarrow t_1 = \frac{1}{600}$ s. Peak: when $\sin(100\pi t_2) = 1 \Rightarrow 100\pi t_2 = \pi/2 \Rightarrow t_2 = \frac{1}{200}$ s. $\Delta t = t_2 - t_1 = \frac{1}{200} - \frac{1}{600} = \frac{3-1}{600} = \frac{2}{600} = \frac{1}{300}$ s $= 3.33$ ms. The answer is Option (2): $\boxed{3.3 \text{ ms}}$ .

Q13JEE Main 2024MCQ4MElectrostatics

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

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📖 Explanation

This question requires matching fundamental laws of electromagnetism with their integral-form equations. Let's analyze each pair:

Gauss's law of magnetostatics (A) states that there are no magnetic monopoles, meaning the net magnetic flux through any closed surface is always zero. This is mathematically expressed as $\oint \vec{B} \cdot d\vec{a} = 0$ , which matches equation (ii).
Faraday's law of induction (B) describes how a time-varying magnetic field creates an electric field. The line integral of the electric field around a closed loop (EMF) equals the negative rate of change of magnetic flux through the surface, $\oint \vec{E} \cdot d\vec{l} = -\frac{d}{dt}\int \vec{B} \cdot d\vec{a}$ . This corresponds to equation (iii) (often written without the minus sign in this context).
Ampere's law (C) relates the magnetic field integrated around a closed loop to the electric current passing through that loop: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$ , which is equation (iv).
Gauss's law of electrostatics (D) relates the net electric flux through a closed surface to the total electric charge enclosed within it: $\oint \vec{E} \cdot d\vec{a} = \frac{Q_{enc}}{\epsilon_0} = \frac{1}{\epsilon_0}\int \rho dV$ . This matches equation (i).

Combining these matches gives us A-II, B-III, C-IV, and D-I.

Q14JEE Main 2024MCQ4MWave Optics

A beam of unpolarised light of intensity $I_0$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $45°$ relative to that of $A$ . The intensity of emergent light is:

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📖 Explanation

Unpolarised light of intensity $I_0$ passes through polaroid A, then polaroid B at $45°$ . After polaroid A (first polarizer), intensity becomes $I_1 = \frac{I_0}{2}$ (Malus's law for unpolarised light through a polarizer). After polaroid B at angle $45°$ to A. By Malus's law: $I_2 = I_1\cos^2 45° = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$ . The correct answer is Option 1 : $\frac{I_0}{4}$ .

Q15JEE Main 2024MCQ4MWork, Power and Energy

If the total energy transferred to a surface in time $t$ is $6.48 \times 10^5$ J, then the magnitude of the total momentum delivered to this surface for complete absorption will be:

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📖 Explanation

The momentum ( $p$ ) of electromagnetic radiation is fundamentally linked to its energy ( $E$ ) by the equation $E = pc$ , where $c$ is the speed of light. For a surface that completely absorbs the radiation, the total momentum delivered is calculated by rearranging this relationship to $p = E/c$ . We are given the total energy transferred, $E = 6.48 \times 10^5$ J. Using the value for the speed of light, $c = 3 \times 10^8$ m/s, we can calculate the momentum.
$p = \frac{E}{c} = \frac{6.48 \times 10^5 \text{ J}}{3 \times 10^8 \text{ m/s}}$
$p = 2.16 \times 10^{-3} \text{ kg m s}^{-1}$
This calculation provides the magnitude of the total momentum delivered to the surface.

Q16JEE Main 2024MCQ4MDual Nature of Matter and Radiation

For the photoelectric effect, the maximum kinetic energy $E_k$ of the photoelectrons is plotted against the frequency $(\nu)$ of the incident photons as shown in figure. The slope of the graph give

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📖 Explanation

According to Einstein's photoelectric equation, the maximum kinetic energy ( $E_k$ ) of a photoelectron is given by:
$E_k = h\nu - \phi_0$
Here, $h$ is Planck's constant, $\nu$ is the frequency of the incident light, and $\phi_0$ is the work function of the metal. This equation is analogous to the equation of a straight line, $y = mx + c$ . When we plot $E_k$ on the y-axis and $\nu$ on the x-axis, the slope ( $m$ ) of the line is equal to Planck's constant ( $h$ ). The y-intercept ( $c$ ) of the graph would be the negative of the work function ( $-\phi_0$ ).

Q17JEE Main 2024MCQ4MDual Nature of Matter and Radiation

An electron revolving in $n^{th}$ Bohr orbit has magnetic moment $\mu_n$ . If $\mu_n \propto n^x$ , the value of $x$ is:

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📖 Explanation

Magnetic moment of electron in $n$ th Bohr orbit: $\mu_n = \frac{eL}{2m_e}$ where $L = n\hbar$ is the angular momentum. $\mu_n = \frac{en\hbar}{2m_e} \propto n$ So $\mu_n \propto n^1$ , meaning $x = 1$ . The correct answer is Option 2 : 1.

Q18JEE Main 2024MCQ4MAtoms and Nuclei

In a nuclear fission reaction of an isotope of mass $M$ , three similar daughter nuclei of same mass are formed. The speed of a daughter nuclei \in terms of mass defect $\Delta M$ will be:

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📖 Explanation

In nuclear fission: $M \rightarrow 3 \times \frac{M}{3}$ . Mass defect = $\Delta M$ . Energy released: $E = \Delta M c^2$ . This energy is shared as kinetic energy among three identical daughter nuclei: $3 \times \frac{1}{2} \cdot \frac{M}{3} \cdot v^2 = \Delta M c^2$ . $\frac{M v^2}{2} = \Delta M c^2$ $v^2 = \frac{2\Delta M c^2}{M}$ $v = c\sqrt{\frac{2\Delta M}{M}}$ The answer is Option (3): $\boxed{c\sqrt{\frac{2\Delta M}{M}}}$ .

Q19JEE Main 2024MCQ4MSemiconductor Electronics

In the given circuit, the voltage across load resistance $(R_L)$ is:

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📖 Explanation

In the given series circuit, the Germanium (Ge) diode ( $D_1$ ) and the Silicon (Si) diode ( $D_2$ ) are forward-biased. The typical voltage drop (knee voltage) across a forward-biased Ge diode is $V_{Ge} = 0.3 \text{ V}$ and for a Si diode is $V_{Si} = 0.7 \text{ V}$ . The total voltage drop across both diodes is $V_D = V_{Ge} + V_{Si} = 0.3 + 0.7 = 1.0 \text{ V}$ .

Applying Kirchhoff's Voltage Law, the net voltage available for the resistors is $V_{net} = V_{source} - V_D = 15 \text{ V} - 1.0 \text{ V} = 14.0 \text{ V}$ . The total resistance \in the circuit is $R_{total} = 1.5 \text{ k}\Omega + 2.5 \text{ k}\Omega = 4.0 \text{ k}\Omega$ . The current flowing through the circuit is $I = \frac{V_{net}}{R_{total}} = \frac{14.0 \text{ V}}{4.0 \text{ k}\Omega} = 3.5 \text{ mA}$ .

The voltage across the load resistance $R_L$ is then found using Ohm's law:
$V_{RL} = I \times R_L = (3.5 \times 10^{-3} \text{ A}) \times (2.5 \times 10^3 \text{ }\Omega) = 8.75 \text{ V}$

Q20JEE Main 2024MCQ4MUnits and Measurements

If 50 Vernier divisions are equal to 49 main scale divisions of a travelling microscope and one smallest reading of main scale is 0.5 mm, the Vernier constant of travelling microscope is:

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📖 Explanation

50 Vernier divisions = 49 main scale divisions. 1 VSD = $\frac{49}{50}$ MSD. Least count = 1 MSD - 1 VSD = $1 - \frac{49}{50} = \frac{1}{50}$ MSD = $\frac{0.5}{50}$ mm = $0.01$ mm. The answer is Option (4): $\boxed{0.01 \text{ mm}}$ .

Q21JEE Main 2024NAT4MMotion in a Plane

A vector has magnitude same as that of $\vec{A} = 3\hat{i} + 4\hat{j}$ and is parallel to $\vec{B} = 4\hat{i} + 3\hat{j}$ . The $x$ and $y$ components of this vector \in first quadrant are $x$ and 3 respectively where $x$ = ____.

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📖 Explanation

$|\vec{A}| = \sqrt{9 + 16} = 5$ . $\vec{B} = 4\hat{i} + 3\hat{j}$ , $|\vec{B}| = 5$ . Unit vector along $\vec{B}$ : $\hat{B} = \frac{4\hat{i} + 3\hat{j}}{5}$ . Required vector = $5\hat{B} = 4\hat{i} + 3\hat{j}$ . The x-component is 4 and y-component is 3. Therefore, $x = \boxed{4}$ .

Q22JEE Main 2024NAT4MRotational Motion

Two discs of moment of inertia $I_1 = 4$ kg m $^2$ and $I_2 = 2$ kg m $^2$ about their central axes & normal to their planes, rotating with angular speeds 10 rad s $^{-1}$ & 4 rad s $^{-1}$ respectively are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is _________ J.

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📖 Explanation

Using conservation of angular momentum: $I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega$ $4(10) + 2(4) = 6\omega \Rightarrow \omega = \frac{48}{6} = 8$ rad/s. Loss \in KE = Initial KE - Final KE: $= \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2 - \frac{1}{2}(I_1+I_2)\omega^2$ $= \frac{1}{2}(4)(100) + \frac{1}{2}(2)(16) - \frac{1}{2}(6)(64)$ $= 200 + 16 - 192 = 24$ J. Therefore, the answer is $\boxed{24}$ .

Q23JEE Main 2024NAT4MProperties of Matter

A big drop is formed by coalescing 1000 small identical drops of water. If $E_1$ be the total surface energy of 1000 small drops of water and $E_2$ be the surface energy of single big drop of water, the $E_1 : E_2$ is $x : 1$ , where $x$ = ________.

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📖 Explanation

Volume of 1000 small drops = volume of big drop. $1000 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow R = 10r$ . $E_1 = 1000 \times 4\pi r^2 \times T$ $E_2 = 4\pi R^2 \times T = 4\pi (10r)^2 T = 400\pi r^2 T$ $\frac{E_1}{E_2} = \frac{1000 \times 4\pi r^2 T}{400\pi r^2 T} = \frac{4000}{400} = 10$ So $E_1 : E_2 = 10 : 1$ , meaning $x = 10$ . Therefore, the answer is $\boxed{10}$ .

Q24JEE Main 2024NAT4MOscillations

A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be _________ s. $[take $g = \pi^2$ m s $^{-2}$ ]$

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📖 Explanation

At height $h = R$ from surface, distance from center = $2R$ . Effective $g' = \frac{g}{(1 + h/R)^2} = \frac{g}{4}$ . $T = 2\pi\sqrt{\frac{l}{g'}} = 2\pi\sqrt{\frac{4}{g/4}} = 2\pi\sqrt{\frac{16}{g}} = 2\pi \times \frac{4}{\sqrt{g}}$ With $g = \pi^2$ : $T = 2\pi \times \frac{4}{\pi} = 8$ s. Therefore, the answer is $\boxed{8}$ .

Q25JEE Main 2024NAT4MWaves

A point source is emitting sound waves of intensity $16 \times 10^{-8}$ W m $^{-2}$ at the origin. The difference \in intensity (magnitude only) at two points located at distances of 2 m and 4 m from the origin respectively will be ________ $\times 10^{-8}$ W m $^{-2}$ .

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📖 Explanation

For a point source, $I\propto\frac{1}{r^2}$ Given intensity at source reference (at 1 m, implied) is $I_0=16\times10^{-8}\ \text{W/m}^2$ At r=2m, $I_1=\frac{16\times10^{-8}}{2^2}$ $=4\times10^{-8}$ At r=4m, $I_2=\frac{16\times10^{-8}}{4^2}$ $=1\times10^{-8}$ Difference \in intensity: $\left|I_1-I_2\right|$ $=(4-1)\times10^{-8}$ $=3\times10^{-8}$

Q26JEE Main 2024NAT4MElectrostatics

Two identical charged spheres are suspended by strings of equal lengths. The string make an angle of $37°$ with each other. When suspended \in a liquid of density $0.7$ g cm $^{-3}$ , the angle remains same. If density of material of the sphere is $1.4$ g cm $^{-3}$ , the dielectric constant of the liquid is _____ ( $\tan 37° = \frac{3}{4}$ )

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📖 Explanation

In air: $\tan(\theta/2) = \frac{F_e}{mg}$ where $F_e$ is the electrostatic force. In liquid: The angle remains the same, so $\tan(\theta/2)$ is unchanged. This means: $\frac{F_e'}{m'g} = \frac{F_e}{mg}$ where $F_e' = \frac{F_e}{K}$ (dielectric constant $K$ ) and $m'g = mg - \text{buoyancy} = mg(1 - \frac{\rho_l}{\rho_s})$ . $\frac{F_e/K}{mg(1 - \rho_l/\rho_s)} = \frac{F_e}{mg}$ $\frac{1}{K} = 1 - \frac{\rho_l}{\rho_s} = 1 - \frac{0.7}{1.4} = 1 - 0.5 = 0.5$ $K = 2$ . Therefore, the answer is $\boxed{2}$ .

Q27JEE Main 2024NAT4MCurrent Electricity

Two resistance of $100\Omega$ and $200\Omega$ are connected \in series with a battery of 4 V and negligible internal resistance. A voltmeter is used to measure voltage across $100\Omega$ resistance, which gives reading as 1 V. The resistance of voltmeter must be _______ $\Omega$ .

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📖 Explanation

$100\Omega$ and $200\Omega$ \in series, battery 4V. Voltmeter across $100\Omega$ reads 1V. The voltmeter (resistance $R_V$ ) is \in parallel with $100\Omega$ . Equivalent = $\frac{100 R_V}{100 + R_V}$ . Total resistance = $\frac{100R_V}{100+R_V} + 200$ . Current = $\frac{4}{\frac{100R_V}{100+R_V} + 200}$ . Voltage across parallel combination = 1V. So voltage across $200\Omega$ = 3V. Current = $\frac{3}{200} = 0.015$ A. For the parallel combination: $V = IR_{eq}$ : $1 = 0.015 \times \frac{100R_V}{100+R_V}$ $\frac{100R_V}{100+R_V} = \frac{1}{0.015} = \frac{200}{3}$ $300R_V = 200(100 + R_V) = 20000 + 200R_V$ $100R_V = 20000 \Rightarrow R_V = 200\Omega$ Therefore, the answer is $\boxed{200}$ .

Q28JEE Main 2024NAT4MMagnetic Effects of Current and Magnetism

The current of 5 A flows in a square loop of sides 1 m is placed in air. The magnetic field at the centre of the loop is $X\sqrt{2} \times 10^{-7}$ T. The value of X is _________.

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📖 Explanation

Magnetic field at center of a square loop of side $a$ with current $I$ : Each side contributes $B_1 = \frac{\mu_0 I}{4\pi d}(\sin\alpha_1 + \sin\alpha_2)$ where $d = a/2$ and $\alpha_1 = \alpha_2 = 45°$ . $B_1 = \frac{\mu_0 I}{4\pi (a/2)} \times 2\sin 45° = \frac{\mu_0 I}{2\pi a} \times \sqrt{2} = \frac{\sqrt{2}\mu_0 I}{2\pi a}$ Total for 4 sides: $B = 4B_1 = \frac{4\sqrt{2}\mu_0 I}{2\pi a} = \frac{2\sqrt{2}\mu_0 I}{\pi a}$ With $I = 5$ A, $a = 1$ m: $B = \frac{2\sqrt{2} \times 4\pi \times 10^{-7} \times 5}{\pi \times 1} = 2\sqrt{2} \times 4 \times 5 \times 10^{-7} = 40\sqrt{2} \times 10^{-7}$ So $X = 40$ . Therefore, the answer is $\boxed{40}$ .

Q29JEE Main 2024NAT4MElectromagnetic Induction

A power transmission line feeds input power at 2.3 kV to a step down transformer with its primary winding having 3000 turns. The output power is delivered at 230 V by the transformer. The current in the primary of the transformer is 5 A and its efficiency is 90%. The winding of transformer is made of copper. The output current of transformer is ____ A.

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📖 Explanation

Step-down transformer: $V_p = 2300$ V, $N_p = 3000$ , $V_s = 230$ V, $I_p = 5$ A, efficiency = 90%. $\frac{N_s}{N_p} = \frac{V_s}{V_p} = \frac{230}{2300} = \frac{1}{10}$ , so $N_s = 300$ . Input power = $V_p I_p = 2300 \times 5 = 11500$ W. Output power = $0.9 \times 11500 = 10350$ W. $I_s = \frac{P_s}{V_s} = \frac{10350}{230} = 45$ A. The output current is 45 A.

Q30JEE Main 2024NAT4MRay Optics and Optical Instruments

In an experiment to measure the focal length $(f)$ of a convex lens, the magnitude of object distance $(x)$ and the image distance $(y)$ are measured with reference to the focal point of the lens. The $y - x$ plot is shown in figure. The focal length of the lens is ____ cm.

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📖 Explanation

This experiment uses Newton's Lens Formula, which relates the object distance ( $x$ ) and image distance ( $y$ ) measured from their respective focal points. The formula is $xy = f^2$ , where $f$ is the focal length of the lens. The given graph is a plot of $y$ versus $x$ , which represents this hyperbolic relationship. To find the focal length, we can pick any point from the experimental curve. Let's choose point C, where the coordinates are clearly defined. At point C, we have $x = 20$ cm and $y = 20$ cm. Substituting these values into Newton's formula gives $f^2 = (20 \text{ cm}) \times (20 \text{ cm}) = 400 \text{ cm}^2$ . Therefore, the focal length is $f = \sqrt{400} = 20$ cm.

Chemistry30 questions

Q31JEE Main 2024MCQ4MClassification of Elements

Given below are two statements: Statement - I: Along the period, the chemical reactivity of the element gradually increases from group 1 to group 18. Statement - II: The nature of oxides formed by group 1 element is basic while that of group 17 elements is acidic. In the light of the above statements, choose the most appropriate from the options given below:

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📖 Explanation

Statement I: Along the period, chemical reactivity does NOT gradually increase from group 1 to group 18. Reactivity first decreases (metals less reactive), then increases for non-metals, and noble gases (group 18) are least reactive. So Statement I is FALSE. Statement II: Group 1 elements form basic oxides (like Na $_2$ O, K $_2$ O) and Group 17 elements form acidic oxides (like Cl $_2$ O $_7$ ). This is TRUE. The answer is Option (3): Statement I is false but Statement II is true.

Q32JEE Main 2024MCQ4MChemical Bonding and Molecular Structure

Given below are two statements: Statement-I: Since fluorine is more electronegative than nitrogen, the net dipole moment of $NF_3$ is greater than $NH_3$ . Statement-II: In $NH_3$ , the orbital dipole due to lone pair and the dipole moment of NH bonds are \in opposite direction, but \in $NF_3$ the orbital dipole due to lone pair and dipole moments of N-F bonds are in same direction. In the light of the above statements, choose the most appropriate from the options given below.

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📖 Explanation

Statement I claims that since fluorine is more electronegative than nitrogen, the net dipole moment of $NF_3$ is greater than $NH_3$ . However, experimental dipole moments are $\mu(NH_3) = 1.47$ D and $\mu(NF_3) = 0.23$ D, so $NF_3$ has a much lower dipole moment than $NH_3$ . This is because \in $NH_3$ , the lone pair dipole and the bond dipoles reinforce each other, while \in $NF_3$ they oppose each other (partially cancelling). Statement I is false. Statement II asserts that \in $NH_3$ the orbital dipole due to lone pair and the dipole moments of N-H bonds are \in opposite direction, but \in $NF_3$ they are \in the same direction. In reality, the situation is reversed: \in $NH_3$ the N-H bond dipoles point from H towards N (since N is more electronegative) and the lone pair on nitrogen also contributes a dipole \in the same direction (away from nitrogen, towards the lone pair region), so they reinforce each other. In $NF_3$ the N-F bond dipoles point from N towards F (since F is more electronegative than N), i.e., away from nitrogen, while the lone pair dipole still points in the direction from the base towards the apex (away from the lone pair), so they oppose each other, causing partial cancellation and a low dipole moment. Statement II is false. Since both statements are false, the correct answer is Option (2): Both Statement I and Statement II are false .

Q33JEE Main 2024MCQ4MAcid-Base Chemistry

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: $H_2Te$ is more acidic than $H_2S$ . Reason R: Bond dissociation enthalpy of $H_2Te$ is lower than $H_2S$ . In the light of the above statements, choose the most appropriate from the options given below.

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📖 Explanation

Assertion A: H $_2$ Te is more acidic than H $_2$ S - TRUE. Acidic strength increases down the group. Reason R: Bond dissociation enthalpy of H-Te bond is lower than H-S bond - TRUE. The larger size of Te makes the bond weaker. R correctly explains A: because the lower bond dissociation enthalpy means easier release of H $^+$ , making H $_2$ Te more acidic. The answer is Option (2): Both A and R are true and R is the correct explanation of A.

Q34JEE Main 2024MCQ4MAmines

IUPAC name of the following compound is:

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📖 Explanation

To determine the IUPAC name, first identify the principal functional group, which is the nitrile ( $-CN$ ) group, as it has higher priority than the amino ( $-NH_2$ ) group. The longest carbon chain including the nitrile carbon consists of four carbons, making the parent name butanenitrile. The numbering of the carbon chain begins at the nitrile carbon, which is assigned position 1. Following this, the amino group ( $-NH_2$ ) is located on the third carbon atom (C-3). Therefore, the correct IUPAC name for the compound is 3-Aminobutanenitrile.

Q35JEE Main 2024MCQ4MPractical Organic Chemistry

Which among the following purification methods is based on the principle of "Solubility" in two different solvents?

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📖 Explanation

The purification method based on the principle of solubility in two different solvents is Differential Extraction. In differential extraction, a substance is partitioned between two immiscible solvents based on its different solubilities in each. The answer is Option (4): $\boxed{\text{Differential Extraction}}$ .

Q36JEE Main 2024MCQ4MOrganic Chemistry - Some Basic Principles

The correct stability order of carbocations is

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📖 Explanation

Carbocation stability depends on the degree of substitution due to hyperconjugation and inductive effects from alkyl groups. The order of carbocation stability is: Tertiary > Secondary > Primary > Methyl $(CH_3)_3C^+ > (CH_3)_2CH^+ > CH_3CH_2^+ > CH_3^+$ The correct answer is Option 3 .

Q37JEE Main 2024MCQ4MAlcohols, Phenols and Ethers

Product A and B formed in the following set of reactions are:

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📖 Explanation

The reaction of 1-methylcyclohexene with acidic water ( $H^+/H_2O$ ) is an acid-catalyzed hydration. This process follows Markovnikov's rule, where the proton adds to the less substituted carbon (C-2) to form a more stable tertiary carbocation at C-1. Water then attacks this carbocation, forming the tertiary alcohol, 1-methylcyclohexan-1-ol, as product A.

Conversely, the reaction with diborane ( $B_2H_6$ ) followed by oxidation ( $H_2O_2, NaOH$ ) is a hydroboration-oxidation. This reaction is an anti-Markovnikov syn-addition of water. The boron atom adds to the less sterically hindered and less substituted carbon (C-2), and after oxidation, the hydroxyl group is placed on this carbon, yielding 2-methylcyclohexanol as product B.

Q38JEE Main 2024MCQ4MSolutions

If a substance ' $A$ ' dissolves \in solution of a mixture of ' $B$ ' and ' $C$ ' with their respective number of moles as $n_A$ , $n_B$ and $n_C$ , mole fraction of $C$ in the solution is:

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📖 Explanation

The mole fraction of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components. The solution contains three components, A, B, and C, with their respective number of moles being $n_A$ , $n_B$ , and $n_C$ .

The total number of moles in the solution is the sum of the moles of all individual components:
$n_{total} = n_A + n_B + n_C$

To find the mole fraction of component C ( $X_C$ ), we divide the number of moles of C by the total number of moles:
$X_C = \frac{\text{moles of C}}{\text{total moles}} = \frac{n_C}{n_A + n_B + n_C}$

Q39JEE Main 2024MCQ4MSolutions

The solution from the following with highest depression in freezing point/lowest freezing point is

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📖 Explanation

We need the solution with highest depression in freezing point. $\Delta T_f = iK_fm$ where $i$ is van't Hoff factor and $m$ is molality. Option 1: 180 g acetic acid (M=60) \in 1 L water. Moles = 3. $m \approx 3$ . Acetic acid dissociates slightly \in water, $i > 1$ . $\Delta T_f \approx 3 \times 1.86 \times i > 5.58$ . Option 2: 180 g acetic acid \in benzene. Acetic acid dimerizes \in benzene, $i \approx 0.5$ . $\Delta T_f$ much lower. Option 3: 180 g benzoic acid (M=122) \in benzene. Moles $\approx 1.48$ . Benzoic acid also dimerizes \in benzene, $i \approx 0.5$ . Option 4: 180 g glucose (M=180) \in water. Moles = 1. Non-electrolyte, $i = 1$ . $\Delta T_f = 1.86$ . Option 1 has the highest molality (3 mol/kg) and \in water, acetic acid ionizes ( $i > 1$ ), giving the largest freezing point depression. The correct answer is Option 1 .

Q40JEE Main 2024MCQ4MElectrochemistry

Reduction potential of ions are given below: $ClO_4^-$ : $E° = 1.19$ V; $IO_4^-$ : $E° = 1.65$ V; $BrO_4^-$ : $E° = 1.74$ V The correct order of their oxidising power is:

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📖 Explanation

Higher reduction potential means stronger oxidizing agent. $BrO_4^-: E° = 1.74$ V (highest), $IO_4^-: E° = 1.65$ V, $ClO_4^-: E° = 1.19$ V (lowest). Oxidising power order: $BrO_4^- > IO_4^- > ClO_4^-$ . The answer is Option (2): $\boxed{BrO_4^- > IO_4^- > ClO_4^-}$ .

Q41JEE Main 2024MCQ4Mp Block Elements

Choose the correct statements about the hydrides of group 15 elements. A. The stability of the hydrides decreases in the order $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$ B. The reducing ability of the hydrides increases \in the order $NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$ C. Among the hydrides, $NH_3$ is strong reducing agent while $BiH_3$ is mild reducing agent. D. The basicity of the hydrides increases \in the order $NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$ Choose the most appropriate from the options given below:

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📖 Explanation

We need to evaluate the statements about hydrides of Group 15 elements ( $NH_3, PH_3, AsH_3, SbH_3, BiH_3$ ). Statement A: The stability of hydrides decreases \in the order $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$ . As we go down the group, the bond length between the central atom and hydrogen increases (N-H < P-H < As-H < Sb-H < Bi-H). Longer bonds are weaker, so the thermal stability decreases down the group. This statement is correct . ✓ Statement B: The reducing ability of hydrides increases \in the order $NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$ . Reducing ability is the tendency to donate electrons or get oxidized. As the stability decreases down the group, the hydride is more easily decomposed and more readily donates its electron pair. Therefore, reducing character increases from $NH_3$ to $BiH_3$ . This statement is correct . ✓ Statement C: Among the hydrides, $NH_3$ is a strong reducing agent while $BiH_3$ is a mild reducing agent. This is the opposite of what is true. $NH_3$ is the weakest reducing agent (most stable hydride), while $BiH_3$ is the strongest reducing agent (least stable). This statement is incorrect . ✗ Statement D: The basicity of hydrides increases \in the order $NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$ . Basicity depends on the availability of the lone pair. As the size of the central atom increases down the group, the electron density on the lone pair decreases (spread over a larger orbital), making it less available for donation. Therefore, basicity decreases down the group: $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$ . This statement claims the opposite, so it is incorrect . ✗ The correct statements are A and B only. The correct answer is Option (3): A and B only .

Q42JEE Main 2024MCQ4Md and f Block Elements

The orange colour of $K_2Cr_2O_7$ and purple colour of $KMnO_4$ is due to

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📖 Explanation

In $K_2Cr_2O_7$ : Cr is \in +6 oxidation state with electronic configuration $d^0$ . Since there are no d-electrons, d-d transitions are impossible. The orange color arises from charge transfer transitions (ligand-to-metal charge transfer, LMCT), where electrons from oxygen ligands are transferred to empty d-orbitals of Cr $^{6+}$ . In $KMnO_4$ : Mn is \in +7 oxidation state with electronic configuration $d^0$ . Again, no d-electrons means no d-d transitions. The intense purple color is due to charge transfer transitions (LMCT from oxygen to Mn $^{7+}$ ). Both colors are due to charge transfer transitions, not d-d transitions. The correct answer is Option 1 : Charge transfer transition in both.

Q43JEE Main 2024MCQ4MCoordination Compounds

$A$ and $B$ formed \in the following reactions are: $CrO_2Cl_2 + 4NaOH \rightarrow A + 2NaCl + 2H_2O$ $A + 2HCl + 2H_2O_2 \rightarrow B + 3H_2O$

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📖 Explanation

Chromyl chloride $CrO_2Cl_2$ contains chromium \in the $+6$ oxidation state. In alkaline medium chloride ions are replaced by oxide ions, converting the molecule into the chromate ion $CrO_4^{2-}$ . Writing the reaction with hydroxide ions first: $CrO_2Cl_2 + 4OH^- \rightarrow CrO_4^{2-} + 2Cl^- + 2H_2O$ On adding sodium ions to both sides, the molecular equation becomes $CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$ Thus, $A = Na_2CrO_4$ (sodium chromate). Next, sodium chromate reacts with dilute acid and hydrogen peroxide. Under mildly acidic conditions the chromate ion forms a blue peroxo complex, commonly written as $CrO_5$ (chromium(VI) peroxide): Ionic form of the reaction: $CrO_4^{2-} + 2H^+ + 2H_2O_2 \rightarrow CrO_5 + 3H_2O$ Adding the spectator sodium and chloride ions gives the required molecular equation: $Na_2CrO_4 + 2HCl + 2H_2O_2 \rightarrow CrO_5 + 2NaCl + 3H_2O$ Hence, $B = CrO_5$ . Therefore $A = Na_2CrO_4,\qquad B = CrO_5$ , which corresponds to Option A.

Q44JEE Main 2024MCQ4MElectrochemistry

Alkaline oxidative fusion of $MnO_2$ gives "A" which on electrolytic oxidation \in alkaline solution produces $B$ . $A$ and $B$ respectively are:

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📖 Explanation

Alkaline oxidative fusion of MnO $_2$ : $2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$ A = MnO $_4^{2-}$ (manganate ion, green color). Electrolytic oxidation \in alkaline solution: $MnO_4^{2-} \rightarrow MnO_4^- + e^-$ B = MnO $_4^-$ (permanganate ion, purple color). The answer is Option (2): $MnO_4^{2-}$ and $MnO_4^-$ .

Q45JEE Main 2024MCQ4MCoordination Compounds

The molecule/ion with square pyramidal shape is:

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📖 Explanation

We need to identify which molecule or ion has a square pyramidal shape. To determine the shape, we use VSEPR theory: first count the total electron pairs (bond pairs + lone pairs) around the central atom, then determine the geometry. Recall: The shape of a molecule depends on the number of bond pairs (BP) and lone pairs (LP) around the central atom. A square pyramidal shape arises when there are 6 electron pairs total: 5 bond pairs and 1 lone pair (derived from octahedral electron geometry). Option 1: $[Ni(CN)_4]^{2-}$ Nickel \in this complex is Ni $^{2+}$ with electronic configuration $[Ar] 3d^8$ . CN $^-$ is a strong field ligand, so it causes pairing of d-electrons. With 4 ligands and a $d^8$ configuration \in a strong field, the complex adopts a square planar geometry ( $dsp^2$ hybridization). This is NOT square pyramidal. Option 2: $PCl_5$ Phosphorus has 5 valence electrons. Each Cl contributes 1 bond pair, giving 5 BP and 0 LP around P. With 5 bond pairs and no lone pairs, the geometry is trigonal bipyramidal ( $sp^3d$ hybridization). This is NOT square pyramidal. Option 3: $BrF_5$ Bromine has 7 valence electrons. Each F contributes 1 bond pair, giving 5 BP. The remaining electrons on Br: $7 - 5 = 2$ electrons = 1 lone pair (LP). So there are 5 BP + 1 LP = 6 total electron pairs around Br. With 6 electron pairs, the electron geometry is octahedral ( $sp^3d^2$ hybridization). When one position of the octahedron is occupied by a lone pair, the molecular shape becomes square pyramidal . Option 4: $PF_5$ Phosphorus has 5 valence electrons, and each F contributes 1 bond pair, giving 5 BP and 0 LP. The geometry is trigonal bipyramidal (same as $PCl_5$ ). This is NOT square pyramidal. Conclusion: Only $BrF_5$ has 5 bond pairs and 1 lone pair arranged \in an octahedral electron geometry, resulting \in a square pyramidal molecular shape. The correct answer is Option (3): $BrF_5$ .

Q46JEE Main 2024MCQ4MCoordination Compounds

The coordination geometry around the manganese in decacarbonyldimanganese(0) is:

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📖 Explanation

Decacarbonyldimanganese(0) is Mn $_2$ (CO) $_{10}$ . Each Mn is bonded to 5 CO ligands and one Mn-Mn bond, giving 6 groups around each Mn. This corresponds to octahedral geometry. The answer is Option (1): Octahedral.

Q47JEE Main 2024MCQ4MHaloalkanes and Haloarenes

Given below are two statements: Statement - I: High concentration of strong nucleophilic reagent with secondary alkyl halides which do not have bulky substituents will follow $S_N2$ mechanism. Statement - II: A secondary alkyl halide when treated with a large excess of ethanol follows $S_N1$ mechanism. In the light of the above statements, choose the most appropriate from the options given below:

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📖 Explanation

We need to evaluate both statements about nucleophilic substitution mechanisms with secondary alkyl halides. Background - Key factors determining S $_N$ 1 vs S $_N$ 2 mechanism: 1. S $_N$ 2 mechanism is favored by: strong nucleophile, high nucleophile concentration, polar aprotic solvent, primary or secondary substrate without bulky groups. 2. S $_N$ 1 mechanism is favored by: weak nucleophile, polar protic solvent, stable carbocation (tertiary > secondary), and the rate depends only on the substrate concentration. Analysis of Statement I: "High concentration of strong nucleophilic reagent with secondary alkyl halides which do not have bulky substituents will follow S $_N$ 2 mechanism." In S $_N$ 2, the rate law is: $\text{Rate} = k[\text{substrate}][\text{nucleophile}]$ . A high concentration of a strong nucleophile increases the rate of the S $_N$ 2 pathway. Secondary alkyl halides can undergo both S $_N$ 1 and S $_N$ 2, but with a strong nucleophile at high concentration and no steric hindrance (no bulky substituents), the S $_N$ 2 pathway dominates because the nucleophile can effectively attack the electrophilic carbon via backside attack. Therefore, Statement I is TRUE . Analysis of Statement II: "A secondary alkyl halide when treated with a large excess of ethanol follows S $_N$ 1 mechanism." Ethanol ( $C_2H_5OH$ ) is a weak nucleophile and a polar protic solvent . In S $_N$ 1, the rate-determining step is the formation of the carbocation: $R\text{-}X \rightarrow R^+ + X^-$ . A polar protic solvent like ethanol stabilizes both the carbocation intermediate and the leaving group through solvation. Since ethanol is a weak nucleophile (it does not strongly drive S $_N$ 2), and it acts as a polar protic solvent that stabilizes the carbocation, the reaction follows the S $_N$ 1 pathway. The large excess of ethanol means it acts primarily as the solvent rather than as a concentrated nucleophile driving a bimolecular mechanism. Therefore, Statement II is TRUE . Since both statements are true, the correct answer is Option (4): Both Statement I and Statement II are true .

Q48JEE Main 2024MCQ4MAldehydes, Ketones and Carboxylic Acids

Salicylaldehyde is synthesized from phenol, when reacted with

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📖 Explanation

The synthesis of salicylaldehyde from phenol is accomplished through the Reimer-Tiemann reaction. In this reaction, phenol is treated with chloroform ( $CHCl_3$ ) \in the presence of a strong base, typically sodium hydroxide (NaOH). The NaOH deprotonates phenol to form the highly reactive phenoxide ion. Simultaneously, NaOH reacts with chloroform to generate the electrophile, dichlorocarbene ( $:CCl_2$ ). The electron-rich phenoxide ring attacks the carbene, and after a series of steps including hydrolysis, a formyl group (-CHO) is introduced, primarily at the ortho-position, yielding salicylaldehyde.

Q49JEE Main 2024MCQ4MAldehydes, Ketones and Carboxylic Acids

m-chlorobenzaldehyde on treatment with 50% KOH solution yields

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📖 Explanation

m-chlorobenzaldehyde is an aldehyde that lacks an $\alpha$ -hydrogen. In the presence of a concentrated base like 50% KOH, it is susceptible to the Cannizzaro reaction. However, under these strong basic conditions, the chlorine atom on the benzene ring first undergoes nucleophilic aromatic substitution by the hydroxide ion $(OH^-)$ , forming m-hydroxybenzaldehyde. This intermediate, which also lacks an $\alpha$ -hydrogen, then undergoes the Cannizzaro reaction. This is a disproportionation reaction where one molecule of the aldehyde is oxidized to a carboxylate salt (m-hydroxybenzoate) and another molecule is reduced to a primary alcohol (m-hydroxybenzyl alcohol).

Q50JEE Main 2024MCQ4MAmines

The products A and B formed in the following reaction scheme are respectively:

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📖 Explanation

The reaction begins with the nitration of benzene using conc. $HNO_3$ /conc. $H_2SO_4$ to form nitrobenzene. Subsequently, the nitro group is reduced to an amino group using $Sn/HCl$ , yielding aniline (A, $C_6H_5NH_2$ ). This two-step process converts benzene into aniline.

Next, aniline (A) undergoes diazotization upon treatment with $NaNO_2$ and $HCl$ at low temperatures (273-278 K), forming benzenediazonium chloride ( $C_6H_5N_2^+Cl^-$ ). This diazonium salt then reacts with phenol in an azo coupling reaction. The hydroxyl group ( $-OH$ ) in phenol is a strong activating and para-directing group, so the coupling occurs at the para-position to form p-hydroxyazobenzene (B), an orange azo dye.

Q51JEE Main 2024NAT4MStructure of Atom

Number of spectral lines obtained in $He^+$ spectra, when an electron makes transition from fifth excited state to first excited state will be

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📖 Explanation

In He $^+$ , transition from 5th excited state (n = 6) to first excited state (n = 2). Number of spectral lines from transitions between levels n = 6 to n = 2: Number of levels involved = 6 - 2 + 1 = 5 (levels 2, 3, 4, 5, 6). Number of spectral lines = $\frac{n(n-1)}{2} = \frac{5 \times 4}{2} = 10$ . Therefore, the answer is $\boxed{10}$ .

Q52JEE Main 2024NAT4MChemical Thermodynamics

Two reactions are given below: $2Fe_{(s)} + \frac{3}{2}O_{2(g)} \rightarrow Fe_2O_{3(s)}$ , $\Delta H° = -822$ kJ/mol $C_{(s)} + \frac{1}{2}O_{2(g)} \rightarrow CO_{(g)}$ , $\Delta H° = -110$ kJ/mol Then enthalpy change for following reaction: $3C_{(s)} + Fe_2O_{3(s)} \rightarrow 2Fe_{(s)} + 3CO_{(g)}$

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📖 Explanation

We need to find the enthalpy change for the reaction: $3C_{(s)} + Fe_2O_{3(s)} \rightarrow 2Fe_{(s)} + 3CO_{(g)}$ We are given two reactions: The first is $2Fe_{(s)} + \frac{3}{2}O_{2(g)} \rightarrow Fe_2O_{3(s)}$ , $\Delta H_1° = -822$ kJ/mol. The second is $C_{(s)} + \frac{1}{2}O_{2(g)} \rightarrow CO_{(g)}$ , $\Delta H_2° = -110$ kJ/mol. Hess's Law states that the enthalpy change of an overall reaction is the \sum of the enthalpy changes of the individual steps, regardless of the pathway taken, so we manipulate the given reactions to arrive at the target reaction. Since the target reaction has $Fe_2O_{3}$ as a reactant and $Fe$ as a product, we reverse the first reaction, which changes the sign of $\Delta H$ : $Fe_2O_{3(s)} \rightarrow 2Fe_{(s)} + \frac{3}{2}O_{2(g)}, \quad \Delta H = +822 \text{ kJ/mol}$ The target reaction requires 3 moles of C and produces 3 moles of CO, so we multiply the second reaction by 3; this also multiplies its enthalpy change by 3: $3C_{(s)} + \frac{3}{2}O_{2(g)} \rightarrow 3CO_{(g)}, \quad \Delta H = 3 \times (-110) = -330 \text{ kJ/mol}$ Adding the reversed iron oxide reaction and the scaled carbon reaction gives: $Fe_2O_{3(s)} + 3C_{(s)} + \frac{3}{2}O_{2(g)} \rightarrow 2Fe_{(s)} + \frac{3}{2}O_{2(g)} + 3CO_{(g)}$ The $\frac{3}{2}O_{2(g)}$ appears on both sides and cancels out, yielding: $3C_{(s)} + Fe_2O_{3(s)} \rightarrow 2Fe_{(s)} + 3CO_{(g)}$ This matches our target reaction. $\Delta H_{total} = (+822) + (-330) = 822 - 330 = 492 \text{ kJ/mol}$ The enthalpy change for the given reaction is $\boxed{492}$ kJ/mol.

Q53JEE Main 2024NAT4MIonic Equilibrium

The pH of an aqueous solution containing 1M benzoic acid ( $pK_a = 4.20$ ) and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is _________ mL.

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📖 Explanation

Henderson-Hasselbalch equation: $pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}$ $4.5 = 4.2 + \log\frac{[\text{benzoate}]}{[\text{benzoic acid}]}$ $\log\frac{[\text{benzoate}]}{[\text{benzoic acid}]} = 0.3 \Rightarrow \frac{[\text{benzoate}]}{[\text{benzoic acid}]} = 2$ In 300 mL buffer: let volume of benzoic acid = $V$ mL (1M), then volume of sodium benzoate = $(300 - V)$ mL (1M). $\frac{300 - V}{V} = 2 \Rightarrow 300 - V = 2V \Rightarrow 300 = 3V \Rightarrow V = 100$ mL. Therefore, the answer is $\boxed{100}$ .

Q54JEE Main 2024NAT4MRedox Reactions

Total number of species from the following which can undergo disproportionation reaction: $H_2O_2$ , $ClO_3^-$ , $P_4$ , $Cl_2$ , $Ag$ , $Cu^{+1}$ , $F_2$ , $NO_2$ , $K^+$

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📖 Explanation

Species that can undergo disproportionation have an element in an intermediate oxidation state: $H_2O_2$ : O is \in -1 state (can go to 0 and -2) - YES $ClO_3^-$ : Cl is \in +5 - YES (can disproportionate) $P_4$ : P is \in 0 state - YES (goes to -3 and +5 \in alkali) $Cl_2$ : Cl is \in 0 state - YES (goes to -1 and +1) $Ag$ : metallic, 0 state - NO (already \in lowest practical state) $Cu^{+1}$ : Can go to Cu(0) and Cu(+2) - YES $F_2$ : F is \in 0 state, but F only goes to -1 (no positive states) - NO $NO_2$ : N is \in +4 - YES (goes to +5 and +3) $K^+$ : Cannot disproportionate - NO Total = 6 (H $_2$ O $_2$ , ClO $_3^-$ , P $_4$ , Cl $_2$ , Cu $^{+1}$ , NO $_2$ ). Therefore, the answer is $\boxed{6}$ .

Q55JEE Main 2024NAT4MHydrocarbons

Number of geometrical isomers possible for the given structure is/are ________.

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📖 Explanation

To determine the number of geometrical isomers, we must identify all stereogenic centers, which are typically double bonds where each carbon atom is attached to two different groups.

The number of geometrical isomers for an unsymmetrical molecule with $n$ stereogenic double bonds is given by the formula $2^n$ .

In the given structure, we can identify two double bonds that act as stereogenic centers:

The leftmost double bond: (H)(D)C=C(CH3)(rest). The left carbon has two different groups (H and D), and the right carbon also has two different groups (CH3 and the rest of the molecule).
The top-right double bond: (rest)(CH3)C=C(H)(D). Similarly, the left carbon has two different groups (CH3 and rest), and the right carbon has H and D.

The third double bond is drawn ambiguously and is considered non-stereogenic in this context. Thus, there are $n=2$ sites for geometrical isomerism. The molecule is unsymmetrical, so the number of possible geometrical isomers is $2^n = 2^2 = 4$ .

Q56JEE Main 2024NAT4MChemical Kinetics

$NO_2$ required for a reaction is produced by decomposition of $N_2O_5$ \in $CCl_4$ as by equation: $2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$ The initial concentration of $N_2O_5$ is 3 mol L $^{-1}$ and it is 2.75 mol L $^{-1}$ after 30 minutes. The rate of formation of $NO_2$ is $x \times 10^{-3}$ mol L $^{-1}$ min $^{-1}$ , value of $x$ is ________.

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📖 Explanation

$2N_2O_5 \rightarrow 4NO_2 + O_2$ Rate of disappearance of N $_2$ O $_5$ : $-\frac{\Delta[N_2O_5]}{\Delta t} = \frac{3 - 2.75}{30} = \frac{0.25}{30}$ mol L $^{-1}$ min $^{-1}$ . Rate of formation of NO $_2$ = $\frac{4}{2} \times$ rate of disappearance of N $_2$ O $_5$ : $= 2 \times \frac{0.25}{30} = \frac{0.5}{30} = \frac{1}{60} \approx 0.01667$ mol L $^{-1}$ min $^{-1}$ . $= 16.67 \times 10^{-3} \approx 17 \times 10^{-3}$ mol L $^{-1}$ min $^{-1}$ . Therefore, $x = \boxed{17}$ .

Q57JEE Main 2024NAT4MCoordination Compounds

Number of complexes which show optical isomerism among the following is _________. $cis-[Cr(ox)_2Cl_2]^{3-}$ , $[Co(en)_3]^{3+}$ , $cis-[Pt(en)_2Cl_2]^{2+}$ , $cis-[Co(en)_2Cl_2]^+$ , $trans-[Pt(en)_2Cl_2]^{2+}$ , $trans-[Cr(ox)_2Cl_2]^{3-}$

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📖 Explanation

We need to determine how many of the given complexes exhibit optical isomerism . Key Concept: Optical isomers (enantiomers) are non-superimposable mirror images. A complex shows optical isomerism if it lacks an improper axis of symmetry (practically, if it has no plane of symmetry and no centre of symmetry). Generally, cis isomers of octahedral complexes with bidentate ligands are optically active, while trans isomers possess a plane of symmetry and are optically inactive. Complex 1: $cis\text{-}[Cr(ox)_2Cl_2]^{3-}$ This is an octahedral complex with two bidentate oxalate (ox) ligands and two monodentate Cl ligands \in the cis arrangement. The cis configuration lacks a plane of symmetry because the two Cl atoms are on the same side, and the bidentate oxalate ligands create an asymmetric arrangement. Its mirror image is non-superimposable. Result: Optically active. Complex 2: $[Co(en)_3]^{3+}$ This complex has three bidentate ethylenediamine (en) ligands arranged around Co $^{3+}$ \in an octahedral geometry. With three identical bidentate ligands, the complex adopts a propeller-like arrangement. The mirror image ( $\Delta$ and $\Lambda$ forms) are non-superimposable. This is a classic example of an optically active complex. Result: Optically active. Complex 3: $cis\text{-}[Pt(en)_2Cl_2]^{2+}$ This octahedral complex has two bidentate en ligands and two Cl ligands \in cis positions. Similar to Complex 1, the cis arrangement with bidentate ligands creates an asymmetric structure with no plane of symmetry. The mirror image is non-superimposable. Result: Optically active. Complex 4: $cis\text{-}[Co(en)_2Cl_2]^+$ This octahedral complex has two bidentate en ligands and two Cl ligands \in cis positions. The cis arrangement produces a structure without a plane of symmetry, and its mirror image is non-superimposable. This is another well-known example of an optically active coordination compound. Result: Optically active. Complex 5: $trans\text{-}[Pt(en)_2Cl_2]^{2+}$ In the trans arrangement, the two Cl ligands are on opposite sides (180 degrees apart). This trans configuration possesses a plane of symmetry that passes through the metal center and the two Cl atoms. The mirror image is superimposable on the original structure. Result: NOT optically active. Complex 6: $trans\text{-}[Cr(ox)_2Cl_2]^{3-}$ In the trans configuration, the two Cl ligands are positioned opposite to each other. This arrangement introduces a plane of symmetry \in the complex, making the mirror image superimposable on the original. Result: NOT optically active. Counting: Complexes 1, 2, 3, and 4 are optically active. Complexes 5 and 6 are not. The total number of complexes showing optical isomerism is $\boxed{4}$ .

Q58JEE Main 2024NAT4MHaloalkanes and Haloarenes

2-chlorobutane + $Cl_2 \rightarrow C_4H_8Cl_2$ (isomers) Total number of optically active isomers shown by $C_4H_8Cl_2$ , obtained in the above reaction is ________.

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📖 Explanation

2-chlorobutane: CH $_3$ CHClCH $_2$ CH $_3$ . Monochlorination gives C $_4$ H $_8$ Cl $_2$ isomers. Possible positions for the second Cl: 1. C1: CH $_2$ ClCHClCH $_2$ CH $_3$ - C2 is chiral, optically active (2 enantiomers) 2. C2: CH $_3$ CCl $_2$ CH $_2$ CH $_3$ - no chiral center, not optically active 3. C3: CH $_3$ CHClCHClCH $_3$ - two chiral centers, can have (R,R), (S,S), and meso. The (R,R) and (S,S) are optically active = 2 enantiomers. Also (R,S) which is meso. 4. C4: CH $_3$ CHClCH $_2$ CH $_2$ Cl - C2 is chiral, 2 enantiomers Total optically active isomers: 2 + 0 + 2 + 2 = 6. Therefore, the answer is $\boxed{6}$ .

Q59JEE Main 2024NAT4MSalt Analysis

Number of metal ions characterized by flame test among the following is _________. $Sr^{2+}$ , $Ba^{2+}$ , $Ca^{2+}$ , $Cu^{2+}$ , $Zn^{2+}$ , $Co^{2+}$ , $Fe^{2+}$

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📖 Explanation

Metal ions characterized by flame test: $Sr^{2+}$ : crimson red - YES $Ba^{2+}$ : apple green - YES $Ca^{2+}$ : brick red - YES $Cu^{2+}$ : blue-green - YES $Zn^{2+}$ : no characteristic flame color - NO $Co^{2+}$ : no characteristic flame color - NO $Fe^{2+}$ : no characteristic flame color - NO Total = 4. Therefore, the answer is $\boxed{4}$ .

Q60JEE Main 2024NAT4MBiomolecules

The total number of correct statements, regarding the nucleic acids is _________. A. RNA is regarded as the reserve of genetic information. B. DNA molecule self-duplicates during cell division. C. DNA synthesizes proteins in the cell. D. The message for the synthesis of particular proteins is present in DNA. E. Identical DNA strands are transferred to daughter cells.

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📖 Explanation

Checking statements about nucleic acids: A. RNA is regarded as the reserve of genetic information - FALSE (DNA is the genetic reserve). B. DNA molecule self-duplicates during cell division - TRUE. C. DNA synthesizes proteins in the cell - FALSE (RNA/ribosomes synthesize proteins). D. The message for synthesis of particular proteins is present in DNA - TRUE. E. Identical DNA strands are transferred to daughter cells - TRUE. Correct statements: B, D, E = 3. Therefore, the answer is $\boxed{3}$ .

Mathematics30 questions

Q61JEE Main 2024MCQ4MComplex Numbers

If $z$ is a complex number, then the number of common roots of the equation $z^{1985} + z^{100} + 1 = 0$ and $z^3 + 2z^2 + 2z + 1 = 0$ , is equal to:

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📖 Explanation

We need common roots of $z^{1985} + z^{100} + 1 = 0$ and $z^3 + 2z^2 + 2z + 1 = 0$ . Factor $z^3 + 2z^2 + 2z + 1 = (z+1)(z^{2+}z+1) = 0$ . Roots: $z = -1$ and $z = \omega, \omega^2$ (cube roots of unity, where $\omega = e^{2\pi i/3}$ ). Check $z = -1$ : $(-1)^{1985} + (-1)^{100} + 1 = -1 + 1 + 1 = 1 \neq 0$ . Not a root. Check $z = \omega$ (where $\omega^3 = 1$ ): $1985 = 3(661) + 2$ , so $\omega^{1985} = \omega^2$ . $100 = 3(33)+1$ , so $\omega^{100} = \omega$ . $\omega^2 + \omega + 1 = 0$ ✓. So $\omega$ is a common root. Similarly $\omega^2$ : $(\omega^2)^{1985} = \omega^{3970} = \omega^{3(1323)+1} = \omega$ . $(\omega^2)^{100} = \omega^{200} = \omega^{3(66)+2} = \omega^2$ . $\omega + \omega^2 + 1 = 0$ ✓. So $\omega^2$ is also a common root. Number of common roots = 2. The answer is Option (2): $\boxed{2}$ .

Q62JEE Main 2024MCQ4MSequences and Series

Let $a$ and $b$ be two distinct positive real numbers. Let 11th term of a GP, whose first term is $a$ and third term is $b$ , is equal to $p^{th}$ term of another GP, whose first term is $a$ and fifth term is $b$ . Then $p$ is equal to

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📖 Explanation

GP1: first term $a$ , third term $b$ . Common ratio $r_1 = \sqrt{b/a}$ . 11th term = $a \cdot r_1^{10} = a(b/a)^5 = b^5/a^4$ . GP2: first term $a$ , fifth term $b$ . Common ratio $r_2 = (b/a)^{1/4}$ . $p$ th term = $a \cdot r_2^{p-1} = a(b/a)^{(p-1)/4}$ . Setting equal: $b^5/a^4 = a(b/a)^{(p-1)/4}$ . $b^5/a^4 = a \cdot b^{(p-1)/4}/a^{(p-1)/4} = b^{(p-1)/4} \cdot a^{1-(p-1)/4}$ Comparing powers of $b$ : $5 = \frac{p-1}{4} \Rightarrow p - 1 = 20 \Rightarrow p = 21$ . Verify powers of $a$ : $-4 = 1 - \frac{p-1}{4} = 1 - 5 = -4$ ✓. The answer is Option (3): $\boxed{21}$ .

Q63JEE Main 2024MCQ4MBinomial Theorem

Suppose $28 - p$ , $p$ , $70 - \alpha$ , $\alpha$ are the coefficients of four consecutive terms \in the expansion of $(1 + x)^n$ . Then the value of $2\alpha - 3p$ equals

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📖 Explanation

Four consecutive coefficients in $(1+x)^n$ : $\binom{n}{r-1} = 28-p$ , $\binom{n}{r} = p$ , $\binom{n}{r+1} = 70-\alpha$ , $\binom{n}{r+2} = \alpha$ . Using the property that consecutive binomial coefficients satisfy: $\binom{n}{r-1} + \binom{n}{r} = \binom{n+1}{r}$ . Sum of 1st and 2nd: $(28-p) + p = 28 = \binom{n+1}{r}$ . Sum of 3rd and 4th: $(70-\alpha) + \alpha = 70 = \binom{n+1}{r+2}$ . Sum of 2nd and 3rd: $p + (70-\alpha) = \binom{n+1}{r+1}$ . Now $\binom{n+1}{r} = 28$ , $\binom{n+1}{r+2} = 70$ . Using $\frac{\binom{n+1}{r+2}}{\binom{n+1}{r}} = \frac{(n+1-r)(n-r)}{(r+1)(r+2)} = \frac{70}{28} = \frac{5}{2}$ . So $2(n+1-r)(n-r) = 5(r+1)(r+2)$ . Try $n = 7, r = 1$ : $2(7)(6) = 84$ , $5(2)(3) = 30$ . No. Try $n = 7, r = 2$ : $2(6)(5) = 60$ , $5(3)(4) = 60$ . Yes! Check: $\binom{8}{2} = 28$ ✓, $\binom{8}{4} = 70$ ✓. $\binom{8}{3} = 56 = p + (70-\alpha)$ , so $p + 70 - \alpha = 56 \Rightarrow \alpha - p = 14$ . Now $\binom{7}{1} = 7 = 28 - p \Rightarrow p = 21$ . $\binom{7}{2} = 21 = p$ ✓. $\binom{7}{3} = 35 = 70 - \alpha \Rightarrow \alpha = 35$ . $\binom{7}{4} = 35 = \alpha$ ✓. $2\alpha - 3p = 70 - 63 = 7$ . The answer is Option (1): $\boxed{7}$ .

Q64JEE Main 2024MCQ4MTrigonometric Ratios and Identities

For $\alpha, \beta \in \left(0, \frac{\pi}{2}\right)$ , let $3\sin(\alpha + \beta) = 2\sin(\alpha - \beta)$ and a real number $k$ be such that $\tan\alpha = k\tan\beta$ . Then the value of $k$ is equal to

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📖 Explanation

Given: $3\sin(\alpha+\beta) = 2\sin(\alpha-\beta)$ . Expanding: $3(\sin\alpha\cos\beta + \cos\alpha\sin\beta) = 2(\sin\alpha\cos\beta - \cos\alpha\sin\beta)$ . $3\sin\alpha\cos\beta + 3\cos\alpha\sin\beta = 2\sin\alpha\cos\beta - 2\cos\alpha\sin\beta$ $\sin\alpha\cos\beta = -5\cos\alpha\sin\beta$ $\frac{\sin\alpha}{\cos\alpha} = \frac{-5\sin\beta}{\cos\beta}$ $\tan\alpha = -5\tan\beta$ So $k = -5$ . The answer is Option (1): $\boxed{-5}$ .

Q65JEE Main 2024MCQ4MStraight Lines and Pair of Straight Lines

If $x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0$ is the locus of a point, which moves such that it is always equidistant from the lines $x + 2y + 7 = 0$ and $2x - y + 8 = 0$ , then the value of $g + c + h - f$ equals

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📖 Explanation

The distance of a point $P(x,y)$ from a straight line $ax+by+c=0$ is $\dfrac{\left|ax+by+c\right|}{\sqrt{a^{2}+b^{2}}}\, .$ For a point that is equidistant from the two given lines $L_1:\;x+2y+7=0$ and $L_2:\;2x-y+8=0$ we must have $\dfrac{\left|x+2y+7\right|}{\sqrt{1^{2}+2^{2}}}= \dfrac{\left|2x-y+8\right|}{\sqrt{2^{2}+(-1)^{2}}}\, .$ Since $1^{2}+2^{2}=5$ and $2^{2}+(-1)^{2}=5$ , the denominators are equal, so squaring both sides eliminates the absolute value and the square roots: $(x+2y+7)^2 = (2x - y + 8)^2\, .$ Expanding the squares: Left side: $(x+2y+7)^2 = x^{2}+4xy+4y^{2}+14x+28y+49.$ Right side: $(2x-y+8)^2 = 4x^{2}-4xy+y^{2}+32x-16y+64.$ Equating and bringing every term to the left: $x^{2}+4xy+4y^{2}+14x+28y+49 -\bigl(4x^{2}-4xy+y^{2}+32x-16y+64\bigr)=0.$ Simplifying term by term: $-3x^{2}+8xy+3y^{2}-18x+44y-15=0.$ Multiplying by $-1$ makes the coefficient of $x^{2}$ positive: $3x^{2}-8xy-3y^{2}+18x-44y+15=0.$ Dividing by $3$ (so that the coefficient of $x^{2}$ becomes $1$ ): $x^{2}-\dfrac{8}{3}xy-y^{2}+6x-\dfrac{44}{3}y+5=0.$ Now compare with the general form $x^{2}-y^{2}+2hxy+2gx+2fy+c=0.$ Thus we read off: $2h=-\dfrac{8}{3}\;\Longrightarrow\;h=-\dfrac{4}{3},$ $2g=6\;\Longrightarrow\;g=3,$ $2f=-\dfrac{44}{3}\;\Longrightarrow\;f=-\dfrac{22}{3},$ $c=5.$ We need $g+c+h-f$ : $g+c+h-f =3+5+\Bigl(-\dfrac{4}{3}\Bigr)-\Bigl(-\dfrac{22}{3}\Bigr) =8+\dfrac{-4+22}{3} =8+\dfrac{18}{3} =8+6 =14.$ Therefore, $g+c+h-f=14,$ which is Option A.

Q66JEE Main 2024MCQ4MEllipse

Let $A(\alpha, 0)$ and $B(0, \beta)$ be the points on the line $5x + 7y = 50$ . Let the point $P$ divide the line segment $AB$ internally \in the ratio $7:3$ . Let $3x - 25 = 0$ be a directrix of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the corresponding focus be $S$ . If from $S$ , the perpendicular on the $x$ -axis passes through $P$ , then the length of the latus rectum of $E$ is equal to

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📖 Explanation

Line $5x + 7y = 50$ : $A(\alpha,0) \Rightarrow 5\alpha = 50 \Rightarrow \alpha = 10$ . $B(0,\beta) \Rightarrow 7\beta = 50 \Rightarrow \beta = 50/7$ . P divides AB \in ratio 7:3 internally: $P = \left(\frac{7(0)+3(10)}{10}, \frac{7(50/7)+3(0)}{10}\right) = (3, 5)$ . Directrix: $3x - 25 = 0 \Rightarrow x = 25/3$ . For ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ : directrix $x = a/e$ . So $a/e = 25/3$ . Focus $S = (ae, 0)$ . Since the perpendicular from S on x-axis passes through P, the x-coordinate of S = x-coordinate of P = 3. So $ae = 3$ . From $a/e = 25/3$ and $ae = 3$ : $a^2 = (a/e)(ae) = (25/3)(3) = 25$ , so $a = 5$ . $e = ae/a = 3/5$ . $b^2 = a^2(1-e^2) = 25(1-9/25) = 16$ . Length of latus rectum = $\frac{2b^2}{a} = \frac{32}{5}$ . The answer is Option (4): $\boxed{\frac{32}{5}}$ .

Q67JEE Main 2024MCQ4MHyperbola

Let $P$ be a point on the hyperbola $H: \frac{x^2}{9} - \frac{y^2}{4} = 1$ , \in the first quadrant such that the area of triangle formed by $P$ and the two foci of $H$ is $2\sqrt{13}$ . Then, the square of the distance of $P$ from the origin is

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📖 Explanation

Hyperbola: $\frac{x^2}{9} - \frac{y^2}{4} = 1$ . So $a = 3, b = 2, c = \sqrt{13}$ . Foci: $(\pm\sqrt{13}, 0)$ . Let $P = (x_0, y_0)$ \in first quadrant on hyperbola. Area of triangle with foci $F_1(-\sqrt{13},0)$ and $F_2(\sqrt{13},0)$ : Area = $\frac{1}{2} \times 2\sqrt{13} \times y_0 = \sqrt{13} \cdot y_0 = 2\sqrt{13}$ . So $y_0 = 2$ . From hyperbola: $\frac{x_0^2}{9} - \frac{4}{4} = 1 \Rightarrow \frac{x_0^2}{9} = 2 \Rightarrow x_0^2 = 18$ . Distance from origin squared: $x_0^2 + y_0^2 = 18 + 4 = 22$ . The answer is Option (3): $\boxed{22}$ .

Q68JEE Main 2024MCQ4MMatrices and Determinants

Let $R = \begin{pmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{pmatrix}$ be a non-zero $3 \times 3$ matrix, where $x\sin\theta = y\sin\left(\theta + \frac{2\pi}{3}\right) = z\sin\left(\theta + \frac{4\pi}{3}\right) \neq 0$ , $\theta \in (0, 2\pi)$ . For a square matrix $M$ , let Trace( $M$ ) denote the \sum of all the diagonal entries of $M$ . Then, among the statements: (I) Trace( $R$ ) = 0 (II) If Trace(adj(adj( $R$ ))) = 0, then $R$ has exactly one non-zero entry.

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📖 Explanation

Given $x\sin\theta = y\sin(\theta + 2\pi/3) = z\sin(\theta + 4\pi/3) \neq 0$ . Let this common value be $k$ . Then $x = k/\sin\theta$ , $y = k/\sin(\theta+2\pi/3)$ , $z = k/\sin(\theta+4\pi/3)$ . Trace(R) = $x + y + z = k\left[\frac{1}{\sin\theta} + \frac{1}{\sin(\theta+2\pi/3)} + \frac{1}{\sin(\theta+4\pi/3)}\right]$ . We know that $\frac{1}{\sin\theta} + \frac{1}{\sin(\theta+2\pi/3)} + \frac{1}{\sin(\theta+4\pi/3)}$ is generally NOT zero. For example, at $\theta = \pi/2$ : $1 + 1/\sin(7\pi/6) + 1/\sin(11\pi/6) = 1 + (-2) + (-2) = -3 \neq 0$ . So statement (I) is NOT always true. For statement (II): adj(adj(R)) for a diagonal matrix has entries related to the (n-1)th powers of cofactors. For 3×3: adj(adj(R)) = det(R) · R (when R is invertible). Trace(adj(adj(R))) = det(R) · Trace(R). If Trace(adj(adj(R))) = 0, then either det(R) = 0 or Trace(R) = 0. This doesn't necessarily mean R has exactly one non-zero entry. So (II) is also not necessarily true. The answer is Option (3): Neither (I) nor (II) is true.

Q69JEE Main 2024MCQ4MMatrices and Determinants

Consider the system of linear equations $x + y + z = 5$ , $x + 2y + \lambda^2 z = 9$ and $x + 3y + \lambda z = \mu$ , where $\lambda, \mu \in R$ . Then, which of the following statement is NOT correct?

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📖 Explanation

System: $x + y + z = 5$ , $x + 2y + \lambda^2 z = 9$ , $x + 3y + \lambda z = \mu$ . Determinant: $D = \begin{vmatrix} 1&1&1\\1&2&\lambda^2\\1&3&\lambda\end{vmatrix} = 1(2\lambda-3\lambda^2) - 1(\lambda-\lambda^2) + 1(3-2) = 2\lambda-3\lambda^{2-}\lambda+\lambda^{2+}1 = -2\lambda^{2+}\lambda+1 = -(2\lambda^{2-}\lambda-1) = -(2\lambda+1)(\lambda-1)$ $D = 0$ when $\lambda = 1$ or $\lambda = -1/2$ . Option (3): "System has unique solution if $\lambda \neq 1$ and $\mu \neq 13$ ." But $D = 0$ also when $\lambda = -1/2$ , not just $\lambda = 1$ . So for unique solution we need $\lambda \neq 1$ AND $\lambda \neq -1/2$ . The statement ignores $\lambda = -1/2$ case. This statement is NOT correct - if $\lambda = -1/2$ and $\mu \neq 13$ , $D = 0$ so no unique solution. The answer is Option (3): The statement that is NOT correct.

Q70JEE Main 2024MCQ4MLogarithms

If the domain of the function $f(x) = \log_e\frac{2x+3}{4x^{2+}x-3} + \cos^{-1}\frac{2x-1}{x+2}$ is $(\alpha, \beta]$ , then the value of $5\beta - 4\alpha$ is equal to

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📖 Explanation

Domain of $f(x) = \log_e\frac{2x+3}{4x^{2+}x-3} + \cos^{-1}\frac{2x-1}{x+2}$ . For log: $\frac{2x+3}{4x^{2+}x-3} > 0$ . Factor denominator: $4x^{2+}x-3 = (4x-3)(x+1)$ . Numerator: $2x+3$ . $\frac{2x+3}{(4x-3)(x+1)} > 0$ . Critical points: $x = -3/2, -1, 3/4$ . Sign analysis: positive on $(-3/2, -1) \cup (3/4, \infty )$ . For $\cos^{-1}$ : $-1 \leq \frac{2x-1}{x+2} \leq 1$ , $x \neq -2$ . $\frac{2x-1}{x+2} \leq 1 \Rightarrow \frac{2x-1-x-2}{x+2} \leq 0 \Rightarrow \frac{x-3}{x+2} \leq 0 \Rightarrow x \in (-2, 3]$ . $\frac{2x-1}{x+2} \geq -1 \Rightarrow \frac{2x-1+x+2}{x+2} \geq 0 \Rightarrow \frac{3x+1}{x+2} \geq 0 \Rightarrow x \in (-\infty , -2) \cup \$[-1/3, \infty )$ . Combined for $\cos^{-1}$ : $(-2,3]\$ \cap [(-\infty ,-2) \cup [-1/3,\infty )] = [-1/3, 3]$ . Intersection with log domain: $[-1/3, 3] \cap [(-3/2,-1) \cup (3/4,\infty )] = (3/4, 3]$ . So domain = $(3/4, 3]$ , $\alpha = 3/4, \beta = 3$ . $5\beta - 4\alpha = 15 - 3 = 12$ . The answer is Option (2): $\boxed{12}$ .

Q71JEE Main 2024MCQ4MDefinite Integrals

Let $f: R \rightarrow R$ be a function defined $f(x) = \frac{x}{(1+x^4)^{1/4}}$ and $g(x) = f(f(f(f(x))))$ then $18\int_0^{\sqrt{2\sqrt{5}}} x^2 g(x) \, dx$

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📖 Explanation

We start with the function $f:\mathbb{R}\rightarrow\mathbb{R}$ defined by $f(x)=\dfrac{x}{(1+x^4)^{1/4}}\,.$ Step 1 : An equation relating $f(x)$ and $x$ . Let $y=f(x)\,.$ Then $y=\dfrac{x}{(1+x^4)^{1/4}} \;\;\Longrightarrow\;\; y^4=\dfrac{x^4}{1+x^4}\;.$ $-(1)$ Re-arranging $-(1)$ to express $x^4$ \in terms of $y^4$ : $y^4(1+x^4)=x^4 \;\;\Longrightarrow\;\; y^4=x^4(1-y^4) \;\;\Longrightarrow\;\; x^4=\dfrac{y^4}{1-y^4}\;.$ $-(2)$ Equation $-(2)$ shows that applying $f$ replaces the quantity $t=x^4$ by the new value $\dfrac{t}{1+t}\,.$ Step 2 : Iterating the transformation. Define $T(t)=\dfrac{t}{1+t}\;.$ If $t_0=x^4$ , then after one application of $f$ we have $t_1=T(t_0)$ , after two applications $t_2=T^2(t_0)$ , and so on. Compute the first four iterates: $T(t)=\dfrac{t}{1+t}\,,$ $T^2(t)=T\!\bigl(T(t)\bigr)=\dfrac{t}{1+2t}\,,$ $T^3(t)=T\!\bigl(T^2(t)\bigr)=\dfrac{t}{1+3t}\,,$ $T^4(t)=T\!\bigl(T^3(t)\bigr)=\dfrac{t}{1+4t}\,.$ Step 3 : Simplifying $g(x)=f^{(4)}(x).$ Setting $t_0=x^4$ , after four applications we get $g(x)^4 = T^4(x^4)=\dfrac{x^4}{1+4x^4}\;.$ Hence $g(x)=\dfrac{x}{(1+4x^4)^{1/4}}\,.$ $-(3)$ Step 4 : Setting up the required integral. The integrand becomes $x^2\,g(x)=x^2\cdot\dfrac{x}{(1+4x^4)^{1/4}}=\dfrac{x^3}{(1+4x^4)^{1/4}}\;.$ Therefore $I=\displaystyle\int_{0}^{\sqrt{2\sqrt{5}}}\dfrac{x^3}{(1+4x^4)^{1/4}}\,dx\;.$ $-(4)$ Step 5 : Substitution to evaluate $I$ . Put $u=1+4x^4\;,$ so that $\dfrac{du}{dx}=16x^3\;\Longrightarrow\;x^3\,dx=\dfrac{du}{16}\,.$ When $x=0$ , $u=1$ ; when $x=\sqrt{2\sqrt{5}}$ , $x^4=(\sqrt{2\sqrt{5}})^4=(2\sqrt{5})^2=4\cdot5=20 \;\Longrightarrow\; u=1+4\cdot20=81\;.$ Substituting \in $-(4)$ : $I=\dfrac{1}{16}\displaystyle\int_{1}^{81}u^{-1/4}\,du\;.$ Use the power-rule integral $\int u^{n}\,du=\dfrac{u^{n+1}}{n+1}+C\;.$ With $n=-\dfrac14$ we get $\int u^{-1/4}\,du=\dfrac{u^{3/4}}{3/4}=\dfrac{4}{3}u^{3/4}\;.$ Hence $I=\dfrac{1}{16}\cdot\dfrac{4}{3}\Bigl\$[u^{3/4}\Bigr]_{1}^{81} =\dfrac{1}{12}\Bigl(81^{3/4}-1^{3/4}\Bigr)\;.$ Because $81=3^4,\;81^{1/4}=3,\;81^{3/4}=3^3=27$ , we have $I=\dfrac{1}{12}(27-1)=\dfrac{26}{12}=\dfrac{13}{6}\;.$ Step 6 : Final multiplication. The question asks for $18I$ : $18I=18\cdot\dfrac{13}{6}=3\cdot13=39\;.$ Answer : 39 (Option D)

Q72JEE Main 2024MCQ4MDefinite Integrals

Let $a$ and $b$ be real constants such that the function $f$ defined by $

f(x) = \begin{cases} x^2 + 3x + a, & x \leq 1 \\ bx + 2, & x > 1 \end{cases}

$ be differentiable on $R$ . Then, the value of $\int_{-2}^{2} f(x) \, dx$ equals

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📖 Explanation

For differentiability at $x = 1$ : continuity and equal derivatives. Continuity: $1 + 3 + a = b + 2 \Rightarrow a + 4 = b + 2 \Rightarrow b = a + 2$ ... (1) Derivatives: $f'(x) = 2x + 3$ for $x \leq 1$ , $f'(x) = b$ for $x > 1$ . At $x = 1$ : $2 + 3 = b \Rightarrow b = 5$ . From (1): $a = 3$ . $

f(x) = \begin{cases} x^2 + 3x + 3, & x \leq 1 \\ 5x + 2, & x > 1 \end{cases}

$ $\int_{-2}^{2} f(x)dx = \int_{-2}^{1}(x^{2+}3x+3)dx + \int_{1}^{2}(5x+2)dx$ $= \left[\frac{x^3}{3}+\frac{3x^2}{2}+3x\right]_{-2}^{1} + \left[\frac{5x^2}{2}+2x\right]_{1}^{2}$ At $x=1$ : $\frac{1}{3}+\frac{3}{2}+3 = \frac{2+9+18}{6} = \frac{29}{6}$ At $x=-2$ : $-\frac{8}{3}+6-6 = -\frac{8}{3}$ First integral: $\frac{29}{6}+\frac{8}{3} = \frac{29+16}{6} = \frac{45}{6}$ At $x=2$ : $10+4 = 14$ . At $x=1$ : $\frac{5}{2}+2 = \frac{9}{2}$ . Second integral: $14 - \frac{9}{2} = \frac{19}{2}$ Total: $\frac{45}{6} + \frac{19}{2} = \frac{45}{6} + \frac{57}{6} = \frac{102}{6} = 17$ . The answer is Option (4): $\boxed{17}$ .

Q73JEE Main 2024MCQ4MDifferential Equations

Let $f: R - \{0\} \rightarrow R$ be a function satisfying $f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$ for all $x, y$ , $f(y) \neq 0$ . If $f'(1) = 2024$ , then

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📖 Explanation

We are given a function $f: \mathbb{R} \setminus \{0\} \rightarrow \mathbb{R}$ satisfying $f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$ for all $x, y$ with $f(y) \neq 0$ , and $f'(1) = 2024$ . We need to find the relationship between $f$ and $f'$ . Determine the form of $f(x)$ . Setting $x = y$ \in the functional equation: $f\left(\frac{x}{x}\right) = \frac{f(x)}{f(x)} \implies f(1) = 1$ Setting $y = x$ and $x = 1$ : $f\left(\frac{1}{y}\right) = \frac{f(1)}{f(y)} = \frac{1}{f(y)}$ The functional equation $f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$ is equivalent to $f(xy) = f(x) \cdot f(y)$ (a multiplicative function). For differentiable functions satisfying this, the solution is $f(x) = x^n$ for some constant $n$ . Determine the value of $n$ using $f'(1) = 2024$ . If $f(x) = x^n$ , then $f'(x) = nx^{n-1}$ . $f'(1) = n \cdot 1^{n-1} = n = 2024$ Therefore, $f(x) = x^{2024}$ . Find the differential equation satisfied by $f$ . With $f(x) = x^{2024}$ : $f'(x) = 2024x^{2023}$ Multiply both sides by $x$ : $xf'(x) = 2024x^{2024} = 2024 \cdot f(x)$ Therefore: $xf'(x) - 2024f(x) = 0$ Verification: We can verify by substituting: $x \cdot 2024x^{2023} - 2024 \cdot x^{2024} = 2024x^{2024} - 2024x^{2024} = 0$ . Confirmed. The correct answer is Option (1): $xf'(x) - 2024f(x) = 0$ .

Q74JEE Main 2024MCQ4MApplication of Derivatives

Let $f(x) = x+3^{2}x-2^3$ , $x \in [-4, 4]$ . If $M$ and $m$ are the maximum and minimum values of $f$ , respectively \in $[-4, 4]$ , then the value of $M - m$ is:

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📖 Explanation

The function appears to be $f(x) = (x+3)^2(x-2)^3$ on $[-4, 4]$ . $f'(x) = 2(x+3)(x-2)^3 + 3(x+3)^2(x-2)^2 = (x+3)(x-2)^2[2(x-2)+3(x+3)] = (x+3)(x-2)^2(5x+5) = 5(x+3)(x-2)^2(x+1)$ Critical points: $x = -3, -1, 2$ . $f(-4) = (-1)^2(-6)^3 = -216$ . $f(-3) = 0$ . $f(-1) = (2)^2(-3)^3 = 4(-27) = -108$ . $f(2) = 0$ . $f(4) = (7)^2(2)^3 = 49(8) = 392$ . Maximum $M = 392$ at $x = 4$ . Minimum $m = -216$ at $x = -4$ . $M - m = 392 - (-216) = 608$ . The answer is Option (3): $\boxed{608}$ .

Q75JEE Main 2024MCQ4MDifferential Equations

Let $y = f(x)$ be a thrice differentiable function \in $(-5, 5)$ . Let the tangents to the curve $y = f(x)$ at $(1, f(1))$ and $(3, f(3))$ make angles $\frac{\pi}{6}$ and $\frac{\pi}{4}$ , respectively with positive x-axis. If $27\int_1^3 \{f'(t)\}^2 + 1\} f''(t) \, dt = \alpha + \beta\sqrt{3}$ where $\alpha, \beta$ are integers, then the value of $\alpha + \beta$ equals

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📖 Explanation

$f'(1) = \tan(\pi/6) = 1/\sqrt{3}$ , $f'(3) = \tan(\pi/4) = 1$ . The integral: $27\int_1^3 \{(f'(t))^2 + 1\}f''(t)dt$ . Let $u = f'(t)$ , $du = f''(t)dt$ . $= 27\int_{f'(1)}^{f'(3)} (u^{2+}1)du = 27\left[\frac{u^3}{3}+u\right]_{1/\sqrt{3}}^{1}$ At $u = 1$ : $\frac{1}{3} + 1 = \frac{4}{3}$ . At $u = 1/\sqrt{3}$ : $\frac{1}{3 \cdot 3\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{1}{9\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{1+9}{9\sqrt{3}} = \frac{10}{9\sqrt{3}}$ . $= 27\left[\frac{4}{3} - \frac{10}{9\sqrt{3}}\right] = 27 \times \frac{4}{3} - 27 \times \frac{10}{9\sqrt{3}} = 36 - \frac{30}{\sqrt{3}} = 36 - 10\sqrt{3}$ So $\alpha = 36, \beta = -10$ . $\alpha + \beta = 26$ . The answer is Option (2): $\boxed{26}$ .

Q76JEE Main 2024MCQ4MIndefinite Integrals

Let $f: R \rightarrow R$ be defined $f(x) = ae^{2x} + be^x + cx$ . If $f(0) = -1$ , $f'(\log_e 2) = 21$ and $\int_0^{\log 4}(f(x) - cx) \, dx = \frac{39}{2}$ , then the value of $|a + b + c|$ equals:

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📖 Explanation

$f(x) = ae^{2x} + be^x + cx$ . $f(0) = a + b = -1$ ... (1) $f'(x) = 2ae^{2x} + be^x + c$ . $f'(\ln 2) = 2a(4) + b(2) + c = 8a + 2b + c = 21$ ... (2) $\int_0^{\ln 4}(f(x)-cx)dx = \int_0^{\ln 4}(ae^{2x}+be^x)dx = \left[\frac{a}{2}e^{2x}+be^x\right]_0^{\ln 4}$ $= \frac{a}{2}(16) + b(4) - \frac{a}{2} - b = \frac{15a}{2} + 3b = \frac{39}{2}$ ... (3) From (3): $15a + 6b = 39 \Rightarrow 5a + 2b = 13$ ... (3') From (1): $b = -1 - a$ . Substituting \in (3'): $5a + 2(-1-a) = 13 \Rightarrow 3a = 15 \Rightarrow a = 5$ . $b = -6$ . From (2): $40 - 12 + c = 21 \Rightarrow c = -7$ . $|a + b + c| = |5 - 6 - 7| = |-8| = 8$ . The answer is Option (4): $\boxed{8}$ .

Q77JEE Main 2024MCQ4MVector Algebra

Let $\vec{a} = \hat{i} + \alpha\hat{j} + \beta\hat{k}$ , $\alpha, \beta \in R$ . Let a vector $\vec{b}$ be such that the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$ and $|\vec{b}|^2 = 6$ . If $\vec{a} \cdot \vec{b} = 3\sqrt{2}$ , then the value of $(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2$ is equal to

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📖 Explanation

$|\vec{a}|^2 = 1 + \alpha^2 + \beta^2$ . $|\vec{b}|^2 = 6$ . $\vec{a} \cdot \vec{b} = 3\sqrt{2}$ . Angle $\pi/4$ between $\vec{a}$ and $\vec{b}$ : $\cos(\pi/4) = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{3\sqrt{2}}{|\vec{a}|\sqrt{6}}$ . $\frac{1}{\sqrt{2}} = \frac{3\sqrt{2}}{|\vec{a}|\sqrt{6}} \Rightarrow |\vec{a}|\sqrt{6} = 3\sqrt{2} \cdot \sqrt{2} = 6 \Rightarrow |\vec{a}| = \sqrt{6}$ . So $1 + \alpha^2 + \beta^2 = 6 \Rightarrow \alpha^2 + \beta^2 = 5$ . $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 = 36 - 18 = 18$ . $(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2 = 5 \times 18 = 90$ . The answer is Option (1): $\boxed{90}$ .

Q78JEE Main 2024MCQ4MVector Algebra

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{b}| = 1$ and $|\vec{b} \times \vec{a}| = 2$ . Then $|(\vec{b} \times \vec{a}) - \vec{b}|^2$ is equal to

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📖 Explanation

$|\vec{b}| = 1$ , $|\vec{b} \times \vec{a}| = 2$ . $|(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 - 2(\vec{b} \times \vec{a}) \cdot \vec{b} + |\vec{b}|^2$ Since $(\vec{b} \times \vec{a}) \perp \vec{b}$ : $(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$ . $= 4 + 0 + 1 = 5$ . The answer is Option (2): $\boxed{5}$ .

Q79JEE Main 2024MCQ4MThree Dimensional Geometry

Let $L_1: \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(\hat{i} - \hat{j} + 2\hat{k})$ , $\lambda \in R$ , $L_2: \vec{r} = (\hat{j} - \hat{k}) + \mu(3\hat{i} + \hat{j} + p\hat{k})$ , $\mu \in R$ and $L_3: \vec{r} = \delta(l\hat{i} + m\hat{j} + n\hat{k})$ , $\delta \in R$ be three lines such that $L_1$ is perpendicular to $L_2$ and $L_3$ is perpendicular to both $L_1$ and $L_2$ . Then the point which lies on $L_3$ is

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📖 Explanation

$L_1$ direction: $(1, -1, 2)$ . $L_2$ direction: $(3, 1, p)$ . $L_1 \perp L_2$ : $3 - 1 + 2p = 0 \Rightarrow p = -1$ . $L_2$ direction: $(3, 1, -1)$ . $L_3$ direction $(l, m, n)$ perpendicular to both $L_1$ and $L_2$ : $(l, m, n) = (1, -1, 2) \times (3, 1, -1) = (1-2, 6+1, 1+3) = (-1, 7, 4)$ . $L_3$ : $\vec{r} = \delta(-\hat{i} + 7\hat{j} + 4\hat{k})$ . Points on $L_3$ : $(-\delta, 7\delta, 4\delta)$ . For $\delta = 1$ : $(-1, 7, 4)$ . The answer is Option (1): $\boxed{(-1, 7, 4)}$ .

Q80JEE Main 2024MCQ4MProbability

Bag $A$ contains 3 white, 7 red balls and bag $B$ contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag $A$ , if the ball drawn is white, is:

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📖 Explanation

$P(A) = P(B) = 1/2$ . $P(W|A) = 3/10$ , $P(W|B) = 3/5$ . $P(W) = P(A) \cdot P(W|A) + P(B) \cdot P(W|B) = \frac{1}{2} \cdot \frac{3}{10} + \frac{1}{2} \cdot \frac{3}{5} = \frac{3}{20} + \frac{3}{10} = \frac{3}{20} + \frac{6}{20} = \frac{9}{20}$ By Bayes' theorem: $P(A|W) = \frac{P(A) \cdot P(W|A)}{P(W)} = \frac{3/20}{9/20} = \frac{3}{9} = \frac{1}{3}$ The answer is Option (3): $\boxed{\frac{1}{3}}$ .

Q81JEE Main 2024NAT4MQuadratic Equation and Inequalities

The number of real solutions of the equation $x(x^2 + 3|x| + 5|x-1| + 6|x-2|) = 0$ is ______.

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📖 Explanation

$x(x^2 + 3|x| + 5|x-1| + 6|x-2|) = 0$ . Either $x = 0$ or $x^2 + 3|x| + 5|x-1| + 6|x-2| = 0$ . For the second factor: all terms $x^2, 3|x|, 5|x-1|, 6|x-2|$ are non-negative. The \sum equals zero only if each term is zero simultaneously, which is impossible ( $|x| = 0$ means $x = 0$ , but then $|x-1| = 1 \neq 0$ ). So the only solution is $x = 0$ . Number of real solutions = 1. Therefore, the answer is $\boxed{1}$ .

Q82JEE Main 2024NAT4MPermutations and Combinations

In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections: A, B and C. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section A has 8 questions, section B has 6 questions and section C has 6 questions, then the total number of ways a student can select 15 questions is _________.

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📖 Explanation

Let $a,\,b,\,c$ denote the numbers of questions chosen from Sections A, B and C respectively. Total questions to be attempted: $a+b+c = 15$ . Minimum requirement: $a \ge 4,\; b \ge 4,\; c \ge 4$ . Upper limits (because of the number of questions present): $a \le 8,\; b \le 6,\; c \le 6$ . Introduce new variables to handle the "at least 4" condition: $a = a' + 4,\; b = b' + 4,\; c = c' + 4,\qquad a',b',c' \ge 0$ . Substituting \in $a+b+c=15$ gives $a' + b' + c' = 15 - 12 = 3 \qquad -(1)$ Equation $(1)$ demands non-negative integer solutions of \sum $3$ . These are: \bullet $\,(3,0,0)$ and its permutations \bullet $\,(2,1,0)$ and its permutations \bullet $\,(1,1,1)$ . Now apply the upper bounds. Case 1: $(a',b',c') = (3,0,0)$ Then $(a,b,c)=(7,4,4)$ which satisfies $a\le 8,\;b\le 6,\;c\le 6$ . The other two permutations give $(a,b,c)=(4,7,4)$ or $(4,4,7)$ , both violating $b\le6$ or $c\le6$ . Hence only one admissible triple: $(7,4,4)$ . Case 2: $(a',b',c') = (2,1,0)$ All six permutations obey the limits, giving $(6,5,4),\;(6,4,5),\;(5,6,4),\;(4,6,5),\;(5,4,6),\;(4,5,6).$ Case 3: $(a',b',c') = (1,1,1)$ This single solution yields $(5,5,5)$ , within the limits. Thus the admissible $(a,b,c)$ triples are $(7,4,4),\;(6,5,4),\;(6,4,5),\;(5,6,4),\;(4,6,5),\;(5,4,6),\;(4,5,6),\;(5,5,5).$ For each triple, multiply the individual combinations: Section A: choose $a$ out of $8 \;\;(\binom{8}{a})$ Section B: choose $b$ out of $6 \;\;(\binom{6}{b})$ Section C: choose $c$ out of $6 \;\;(\binom{6}{c})$ . Compute every case:

\begin{aligned} (7,4,4):&\;\binom{8}{7}\binom{6}{4}\binom{6}{4}=8\cdot15\cdot15=1800\$$2pt] (6,5,4):&\;\binom{8}{6}\binom{6}{5}\binom{6}{4}=28\cdot6\cdot15=2520\$$2pt] (6,4,5):&\;28\cdot15\cdot6=2520\$$2pt] (5,6,4):&\;\binom{8}{5}\binom{6}{6}\binom{6}{4}=56\cdot1\cdot15=840\$$2pt] (4,6,5):&\;\binom{8}{4}\cdot1\cdot6=70\cdot6=420\$$2pt] (5,4,6):&\;56\cdot15\cdot1=840\$$2pt] (4,5,6):&\;70\cdot6\cdot1=420\$$2pt] (5,5,5):&\;56\cdot6\cdot6=2016 \end{aligned}

$ Add the eight results: $1800+2520+2520+840+420+840+420+2016 = 11376.$ Hence, a student can select the required 15 questions \in $\mathbf{11376}$ different ways.

Q83JEE Main 2024NAT4MSequences and Series

Let $S_n$ be the \sum to n-terms of an arithmetic progression $3, 7, 11, \ldots$ , if $40 < \frac{6}{n(n+1)}\sum_{k=1}^{n} S_k < 42$ , then $n$ equals ____________.

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📖 Explanation

The given arithmetic progression is $3,\,7,\,11,\ldots$ , so first term $a = 3$ and common difference $d = 4$ . Sum of the first $n$ terms of an A.P. is $S_n = \frac{n}{2}\Bigl[2a + (n-1)d\Bigr]$ $= \frac{n}{2}\Bigl[2\cdot3 + (n-1)\cdot4\Bigr]$ $= \frac{n}{2}\Bigl[6 + 4n - 4\Bigr]$ $= \frac{n}{2}\,(4n + 2)$ $= n(2n + 1)\quad -(1)$ Next, we need the cumulative \sum $\sum_{k=1}^{n} S_k$ . Using $(1)$ , $S_k = k(2k + 1) = 2k^2 + k$ . Hence $\sum_{k=1}^{n} S_k = \sum_{k=1}^{n} (2k^2 + k)$ $= 2\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$ Using the standard formulas $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \qquad \sum_{k=1}^{n} k = \frac{n(n+1)}{2},$ we get $\sum_{k=1}^{n} S_k = 2\cdot\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$ $= \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$ $= n(n+1)\left[\frac{2n+1}{3} + \frac{1}{2}\right]$ Common denominator $6$ gives $\frac{2n+1}{3} + \frac{1}{2} = \frac{4n+2}{6} + \frac{3}{6} = \frac{4n + 5}{6}.$ Therefore $\sum_{k=1}^{n} S_k = \frac{n(n+1)(4n + 5)}{6}\quad -(2)$ The problem states $40 \lt \frac{6}{n(n+1)}\sum_{k=1}^{n} S_k \lt 42.$ Substituting from $(2)$ : $\frac{6}{n(n+1)}\cdot\frac{n(n+1)(4n + 5)}{6} = 4n + 5.$ Hence the inequality becomes $40 \lt 4n + 5 \lt 42.$ Subtract $5$ everywhere: $35 \lt 4n \lt 37.$ The only integer multiple of $4$ between $35$ and $37$ is $36$ , so $4n = 36 \;\Rightarrow\; n = 9.$ Therefore, $n = 9$ .

Q84JEE Main 2024NAT4MSequences and Series

Let $\alpha = \sum_{k=0}^{n} \frac{{}^nC_k^2}{k+1}$ and $\beta = \sum_{k=0}^{n-1} \frac{{}^nC_k \cdot {}^nC_{k+1}}{k+2}$ . If $5\alpha = 6\beta$ , then $n$ equals

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📖 Explanation

We start with the two sums $\alpha = \sum_{k=0}^{n} \frac{\binom{n}{k}^{2}}{k+1},\qquad \beta = \sum_{k=0}^{\,n-1} \frac{\binom{n}{k}\,\binom{n}{k+1}}{k+2}.$ Case 1: Closed form for $\alpha$ First use the relation $\frac{\binom{n}{k}}{k+1}=\frac{1}{\,n+1\,}\binom{\,n+1\,}{\,k+1\,}.$ Hence $\alpha =\frac{1}{n+1}\sum_{k=0}^{n}\binom{n}{k}\binom{\,n+1\,}{\,k+1\,}.$ Notice that $\binom{\,n+1\,}{\,k+1\,}=\binom{\,n+1\,}{\,n-k\,}.$ Putting $j=n-k$ the summation becomes $\alpha=\frac{1}{n+1}\sum_{j=0}^{n}\binom{n}{j}\binom{\,n+1\,}{\,j\,}.$ By Vandermonde's identity $\sum_{j=0}^{n}\binom{n}{j}\binom{\,n+1\,}{\,j\,} =\binom{\,2n+1\,}{\,n\,}.$ Therefore $\boxed{\alpha=\dfrac{1}{\,n+1\,}\binom{\,2n+1\,}{\,n\,}.}$ Case 2: Closed form for $\beta$ Rewrite the general term of $\beta$ with a shifted index $r=k+1$ ( $1\le r\le n$ ): $\beta=\sum_{r=1}^{n}\frac{\binom{n}{r-1}\,\binom{n}{r}}{r+1}.$ Use $(r+1)\binom{n}{r}=n\binom{\,n-1\,}{\,r-1\,}.$ Multiplying numerator and denominator of each term by $r+1$ gives $\beta =\frac{n}{(n+1)(n+2)} \sum_{r=1}^{n}\binom{\,n-1\,}{\,r-1\,}\binom{\,n+2\,}{\,r+1\,}.$ Put $s=r-1$ (so $0\le s\le n-1$ ): $\beta=\frac{n}{(n+1)(n+2)} \sum_{s=0}^{n-1}\binom{\,n-1\,}{\,s\,}\binom{\,n+2\,}{\,s+2\,}.$ Since $\binom{\,n+2\,}{\,s+2\,}=\binom{\,n+2\,}{\,n-s\,},$ the \sum equals the coefficient of $x^{\,n}$ \in $(1+x)^{\,n-1}(1+x)^{\,n+2}=(1+x)^{\,2n+1},$ which by the binomial theorem is $\binom{\,2n+1\,}{\,n\,}.$ Hence $\boxed{\beta=\dfrac{n}{(n+1)(n+2)}\binom{\,2n+1\,}{\,n\,}.}$ Case 3: Using the given relation The problem states $5\alpha = 6\beta.$ Insert the closed forms: $5\left[\frac{1}{n+1}\binom{\,2n+1\,}{\,n\,}\right] =6\left[\frac{n}{(n+1)(n+2)}\binom{\,2n+1\,}{\,n\,}\right].$ Because $\binom{\,2n+1\,}{\,n\,}\neq 0$ , it cancels out, and so does the common factor $n+1$ , giving $5 = 6\,\frac{n}{n+2}.$ Solve for $n$ : $5(n+2)=6n \;\;\Longrightarrow\;\; 5n+10=6n \;\;\Longrightarrow\;\; n=10.$ Hence $n=10$ .

Q85JEE Main 2024NAT4MCircles

Consider two circles $C_1: x^2 + y^2 = 25$ and $C_2: (x-\alpha)^2 + y^2 = 16$ , where $\alpha \in (5, 9)$ . Let the angle between the two radii (one to each circle) drawn from one of the intersection points of $C_1$ and $C_2$ be $\sin^{-1}\frac{\sqrt{63}}{8}$ . If the length of common chord of $C_1$ and $C_2$ is $\beta$ , then the value of $(\alpha\beta)^2$ equals _________.

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📖 Explanation

$C_1: x^{2+}y^2=25$ (center O₁(0,0), r₁=5). $C_2: (x-\alpha)^{2+}y^2=16$ (center O₂(\alpha ,0), r₂=4). The angle between radii at intersection point is $\sin^{-1}\frac{\sqrt{63}}{8}$ . Let the angle be $\theta$ at the intersection point P. Using the law of cosines \in triangle O₁PO₂: $\alpha^2 = 25 + 16 - 2(5)(4)\cos\theta = 41 - 40\cos\theta$ where $\theta$ is the angle at P. $\sin\theta = \frac{\sqrt{63}}{8}$ , $\cos\theta = \pm\frac{1}{8}$ . Since $\alpha \in (5,9)$ , $\alpha^2 \in (25,81)$ . With $\cos\theta = 1/8$ : $\alpha^2 = 41-5 = 36$ , $\alpha = 6$ ✓. Common chord length: The perpendicular from O₁ to the radical axis. Radical axis: subtract circles: $-2\alpha x + \alpha^2 = -9 \Rightarrow x = \frac{\alpha^{2+}9}{2\alpha} = \frac{45}{12} = \frac{15}{4}$ . $\beta = 2\sqrt{r_1^2 - x^2} = 2\sqrt{25 - 225/16} = 2\sqrt{175/16} = 2 \cdot \frac{5\sqrt{7}}{4} = \frac{5\sqrt{7}}{2}$ . $(\alpha\beta)^2 = (6 \cdot \frac{5\sqrt{7}}{2})^2 = (15\sqrt{7})^2 = 225 \times 7 = 1575$ . Therefore, the answer is $\boxed{1575}$ .

Q86JEE Main 2024NAT4MStatistics

If the variance $\sigma^2$ of the data is $k$ , then the value of $k$ is ______ (where $.$ denotes the greatest integer function)

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📖 Explanation

To find the variance $\sigma^2 = k$ , we first calculate the mean of the data, $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ . The total frequency is $\sum f_i = 3+2+3+2+6+3+3 = 22$ . The sum of the products of the data points and their frequencies is $\sum f_i x_i = (0)(3) + (1)(2) + (5)(3) + (6)(2) + (10)(6) + (12)(3) + (17)(3) = 0+2+15+12+60+36+51 = 176$ . Thus, the mean is $\bar{x} = \frac{176}{22} = 8$ .

Next, we calculate the variance using the formula $\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \bar{x}^2$ .
The sum $\sum f_i x_i^2 = 3(0^2) + 2(1^2) + 3(5^2) + 2(6^2) + 6(10^2) + 3(12^2) + 3(17^2) = 0+2+75+72+600+432+867 = 2048$ .
The variance is $k = \sigma^2 = \frac{2048}{22} - (8)^2 = \frac{1024}{11} - 64$ .
This simplifies to $k = \frac{1024 - 704}{11} = \frac{320}{11} \approx 29.09$ .
The question asks for the greatest integer value of $k$ , so we find $[k] = [29.09] = 29$ .

Q87JEE Main 2024NAT4MSets and Relations

The number of symmetric relations defined on the set $\{1, 2, 3, 4\}$ which are not reflexive is _______.

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📖 Explanation

We need to find the number of symmetric relations on the set $\{1, 2, 3, 4\}$ that are NOT reflexive. A relation $R$ on a set $A$ is symmetric if whenever $(a, b)\in R$ then $(b, a)\in R$ . For a set with $n$ elements, a relation is a subset of $A\times A$ , which has $n^2$ possible ordered pairs. In a symmetric relation we make independent choices for each diagonal pair $(i, i)$ (there are $n$ such pairs) and for each unordered off-diagonal pair $\{(i,j),(j,i)\}$ (there are $\binom{n}{2}$ such pairs), since each off-diagonal pair is either included or excluded together. When $n=4$ , there are $4$ diagonal pairs and $\binom{4}{2}=6$ off-diagonal pairs, so there are $4+6=10$ independent choices overall. Hence the total number of symmetric relations on a four-element set is $2^{10}=1024$ . A relation is reflexive if $(i,i)\in R$ for all $i\in A$ . To count symmetric relations that are also reflexive, we must include all four diagonal pairs, leaving only the six off-diagonal pairs as free choices. Thus there are $2^6=64$ reflexive symmetric relations. Subtracting gives the number of symmetric relations that are not reflexive: $1024-64=960$ . The answer is $\boxed{960}$ .

Q88JEE Main 2024NAT4MArea Under The Curves

The area of the region enclosed by the parabola $(y-2)^2 = x - 1$ , the line $x - 2y + 4 = 0$ and the positive coordinate axes is __________.

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📖 Explanation

The given curves are the parabola $(y-2)^2 = x - 1$ and the line $x - 2y + 4 = 0$ . Rewriting them \in $x = \dots$ form: Parabola : $x = (y-2)^2 + 1 = y^2 - 4y + 5$ $-(1)$ Line : $x = 2y - 4$ $-(2)$ To find their intersection, equate $(1)$ and $(2)$ : $(y-2)^2 + 1 = 2y - 4$ $y^2 - 4y + 4 + 1 = 2y - 4$ $y^2 - 6y + 9 = 0 \;\;\Rightarrow\;\; (y-3)^2 = 0$ Hence $y = 3$ and, from $(2)$ , $x = 2(3) - 4 = 2$ . The curves meet at $P(2,\,3)$ . Intercepts with the positive axes: \bullet Line with $y$ -axis: put $x = 0$ \in $(2)$ \to $-2y + 4 = 0 \Rightarrow y = 2$ \Rightarrow point $A(0,\,2)$ . \bullet Parabola with $x$ -axis: put $y = 0$ \in $(1)$ \to $x = (-2)^2 + 1 = 5$ \Rightarrow point $B(5,\,0)$ . Tracing the boundary \in the first quadrant gives the closed path $O(0,0) \rightarrow A(0,2) \rightarrow P(2,3) \rightarrow B(5,0) \rightarrow O(0,0)$ . A horizontal slice at height $y$ therefore starts at the left boundary and ends at the right boundary as follows: Case 1: For $0 \le y \le 2$ , the left boundary is the $y$ -axis $x = 0$ and the right boundary is the parabola $(1)$ . Area $A_1 = \displaystyle\int_{0}^{2} \bigl(y^2 - 4y + 5\bigr)\,dy$ Integrating term by term, $\int y^2\,dy = \frac{y^3}{3},\;\; \int (-4y)\,dy = -2y^2,\;\; \int 5\,dy = 5y$ Evaluated from $0$ to $2$ : $A_1 = \left[\frac{y^3}{3} - 2y^2 + 5y\right]_{0}^{2} = \left(\frac{8}{3} - 8 + 10\right) - 0 = \frac{14}{3}$ Case 2: For $2 \le y \le 3$ , the left boundary is the line $(2)$ and the right boundary is the parabola $(1)$ . Area $A_2 = \displaystyle\int_{2}^{3} \Bigl\$[(y^2 - 4y + 5) - (2y - 4)\Bigr]\$\,dy$ Simplify the integrand: $(y^2 - 4y + 5) - (2y - 4) = y^2 - 6y + 9 = (y-3)^2$ Integrate: $A_2 = \int_{2}^{3} (y^2 - 6y + 9)\,dy = \left[\frac{y^3}{3} - 3y^2 + 9y\right]_{2}^{3}$ At $y = 3$ : $\frac{27}{3} - 27 + 27 = 9$ At $y = 2$ : $\frac{8}{3} - 12 + 18 = \frac{26}{3}$ Hence $A_2 = 9 - \frac{26}{3} = \frac{1}{3}$ Total required area: $A = A_1 + A_2 = \frac{14}{3} + \frac{1}{3} = \frac{15}{3} = 5$ Area of the enclosed region = 5 square units.

Q89JEE Main 2024NAT4MArea Under The Curves

Let $Y = Y(X)$ be a curve lying \in the first quadrant such that the area enclosed by the line $Y - y = Y'(x)(X - x)$ and the co-ordinate axes, where $(x, y)$ is any point on the curve, is always $\frac{-y^2}{2Y'(x)} + 1$ , $Y'(x) \neq 0$ . If $Y(1) = 1$ , then $12Y(2)$ equals ________.

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📖 Explanation

The tangent line at $(x,y)$ is $Y - y = Y'(x)(X - x)$ . This meets the axes at: X-axis ( $Y=0$ ): $X = x - y/Y'(x)$ . Y-axis ( $X=0$ ): $Y = y - xY'(x)$ . Area of triangle with these intercepts and origin = $\frac{1}{2}|x - y/Y'||y - xY'|$ . Given this equals $-\frac{y^2}{2Y'} + 1$ . Since the curve is \in the first quadrant with the tangent having appropriate orientation: $\frac{1}{2}\left(x - \frac{y}{Y'}\right)(y - xY') = -\frac{y^2}{2Y'} + 1$ $\frac{1}{2}\left(xy - x^2Y' - \frac{y^2}{Y'} + xy\right) = -\frac{y^2}{2Y'} + 1$ $xy - \frac{x^2Y'}{2} - \frac{y^2}{2Y'} = -\frac{y^2}{2Y'} + 1$ $xy - \frac{x^2Y'}{2} = 1$ $Y' = \frac{2(xy-1)}{x^2} = \frac{2y}{x} - \frac{2}{x^2}$ This is linear: $Y' - \frac{2}{x}Y = -\frac{2}{x^2}$ . IF = $e^{-2\ln x} = x^{-2}$ . $\frac{d}{dx}(Yx^{-2}) = -\frac{2}{x^4}$ . $Yx^{-2} = \frac{2}{3x^3} + C$ . $Y = \frac{2}{3x} + Cx^2$ . $Y(1) = 1$ : $1 = 2/3 + C \Rightarrow C = 1/3$ . $Y = \frac{2}{3x} + \frac{x^2}{3}$ . $Y(2) = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}$ . $12Y(2) = 12 \times 5/3 = 20$ . Therefore, the answer is $\boxed{20}$ .

Q90JEE Main 2024NAT4MThree Dimensional Geometry

Let a line passing through the point $(-1, 2, 3)$ intersect the lines $L_1: \frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{-2}$ at $M(\alpha, \beta, \gamma)$ and $L_2: \frac{x+2}{-3} = \frac{y-2}{-2} = \frac{z-1}{4}$ at $N(a, b, c)$ . Then the value of $\frac{(\alpha + \beta + \gamma)^2}{(a + b + c)^2}$ equals ________________.

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📖 Explanation

Line through $(-1,2,3)$ intersects $L_1: \frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}$ at $M$ and $L_2: \frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}$ at $N$ . $M = (3s+1, 2s+2, -2s-1)$ on $L_1$ . $N = (-3t-2, -2t+2, 4t+1)$ on $L_2$ . Direction from $(-1,2,3)$ to $M$ : $(3s+2, 2s, -2s-4)$ . Direction from $(-1,2,3)$ to $N$ : $(-3t-1, -2t, 4t-2)$ . These must be proportional: $\frac{3s+2}{-3t-1} = \frac{2s}{-2t} = \frac{-2s-4}{4t-2}$ . From $\frac{2s}{-2t} = \frac{s}{-t}$ : Let ratio = $k = s/(-t)$ , i.e., $s = -kt$ . From $\frac{3s+2}{-3t-1} = \frac{s}{-t}$ : $-t(3s+2) = s(-3t-1) \Rightarrow -3st-2t = -3st-s \Rightarrow -2t = -s \Rightarrow s = 2t$ . From $\frac{-2s-4}{4t-2} = \frac{s}{-t}$ : $-t(-2s-4) = s(4t-2) \Rightarrow 2st+4t = 4st-2s \Rightarrow 4t+2s = 2st$ . With $s = 2t$ : $4t + 4t = 4t^2 \Rightarrow 8t = 4t^2 \Rightarrow t = 2$ , $s = 4$ . $M = (13, 10, -9)$ : $\alpha+\beta+\gamma = 14$ . $N = (-8, -2, 9)$ : $a+b+c = -1$ . $\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2} = \frac{196}{1} = 196$ . Therefore, the answer is $\boxed{196}$ .

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