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Physics30 questions

Q1JEE Main 2021MCQ4MOscillations

When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is

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📖 Explanation

Since, the particle is executing SHM. Therefore, displacement equation of wave will be $y \;\; =A \sin \omega t$ $\Rightarrow \;\; y / A \;\; =\sin \omega t$ and wave velocity equation will be $\;\;\;\;\; v_{y}=\;\frac{d y}{d t}=A \omega \cos \omega t$ $\Rightarrow \;\; v_{y} / A \omega=\cos \omega t$ $\;\text{ Now, }\; \;\; \sin ^{2} \omega t+\cos ^{2} \omega t=1$ $\therefore \;\; (y / A)^{2}+(v_{y} / A \omega)^{2}=1$ This equation is similar to the equation of ellipse.

Q2JEE Main 2021MCQ4MOscillations

Two electrons each are fixed at a distance $2 d$ . A third charge proton placed at the mid-point is displaced slightly by a distance $x(x < < d)$ perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency? ( $m=$ mass of charged particle)

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📖 Explanation

The arrangement of charges is shown belowAs we know that, Coulomb's force between two charges. i.e. $q_{1}$ and $q_{2}$ , $F=\;\frac{1}{4 \pi \epsilon_{0}} \;\frac{q_{1} q_{2}}{r^{2}}=\;\frac{1}{4 \pi \epsilon_{0}} \;\frac{q_{1} q_{2}}{(d^{2}+x^{2})}$ . . . (i) Here, $q_{1}=q_{2}=q$ Force in ${SHM}, F=m \omega^{2} x$ ...(ii) Since, in order to have ${SHM}+q$ should move downwards and force responsible for this will be only $F^{'}=F \sin \theta+F \sin \theta=2 F \sin \theta$ . . . (iii) Using Eqs. (ii) and (iii), we get $2 F \sin \theta=m \omega^{2} x$ $\Rightarrow \;\; \;\frac{2}{4 \pi \epsilon_{0}} \;\frac{q^{2}}{(d^{2}+x^{2})} \sin \theta=m \omega^{2} x$ $\Rightarrow \;\; \;\frac{2}{4 \pi \epsilon_{0}} \;\frac{q^{2}}{(d^{2}+x^{2})} \cdot \;\frac{x}{(d^{2}+x^{2})^{1 / 2}}=m \omega^{2} x$ $\Rightarrow \;\; \omega=(\;\frac{1}{2 \pi \epsilon_{0}} \;\frac{q^{2}}{(d^{2}+x^{2})^{3 / 2} m})^{1 / 2}$ As, $x < < d$ $\therefore \;\; \omega=(\;\frac{1}{2 \pi \epsilon_{0}} \;\frac{q^{2}}{m d^{3}})^{1 / 2}$

Q3JEE Main 2021MCQ4MThermodynamics

On the basis of kinetic theory of gases, the gas exerts pressure because its molecules

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Q4JEE Main 2021MCQ4MMagnetism and Matter

A soft ferromagnetic material is placed in an external magnetic fiel

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📖 Explanation

The magnetic susceptibility $(χ)$ of ferromagnetic material is in the range of $10^{3}$ to $10^{4}$ , so it gets easily magnetised and demagnetised in the presence of external field region. Therefore, when a soft ferromagnetic material is placed in an external magnetic field region, the magnetic domains may increase or decrease in size and change its orientation.

Q5JEE Main 2021MCQ4MSemiconductor Electronics

The logic circuit shown above is equivalent to

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📖 Explanation

The logic circuit is given as By using De-morgan's theorem, $C=\overline{A+\overline{B}}=\overline{A} \cdot \overline{\overline{B}}=\overline{A} \cdot B$ This relation can be shown by the circuit drawn below

Q6JEE Main 2021MCQ4MUnits and Measurements

The period of oscillation of a simple pendulum is $T=2 \pi \sqrt{\;\frac{L}{g}} .$ Measured value of $L$ is $1.0 {m}$ from metre scale having a minimum division of $1 {mm}$ and time of one complete oscillation is $1.95 {s}$ measured from stopwatch of $0.01 {s}$ resolution. The percentage error in the determination of $g$ will be

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📖 Explanation

Given, $T=2 \pi \sqrt{\;\frac{L}{g}}$ $. . .$ (i) where, time period, $T=1.95 {s}$ Length of string, $I=1 {m}$ Acceleration due to gravity $=g$ Error in time period, $\Delta T=0.01 {s}=10^{-2} {s}$ Error in length, $\Delta L=1 {mm}=1 \times 10^{-3} {m}$ Squaring Eq. (i) on both sides, we get $T^{2} \;\; =4 \pi^{2} \;\frac{L}{g}$ $g \;\; =4 \pi^{2} \;\frac{L}{T^{2}}$ $\Rightarrow \;\; \;\frac{\Delta g}{g} \;\; =\;\frac{\Delta L}{L}+\;\frac{2 \Delta T}{T}=\;\frac{10^{-3}}{1}+\;\frac{2 \times 10^{-2}}{1.95}$ $\;\; =10^{-3}+1.025 \times 10^{-2}$ $\;\; =10^{-3}+10.25 \times 10^{-3}$ $\;\; =11.25 \times 10^{-3}$ $\because \;\; \Delta g / g \times 100 \;\; =11.25 \times 10^{-3} \times 10^{2}$ $\;\; =1.125 \%=1.13 \%$

Q7JEE Main 2021MCQ4MSemiconductor Electronics

Given below are two statements: Statement I $p-n$ junction diodes can be used to function as transistor, simply by connecting two diodes, back to back, which acts as the base terminal. Statement II In the study of transistor, the amplification factor $\beta$ indicates ratio of the collector current to the base current. In the light of the above statements, choose the correct answer from the options given below.

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📖 Explanation

Transistor is a device used as switch or amplifier. It is made by sandwiching three semiconductors, i.e. $p-n-p$ and $n-p-n$ . Hence, Statement I is false, because we cannot make transistor from diode. As we know that, amplification factor $(\beta)$ is the ratio of collector current to base current. $\therefore$ $\beta=\;\frac{I_{C}}{I_{B}}$ Hence, Statement II is true. Therefore, option (a) is correct.

Q8JEE Main 2021MCQ4MOscillations

In the given figure, a body of mass $M$ is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant $k$ , then the frequency of oscillation of given body is

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📖 Explanation

Let $T$ be the time period of oscillation, then $T \;\; =2 \pi \sqrt{\;{M}/{k_{ {eq}}}}\;$ $\therefore \;\; T=2 \pi \sqrt{\;\frac{M}{2 k}} \;[ \because k_{ {eq}}=k+k]$ and frequency $(f)=\;\frac{1}{T}=\;\frac{1}{2 \pi} \sqrt{\;\frac{2 k}{M}}$

Q9JEE Main 2021MCQ4MCurrent Electricity

Figure shows a circuit that contains four identical resistors with resistance $R=2.0 Ω$ , two identical inductors with inductance $L=2.0 {mH}$ and an ideal battery with electromotive force $E=9 {V}$ . The current $i$ just after the switch $S$ is closed will be

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📖 Explanation

Given, resistance, $R=2 Ω$ , Inductance, $L=2 {mH}$ , emf, $E=9 {V}$ and $i$ be the current. $\because$ At $t=0$ when switch is closed, inductors behave as open circuit. $\therefore$ Effective circuit will be By using Ohm's law, $V=i R_{\;\text{eq }\;}$ $\Rightarrow$ $i=V / R_{\;\text{eq }\;}$ where, $R_{\;\text{eq }\;}$ is equivalent resistance of series resistors, i.e., $R_{ {eq}} \;\; =R+R=2 R=2 \times 2=4 Ω$ $i \;\; =\;\frac{9}{4}=2.25 {A}$

Q10JEE Main 2021MCQ4MDual Nature of Matter and Radiation

The de-Broglie wavelength of a proton and $\alpha$ -particle are equal. The ratio of their velocities is

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📖 Explanation

Let $\lambda_{p}, \lambda_{\alpha}, m_{p}, m_{\alpha}, v_{p}, v_{\alpha '} p_{p}$ and $p_{\alpha}$ be the wavelength, mass, velocity and momentum of proton and $\alpha$ -particle, respectively. Given, $\lambda_{p}=\lambda_{\alpha}$ As we know that, $\lambda \;\; =h / p\;$ $\;\;\frac{h}{p_{p}} \;\; =\;\frac{h}{p_{\alpha}}$ $\; \therefore \;p_{p} \;\; =p_{\alpha}$ $\Rightarrow \;m_{p} v_{p} \;\; =m_{\alpha} v_{\alpha}$ $\Rightarrow \;m_{p} v_{p} \;\; =4 m_{p} v_{\alpha}$ $\Rightarrow \;\;\frac{v_{p}}{v_{\alpha}} \;\; =\;\frac{4}{1} \;\text{ or }\; 4: 1$

Q11JEE Main 2021MCQ4MThermodynamics

If one mole of an ideal gas at $(p_{1}, V_{1})$ is allowed to expand reversibly and isothermally ( $A$ to $B$ ), its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value $(B \to C)$ . Then, it is restored to its initial state by a reversible adiabatic compression ( $C$ to $A$ ). The net work done by the gas is equal to

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📖 Explanation

Let $p_{i}, p_{f}, V_{i}$ and $V_{f}$ be the initial and final pressure and volume. Given, $A B$ is isothermal $(\Delta T=0)$ , $B C$ is isochoric $(\Delta V=0)$ and $C A$ is adiabatic $(\Delta Q=0)$ Since, isothermal work $(W_{A B})=p_{1} V_{1} \ln \;\frac{V_{f}}{V_{i}}$ where, $V_{i}$ and $V_{f}$ are volume at $A$ and $B$ , respectively. $\therefore \;\; W_{A B}=p_{1} V_{1} \ln \;\frac{2 V_{1}}{V_{1}}=p_{1} V_{1} \ln 2$ Since, at constant volume, work done is zero. $\therefore \;\; W_{B C}=0$ Since, $W_{C A}$ is an adiabatic work done, i.e. $W_{C A} \;\; =\;\frac{1}{1-\gamma}(p_{f} V_{f}-p_{i} V_{i})$ $\Rightarrow \;\; W_{C A} \;\; =\;\frac{1}{1-\gamma}(p_{1} V_{1}-\;\frac{p_{1}}{4} \times 2 V_{1})$ $\;\; =\;\frac{1}{1-\gamma}(p_{1} V_{1}-p_{1} V_{1} / 2)=\;\frac{1}{1-\gamma} \;\frac{p_{1} V_{1}}{2}$ $\therefore$ Net work done, $W_{\;\text{net }\;}=W_{A B}+W_{B C}+W_{C A}$ $\;\; =p_{1} V_{1} \ln 2+0+\;\frac{1}{1-\gamma} \;\frac{p_{1} V_{1}}{2}$ $\;\; =p_{1} V_{1}[\ln 2+1 / 2(1-\gamma)]$ From ideal gas law, $p V=n R T$ $\therefore W_{\;\text{net }\;}=R T[\ln 2-1 / 2(\gamma-1)]$ $( \because n=1)$

Q12JEE Main 2021MCQ4MDual Nature of Matter and Radiation

An X-ray tube is operated at $1.24$ million volt. The shortest wavelength of the produced photon will be

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📖 Explanation

Given, $V=1.24$ million volt $=1.24 \times 10^{6}$ volt Since, energy $(E)= {eV}$ where, ${e}$ is the charge of electron $=1.6 \times 10^{-19} {C}$ $\therefore \;\; E=1.6 \times 10^{-19} \times 1.24 \times 10^{6}$ ... (i) As we know that, Energy of photon, $E=\;\frac{h c}{\lambda}$ ...(ii) Here, Planck's constant, $h=6.67 \times 10^{-34} {J}- {s}$ , ${C}=$ speed of light in free space, ${C}=3 \times 10^{8} {ms}^{-1}$ Equating Eqs. (i) and (ii), we get $1.6 \times 10^{-19} \times 1.24 \times 10^{6} \;\; =\;\frac{6.67 \times 10^{-34} \times 3 \times 10^{8}}{\lambda}$ $\Rightarrow \;\; \lambda \;\; =\;\frac{20.01 \times 10^{-13}}{1.6 \times 1.24}=10.09 \times 10^{-13}$ $\;\; =1.009 \times 10^{-12} ∼eq 10^{-3} \times 10^{-9}$ $\;\; =10^{-3} {nm}$

Q13JEE Main 2021MCQ4MWaves

Which of the following equations represents a travelling wave?

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📖 Explanation

The equation of a travelling wave in standard form, $y=A \sin (\omega t \pm k x)$ Only option (a), i.e. $y=A \sin (15 x-2 t)$ satisfies this equation.

Q14JEE Main 2021MCQ4MAtoms and Nuclei

According to Bohr atom model, in which of the following transitions will the frequency be maximum ?

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📖 Explanation

Let $n_{f}, n_{i}$ be the final and initial orbit. As we know that, $\;\frac{1}{\lambda}=109 \times 10^{7}[\;\frac{1}{n_{f}^{2}}-\;\frac{1}{n_{i}^{2}}]$ Now, checking for each option, we get (a) $\;\frac{1}{\lambda} \propto \$ [;\frac{1}{3^{2}}-;\frac{1}{4^{2}}]$=[;\frac{1}{9}-;\frac{1}{16}]=0.05. . . $(i) (b)$ ;\frac{1}{\lambda} \propto [;\frac{1}{1}-;\frac{1}{4}]=0.75 $... (ii) (c)$ ;\frac{1}{\lambda} \propto [;\frac{1}{16}-;\frac{1}{25}]=0.0225 $... (iii) (d)$ ;\frac{1}{\lambda} \propto [;\frac{1}{4}-;\frac{1}{9}]=0.14 $... (iv) The option (b) has highest value. Since, frequency,$ f=;\frac{C}{\lambda} \Rightarrow f \propto ;\frac{1}{\lambda}\therefore $Frequency will be maximum for transition$ n=2 $to$ n=1$.

Q15JEE Main 2021MCQ4MWave Optics

If the source of light used in a Young's double slit experiment is changed from red to violet, then

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📖 Explanation

According to Young's double slit experiment, The distance of $n$ th bright fringe from the centre, $y_{n}=\;\frac{n \lambda D}{d}$ Since, $\;\lambda_{\;\text{violet }\;}< \lambda_{\;\text{red }\;}$ $\therefore$ $\;y_{\;\text{violet }\;}< y_{\;\text{red }\;}$ $\therefore$ Consecutive fringe lines will come closer.

Q16JEE Main 2021MCQ4MCentre of Mass and Collision

A circular hole of radius $(\;\frac{a}{2})$ is cut out of a circular disc of radius $a$ as shown in figure The centroid of the remaining circular portion with respect to point $O$ will be

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📖 Explanation

Given, radius of hole, $r= {a} / 2$ and radius of disc, $R= {a}$ Let $x_{C M}$ be the centre of mass of system, $m_{1}, x_{1}$ be the mass and centre of mass of disc $m_{2}, x_{2}$ be the mass and centre of mass of circular hole. $\therefore \;m_{2}=\;\frac{m_{1}}{\pi R^{2}} \pi r^{2}$ $\Rightarrow \;m_{2}=\;\frac{m_{1} r^{2}}{R^{2}}=\;\frac{m_{1}(a / 2)^{2}}{a^{2}}=\;\frac{m_{1}}{4}$ and $x_{ {CM}}=\;\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}$ $\therefore \;\; {x}_{ {CM}}=\;\frac{m_{1} {a}-(m_{1} / 4)(3 {a} / 2)}{m_{1}-(m_{1} / 4)}$ $\Rightarrow \;\; {x}_{ {CM}}=\;\frac{m_{1} {a}(1-3 / 8)}{3 m_{1} / 4} \Rightarrow {x}_{ {CM}}=\;\frac{5 {a}}{6}$

Q17JEE Main 2021MCQ4MSemiconductor Electronics

Zener breakdown occurs in a $p-n$ junction having $p$ and $n$ both

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📖 Explanation

As we know that, Zener breakdown takes place, when we supply reverse bias voltage to Zener diode. Due to heavily dopping, the electrons in the valence band of $p$ -type region can jump easily to the conduction band of $n$ -type region, hence due to high electric field, zener breakdown occur. Thus, there is very high sudden increase in Zener current $(I_{Z})$ that is caused by reverse breakdown voltage $(V_{Z})$ . Hence, Zener breakdown is easily observed in Zener diode which is heavily doped and having narrow depletion region.

Q18JEE Main 2021MCQ4MElectromagnetic Waves

List-I List-Il A. Source of microwave frequency 1. Radioactive decay of nucleus B. Source of infrared frequency 2. Magnetron B. Source gamma rays 3. Inner shell electrons C. Source of X-rays Vibration of atoms and molecules 5.LASER 6. R-C circuit choose the correct answer from teh options given below

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📖 Explanation

As we know that; A. Source of microwave frequency is magnetron. B. Source of infrared frequency is vibration of atoms and molecules. C. Source of gamma ray is radioactive decay of nucleus. D. Source of X-ray is transition of electron in inner shells. $\therefore$ The correct match is ${A} \to (2), {B} \to (4), {C} \to (1)$ , ${D} \to$ (3).

Q19JEE Main 2021MCQ4MLaws of Motion

A particle is projected with velocity $v_{0}$ along $x$ -axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin, i.e. $m a=-\alpha x^{2}$ . The distance at which the particle stops is

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📖 Explanation

Given, speed of projection $=v_{0}$ Damping force, $F=m a=-\alpha x^{2}$ $\Rightarrow \;\; {a}=-\alpha x^{2} / m$ Also, $\Rightarrow \;\; v d v=a d x=-\;\frac{\alpha}{m} x^{2} d x$ Integrating both sides, we get $int_{v_{0}}^{v} v d v \;\; =int_{0}^{x}-\;\frac{\alpha}{m} x^{2} d x$ $\Rightarrow \;\; (\;\frac{v^{2}}{2})_{v_{0}}^{0} \;\; =-\;\frac{\alpha}{m}(\;\frac{x^{3}}{3})_{0}^{x}$ $\Rightarrow \;\; 0-v_{0}^{2} / 2 \;\; =-\;\frac{\alpha}{m} \;\frac{x^{3}}{3} \Rightarrow x=(\;\frac{3 m}{2} \;\frac{v_{0}^{2}}{\alpha})^{1 / 3}$

Q20JEE Main 2021MCQ4MGravitation

A body weights $49 {N}$ on a spring balance at the North pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator ? (Use, $g=\;\frac{G M}{R^{2}}=9.8 {ms}^{-2}$ and radius of earth, $R=6400 {km}$ )

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📖 Explanation

Given, weight of body at North pole, $w_{p}=m g=49 {N}$ Radius of Earth, $R=6400 {km}$ Let weight of body at equator be $w_{e}$ . At equator, $g_{e}=g-R \omega^{2}$ $\therefore \;\; w_{e}=m g_{e}=m(g-R \omega^{2})$ Since, $w_{p} > w_{e} \Rightarrow w_{e} < 49 {N}$ Hence, above condition is satisfied by only option (b).

Q21JEE Main 2021NAT4MProperties of Matter

A uniform metallic wire is elongated by $0.04 {m}$ when subjected to a linear force $F$ . The elongation, if its length and diameter is doubled and subjected to the same force will be ........... cm

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📖 Explanation

Let initial length and diameter be $l_{1}$ and $d_{1}$ , whereas final length and diameter be $I_{2}$ and $d_{2}$ . Given, $I_{2}=2 I_{1}, d_{2}=2 d_{1}, \Delta I_{1}=0.04 {m}$ By using formula of Young's modulus of elasticity, $Y \;\; =\;\frac{F \cdot l}{A \Delta l}\;$ $\because \;Y_{1} \;\; =Y_{2}$ $\Rightarrow \;\;\frac{F I_{1}}{A_{1} \times \Delta I_{1}}=\;\frac{F I_{2}}{A_{2} \times \Delta I_{2}}$ $\Rightarrow \;\;\frac{F I_{1}}{\pi(d_{1} / 2)^{2} \times 0.04}=\;\frac{F 2 I_{1}}{\pi(2 d_{1} / 2)^{2} \Delta I_{2}}$ $\Rightarrow \;\;\frac{1}{1 / 4 \times 0.04}=\;\frac{2}{\Delta I_{2}}$ $\Rightarrow \;\Delta I_{2}=0.02 {m}=2 {cm}$

Q22JEE Main 2021NAT4MCurrent Electricity

A cylindrical wire of radius $0.5 {mm}$ and conductivity $5 \times 10^{7} {S} / {m}$ is subjected to an electric field of $10 {mV} / {m}$ . The expected value of current in the wire will be $x^{3} \pi {mA}$ . The value of $x$ is .......... .

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📖 Explanation

Given, radius of cylindrical wire, $r=0.5 {mm}=0.5 \times 10^{-3} {m}$ Conductivity, $\sigma=5 \times 10^{7} {S} / {m}$ Electric field, $E=10 {mV} / {m}=10 \times 10^{-3} {V} / {m}$ We know that current density, $\therefore J \;\; =\sigma E$ $\;\; =5 \times 10^{7} \times 10 \times 10^{-3}=5 \times 10^{5} {A} / {m}^{2}$ Also, $J=I / {A} \Rightarrow I=J {A}$ $\Rightarrow$ $I \;\; =5 \times 10^{5} \times \pi \times (0.5 \times 10^{-3})^{2}$ $\;\; =5 \times 10^{5} \times \pi \times 25 \times 10^{-8}=125 \pi \times 10^{-3}$ $\Rightarrow \;\; x^{3} \pi {mA}=125 \pi {mA} \Rightarrow x^{3}=5^{3}$ $\Rightarrow$ $x=5$

Q23JEE Main 2021NAT4MRotational Motion

A uniform thin bar of mass $6 {kg}$ and length $2.4 {m}$ is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is $.......... \times 10^{-1} {kg}- {m}^{2}$ .

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📖 Explanation

Given, mass of uniform bar, $m=6 {kg}$ Length of bar, $I=2.4 {m}$ Side of hexagon, ${a}=\;\frac{2.4}{6}=0.4 {m}$ Mass of each side, $m^{'}=6 / 6=1 {kg}$ and $O P=\sqrt{a^{2}-a^{2} / 4}=a \sqrt{3} / 2$ $=0.2 \sqrt{3} {m}$ Now, by using parallel axes theorem, $I_{O P} \;\; =\;\frac{m^{'} a^{2}}{12}+m^{'}(\;\frac{\sqrt{3} a}{2})^{2}$ $\;\; =\;\frac{a^{2}}{12}+\;\frac{3 a^{2}}{4}=\;\frac{10 a^{2}}{12}=\;\frac{5 a^{2}}{6}$ $\Rightarrow \;\; I_{O P} \;\; =\;\frac{5}{6} \times 0.4 \times 0.4$ and $I_{\;\text{net }\;}$ (net moment of inertia at $O$ ) $=6 \times \;\frac{5}{6} \times 0.4 \times 0.4=8 \times 10^{-1} {kg}- {m}^{2}$

Q24JEE Main 2021NAT4MWork, Power and Energy

Two solids $A$ and $B$ of mass $1 {kg}$ and $2 {kg}$ respectively are moving with equal linear momentum. The ratio of their kinetic energies $(K E)_{A}:(K E)_{B}$ will be $\;\frac{A}{1}$ , so the value of A will be .......... .

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📖 Explanation

Given, $m_{A}=1 {kg}, m_{B}=2 {kg},( {KE})_{A}:( {KE})_{B}=A: 1$ Linear momentum of $A$ and $B$ are equal. $\Rightarrow \;\; p_{A}=p_{B}$ $\because$ Kinetic energy $( {KE})=p^{2} / 2 m$ $\therefore \;\; \;\frac{ {KE}_{A}}{ {KE}_{B}}=\;\frac{m_{B}}{m_{A}}=\;\frac{2}{1}=\;\frac{A}{1} \Rightarrow A=2$

Q25JEE Main 2021NAT4MThermodynamics

The root mean square speed of molecules of a given mass of a gas at $27^{\circ} {C}$ and 1 atmosphere pressure is $200 {ms}^{-1}$ . The root mean square speed of molecules of the gas at $127^{\circ} {C}$ and 2 atmosphere pressure is $\;\frac{x}{\sqrt{3}} {ms}^{-1}$ . The value of $x$ will be

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📖 Explanation

Given, $T_{1}=27^{\circ} {C}=27+273=300 {K}$ , $p_{1}=1 {atm}, v_{1}=200 {ms}^{-1}, T_{2}=127^{\circ} {C}=400 {K}$ , $p_{2}=2 {atm}, v_{2}=$ ? As we know that, Root mean square speed, $v_{ {rms}}=\sqrt{\;\frac{3 R T}{m}}$ $\therefore \;\; \;\frac{v_{1}}{v_{2}}=\sqrt{\;\frac{T_{1}}{T_{2}}}=\sqrt{\;\frac{300}{400}}=\sqrt{\;\frac{3}{4}}$ $\Rightarrow \;\; v_{2}=\sqrt{\;\frac{4}{3}} v_{1}=\;\frac{2}{\sqrt{3}} \times 200=\;\frac{400}{\sqrt{3}} {ms}^{-1}$ $\Rightarrow \;\; \;\frac{x}{\sqrt{3}}=\;\frac{400}{\sqrt{3}} \Rightarrow x=400$

Q26JEE Main 2021NAT4MElectrostatics

A point charge of $+12 \mu {C}$ is at a distance $6 {cm}$ vertically above the centre of a square of side $12 {cm}$ as shown in figure. The magnitude of the electric flux through the square will be .......... $\times 10^{3} {N}- {m}^{2} / {C}$

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📖 Explanation

Given, charge, $q=12 \mu {C}=12 \times 10^{-6} {C}$ Height of charge from surface, $h=6 {cm}=6 \times 10^{-2} {m}$ and side of square, $a=12 {cm}=12 \times 10^{-2} {m}$ From figure, it is clear that the given square is one of the face of a cube of side $12 {cm}$ and $+12 \mu {C}$ charge is placed at its centre. Then, by Gauss's theorem,Flux through any face, $\varphi =\;\frac{q}{6 \epsilon_{0}}$ $\Rightarrow \;\; \varphi \;\; =\;\frac{12 \times 10^{-6}}{6 \times 8.854 \times 10^{-12}}=0.226 \times 10^{6} {N}- {m}^{2} / {C}$ $\;\; =226 \times 10^{3} {N}- {m}^{2} / {C}$

Q27JEE Main 2021NAT4MElectromagnetic Waves

A signal of $0.1 {kW}$ is transmitted in a cable. The attenuation of cable is $-5 {dB}$ per ${km}$ and cable length is $20 {km}$ . The power received at receiver is $10^{-x} {W}$ . The value of $x$ is .......... . [Gain in dB $=10 \log _{10}(\;\frac{P_{0}}{P_{i}})$ ]

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📖 Explanation

Given, power of transmitted signal, $P_{i}=0.1 {kW}=0.1 \times 10^{3} {W}$ $=10^{2} {W}$ Rate of attenuation, $R=-5 {dB} / {km}$ Length of cable, $I=20 {km}$ Power received at receiver, $P_{x}=10^{-x} {W}$ Total loss, $\beta=R \times I=-5 \times 20=-100 {dB}$ $\because \;\;$ Gain $(\beta)=10 \log _{10} \;\frac{P_{0}}{P_{i}}$ $\therefore \;\; \beta=-100=-10 \log _{10} \;\frac{P_{0}}{P_{i}}$ $\Rightarrow \;\; -10=\log _{10} \;\frac{P_{0}}{P_{i}} \Rightarrow 10^{-10}=\;\frac{P_{0}}{P_{i}}$ $\Rightarrow \;\; P_{0}=10^{-10} P_{i}=10^{-10} \times 10^{2}=10^{-8} \Rightarrow P_{0}=10^{-8} {W}$ Hence, $\;\; x=8$

Q28JEE Main 2021NAT4MAlternating Current

A series $L-C-R$ circuit is designed to resonate at an angular frequency $\omega_{0}=10^{5} {rad} / {s}$ . The circuit draws $16 {W}$ power from $120 {V}$ source at resonance. The value of resistance $R$ in the circuit is .......... $Ω$ .

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📖 Explanation

Given, angular frequency at resonance, $\omega_{0}=10^{5} {rads}^{-1}$ Power drawn from circuit, $P=16 {W}$ and supply voltage, $V=120 {V}$ Let resistance of circuit $=R$ . As, $\;\; P=V^{2} / R$ $\Rightarrow \;\; R=V^{2} / P=\;\frac{120 \times 120}{16}$ $=30 \times 30=900 Ω$

Q29JEE Main 2021NAT4MWaves

Two cars are approaching each other at an equal speed of $7.2 {km} / {h}$ . When they see each other, both blow horns having frequency of $676 {Hz}$ . The beat frequency heard by each driver will be ........ ${Hz}$ . [Velocity of sound in air is $340 {m} / {s}$ .]

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📖 Explanation

Given, $v_{A}=v_{B}=7.2 {kmh}^{-1}$ $=\;\frac{72}{10} \times \;\frac{5}{18}=2 {ms}^{-1}$ Frequency of source, $f_{ {s}}=676 {Hz}$ Speed of sound in air, $v=340 {ms}^{-1}$ Let $f_{0}$ be the frequency heard by each driver. By using Doppler effect for $A$ , $(v-v_{A}) f_{s} \;\; =(v+v_{B}) f_{0}$ $\Rightarrow f_{0}=(\;\frac{v+v_{A}}{v-v_{B}}) f_{s} \;\; =(\;\frac{340+2}{340-2}) 676=\;\frac{342}{338} \times 676=684 {Hz}$ Now, beat frequency $=f_{0}-f_{s}=684-676=8 {Hz}$

Q30JEE Main 2021NAT4MElectromagnetic Waves

An electromagnetic wave of frequency $3 {GHz}$ enters a dielectric medium of relative electric permittivity $2.25$ from vacuum. The wavelength of this wave in that medium will be $. . . . . . . . . \times 10^{-2} {cm}$ .

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📖 Explanation

Given, frequency of wave, $f=3 {GHz}=3 \times 10^{9} {Hz}$ Relative permittivity, $\epsilon_{r}=2.25$ Since, $f \;\; =C / \lambda$ $\Rightarrow \;\; \lambda \;\; =\;\frac{C}{f}=\;\frac{3 \times 10^{8}}{3 \times 10^{9}}=0.1 {m}$ $\because \lambda_{m}$ (wavelength of wave in a medium) $=\lambda / \mu$ and as we know that, $\mu =\sqrt{\mu _{r} \epsilon_{r}}$ As, dielectric is non-magnetic, $\mu _{r}=1$ $\Rightarrow$ $\mu =\sqrt{2.25}=1.5$ $\Rightarrow$ $\lambda_{m} \;\; =\;\frac{0.1}{1.5}=\;\frac{1}{15}=0.0667 {m}$ $\;\; =6.67 {cm}=667 \times 10^{-2} {cm}$

Chemistry30 questions

Q31JEE Main 2021MCQ4MAmines

What is the correct sequence of reagents used for converting nitrobenzene into $m$ -dibromobenzene?

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📖 Explanation

In first step, ${NO}_{2}$ group is electron withdrawing group, it decreases the electron density in meta position and bromination occur in meta position. In second step, ${Sn} / {HCl}$ used for reduction to convert $- {NO}_{2}$ to $- {NH}_{2}$ and form aniline. In next step, diazonium salt is formed and at last bromine replace ${N}_{2}^{+} {Cl}^{-}$ group to form 1 , 3-dibromobenzene. Complete reaction is as follows

Q32JEE Main 2021MCQ4MBiomolecules

Most suitable salt which can be used for efficient clotting of blood will be

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📖 Explanation

According to Hardy-Schulze rule, for negatively charged sol, most (+ve) charged ion is needed for efficient coagulation. Blood is a negatively charged sol. Hence ${FeCl}_{3}$ can be used for blood clotting and it from ${Fe}^{3+}$ ion.

Q33JEE Main 2021MCQ4MHaloalkanes and Haloarenes

The correct order of the following compounds showing increasing tendency towards nucleophilic substitution reaction is

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📖 Explanation

Rate of nucleophilic substitution in aromatic halide are as follows Greater the number of nitro $(- {NO}_{2})$ group in chlorobenzene, greater the rate of nucleophilic substitution reaction in aromatic halides. When a nitro group is present at orthoand para position, it can withdraw electrons from the benzene ring. This facilitates, the attack of nucleophiles on haloarenes.

Q34JEE Main 2021MCQ4MStructure of Atom

According to Bohr's atomic theory, I. kinetic energy of electron is $\propto \;\frac{Z^{2}}{n^{2}}$ II. the product of velocity $(v)$ of electron and principal quantum number $(n), ' v n^{'} \propto Z^{2} .$ III. frequency of revolution of electron in an orbit is $\propto \;\frac{Z^{3}}{n^{3}}$ IV. coulombic force of attraction on the electron is $\propto \;\frac{Z^{3}}{n^{4}}$ Choose the most appropriate answer from the options given below.

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📖 Explanation

According to Bohr's theory, I. ${KE} \propto \;\frac{ {Z}^{2}}{n^{2}}$ or $13.6 \propto \;\frac{ {Z}^{2}}{n^{2}} \;\frac{( {eV})}{(\;\text{ atom }\;)}$ ( ∴ Correct) II. Speed of electron $\propto \;\frac{Z}{n}$ (Here, Z = atomic number, $n=$ number of shells) $\therefore \;\; v \times n \propto Z$ ( $\therefore$ Incorrect) III. Frequency of revolution of electron $=\;\frac{v}{2 \pi r}$ Frequency $\propto \;\frac{ {Z}^{2}}{n^{3}} \;\; ( \because v \propto \;\frac{z}{n}, r \propto \;\frac{n^{2}}{z})$ ( ∴ Incorrect) IV. $\;\; F=\;\frac{k q_{1} q_{2}}{r^{2}}=\;\frac{k Z e^{2}}{r^{2}}$ $F=\;{Z}/{(\;\frac{n^{2}}{Z})^{2}}$ $F \propto \;\frac{Z^{3}}{n^{4}}$ ( ∴ Correct) Hence, only I, and IV are correct.

Q35JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Match List-I with List-II. List-I List-II A. $R- {C}^{||}^{\text{O}}- {Cl} \longrightarrow R- {CHO}$ 1. ${Br}_{2} / {NaOH}$ B. $\;\; {R}- {CH}_{2}- {COOH}\longrightarrow R- {C_{|}_{\text{CI}}H}- {COOH}$ 2. ${H}_{2} / {Pd}- {BaSO}_{4}$ C. $R- {C}^{||}^{\text{O}}- {NH}_{2} \longrightarrow R- {NH}_{2}$ 3. ${Zn}( {Hg}) /$ Conc. ${HCl}$ D. ${R}- {C}- {CH}_{3} \longrightarrow {R}- {CH}_{2}- {CH}_{3}$ 4. ${Cl}_{2} /$ Red ${P}, {H}_{2} {O}$ Choose the correct answer from the options given below.

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📖 Explanation

(A) Alkyl chloride reacts with ${H}_{2} / {Pd}- {BaSO}_{4}$ and reduced to alkyl aldehyde. This is known as Rosenmund reduction. (B) Carboxylic acid reacts with ${Cl}_{2} / {P}$ in aqueous medium to form 2-chlorocarboxylic acid. This reaction is known as HVZ reaction. [Hell-Volhard-Zelinsky (HVZ) reaction] (C) ${Alkyl}$ amide reacts with ${Br}_{2}$ in presence of ${NaOH}$ to give alkyl amine. This reaction is known as Hofmann bromide reaction, (D) Ketone reacts with ${Zn}- {Hg}$ in presence of conc. ${HCl}$ to give alkane. This reaction is known as clemmensen reduction. Hence, correct match is (A)-2, (B)-4, (C)-1, (D)-3.

Q36JEE Main 2021MCQ4MCoordination Compounds

The calculated magnetic moments (spin only value) for species $[ {FeCl}_{4}]^{2-},[ {Co}( {C}_{2} {O}_{4})_{3}]^{3-}$ and ${MnO}_{4}^{2-}$ respectively are

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📖 Explanation

(i) $[ {FeCl}_{4}]^{2-} \longrightarrow {Fe}^{2+} \longrightarrow [ {Ar}] 3 d^{6}$ $\therefore {Cl}$ is weak field ligand so does not pairing occurSo, magnetic moment $(\mu )=\sqrt{n(n+2)}$ BM $=\sqrt{4(4+2)}$ BM $\;\; (n=$ Number of total unpaired ${e}^{-}=4)$ $=\sqrt{24} {BM}=4.90 {BM}$ (ii) $[ {Co}( {C}_{2} {O}_{4})_{3}]^{3-} \longrightarrow {Co}^{3+} \longrightarrow [ {Ar}] 3 d^{6}$ ${C}_{2} {O}_{4}$ is strong field ligand so pairing occur.All electrons are paired, $n=0$ hence, $\mu =0$ (iii) ${MnO}_{4}^{2-} \to {Mn}^{+6} \to [ {Ar}] {Bd}^{1}$ $n=1$ $=\sqrt{n(n+2)} {BM}$ $=\sqrt{1(1+2)} {BM}$ $=\sqrt{3} {BM}$ $=1.73 {BM}$

Q37JEE Main 2021MCQ4MSalt Analysis

Match List-I with List-II. List-I(Salt) List-II (Flame colour wavelength) A. ${LiCl}$ 1. $455.5 {nm}$ B. ${NaCl}$ 2. $670.8 {nm}$ C. ${RbCl}$ 3. $780.0 {nm}$ D. ${CsCl}$ 4. $589.2 {nm}$ Choose the correct answer from the options given below.

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📖 Explanation

Alkali metal Colour (Flame) $\lambda( {nm})$ Li Crimson red $670.8$ Na Yellow 589.2 Rb Red, violet 780.0 Cs Blue 455.5Alkali metals have very low value of ionisation energy as compared to other metals. So, alkali metals easily get excited and impart colour to flame. Hence, $R b$ is most excited and having high value of wavelength in all alkali metals.

Q38JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Which one of the following carbonyl compounds cannot be prepared by addition of water on an alkyne in the presence of ${HgSO}_{4}$ and ${H}_{2} {SO}_{4}$ ?

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📖 Explanation

Reaction of ${HgSO}_{4} / {dil} . {H}_{2} {SO}_{4}$ with alkyne result in addition of water as per Markownikoff's rule.

Q39JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

In polymer buna-S: ' $S$ ' stands for

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📖 Explanation

Buna-S is the co-polymer of but 1,3 -diene and styrene as follows.Styrene butadiene rubber is also called buna- ${S}$ , in which ' ${Bu}$ ' stands for butadiene, 'na' stands for sodium and ' ${S}$ ' stands for styrene. Since, butadiene and styrene is one of the constituent monomer of given polymer.

Q40JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Which of the following reagent is suitable for the preparation of the product in the above reaction?

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📖 Explanation

To reduce the carbonyl groups into alkane, wolff-Kishner reduction is used, without affecting the double bond. Wolff-Kishner reagent It utilises hydrazine $( {NH}_{2}- {NH}_{2})$ as the reducing agent in the presence of strong base ${KOH}$ or ${C}_{2} {H}_{5} \overline{ {O}} {Na}^{+}$ in a high boiling protic solvent. The driving force for the reaction is the conversion of hydrazine to nitrogen gas.

Q41JEE Main 2021MCQ4MBiomolecules

Match List-I and List-II. List-I List-II A. Valium 1. Antifertility drug B. Morphine 2. Pernicious anaemia C. Norethindrone 3. Analgesic D. Vitamin ${B}_{12}$ 4. TranquiliserChoose the correct answer from the option given below.

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📖 Explanation

(A) Valium - (4) Tranquilizer A tranquilizer drug became a standard drug for the treatment of anxiety and one of most commonly prescribed drugs of all time. (B) Morphine - (3) Analgesic Morphine is effective for both acute and chronic pain and often usedbefore and after surgery. (C) Norethindrone - (1) Antifertility drug It is a form of progesterone, a female hormone important for regulating ovulation and menstruation. (D) Vitamin ${B}_{12}-$ (2) Pernicious anaemia It is a nutrient that helps to keepour body blood cells and nerve cells healthy and help in making DNA.

Q42JEE Main 2021MCQ4Md and f Block Elements

Match List-I with List-II. List-I (Metal) List-II (Ores) A. Aluminium 1. Siderite B. Iron 2. Calamine C. Copper 3. Kaolinite D. Zinc 4. MalachiteChoose the correct answer from the options given below.

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📖 Explanation

(A) Siderite $\to {FeCO}_{3}$ Siderite is a mineral composed of iron (II) carbonate $( {FeCO}_{3})$ . (B) Calamine $\to {ZnCO}_{3}$ Calamine is used to treat mild itchiness and contain zinc in its formula. (C) Kaolinite $\to {Al}_{2}( {OH})_{4} \cdot {Si}_{2} {O}_{5}$ Kaolinite is a type of clay found in nature. It is a dioctahedral pyrosilicate clay containing ${Si}_{2} {O}_{5}$ . (D) Malachite $\to {Cu}( {OH})_{2} \cdot {CuCO}_{3}$ Malachite is a copper carbonate hydroxide mineral, with the formula ${Cu}_{2} {CO}_{3}( {OH})_{2}$ .

Q43JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

Which one of the following compounds is non-aromatic ?

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📖 Explanation

Compound, which obey Huckel rule, $(4 n+2) \pi$ is called aromatic compound. Compound, which obey $4 n \pi$ rule, is called anti-aromatic compound. Compound which contain one or more $s p^{3}$ carbon in its structure is called non-aromatic compound. The nature of following ions/compounds are as follows :Only (a) is non-aromatic and rest all are aromatic.

Q44JEE Main 2021MCQ4Md and f Block Elements

What is the correct order of the following elements with respect to their density ?

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📖 Explanation

Generally, due to decrease in metallic radius and increase in atomic mass density increase across the period from left to right. Metal Density $( {g} / {cm}^{3})$ Zn 7.13 Cr 7.19 Fe 7.8 Co 8.7 Cu 8.9Correct order is ${Cu} > {Co} > {Fe} > {Cr} > {Zn}$ .

Q45JEE Main 2021MCQ4MSome Basic Concepts of Chemistry

Given below are two statements. Statement I The value of the parameter "Biochemical Oxygen Demand (BOD)" is important for survival of aquatic life. Statement II The optimum value of BOD is $6.5 {ppm}$ . In the light of the above statements, choose the most appropriate answer from the options given below.

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📖 Explanation

Statement I is true but statement II is false. Clean water would have BOD value of less than $5 {ppm}$ whereas highly polluted water could have a BOD value of 17 ppm or more. Hence, the value of parameter ' $B O D$ ' is important for survival of aquatic life but optimum value of $B O D$ is $17 {ppm}$ or more. So, statement II is incorrect.

Q46JEE Main 2021MCQ4Md and f Block Elements

The incorrect statement among the following is

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📖 Explanation

Red colour of ruby is due to presence of ${Cr}^{3+}$ ions in ${Al}_{2} {O}_{3}$ . Chromium is the trace element that causes ruby's red colour, which ranges from an orange red to a publish red. The strength of ruby's red depends on how much chromium is present.

Q47JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

The correct shape and I-I-I bond angles respectively in ${I}_{3}^{-}$ , ion are

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📖 Explanation

Hybridisation of central ${I}$ in ${I}_{3}^{-}$ is $s p^{3} d$ with 3 lone pair and 2 bond pair. Shape Linear Lone pair 3 lone pair Bond angle $180^{\circ}$ (for linear molecule)

Q48JEE Main 2021MCQ4MClassification of Elements

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason ${R}$ . Assertion A Hydrogen is the most abundant element in the universe, but it is not the most abundant gas in the troposphere. Reason R Hydrogen is the lightest element. In the light of the above statements, choose the correct answer from the options given below.

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📖 Explanation

Both $A$ and $R$ are true and $R$ is the correct explanation of $A$ . As we know, hydrogen is most abundant element in the universe but the most abundant gas in the troposphere is nitrogen, because hydrogen is lightest element. Troposphere contains 3 quarters of mass of the entire atmosphere. The air here contain $78 \%$ nitrogen and $21 \%$ oxygen. The last $1 \%$ is made of argon, water vapour and carbon dioxide.

Q49JEE Main 2021MCQ4MAmines

The diazonium salt of which of the following compounds will form a coloured dye on reaction with $\beta$ -naphthol in ${\text{NaOH}}$ ?

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📖 Explanation

Initially aniline reacts with diazotisation reagent to form diazonium salt. Then $\beta$ -naphthol react with salt and orange-red dye is obtained. So, diazonium salt of aniline is used to prepare orange-red dye.

Q50JEE Main 2021MCQ4MSolid State

The correct set from the following in which both pairs are in correct order of melting point is

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📖 Explanation

Correct option is (a) i.e. ${LiF} > {LiCl} ; {MgO} > {NaCl}$ . Melting point is directly proportional to lattice energy. Lattice energy is the energy required to separate a mole of an ionic solid into gaseous ions. It depends upon charge of ions and size of ions. ${M} . {P} \propto {L} . {E}\;\propto \;\frac{\;\text{ Charge }\;}{\;\text{ Size }\;}\;$ ${Li} \to +1\; {Li} \to +1$ ${F}=-1\; {Cl} \to -1$ Both LiF and LiCl having same charge, so melting point will depend on size. Larger the size of anion, lesser the lattice energy and hence, melting point order is ${LiF} > {LiCl}$ . Similarly, ${MgO} {NaCl}$ ${Mg} \to 2+\; {Na} \to 1+$ ${O} \to 2-\; {Cl} \to 1-$ ${MgO}$ having $+2$ charge which is greater than ${NaCl}(+1)$ charge. So, greater the charge on the ions greater will be lattice energy and hence, melting point order is ${MgO} > {NaCl}$ .

Q51JEE Main 2021MCQ4MAmines

The total number of amines among the following which can be synthesised by Gabriel synthesis is

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📖 Explanation

Gabriel phthalimide synthesis is used to prepare $1^{\circ}$ aliphatic or alicyclic amine. Hence, amine which can be synthesised by Gabriel phthalimide synthesis method contains $\alpha$ -carbon. Aniline $( {C}_{6} {H}_{5} {NH}_{2})$ does not contain $\alpha- {C}$ cannot be prepared by Gabriel reaction. Rest amines all contain $\alpha- {C}$ in its respective position, hence they can easily give Gabriel phthalimide reaction. $\therefore$ Three amins out of given four anions can be prepared by Gabriel synthesis.

Q52JEE Main 2021NAT4Mp Block Elements

Among the following allotropic forms of sulphur, the number of allotropic forms, which will show paramagnetism is .......... a. $\alpha$ -sulphur b. $\beta$ -sulphur c. $S_{2}$ -form

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📖 Explanation

Only ${S}_{2}$ -form of sulphur is paramagnetic in nature. Because ${S}_{2}$ is like ${O}_{2}$ i.e. paramagnetic as per molecular orbital theory. It contains unpaired electron. While $\alpha$ -sulphur and $\beta$ -sulphur are diamagnetic as they do not have unpaired electron.

Q53JEE Main 2021NAT4MHydrocarbons

The formula of a gaseous hydrocarbon, which requires 6 times of its own volume of ${O}_{2}$ for complete oxidation and produces 4 times its own volume of ${CO}_{2}$ is ${C}_{x} {H}_{y}$ . The value of $y$ is ........... .

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📖 Explanation

Combustion reaction: ${C}_{x} {H}_{y}(g)+(x+\;\frac{y}{4}) {O}_{2}(g) \longrightarrow x {CO}_{2}(g)+\;\frac{y}{z} {H}_{2} {O}(l)$ Suppose, volume of ${C}_{x} {H}_{y}$ is $V$ and volume of ${O}_{2}$ is 6 times greater than ${C}_{x} {H}_{y}=6 {V}$ then volume of $x{CO}_{2} \Rightarrow V x=4 {V}$ $x=4$ Since, $V_{ {O}_{2}} \;\; =6 \times V_{ {C}_{x} {H}_{y}}$ $V(x+\;\frac{y}{4}) \;\; =6 {V}$ $(x+\;\frac{y}{4}) \;\; =6$ Put value of $x=4$ in Eq. (i) We get, $4+\;\frac{y}{4}=6 \Rightarrow y=8$

Q54JEE Main 2021NAT4MStates of Matter

The volume occupied by $4.75 {g}$ of acetylene gas at $50^{\circ} {C}$ and $740 {mm} {Hg}$ pressure is ...... L (Rounded off to the nearest integer). $[$ Given, $R=0.0826 {L}$ atm ${K}^{-1} {mol}^{-1}]$

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📖 Explanation

Given, mass of ${C}_{2} {H}_{2}( {g})=4.75 {g}$ Molecular weight $=26 {g} / {mol}$ Temperature $=50+273=323 {K}$ Pressure $=740$ torr $/ {mm}$ of ${Hg}$ $\;\text{ Pressure }\; \;\; =\;\frac{740}{760} {atm}$ $R \;\; =0.0821 {L} {atm} {mol}^{-1} {K}^{-1}$ Hence, no. of mole $n=\;\frac{4.75}{26} {mol}$ Formula used, $p V=n R T$ (ideal gas) $\Rightarrow \;\; V \;\; =\;\frac{n R T}{p}=\;\frac{4.75}{26} \times \;\frac{0.0821 \times 323}{(740 / 760)}$ $\;\; =\;\frac{96314.078}{19240}=5.0059 {L}=5 {L}$

Q55JEE Main 2021NAT4MSolutions

${C}_{6} {H}_{6}$ freezes at $5.5^{\circ} {C}$ . The temperature at which a solution $10 {g}$ of ${C}_{4} {H}_{10}$ in $200 {g}$ of ${C}_{6} {H}_{6}$ freeze is ........ ${ }^{\circ} {C}$ . (The molal freezing point depression constant of ${C}_{6} {H}_{6}$ is $5.12^{\circ} {C} / {m}$ .)

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📖 Explanation

Pure solvent : ${C}_{6} {H}_{6}( {l})$ Given, $T_{f}^{\circ} \;\; =5.5^{\circ} {C}$ $K_{f} \;\; =5.12^{\circ} {C} / {m} \Rightarrow m=200 {g}$ $m_{\;\text{solute }\;} \;\; =10 {g}$ Molar mass of solute ${C}_{4} {H}_{10}=12 \times 4+10=58$ Solute $( {C}_{4} {H}_{10})$ is non-dissociative; $\therefore \;i=1$ $\therefore \;\Delta T_{f}=i K_{f} m$ $\Rightarrow \;(T_{f}^{\circ}-T_{f}^{1})=1 \times 5.12 \times \;\frac{(10 / 58)}{(200 / 1000)}$ $5.5-T_{f}^{1}=\;5.12 \times 5 \times 10$ $\;58$ $\;\text{ or }\;\;T_{f}^{1} \approx 1^{\circ} {C}$

Q56JEE Main 2021NAT4MElectrochemistry

The magnitude of the change in oxidising power of the ${MnO}_{4}^{-} / {Mn}^{2+}$ couple is $x \times 10^{-4}$ ${V}$ , if the ${H}^{+}$ concentration is decreased from $1 {M}$ to $10^{-4} {M}$ at $25^{\circ} {C}$ . (Assume concentration of ${MnO}_{4}^{-}$ and ${Mn}^{2+}$ to be same on change in ${H}^{+}$ concentration). The value of $x$ is ......... (Rounded off to the nearest integer). [ Given, $\;\frac{2.303 R T}{F}=0.059]$

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📖 Explanation

Reaction, ${MnO}_{4}^{-}+ {H}^{+}+5 {e}^{-} \longrightarrow {Mn}^{2+}+4 {H}_{2} {O}$ Applying Nernst equation, $E_{c e l l}=E_{c e l l}^{\circ}-\;\frac{0.0591}{n} \log \;\frac{[P]}{[R]}$ or $E_{c e l l}=E_{c e l l}^{\circ}-\;\frac{0.0591}{5} \log \;\frac{[ {Mn}^{2+}]\$ }{[ {MnO}{4}^{-}]$}$[;\frac{1}{ {H}^{+}}]^{8} $(I) Given,$ [ {H}^{+}]=1 {M}E{1}=E^{\circ}-;\frac{0.0591}{5} \log ;\frac{[ {Mn}^{2+}]$}{[ {MnO}{4}^{-}]$} $(II)$ {Now},$[ {H}^{+}]$=10^{-4} {M}E{2}=E^{\circ}-;\frac{0.0591}{5} \log ;\frac{[ {Mn}^{2+}]$}{[ {MnO}{4}^{-}]$} × ;\frac{1}{(10^{-4})^{8}}∴ |E{1}-E_{2}||E_{1}-E_{2}| ;; =;\frac{0.0591}{5} × 32=0.3776 {V}=3776 × 10^{-4}x ;; =3776$

Q57JEE Main 2021NAT4MIonic Equilibrium

The solubility product of ${Pbl}_{2}$ is $8.0 \times 10^{-9}$ . The solubility of lead iodide in $0.1$ molar solution of lead nitrate is $x \times 10^{-6} {mol} / {L}$ . The value of $x$ is ....... (Rounded off to the nearest integer). [Given, : $\sqrt{2}=1.41$ ]

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📖 Explanation

Given, $[K_{ {sp}}]_{ {Pbl}_{2}}=8 \times 10^{-9}$ To calculate solubility of ${PbI}_{2}$ in $0.1 {M}$ solution of ${Pb}( {NO}_{3})_{2}$ . (I) ${ {Pb}( {NO}_{3})_{2} }_{0 . \overline{1} {M}}\longrightarrow {{Pb}^{2+}( {aq})}_{0 . \overline{1} {M}}+{2 {NO}_{3}^{-}( {aq})}_{0 . \overline{2} {M}}$ (II) ${Pbl}_{2}(s) ⇌ {Pb}^{2+}( {aq})+2 {l}^{-}( {aq})$ $S\;2 S$ $\therefore \;\; {[ {Pb}^{2+}]\$ =S+0.1} ;; ≈ 0.1\because ;; S < < 0.1 $Now,$ K_{ {sp}}=8 × 10^{-9}[ {Pb}^{2}]$[I^{-}]^{2}=8 \times 10^{-9}0.1 \times (2 S)^{2}=8 \times 10^{-9}4 S^{2}=8 \times 10^{-8} \Rightarrow S=141 \times 10^{-6} {M}x=141$

Q58JEE Main 2021NAT4MChemical Kinetics

Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of $3.33 {h}$ at $25^{\circ} {C}$ . After $9 {h}$ , the fraction of sucrose remaining is $f$ . The value of $\log _{10}(\;\frac{1}{f})$ is ......... $\times 10^{-2}$ (Rounded off to the nearest integer). [Assume, $\ln 10=2.303, \ln 2=0.693$ ]

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📖 Explanation

Given, ${ {C}_{12} {H}_{22} {O}_{11}}_{\text{Sucrose}}+ {H}_{2} {O} {\to }^{\;\text{ 1st order }\;} {{C}_{6} {H}_{12} {O}_{6}}_{\text{Glucose}}+ {{C}_{6} {H}_{12} {O}_{6} }_{\text{Fructose}}$ $t_{1 / 2}\;=\;\frac{10}{3} {h}$ $t\;=0, \;\; {a}=[ {A}]_{0} \;\; \;\text{ (initial conc.) }\;$ $\;\text{ At }\; \;\; t\;=9 {h} \;\; {a}- {x}=[ {A}]_{t}$ [conc. at time $t]$ For using 1st order equation, $K \;\; =\;\frac{2.303}{t} \log \;\frac{[A]_{0}}{[A]_{t}} \Rightarrow \;\frac{K \times t}{2.303}=\log \;\frac{[A]_{0}}{[A]_{t}}$ $\;\frac{\ln 2 \times 9}{10 / 3 \times 2303} \;\; =\log (\;\frac{1}{F}) \Rightarrow \log (\;\frac{1}{F})=0.8124 \;\; ( \because k=\;\frac{\ln 2}{t_{1 / 2}})$ $\log (\;\frac{1}{F}) \;\; =81.24 \times 10^{-2}$ $x \;\; =81.24 \;\text{ or }\; x \approx 81$

Q59JEE Main 2021NAT4MAmines

$1.86 {g}$ of aniline completely reacts to form acetanilide. $10 \%$ of the product is lost during purification. Amount of acetanilide obtained after purification (in g) is .......... $\times 10^{-2}$ .

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📖 Explanation

Given, weight $=1.86 {g}$ Here, 1 mole of aniline gives 1 mole of acetanilide $\therefore$ mole of aniline $=$ mole of acetanilide $\Rightarrow \;\; \;\frac{1.86}{93}=\;{W_{\;\text{Actanilide }\;}}/{135}$ $\;W_{\;\text{Acetanilide }\;}=\;\frac{1.86 \times 135}{93} {g}=2.70 {g}$ But efficiency of reaction is $90 \%$ only. Hence, mass of acetanilide produced $\;\; =2.70 \times \;\frac{90}{100} {g}=2.43 {g}=243 \times 10^{2} {g}$ $x \;\; =243$

Q60JEE Main 2021NAT4MChemical Equilibrium

Assuming ideal behaviour, the magnitude of log $K$ for the following reaction at $25^{\circ} {C}$ is $x \times 10^{-1} .$ The value of $x$ is .......... (Integer answer) 3HC $\equiv {CH}( {g}) ⇌ {C}_{6} {H}_{6}(l)$ $[ {Given}, \Delta _{f} G^{\circ}( {HC} \equiv {CH})=-2.04 \times 10^{5}{J} {mol}^{-1},$ , $\Delta _{f} G^{\circ}( {C}_{6} {H}_{6})=-1.24 \times 10^{5} {J} {mol}^{-1}$ , $R=8.314 {J} {K}^{-1} {mol}^{-1}]$

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📖 Explanation

Reaction, ${3 {HC} \equiv {CH}( {g})}_{\text{Acetylene} } \longrightarrow {{C}_{6} {H}_{6}( {l})}_{\text{Benzene} }$ Given, $\Delta {G}_{f}^{\circ}=( {CH} \equiv {CH})=-2.04 \times 10^{5} {J} {mol}^{-1}$ $\Delta {G}_{f}^{\circ}( {C}_{6} {H}_{6})=-1.24 \times 10^{5} {J} {mol}^{-1}$ Gibb's free energy, $\Delta {G}^{\circ}=-n R T \text{lnK}$ $\Delta {G}^{\circ}=\sum(\Delta {G}_{f}^{\circ})_{p}-\sum(\Delta {G}_{f}^{\circ})_{R}$ $-n R T \ln K=-n^{'} R T \ln K_{p}-(-n^{' '} R T \ln K_{f})$ $\Rightarrow \;\; -R T \ln K=1 \times (-1.24 \times 10^{5})-(-3 \times 2.04 \times 10^{5})$ $-2.303 \times R \times T \log K=4.88 \times 10^{5}$ $\log K=-\;\frac{4.88 \times 10^{5}}{2.303 \times 8.314 \times 273}$ $n \ln K=n^{'} \ln K_{p}-(-n^{' '} \ln K_{f})$ $K=85.52 \Rightarrow K=855 \times 10^{-1}$ $x=855$

Mathematics30 questions

Q61JEE Main 2021MCQ4MSets and Relations

For the statements $p$ and $q$ , consider the following compound statements A. $[∼ q ∧(p \to q)] \to ∼ p$ B. $[(p ∨ q) ∧ ∼ p] \to q$ Then, which of the following statement(s) is/are correct?

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📖 Explanation

Given statements, (A) $[∼ q ∧(p \to q] \to ∼ p$ (B) $[(p ∨ q) ∧ ∼ p] \to q$ For statement (A), $\therefore$ Statement (B) is tautology. $\therefore$ (A) and (B) both are tautologies.

Q62JEE Main 2021MCQ4MThree Dimensional Geometry

Let $a, b \in R$ . If the mirror image of the point $P(a, 6,9)$ with respect to the line $\;\frac{x-3}{7}=\;\frac{y-2}{5}=\;\frac{z-1}{-9}$ is $(20, b,-a-9)$ , then $|a+b|$ is equal to

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📖 Explanation

Given, $P(a, 6,9)$ Equation of line $\;\frac{x-3}{7}=\;\frac{y-2}{5}=\;\frac{z-1}{-9}$ Image of point $P$ with respect to line is point $Q(20, b,-a-9)$ Mid-point of $P$ and $Q=(\;\frac{a+20}{2}, \;\frac{6+b}{2}, \;\frac{-a}{2})$ This point lies on line $\therefore \;\; \;\frac{\;\frac{a+20}{2}-3}{7}=\;\frac{\;\frac{6+b}{2}-2}{5}=\;\frac{\;\frac{-a}{2}-1}{-9}$ $\Rightarrow \;\; \;\frac{a+14}{14}=\;\frac{b+2}{10}=\;\frac{a+2}{18}$ $\Rightarrow \;\; \;\frac{a+14}{14}=\;\frac{a+2}{18}$ and $\;\frac{b+2}{10}=\;\frac{a+2}{18}$ Solving, we get $a=-56, b=-32$ $\therefore \;\; |a+b|=|-56-32|=88$

Q63JEE Main 2021MCQ4MThree Dimensional Geometry

The vector equation of the plane passing through the intersection of the planes $r \cdot (\hat{i}+j+\hat{k})=1$ and $r \cdot (\hat{i}-2 \hat{j})=-2$ and the point $(1,0,2)$ is

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📖 Explanation

Given, point $(1,0,2)$ Equation of plane $=$ $r \cdot (\hat{i}+\hat{j}+\hat{k})=1 \;\text{ and }\; r \cdot (\hat{i}-2 \hat{j})=-2$ Equation of plane passing through the intersection of given planes is $[r \cdot (\hat{i}+\hat{j}+\hat{k})-1]+\lambda[r \cdot (\hat{i}-2 \hat{j})+2]=0$ $\because$ This plane passes through point $(1,0,2)$ i.e., $vector(\hat{i}+2 \hat{k})$ $\therefore \;\; [(\hat{i}+2 \hat{k}) \cdot (\hat{i}+\hat{j}+\hat{k})-1]+\lambda[(\hat{i}+2 \hat{k}) \cdot (\hat{i}-2 \hat{j})+2]=0$ $\Rightarrow \;\; (3-1)+\lambda(1+2)=0$ $\Rightarrow \;2+\lambda \times 3 \;\; =0$ $\Rightarrow \;\lambda \;\; =-2 / 3$ Hence, equation of required plane is ${[r \cdot (\hat{i}+\hat{j}+\hat{k})-1]+(\;\frac{-2}{3})[r \cdot (\hat{i}-2 \hat{j})+2]}\;=0$ $\;\text{ or }\; \;\; 3[r \cdot (\hat{i}+\hat{j}+\hat{k})-1]-2[r \cdot (\hat{i}-2 \hat{j})+2]\;=0$ $\;\text{ or }\; \;\; r \cdot (\hat{i}+7 \hat{j}+3 \hat{k})\;=7$

Q64JEE Main 2021MCQ4MApplication of Derivatives

If $P$ is a point on the parabola $y=x^{2}+4$ which is closest to the straight line $y=4 x-1$ , then the coordinates of $P$ are

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📖 Explanation

Given, curve $y=x^{2}+4$ and, line $y=4 x-1$ Here, $y=x^{2}+4$ $\therefore \;\frac{d y}{d x} \;\; =2 x$ . . . (i) $\;\text{ and }\; y \;\; =4 x-1$ $\;\frac{d y}{d x} \;\; =4$ . . . (ii) Let the required point be $P(x_{1}, y_{1})$ . $\therefore \;\; \;\frac{d y}{d x}|_{p}=2 x_{1}$ . . . (iii) $\because$ Slopes will be equal. $\therefore \;\; 2 x_{1}=4 \;\;$ [from Eqs. (ii) and (iii)] $\Rightarrow$ $2 x_{1}=4$ $4=2$ Now, the given point $P(x_{1}, y_{1})$ lies on curve $y=x^{2}+4$ , $\therefore \;y_{1}=x_{1}^{2}+4$ $\Rightarrow \;y_{1}=2^{2}+4=8$ Hence, required coordinate of $P=(2,8)$

Q65JEE Main 2021MCQ4MTrigonometric Ratios and Identities

The angle of elevation of a jet plane from a point $A$ on the ground is $60^{\circ}$ . After a flight of $20 {s}$ at the speed of $432 {km} / {h}$ , the angle of elevation changes to $30^{\circ}$ . If the jet plane is flying at a constant height, then its height is

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📖 Explanation

Given, angle of elevation are $60^{\circ}$ and then $30^{\circ}$ . Also, in 20 sec plane covers the distance from $A$ to $B$ with speed $432 {km} / {h}$ . $\because \;\; 432 \times \;\frac{5}{18} {m} / {sec} \;\; =120 {m} / {sec}$ $\therefore \;\; A B=\;\text{ distance }\; \;\; =S \times T=120 \times 20$ $\;\; =2400 {m}$ In $\triangle P B D$ $\tan 30^{\circ} \;\; =\;\frac{B D}{P D} \Rightarrow \;\frac{1}{\sqrt{3}}=\;\frac{h}{P D}$ $\Rightarrow \;\; P D \;\; =h \sqrt{3}$ In $\triangle P A C$ $\tan 60^{\circ} \;\; =\;\frac{A C}{P C} \Rightarrow \sqrt{3}=\;\frac{h}{P C}$ $\Rightarrow \;\; P C \;\; =h / \sqrt{3}$ $\;\text{ Now, }\; C D \;\; =P D-P C$ $2400 \;\; =h \sqrt{3}-\;\frac{h}{\sqrt{3}}$ $\Rightarrow \;\; 2400 \;\; =h(\;\frac{3-1}{\sqrt{3}})$ $\therefore \;\; h \;\; =\;\frac{2400 \times \sqrt{3}}{2}=1200 \times \sqrt{3}$ $\therefore \;\text{ Required height }\; \;\; =1200 \sqrt{3} {m}$

Q66JEE Main 2021MCQ4MPermutations and Combinations

If $n \ge 2$ is a positive integer, then the sum of the series ${ }^{n+1} C_{2}+2({ }^{2} C_{2}+{ }^{3} C_{2}+{ }^{4} C_{2}+. . .+{ }^{n} C_{2})$ is

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📖 Explanation

Given, $n \ge 2$ Let $S={ }^{2} C_{2}+{ }^{3} C_{2}+. . .+{ }^{n} C_{2}={ }^{n+1} C_{3}$ Now, $\;{ }^{n+1} C_{2}+2 \times ({ }^{2} C_{2}+{ }^{3} C_{2}+. . .+{ }^{n} C_{2})$ $\;\; ={ }^{n+1} C_{2}+2 \times { }^{n+1} C_{3}$ $\;\; =({ }^{n+1} C_{2}+{ }^{n+1} C_{3})+{ }^{n+1} C_{3}$ $\;\; ={ }^{n+2} C_{3}+{ }^{n+1} C_{3}=\;\frac{(n+2) !}{3 !(n-1) !}+\;\frac{(n+1) !}{3 !(n-2) !}$ $\;\; =\;\frac{(n+2)(n+1) n(n-1) !}{3 \times 2 \times 1 \times (n-1) !}+\;\frac{(n+1) \times n \times (n-1) \times (n-2) !}{3 \times 2 \times 1 \times (n-2) !}$ $\;\; =\;\frac{n(n+1)}{6}[n+2+n-1]$ $\;\; =\;\frac{n(n+1)(2 n+1)}{6}$

Q67JEE Main 2021MCQ4MApplication of Derivatives

Let $f: R \to R$ be defined as $f(x)=\begin{cases}-55 x & \\ \\ \text{ if }\\ \\ x< -5 \\ 2 x^{3}-3 x^{2}-120 x & \\ \\ \text{ if }\\ \\ -5 \le x< 4 \\ 2 x^{3}-3 x^{2}-36 x-336 & \\ \\ \text{ if }\\ \\ x \ge 4\end{cases}$ Let $A=\{x \in R$ : $f$ is increasing $\}$ . Then, $A$ is equal to

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📖 Explanation

Given, $f(x) \;\; =\begin{cases}-55 x & \\ \\ \\ \\ x < -5 \\ 2 x^{3}-3 x^{2}-120 x & -5 \le x< 4 \\ 2 x^{3}-3 x^{2}-36 x+10 & x \ge 4\end{cases}$ $\therefore \;\; f^{'}(x) \;\; =\begin{cases}-55 & x < -5 \\ 6(x^{2}-x-20) & -5 \le x < 4 \\ 6(x^{2}-x-6) & x \ge 4\end{cases}$ $f^{'}(x) \;\; =\begin{cases}-55 & x < -5 \\ 6(x-5)(x+4) & -5 \le x < 4 \\ 6(x-3)(x+2) & x \ge 4\end{cases}$ For $f$ to be increasing, $f^{'}(x) > 0$ Now, $f^{'}(x)=-55$ is always less than zero. $f^{'}(x)=6(x-5)(x+4) < 0,-5 \le x < 4$ Critical points $=5,-4$ $\;\; [x \in(-5,-4)$ and $f^{'}(x)=6(x-3)(x+2) < 0, x \ge 4$ Critical point, $=3,-2$ $x \in(4, \infty )$ ... (ii) From Eqs. (i) and (ii), $f(x)$ is increasing in $x \in(-5,-4) \cup(4, \infty )$

Q68JEE Main 2021MCQ4MDifferential Equations

Let $f$ be a twice differentiable function defined on $R$ , such that $f(0)=1, f^{'}(0)=2$ and $f^{' '}(x) \ne 0$ for all $x \in R$ . If $\begin{vmatrix}f(x) & f^{'}(x) \\ f^{'}(x) & f^{' '}(x)\end{vmatrix}=0$ , for all $x \in R$ , then the value of $f(1)$ lies in the interval

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📖 Explanation

Given, $f(0)=1$ , $\;f^{'}(0) \;\; =2,$ $\;f^{' '}(x)\;\ne 0\;$ $\;\;\begin{vmatrix}f(x) & f^{'}(x) \\ f^{'}(x) & f^{' '}(x)\end{vmatrix} \;\; =0$ $\Rightarrow \;f(x) f^{' '}(x)-f^{'}(x) f^{'}(x) \;\; =0$ $\Rightarrow \;\;\frac{f^{' '}(x)}{f^{'}(x)} \;\; =\;\frac{f^{'}(x)}{f(x)}$ $\Rightarrow \;\int \;\frac{f^{' '}(x)}{f^{'}(x)} d x \;\; =\int \;\frac{f^{'}(x)}{f(x)} d x$ $\Rightarrow \;\log f^{'}(x) \;\; =\log f(x)+\log c$ $\Rightarrow \;f^{'}(x) \;\; =c f(x)$ Now, put $x=0$ , we get $\;f^{'}(0) \;\; =c f(0)$ $\Rightarrow \;2 \;\; =c \times 1$ $\Rightarrow \;c \;\; =2$ Putting the value of $c=2$ in Eq. (i), we get $\;\log f^{'}(x) \;\; =\log f(x)+\log 2$ $\Rightarrow \;f^{'}(x) \;\; =2 f(x) \Rightarrow \int \;\frac{f^{'}(x)}{f(x)} d x=\int 2 d x$ $\Rightarrow \;\log f(x) \;\; =2 x+D \Rightarrow f(x)= {e}^{2 x+D}$ $\Rightarrow \;f(x) \;\; = {e}^{D} \cdot {e}^{2 x}$ $\Rightarrow \;f(x) \;\; =k \cdot {e}^{2 x}$ $[Let $k=e^{D}$ ]$ Put $x=0$ , we get $\;f(0) \;\; =k \cdot e^{0}$ $\Rightarrow \;1 \;\; =k \Rightarrow f(x)=k \cdot e^{2 x}$ $\; \therefore \;f(x) \;\; =e^{2 x}$ Put $x=1$ , we get $f(1)=e^{2}$ Clearly, ${e}^{2}$ lies in $(6,9)$ .

Q69JEE Main 2021MCQ4MApplication of Derivatives

For which of the following curves, the line $x+\sqrt{3} y=2 \sqrt{3}$ is the tangent at the point $(\;\frac{3 \sqrt{3}}{2}, \;\frac{1}{2}) ?$

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📖 Explanation

Given, line $x+\sqrt{3} y=2 \sqrt{3}$ and point $(\;\frac{3 \sqrt{3}}{2}, \;\frac{1}{2})$ From options, we take the conic $x^{2}+9 y^{2}=9$ Equation of any tangent at $(x_{1}, y_{1})$ is $x x_{1}+9 \cdot y y_{1} \;\; =9$ $\Rightarrow \;\;\frac{3 \sqrt{3}}{2} x+9 \times \;\frac{1}{2} \times y=9 \;\; [ \because (x_{1}, y_{1})=(\;\frac{3 \sqrt{3}}{2}, \;\frac{1}{2})]$ $\Rightarrow \;\;\frac{3 \sqrt{3}}{2}(x+\sqrt{3} y)=9$ $\Rightarrow \;x+\sqrt{3} y=\;\frac{9 \times 2}{3 \sqrt{3}}$ $\Rightarrow \;x+\sqrt{3} y=2 \sqrt{3}$ Clearly, $x+\sqrt{3} y=2 \sqrt{3}$ is a tangent to the curve $x^{2}+9 y^{2}=9$ .

Q70JEE Main 2021MCQ4MDefinite Integrals

The value of the integral $int_{1}^{3}\$ [x^{2}-2 x-2]$ d x $, where$ [x] $denotes the greatest integer less than or equal to$ x$, is

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📖 Explanation

Let $I \;\; =int_{1}^{3}\$ [x^{2}-2 x-2]$ d x;=int_{1}^{3}$[x^{2}-2 x+1-3]$ d x=int_{1}^{3}(x-1)^{2}-3] d x;; =int_{1}^{3}$[(x-1)^{2}]$ d x+int_{1}^{3}-3 d x $Put$ x-1=t ; d x=d t $, when$ x=1, t=0 $and$ x=3, t=2\therefore ;; I=-3[x]{1}^{3}+int{0}^{2}$[t^{2}]$ d t=-6+int_{0}^{1} 0 d t+int_{1}^{\sqrt{2}} 1 d t+int_{\sqrt{2}}^{\sqrt{3}} 2 d t+int_{\sqrt{3}}^{2} 3 d t=-6+(0)+(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3})=-6+\sqrt{2}-1+2 \sqrt{3}-2 \sqrt{2}+6-3 \sqrt{3}I=-1-\sqrt{2}-\sqrt{3}$

Q71JEE Main 2021MCQ4MInverse Trigonometric Functions

A possible value of $\tan (\;\frac{1}{4} \sin ^{-1} \;\frac{\sqrt{63}}{8})$ is

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📖 Explanation

Given, $\tan (\;\frac{1}{4} \sin ^{-1} \;\frac{\sqrt{63}}{8})$ Let $\sin ^{-1} \;\frac{\sqrt{63}}{8}=\theta$ $\Rightarrow \;\; \sin \theta \;\; =\;\frac{\sqrt{63}}{8} \Rightarrow \cos \theta=\;\frac{1}{8}$ $\;\text{ Also, }\; \cos \;\frac{\theta}{2} \;\; =\sqrt{\;\frac{1+\cos \theta}{2}}$ $\;\; =\sqrt{\;\frac{1+\frac{1}{8}}{2}}=\sqrt{\;\frac{9}{16}}=\;\frac{3}{4}$ $\therefore \tan (\;\frac{1}{4} \sin ^{-1} \;\frac{\sqrt{63}}{8}) \;\; =\tan (\;\frac{\theta}{4})$ $\;\; =\sqrt{\;\frac{1-\cos \frac{\theta}{2}}{1+\cos \;\frac{\theta}{2}}}=\sqrt{\;\frac{1-\frac{3}{4}}{1+\;\frac{3}{4}}}=\;\frac{1}{\sqrt{7}}$

Q72JEE Main 2021MCQ4MSets and Relations

The negative of the statement $∼ p ∧(p ∨ q)$ is

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📖 Explanation

Given, statement: $∼ p ∧(p ∨ q$ Negative of given statement $∼\;∼[∼ p ∧(p ∨ q)$ $=p ∨ ∼[p ∨ q)$ [by De Morgan's law] $=p ∨(∼ p ∧ ∼ q)$ [by De Morgan's law] $=(p ∨ ∼ p) ∧(p ∨ ∼ q)$ [Using distributive property] $=t ∧(p ∨ ∼ q)$ $=p ∨ ∼ q$

Q73JEE Main 2021MCQ4MApplication of Derivatives

If the curve $y=a x^{2}+b x+c, x \in R$ , passes through the point $(1,2)$ and the tangent line to this curve at origin is $y=x$ , then the possible values of $a, b, c$ are

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📖 Explanation

Given, curve $\Rightarrow y=a x^{2}+b x+c, x \in R$ and point $(1,2)$ $\because$ The given curve passes through $(1,2)$ . $\therefore \;\; 2=a+b+c$ Also, slope of tangent of $y=a x^{2}+b x+c$ is $\;\frac{d y}{d x}=2 a x+b$ $\because$ Tangent passes through origin $(0,0)$ . ` $\therefore \;\;\;\frac{d y}{d x}|_{(0,0)}=2 a \times 0+b=b$ $. . .$ (i) According to the question, tangent at origin is $y=x$ $\therefore$ Its slope is 1 . ... (ii) From Eqs. (i) and (ii), $b=1$ Also, $a+b+c=2$ $\Rightarrow \;\; a+c+1=2 \Rightarrow a+c=1$ From the option look for $b=1$ and $a+c=1$ The only correct order triplet is $a=1, b=1, c=0$ .

Q74JEE Main 2021MCQ4MArea Under The Curves

The area of the region $R=\{(x, y): 5 x^{2} \le y \le 2 x^{2}+9\}$ is

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📖 Explanation

Given, $R=\{(x, y): 5 x^{2} \le y \le 2 x^{2}+9\}$ Here, we have two curves $y=5 x^{2}$ and $y=2 x^{2}+9$ , point of intersection of both curves is find by solving both equations i.e. $5 x^{2} \;\; =2 x^{2}+9$ $\Rightarrow \;\; x^{2} \;\; =3 \Rightarrow x=\pm \sqrt{3}$ $\therefore \;\; \;\text{ Area }\; \;\; =int_{-\sqrt{3}}^{\sqrt{3}}(2 x^{2}+9-5 x^{2}) d x$ $\;\; =2 int_{0}^{\sqrt{3}}(9-3 x^{2}) d x$ $\;\; =2\$ [9 x-x^{3}]_{0}^{\sqrt{3}};; =2[9 \sqrt{3}-3 \sqrt{3}];; =12 \sqrt{3} ;\text{ sq units };$

Q75JEE Main 2021MCQ4MDifferential Equations

If a curve $y=f(x)$ passes through the point $(1,2)$ and satisfies $x \;\frac{d y}{d x}+y=b x^{4}$ , then for what value of $b, int_{1}^{2} f(x) d x=\;\frac{62}{5}$ ?

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📖 Explanation

Given, curve $y=f(x)$ passes through $(1,2)$ and satisfies $\;x \;\frac{d y}{d x}+y \;\; =b x^{4}$ $\Rightarrow \;x \;\frac{d y}{d x}+y \;\; =b x^{4}$ $\Rightarrow \;\;\frac{d y}{d x}+\;\frac{y}{x} \;\; =b x^{3}$ $\; \;\; \;\text{ IF }\; \;\; =e^{j \;\frac{1}{x} \cdot d x}=x$ $\; \;\; y x \;\; =\int b x^{4} d x=\;\frac{b x^{5}}{5}+C$ $\; \therefore \;\; y\; \;\; =\;\frac{b x^{4}}{5}+\;\frac{C}{x} \;\; =f(x)$ . . . (i) $\because$ This curve passes through $(1,2)$ . $\therefore \;\; 2 \times 1=\;\frac{b \times (1)^{5}}{5}+C$ $\Rightarrow \;\; 2=\;\frac{b}{5}+C$ . . . (ii) Also, $int_{1}^{2} f(x) d x=\;\frac{62}{5}$ $\Rightarrow \;\; int_{1}^{2}(\;\frac{b x^{4}}{5}+\;\frac{C}{x}) d x=\;\frac{62}{5}$ $\Rightarrow \;\; \$ [b × ;\frac{x^{5}}{25}+C \log x]_{1}^{2}=;\frac{62}{5}\Rightarrow ;; [(;\frac{b × 32}{25}+C \log 2)-(;\frac{b}{25}+c \log 1)]=;\frac{62}{5}\Rightarrow ;\frac{b × 32}{25}+C \log 2-;\frac{b}{25}=;\frac{62}{5}+0 \log 2 $[from Eq. (i)] On comparing, we get$ ;\frac{b}{25} × 31 ;; =;\frac{62}{5} ;\text{ and }; c=0b ;; =;\frac{62 × 25}{31 × 5}b ;; =10 $Hence, the required value of$ b=10$.

Q76JEE Main 2021MCQ4MDefinite Integrals

Let $f(x)$ be a differentiable function defined on $[0,2]$ , such that $f^{'}(x)=f^{'}(2-x)$ , for all $x \in(0,2), f(0)=1$ and $f(2)=e^{2}$ . Then, the value of $int_{0}^{2} f(x) d x$ is

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📖 Explanation

Given, $f(0)=1$ . . . (i) $f(2) \;\; = {e}^{2}$ . . . (ii) $f^{'}(x) \;\; =f^{'}(2-x)$ Integrating w.r.t. $x$ , $f(x)=-f(2-x)+C$ Put $x=0$ $\;f(0) \;\; =-f(2)+C$ $\Rightarrow \;1 \;\; =- {e}^{2}+C$ [from Eqs. (i) and (ii)] $\Rightarrow \;C \;\; =1+ {e}^{2}$ $\; \therefore \;f(x) \;\; =-f(2-x)+1+ {e}^{2}$ $\;\text{ or }\; f(x)+f(2-x)=1+ {e}^{2}$ . . . (iii) $\;\text{ Let }\; I=int_{0}^{2} f(x) d x$ . . . (iv) $\;\text{ Also, }\; I=int_{0}^{2} f(2-x) d x$ . . . (v) Now, adding Eqs. (iv) and (v), $2 I \;\; =int_{0}^{2}[f(x)+f(2-x)] d x$ $2 I \;\; =int_{0}^{2}(1+e^{2}) d x$ [from Eq. (iii)] $2 I \;\; =2(1+e^{2})$ $\therefore \;\; I \;\; =(1+e^{2})$

Q77JEE Main 2021MCQ4MMatrices and Determinants

Let $A$ and $B$ be $3 \times 3$ real matrices, such that $A$ is symmetric matrix and $B$ is skew-symmetric matrix. Then, the system of linear equations $(A^{2} B^{2}-B^{2} A^{2}) X=0$ , where $X$ is a $3 \times 1$ column matrix of unknown variables and $O$ is a $3 \times 1$ null matrix, has

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📖 Explanation

Given, $A$ be a $3 \times 3$ matrix. $A$ is symmetric and $B$ is skew-symmetric. $\therefore \;\; A^{T}=A, B^{T}=-B$ $\;\text{ Let }\; A^{2} B^{2}-B^{2} A^{2}=P$ $\;P^{T}=(A^{2} B^{2}-B^{2} A^{2})^{T}=(A^{2} B^{2})^{T}-(B^{2} A^{2})^{T}$ $\;\; =(B^{2})^{T}(A^{2})^{T}-(A^{2})^{T}(B^{2})^{T}$ $\;\; =B^{2} A^{2}-A^{2} B^{2}=-(A^{2} B^{2}-B^{2} A^{2})=-P$ $\;P^{T}=-P$ $P$ is skew-symmetric. $\therefore |P|=0$ Hence, $P X=0$ have infinite solutions.

Q78JEE Main 2021MCQ4MQuadratic Equation and Inequalities

Let $a, b, c$ be in an arithmetic progression. Let the centroid of the triangle with vertices $(a, c),(2, b)$ and $(a, b)$ be $(\;\frac{10}{3}, \;\frac{7}{3})$ . If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+1=0$ , then the value of $\alpha^{2}+\beta^{2}-\alpha \beta$ is

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📖 Explanation

Given, $a, b$ and $c$ are in AP. $(a, c),(2, b)$ and $(a, b)$ are vertices of triangle. Centroid $=(\;\frac{10}{3}, \;\frac{7}{3})$ $\alpha$ and $\beta$ are the roots of equation $a x^{2}+b x+1=0$ $\because a, b, c$ are in AP. $\therefore$ $2 b=a+c$ . . . (i) $\Rightarrow \;\;\frac{2 a+2}{3} \;\;=\;\frac{10}{3} \;\text{ and }\; \;\frac{c+2 b}{3}=\;\frac{7}{3}\;$ $a \;\; =4\;$ $\;\text{ and }\; \;\; c+a+c \;\; =7\;$ $\Rightarrow \;\; 2 c \;\; =7-4\;\;[ \because a=a+c]$ $c=3 / 2$ $\;\text{ Also, }\; 2 b=a+c=4+\;\frac{3}{2}$ $\Rightarrow \;\; b=11 / 4$ Now, $\alpha$ and $\beta$ are roots of $a x^{2}+b x+1=0$ $\therefore \;\; \alpha+\beta=\;\frac{-b}{a}=\;\frac{-11 / 4}{4}$ $\Rightarrow \;\; \alpha+\beta=\;\frac{-11}{16} \Rightarrow \alpha \beta=\;\frac{1}{ {a}}=\;\frac{1}{4}$ $\Rightarrow \;\; \alpha \beta=\;\frac{1}{4}$ Now, $\alpha^{2}+\beta^{2}-\alpha \beta$ $=(\alpha+\beta)^{2}-3 \alpha \beta=(\;\frac{-11}{16})^{2}-3 \times \;\frac{1}{4}$ $=\;\frac{121-192}{256}=\;\frac{-71}{256}$

Q79JEE Main 2021MCQ4MMatrices and Determinants

For the system of linear equations $x-2 y=1, x-y+k z=-2, k y+4 z=6, k \in R$ , consider the following statements (A) The system has unique solution, if $k \ne 2$ , $k \ne -2$ (B) The system has unique solution, if $k=-2$ (C) The system has unique solution, if $k=2$ (D) The system has no solution, if $k=2$ (E) The system has infinite number of solutions, if $k \ne -2$ Which of the following statements are correct?

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📖 Explanation

Given, $x-2 y+0 z=1$ $\;x-y+k x=-2$ $\;0 x+k y+4 z=6$ $\;\text{ Here, }\; \Delta \;\; ={\begin{vmatrix}1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4\end{vmatrix}}=1(-4-k^{2})+2(4)$ $\;\; =-4-k^{2}+8=4-k^{2}$ $\;\Delta _{x}={\begin{vmatrix}1 & -2 & 0 \\ -2 & -1 & k \\ 6 & k & 4\end{vmatrix}}=1(-4-k^{2})+2(-8-6 k)$ $\;\; =-4-k^{2}-16-12 k=-k^{2}-12 k-20$ $\;\text{ If }\; \Delta \ne 0 \;\text{, then it has unique solution i.e. }\; 4-k^{2} \ne 0$ $\Rightarrow k \ne \pm 2 \;\text{ for unique solution. }\;$ $\;\text{ Also at }\; k=2$ $\Delta _{x} \;\; =-2^{2}-12 \times 2-20=-48 \ne 0$ $\;\text{ Then, \in this case it has no solution. }\;\;$ $\;\text{ Hence, statement (A) and statement (D) both are correct. }\;$ $\Rightarrow k \ne \pm 2$ for unique solution. Also at $k=2$ Then, in this case it has no solution. Hence, statement (A) and statement (D) both are correct.

Q80JEE Main 2021MCQ4MProbability

The probability that two randomly selected subsets of the set $\{1,2,3,4,5\}$ have exactly two elements in their intersection, is

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📖 Explanation

Given, set $=\{1,2,3,4,5\}$ Let the two subsets be $A$ and $B$ . Then, $n(A \cap B)=2$ (as given in question) $\therefore$ Required probability $=\;\frac{{ }^{5} C_{2} \times 3^{3}}{4^{5}}=\;\frac{10 \times 27}{2^{10}}=\;\frac{135}{2^{9}}$

Q81JEE Main 2021NAT4MPermutations and Combinations

For integers $n$ and $r$ , let $(\text{table} n ; r)=\begin{cases}{ }^{n} C_{r^{'}} & {,}\ \\ \text{if} \ \\ n \ge r \ge 0 \\ 0 {,} & \text{otherwise}$ The maximum value of $k$ for which the sum $sum_{i=0}^{k}(\text{table} 10 \\ i)(\text{table} 15 \\ k-i)+sum_{i=0}^{k+1}(\text{table}12 \\ i)(\text{table} 13 \\ k+1-i)$ exists & is equal to\end{cases}.........

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📖 Explanation

Given, $(\text{table} n ; r)=\begin{cases}{ }^{n} C_{r} & \\ \\ \text{, if }\\ \\ n \ge r \ge 0 \\ 0 & \\ \\ \text{ otherwise }\\ \\ \end{cases}$ and $sum_{i=0}^{k}(\text{table} 10 ; i)(\text{table} 15 ; k-i)+sum_{i=0}^{k+1}(\text{table} 12 ; i)(\;\frac{13}{k+1-i})$ $\because \;\; (1+x)^{10}={ }^{10} C_{0}+{ }^{10} C_{1} x+{ }^{10} C_{2} x^{2}+. . .+{ }^{10} C_{10} x^{10}$ and $(1+x)^{15}={ }^{15} C_{0}+{ }^{15} C_{1} x+{ }^{15} C_{2} x^{2}+. . .+{ }^{15} C_{15} x^{15}$ $\therefore sum_{i=0}^{k}({ }^{10} C_{i})({ }^{15} C_{k-i})={ }^{10} C_{0} \cdot { }^{15} C_{k}+{ }^{10} C_{1} \cdot { }^{15} C_{k-1}+. . .+{ }^{10} C_{k}{ }^{15} C_{0}$ $\Rightarrow$ Coefficient of $x^{k}$ in $(1+x)^{25}={ }^{25} C_{k}$ Also, $sum_{i=0}^{k+1}({ }^{12} C_{i})({ }^{13} C_{k+1-i})={ }^{12} C_{0} \cdot { }^{13} C_{k+1}$ $+{ }^{12} C_{1} \cdot { }^{13} C_{k}+. . .+{ }^{12} C_{k+1} \cdot { }^{13} C_{0}$ $\Rightarrow$ Coefficient of $x^{k+1}$ in $(1+x)^{25}={ }^{25} C_{k+1}$ $\Rightarrow \;\; { }^{25} C_{k}+{ }^{25} C_{k+1}={ }^{26} C_{k+1}$ So, ${ }^{26} C_{k+1}$ always exists. Now $k$ can be as larger as possible.

Q82JEE Main 2021NAT4MThree Dimensional Geometry

Let $\lambda$ be an integer. If the shortest distance between the lines $x-\lambda=2 y-1=-2 z$ and $x=y+2 \lambda=z-\lambda$ is $\;\frac{\sqrt{7}}{2 \sqrt{2}}$ , then the value of $|\lambda|$ is ......... .

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📖 Explanation

Given, equation of line $\Rightarrow x-\lambda=2 y-1=$ $\Rightarrow \;\; \;\frac{x-\lambda}{1}=\;\frac{y-1 / 2}{\frac{1}{2}}=\;\frac{z}{-\;\frac{1}{2}}$ or $\;\; \;\frac{x-\lambda}{2}=\;\frac{y-1 / 2}{1}=\;\frac{z}{-1}$ Point on this line through which it passes is $(\lambda, \;\frac{1}{2}, 0)$ . Equation of another line $\Rightarrow x=y+2 \lambda=z-\lambda$ $\Rightarrow \;\; \;\frac{x}{1}=\;\frac{y-(-2 \lambda)}{1}=\;\frac{z-\lambda}{1}$ ... (ii) A point through which this line passes is $(0,-2 \lambda, \lambda)$ . Now, distance between two skew lines $=\;\frac{|(a_{2}-a_{1}) \cdot (b_{1} \times b_{2})|}{|b_{1} \times b_{2}|}$ According to the question, $\;\frac{|-5 \lambda-\frac{3}{2}|}{\sqrt{14}}=\;\frac{\sqrt{7}}{2 \sqrt{2}}$ $\Rightarrow \;\; |10 \lambda+3| \;\; =7$ $10 \lambda+3 \;\; =\pm 7$ $\Rightarrow \;\; 10 \lambda \;\; =4,-10$ $\Rightarrow \;\; \lambda \;\; =\;\frac{2}{5} \;\text{ and }\; \lambda=-1$ $\therefore \;\; \lambda=-1$ $\;\text{ ( }\; \lambda=\;\frac{2}{5} \;\text{ is not possible as }\; \lambda \;\text{ is an integer) }\;\;$ $\therefore \;$ $\;\text{ Hence, }\;|\lambda|=|-1|=1\;$

Q83JEE Main 2021NAT4MFunctions

If $a+\alpha=1, b+\beta=2$ and $a f(x)+\alpha f(\;\frac{1}{x})=b x+\;\frac{\beta}{x}, x \ne 0$ , then the value of expression $\;\frac{f(x)+f(\frac{1}{x})}{x+\;\frac{1}{x}}$ is .......... .

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📖 Explanation

Given, $a+\alpha=1$ $b+\beta=2$ $\because \;\; a \cdot f(x)+\alpha \cdot f(\;\frac{1}{x})=b x+\;\frac{\beta}{x}$ Replace $x$ by $\;\frac{1}{x^{'}}$ $a f(\;\frac{1}{x})+\alpha f(x)=\;\frac{b}{x}+\beta x$ Adding Eqs. (i) and (ii), we get $(a+\alpha)[f(x)+f(\;\frac{1}{x})]=(x+\;\frac{1}{x})(b+\beta)$ $\Rightarrow \;\; \;\frac{f(x)+f(\frac{1}{x})}{x+\;\frac{1}{x}}=\;\frac{b+\beta}{a+\alpha}=\;\frac{2}{1}=2$

Q84JEE Main 2021NAT4MCircles

Let a point $P$ be such that its distance from the point $(5,0)$ is thrice the distance of $P$ from the point $(-5,0)$ . If the locus of the point $P$ is a circle of radius $r$ , then $4 r^{2}$ is equal to ......... .

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📖 Explanation

Let $P$ be $(h, k), A(5,0)$ and $B(-5,0)$ . Given, $\Rightarrow \;\; P A^{2}=9 P B^{2}$ $\Rightarrow (h-5)^{2}+k^{2}=9[(h+5)^{2}+k^{2}]$ $\Rightarrow 8 h^{2}+8 k^{2}+100 h+200=0$ $\therefore \;\text{ Locus of }\; P \;\text{ is }\; x^{2}+y^{2}+\;\frac{25}{2} x+25=0$ $\;\text{ Centre }\;=(\;\frac{-25}{4}, 0)$ $\;\text{ and }\; r^{2}=(\;\frac{-25}{4})^{2}+0^{2}-25$ $\;\; =\;\frac{625}{16}-25=\;\frac{225}{16}$ $\therefore \;\; 4 r^{2} \;\; =4 \times \;\frac{225}{16}=\;\frac{225}{4}=56.25$

Q85JEE Main 2021NAT4MCircles

If the area of the triangle formed by the positive $x$ -axis, the normal and the tangent to the circle $(x-2)^{2}+(y-3)^{2}=25$ at the point $(5,7)$ is $A$ , then $24 A$ is equal to ......... .

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📖 Explanation

Given, circle $(x-2)^{2}+(y-3)^{2}=5^{2}$ $\;c=(2,3)$ $\;r=5$ Equation of normal at $P$ (i.e. $P A$ line) $\Rightarrow \;\; (y-7)=(\;\frac{7-3}{5-2})(x-5)$ $\Rightarrow \;\; 3 y-21=4 x-20$ $\Rightarrow \;\; 4 x-3 y+1=0$ Therefore, $M=(\;\frac{-1}{4}, 0)$ [Put $y=0$ in above equation] Now, equation of tangent at $P$ . $y-7 \;\; =\;\frac{-3}{4}(x-5) \;\; [ \because \;\text{ slope of }\; P N=\;\frac{-1}{\;\text{ Slope of }\; P A}]$ $\Rightarrow 4 y-28 \;\; =-3 x+15$ $\Rightarrow 3 x+4 y \;\; =43$ Therefore, $N=(\;\frac{43}{3}, 0)$ [Put $y=0$ in above equation] $\therefore \;\text{ Area }\;(A) \;\; =\;\frac{1}{2} \times M N \times P Q$ $\;\; =\;\frac{1}{2} \times (\;\frac{43}{3}+\;\frac{1}{4}) \times 7$ $\;\; =\;\frac{1}{2} \times \;\frac{175}{12} \times 7$ $\therefore \;\; 24 A \;\; =24 \times \;\frac{1}{2} \times \;\frac{175}{12} \times 7=1225$ But this question is wrong as in question. It is mentioned that the triangle is formed with the positive $X$ -axis which contradicts the solution.

Q86JEE Main 2021NAT4MStatistics

If the variance of 10 natural numbers 1,1 , $1, . . . . .1, k$ is less than 10 , then the maximum possible value of $k$ is ............ .

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📖 Explanation

Given, 10 natural numbers $=1,1,1, . . . 1, k$ According to the question, variance $(\sigma^{2}) < 10$ $\because \;\sigma^{2}=\;\frac{\sum x^{2}}{n}-(\;\frac{\sum x}{n})^{2}$ $\;\sigma^{2}=\;\frac{(9+k^{2})}{10}-(\;\frac{9+k}{10})^{2} < 10$ $\Rightarrow \;10(9+k^{2})-(81+k^{2}+18 k) < 1000$ $\Rightarrow \;90+10 k^{2}-81-k^{2}-18 k < 1000$ $\Rightarrow \;9 k^{2}-18 k+9 < 1000$ $\Rightarrow \;(k-1)^{2} < \;\frac{1000}{9}$ $\Rightarrow \;k-1 < \;\frac{10 \sqrt{10}}{3}$ and $\;\; k-1 > \;\frac{-10 \sqrt{10}}{3} \Rightarrow k < \;\frac{10 \sqrt{10}}{3}+1$ It is not possible because $k \in N$ . $\therefore$ Maximum possible integral value of $k$ is 11 .

Q87JEE Main 2021NAT4MSequences and Series

The sum of first four terms of a geometric progression (G.P.) is $\;\frac{65}{12}$ and the sum of their respective reciprocals is $\;\frac{65}{18}$ . If the product of first three terms of the G.P. is 1 and the third term is $\alpha$ , then $2 \alpha$ is .......... .

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📖 Explanation

Let four numbers in GP be $a, a r, a r^{2}, a r^{3}$ . According to the question, $a+a r+a r^{2}+a r^{3}=\;\frac{65}{12} \;\; . . .$ (i) and $\;\frac{1}{a}+\;\frac{1}{a r}+\;\frac{1}{a r^{2}}+\;\frac{1}{a r^{3}}=\;\frac{65}{18}$ $\Rightarrow \;\; \;\frac{1}{a}(\;\frac{1+r+r^{2}+r^{3}}{r^{3}})=\;\frac{65}{18}$ Dividing Eq. (i) by Eq. (ii), we get $\;\frac{a(1+r+r^{2}+r^{3})}{1} \;\frac{(1+r+r^{2}+r^{3})}{r^{3}} \;\; =\;\frac{65 / 12}{65 / 18}$ $\Rightarrow \;\; a^{2} r^{3} \;\; =\;\frac{18}{12} \Rightarrow a^{2} r^{3}=\;\frac{3}{2}$ Also, product of first three terms $=1$ $a \times a r \times a r^{2} \;\; =1$ $\Rightarrow \;\; a^{3} r^{3} \;\; =1$ $\Rightarrow \;\; a^{3} \times \;\frac{3}{2 a^{2}} \;\; =1$ $a \;\; =\;\frac{2}{3}$ $\Rightarrow \;\; \;\text{ and }\; r^{3}=\;\frac{3 / 2}{(2 / 3)^{2}}=(\;\frac{3}{2})^{3} \Rightarrow r=\;\frac{3}{2}$ According to the question, third term $=\alpha=a r^{2}=\;\frac{2}{3} \times \;\frac{3}{2} \times \;\frac{3}{2}=\;\frac{3}{2}$ $\therefore \;\; 2 \alpha=2 \times \;\frac{3}{2}=3$

Q88JEE Main 2021NAT4MPermutations and Combinations

The students $S_{1}, S_{2}, . . ., S_{10}$ are to be divided into 3 groups $A, B$ and $C$ such that each group has at least one student and the group $C$ has at most 3 students. Then, the total number of possibilities of forming such groups is ........... .

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📖 Explanation

Given, total students $=10$ Number of groups $=3$ (i.e. $A, B$ and $C$ ) Each group has atleast one student but group $C$ has atmost 3 students. $\therefore$ There are 3 cases depending on number of students in group $C$ . Case I C has 1 student, then $← 9$ students. $\therefore$ Number of ways $={ }^{10} C_{1} \times [2^{9}-2]$ Case II $C$ has 2 students, then $\;\frac{A}{B}] ← 8$ Students. $\therefore$ Number of ways $={ }^{10} C_{2} \times [2^{8}-2]$ Case III $C$ has 3 students, then $← 7$ Students. $\therefore$ Number of ways $={ }^{10} C_{3} \times [2^{7}-2]$ $\therefore$ Required number of possibilities $\;\; ={ }^{10} C_{1}(2^{9}-2)+{ }^{10} C_{2}(2^{8}-2)+{ }^{10} C_{3}(2^{7}-2)$ $\;\; =2^{7}\$ [{ }^{10} C_{1} × 4+{ }^{10} C_{2} × 2+{ }^{10} C_{3}]$-20-90-240;; =128[40+90+120]-350;; =(128 \times 250)-350=31650$

Q89JEE Main 2021NAT4MComplex Numbers

Let $i=\sqrt{-1}$ . If $\;\frac{(-1+i \sqrt{3})^{21}}{(1-i)^{24}}+\;\frac{(1+i \sqrt{3})^{21}}{(1+i)^{24}}=k$ and $n=[|k|]$ be the greatest integral part of $|k|$ . Then, $sum_{j=0}^{n+5}(j+5)^{2}-sum_{j=0}^{n+5}(j+5)$ is equal to ............. .

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📖 Explanation

Given, $\;\frac{(-1+i \sqrt{3})^{21}}{(1-i)^{24}}+\;\frac{(1+i \sqrt{3})^{21}}{(1+i)^{24}}=k$ $\because \;\; -1+i \sqrt{3}=2 {e}^{i 2 \pi / 3}$ $1+i \sqrt{3}=2 {e}^{i \pi / 3}$ $1-i=\sqrt{2} e^{-i \pi / 4}$ $1+i=\sqrt{2} e^{i \pi / 4}$ Now, $\;{(2 e^{i \frac{2 \pi}{3}})^{21}}/{(\sqrt{2} e^{-i \pi / 4})^{24}}+\;\frac{(2 e^{i \pi / 3})^{21}}{(\sqrt{2} e^{i \pi / 4})^{24}}$ $=\;\frac{2^{21} \cdot e^{i 14 \pi}}{2^{12} \cdot e^{-i 6 \pi}}+\;\frac{2^{21} \cdot e^{i 7 \pi}}{2^{12} \cdot e^{i 6 \pi}}$ $=2^{9} \cdot e^{i 20 \pi}+2^{9} \cdot e^{i \pi}$ $=2^{9}(1)+2^{9}(-1)$ $n=\begin{bmatrix}2^{9}-2^{9} & =0=k \\ n & =[|k|\end{bmatrix}=0 ; \;\text{ Now, }\; sum_{i=0}^{5}(j+5)^{2}-sum_{i=0}^{5}(j+5) , ; = , {[5^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2}]\$ -[5+6+7+8+9+10]} ; = , {[(1^{2}+2^{2}+3^{2}+. . .+10^{2})-(1^{2}+2^{2}+. . .+4^{2})]$} ; , -[(1+2+3+.. .+10)-(1+2+3+4)] ; = , {[;\frac{10 \times 11 \times 21}{6}-;\frac{4 \times 5 \times 9}{6}]-[;\frac{10 \times 11}{2}-;\frac{4 \times 5}{2}]} ; = , (385-30)-(55-10) ; = , 355-45=310].$

Q90JEE Main 2021NAT4MQuadratic Equation and Inequalities

The number of the real roots of the equation $(x+1)^{2}+|x-5|=\;\frac{27}{4}$ is ............. .

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📖 Explanation

Given, equation $(x+1)^{2}+|(x-5)|=\;\frac{27}{4}$ Case I For $x \ge 5$ $\Rightarrow \;\; (x+1)^{2}+(x-5) \;\; =\;\frac{27}{4}$ $\Rightarrow \;\; x^{2}+3 x-4 \;\; =\;\frac{27}{4}$ $\Rightarrow \;\; 4 x^{2}+12 x-43 \;\; =0$ $x \;\; x \;\; =\;\frac{-12 \pm \sqrt{144+688}}{8}$ $\therefore \;\; \;\; =\;\frac{-12 \pm \sqrt{832}}{8}$ $\;\; =\;\frac{-12 \pm 28.8}{8}$ $x \;\; =\;\frac{-3 \pm 7.2}{2}$ $x \;\; =\;\frac{-3+7.2}{2}, \;\frac{-3-7.2}{2}$ Both the values are less than $5 .$ $\therefore$ No solution from here. Case II $x < 5$ $\Rightarrow \;\; \;\; (x+1)^{2}-(x-5)=\;\frac{27}{4}$ $\Rightarrow \;\; x^{2}+x+6-\;\frac{27}{4}=0$ $\Rightarrow \;\; 4 x^{2}+4 x-3=0$ $\Rightarrow \;\; x=\;\frac{-4 \pm \sqrt{16+48}}{8}$ $\Rightarrow \;\; x=\;\frac{-12}{8}, \;\frac{4}{8}, \;\text{ both are less than }\; 5 .$ $\therefore$ These values must be the solution. Hence, here 2 real roots are possible.

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