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Physics30 questions

Q1JEE Main 2021MCQ4MUnits and Measurements

One main scale division of a vernier callipers is $a {cm}$ and $n$ th division of the vernier scale coincide with ( $n-1)$ th division of the main scal

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📖 Explanation

According to the question, One division of main scale reading = a cm nth vernier scale division $=(n-1)$ th main scale division $\therefore$ One division of vernier scale reading $=\;\frac{(n-1) \times a}{n}$ ...(i) We know that, Least count $(L C)=$ [1 main scale division $-1$ vernier scale division] ${cm}$ $=a-\;\frac{(n-1) a}{n}$ [using Eq. (i)] $=\;\frac{a(n-n+1)}{n}=\;\frac{a}{n} {cm}=\;\frac{a}{n} \times 10 {mm}$ $\Rightarrow \;\; L C=\;\frac{10 a}{n} {mm}$

Q2JEE Main 2021MCQ4MElectrostatic Potential and Capacitance

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\;\frac{3}{4} d$ , where $d$ is the separation between the plates of parallel plate capacitor. The new capacitance $(C^{'})$ in terms of original capacitance $(C_{0})$ is given by the following relation

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📖 Explanation

As per question, the figure can be shown for a parallel plate capacitorsInitially, capacitance, $C_{0}=\;\frac{\epsilon_{0} A}{d}$ ...(i) where, $\epsilon_{0}=$ permittivity of free space, $A=$ area of plates and $d=$ distance between plates. In presence of a dielectric medium between the plates, the capacitance will be $C=\;\frac{\epsilon_{0} K A}{d}$ ...(ii) Also, from the figure capacitors $C_{1}$ and $C_{2}$ are in series. $\therefore$ Equivalent capacitance is given by $\;{1}/{C_{ {eq}}}=\;\frac{1}{C_{1}}+\;\frac{1}{C_{2}}$ $\Rightarrow \;\; \;\frac{1}{C_{e q}}=(\;\frac{3 d {/} 4}{\epsilon_{0} K A})+(\;\frac{d {/} 4}{\epsilon_{0} A})$ [using Eqs. (i) and (ii)] $\Rightarrow \;\; \;{1}/{C_{ {eq}}}=\;\frac{d}{4 \epsilon_{0} A}(\;\frac{3+K}{K})$ $\Rightarrow \;\; C_{\;\text{eq }\;}=\;\frac{4 K C_{0}}{(3+K)}$ [using Eq.(i)]

Q3JEE Main 2021MCQ4MLaws of Motion

A block of mass $m$ slides along a floor, while a force of magnitude $F$ is applied to it at an angle $\theta$ as shown in figure. The coefficient of kinetic friction is $\mu _{k}$ . Then, the block's acceleration $a$ is given by ( $g$ is acceleration due to gravity)

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📖 Explanation

The diagram and the required components of force on given block are shown belowFrom the above diagram, $N=m g-F \sin \theta$ ...(i) where, $N=\;\text{ normal force }\;$ and $F \cos \theta-f_{k}=m a$ $\Rightarrow \;F \cos \theta-\mu _{k} N=m a$ ...(ii) where, $f_{k}=$ kinetic friction force. From Eq. (i) and Eq. (ii), we get $F \cos \theta-\mu _{k}(m g-F \sin \theta)=m a$ $\Rightarrow \;\; a=\;\frac{F}{m} \cos \theta-\mu _{k}(g-\;\frac{F}{m} \sin \theta)$ This is the required acceleration of the block.

Q4JEE Main 2021MCQ4MProperties of Matter

The pressure acting on a submarine is $3 \times 10^{5} {Pa}$ at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be (Assume that atmospheric pressure is $1 \times 10^{5} {Pa}$ , density of water is $10^{3} {kg} {m}^{-3}$ , $g=10 {ms}^{-2}$ )

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📖 Explanation

Given, pressure on submarine at a certain depth, $p_{i}=3 \times 10^{5} {Pa}$ Since, we know that at a certain depth or below the surface of a liquid of density $\rho$ , pressure is given as $p_{i}=p_{0}+\rho g h$ $\Rightarrow \;\; 3 \times 10^{5}=p_{0}+\rho g h$ $\Rightarrow \;\; 3 \times 10^{5}=1 \times 10^{5}+\rho g h$ $( \because p_{0}=1 \times 10^{5} {Pa})$ $\Rightarrow \;\; 3 \times 10^{5}-1 \times 10^{5}=\rho g h$ $\Rightarrow \;\; 2 \times 10^{5}=\rho g h$ ....(i) When the depth is doubled, then final pressure will be $p_{f}=p_{0}+\rho g 2 h=1 \times 10^{5}+2(\rho g h)$ $=1 \times 10^{5}+2(2 \times 10^{5})$ [using Eq. (i)] $=1 \times 10^{5}+4 \times 10^{5}$ $\Rightarrow \;\; p_{f}=5 \times 10^{5} {Pa}$ $\therefore$ Percentage increase in pressure $=\;\frac{p_{f}-p_{i}}{p_{i}} \times 100=\;\frac{(5-3) \times 10^{5}}{3 \times 10^{5}} \times 100=\;\frac{200}{3} \%$

Q5JEE Main 2021MCQ4MRay Optics and Optical Instruments

The angle of deviation through a prism is minimum when A. incident ray and emergent ray are symmetric to the prism B. the refracted ray inside the prism becomes parallel to its base C. angle of incidence is equal to that of the angle of emergence D. angle of emergence is double the angle of incidence Choose the correct answer from the options given below.

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Q6JEE Main 2021MCQ4MElectromagnetic Waves

A plane electromagnetic wave of frequency $500 {MHz}$ is travelling in vacuum along $y$ -direction. At a particular point in space and time, $B=8.0 \times 10^{-8} \hat{z} T$ . The value of electric field at this point is (speed of light $=3 \times 10^{8} {ms}^{-1}$ ; $\hat{x}, \hat{y}, \hat{z}$ are unit vectors along $x, y$ and $z$ -direction).

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📖 Explanation

Given, frequency, $f=500 {MHz}=5 \times 10^{8} {Hz}$ $B=8.0 \times 10^{-8} \hat{Z} {T}$ $\therefore$ Magnitude of peak value of magnetic field is given by $B_{0}=8 \times 10^{-8} {T}$ We know that, $\;\frac{E_{0}}{B_{0}}=C$ where, $E_{0}$ is the magnitude of peak value of electric field and $c$ is the speed of electromagnetic wave in air (or vacuum). $\Rightarrow \;\; E_{0}=c B_{0}$ $=3 \times 10^{8} \times 8 \times 10^{-8}=24 \frac{V}{m}$ Since, the direction of propagation of electromagnetic wave is perpendicular to the direction of $E$ and $B$ both. $\therefore$ Direction of propagation is given by $\hat{E} \times \hat{B}$ . As, the wave is travelling in $y$ -direction, and the magnetic field is in z-direction. $\Rightarrow \hat{E} \times \hat{z} \;\; =\hat{y}$ $\hat{E} \;\; =-\hat{x}$ $\therefore$ The value of electric field will be $-24 \hat{x} {V} / {m}$ .

Q7JEE Main 2021MCQ4MGravitation

The maximum and minimum distance of a comet from the Sun are $1.6 \times 10^{12} {m}$ and $8.0 \times 10^{10} {m}$ , respectively. If the speed of the comet at the nearest point is $6 \times 10^{4} {ms}^{-1}$ , then the speed at the farthest point is

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📖 Explanation

Given, maximum distance of comet from the Sun, $r_{1}=1.6 \times 10^{12} {m}$ Minimum distance of comet from the Sun, $r_{2}=8.0 \times 10^{10} {m}$ Speed of the comet at nearest point, $v_{2}=6 \times 10^{4} {ms}^{-1}$ Applying law of conservation of angular momentum, $m v_{1} r_{1}=m v_{2} r_{2}( \because$ mass of comet will remain same $)$ $\Rightarrow \;\; V_{1}=\;\frac{V_{2} r_{2}}{r_{1}}$ $=\;\frac{6 \times 10^{4} \times 8 \times 10^{10}}{1.6 \times 10^{12}}=\;\frac{48 \times 10^{14}}{1.6 \times 10^{12}}$ $=3 \times 10^{3} \frac{m}{s}$

Q8JEE Main 2021MCQ4MMagnetism and Matter

A bar magnet of length $14 {cm}$ is placed in the magnetic meridian with its North pole pointing towards the geographic North pole A neutral point is obtained at a distance of $18 {cm}$ from the centre of the magnet. If $B_{H}=0.4 {G}$ , then the magnetic moment of the magnet is $(1 {G}=10^{-4} {T})$

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📖 Explanation

The given situation can be shown as belowFrom the above figure, it is clear that the neutral point will lie on equitorial plane. $B_{0}=\;\frac{\mu _{0} m}{4 \pi} \;\frac{1}{r^{2}}$ $B_{H}=2 B_{0} \cos \theta$ $\Rightarrow \;\; 0.4 \times 10^{-4}=\;\frac{2 \mu _{0} m}{4 \pi r^{2}} \cdot \;\frac{7 \times 10^{-2}}{r}$ $( \because B_{H}=0.4 {G}=0.4 \times 10^{-4} {T})$ $[ \because \cos \theta=\;\frac{7}{r}$ and $r=(7^{2}+18^{2})^{1 {/} 2}]$ $\Rightarrow \;\; 0.4 \times 10^{-4}=2 \times 10^{-7} \times \;{m \times 7}/{(7^{2}+18^{2})^{3 {/} 2}} \times 10^{4}$ $\Rightarrow \;\; m=\;{4 \times 10^{-2} \times (373)^{3 {/} 2}}/{14}$ ...(i) $\because$ Magnetic moment, $M=m \times 2 l=m \times \;\frac{14}{100}$ ...(ii) From Eqs. (i) & (ii), we get $M \;\; =\;{4 \times 10^{-2} \times (373)^{3{/} 2}}/{14} \times \;\frac{14}{100}$ $\;\; =4 \times 10^{-4} \times 7203.82=2.88 \frac{J}{T}$

Q9JEE Main 2021MCQ4MThermodynamics

The volume $V$ of an enclosure contains a mixture of three gases, $16 {g}$ of oxygen, $28 {g}$ of nitrogen and $44 {g}$ of carbon dioxide at absolute temperature $T$ . Consider $R$ as universal gas constant. The pressure of the mixture of gases is

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📖 Explanation

From ideal gas equation, $p V=(n_{1}+n_{2}+n_{3}) R T$ where, $n_{1}=$ number of moles of oxygen $=\;\frac{16}{32}$ $n_{2}=$ number of moles of nitrogen $=\;\frac{28}{28}$ $n_{3}=$ number of moles of carbon dioxide $=\;\frac{44}{44}$ $\Rightarrow \;\; p V=[\;\frac{16}{32}+\;\frac{28}{28}+\;\frac{44}{44}] R T=[\;\frac{1}{2}+1+1]_{R T}$ $=\;\frac{5}{2} R T \Rightarrow p=\;\frac{5}{2} \;\frac{R T}{V}$ This is the required pressure of the mixture of the gases.

Q10JEE Main 2021MCQ4MThermodynamics

In thermodynamics, heat and work are

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📖 Explanation

According to first law of thermodynamics dQ = dU + dW.......(i) As we know that, work done by the gas depends on the type of process, i.e. path and $d U$ depends on the initial and final states. So, considering Eq. (i), dQ will also be dependent on path. It means that, in thermodynamics, heat and work are path functions.

Q11JEE Main 2021MCQ4MRotational Motion

Four equal masses, $m$ each are placed at the corners of a square of length $(I)$ as shown in the figure

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📖 Explanation

The given situation can be represented as followsIn the above figure, XX′ be the axis which passes through A and is parallel to DB. ∴Moment of inertia of the system $=$ Mass $\times$ (Perpendicular distance from axis $)^{2}$ $I=m(A C)^{2}+m(E D)^{2}+m(F B)^{2}$ $=m(0)^{2}+m(I \sqrt{2})^{2}+m(\;\frac{1}{\sqrt{2}})^{2}+m(\;\frac{1}{\sqrt{2}})^{2}$ $=3 m l^{2}$

Q12JEE Main 2021MCQ4MCurrent Electricity

A conducting wire of length $l$ , area of cross-section $A$ and electric resistivity $\rho$ is connected between the terminals of a battery. A potential difference $V$ is developed between its ends, causing an electric current. If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be

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📖 Explanation

Initially, the resistance of wire is $R_{1}=\rho L {/} A$ In second case, Length, $I^{'}=2 I$ Area, $A^{'}=\;\frac{A}{2}$ $\therefore \;\; R_{2}=\;\frac{\rho l^{'}}{A^{'}}=\;\frac{\rho (2 l)}{(A {/}2)}=\;\frac{4 \rho l}{A}$ According to Ohm's law, $I=\;\frac{V}{R_{2}}$ $\Rightarrow \;\; l=\;\frac{V}{4 \rho l{/}A}=\;\frac{1}{4} \;\frac{V A}{\rho l}$ This is the required value of resultant current.

Q13JEE Main 2021MCQ4MOscillations

Time period of a simple pendulum is $T$ inside a lift, when the lift is stationary. If the lift moves upwards with an acceleration $g {/} 2$ , then the time period of pendulum will be

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📖 Explanation

Time period of a simple pendulum can be given as $T=2 \pi \sqrt{\;\frac{I}{g}}$ ...(i) When the lift moves upwards, then effective acceleration is $\Rightarrow$ $\Rightarrow g_{\;\text{eff }\;}=g+a=g+\;\frac{g}{2}=\;\frac{3 g}{2}$ $\therefore$ New time period, $T_{1}=2 \pi \sqrt{\;{1}/{g_{\;\text{eff }\;}}}=2 \pi \sqrt{\;\frac{2 L}{3 g}}$ $\Rightarrow \;\; T_{1}=\sqrt{\;\frac{2}{3}} T$ [From Eq. (i)]

Q14JEE Main 2021MCQ4MMotion in a Straight Line

The velocity-displacement graph describing the motion of a bicycle is shown in the following figure.The acceleration-displacement graph of the bicycle's motion is best described by

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📖 Explanation

From the given velocity-displacement graph, for $0 \le x \le 200$ , the graph is a straight line i.e., varying linearly. As, the equation of straight line is $y=m x+c$ Here, it can be given as $v=m x+c$ where, $m=$ slope $=\;\frac{(50-10)}{200-0}=\;\frac{1}{5}$ and $c=10$ $\Rightarrow \;\; v=\;\frac{x}{5}+10$ As, acceleration, $a=\;\frac{d v}{d t}=\;\frac{v d v}{d x}$ $a=(\;\frac{x}{5}+10) \;\frac{d}{d x}(\;\frac{x}{5}+10)$ [using Eq. (i)] $=(\;\frac{x}{5}+10)(\;\frac{1}{5})=2+\;\frac{x}{25}$ For $x=0, a=2+\;\frac{0}{25}=2 {ms}^{-2}$ and for $x=200, a=2+\;\frac{200}{25}=10 {ms}^{-2}$ Therefore, the graph will have straight line till $x=200$ and for $x>200 ; v=$ constant and $a=0$ So, from the graphs given in the options, the above conditions are satisfied in option (a) only.

Q15JEE Main 2021MCQ4MElectromagnetic Waves

A $25 {m}$ long antenna is mounted on an antenna tower. The height of the antenna tower is $75 {m}$ . The wavelength (in ${m}$ ) of the signal transmitted by this antenna would be

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📖 Explanation

Since, height of antenna = Wavelength of the signal $\;\frac{\;\text{ transmitted by antenna }\;(\lambda)}{4}$ $\Rightarrow 25 \;\; =\;\frac{\lambda}{4}$ $\Rightarrow \lambda \;\; =25 \times 4$ $\Rightarrow \lambda \;\; =100 {m}$

Q16JEE Main 2021MCQ4MElectromagnetic Waves

For an electromagnetic wave travelling in free space, the relation between average energy densities due to electric $(U_{e})$ and magnetic $(U_{m})$ fields is

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📖 Explanation

We know that, $U_{e}=\;\frac{1}{2} \epsilon_{0} E^{2}$ ....(1) where, $U_{e}=$ average energy density due to electric field, $\epsilon_{0}=$ electrical permittivity of free space and $E=$ electric field. and $U_{m}=\;\frac{B^{2}}{2 \mu _{0}}$ .....(2) where, $U_{m}=$ average energy density due to magnetic field, $\mu _{0}=$ magnetic permeability of free space and $B=$ magnetic field. Also, $E=C B$ $( \because C=\;{1}/{\sqrt{\mu _{0} \epsilon_{0}}})$ $\Rightarrow \;\; E=\;{B}/{\sqrt{\mu _{0} \epsilon_{0}}}$ Put this value in Eq. (1), we get $U_{e}=\;\frac{1}{2} \epsilon_{0} \times \;\frac{1}{\mu _{0} \epsilon_{0}} B^{2}$ $\Rightarrow \;U_{e}=\;\frac{B^{2}}{2 \mu _{0}}$ $\Rightarrow \;U_{e}=U_{m}$ [using Eq. (2)]

Q17JEE Main 2021MCQ4MAlternating Current

An $R-C$ circuit as shown in the figure is driven by an AC source generating a square wav The output wave pattern monitored by CRO would look close to

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📖 Explanation

When square wave is applied at the input, then For charging, the capacitor $Q_{1}=Q(1-e^{\;\frac{-t}{R C}})$ Similarly, for discharging the capacitor, $Q_{2}=Q_{\max }(e^{\;\frac{-t}{R C}})$ In this manner, charging and discharging exponentially with time will keep on happening alternatively. Therefore, the output wave pattern monitored by CRO would look close to

Q18JEE Main 2021MCQ4MDual Nature of Matter and Radiation

The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation

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📖 Explanation

The stopping potential $V_{0}$ is related to the frequency of incident radiations by following relation ${eV}_{0}=h v-\varphi \;\; (\varphi =$ work-function $)$ The stopping potential in the context of photoelectric effect depends only on the frequency $v$ of incident electromagnetic radiation.

Q19JEE Main 2021MCQ4MCircular Motion

A block of $200 {g}$ mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius $20 {cm}$ . If the block takes 40 s to complete one round, the normal force by the side walls of the groove is

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📖 Explanation

The normal force by the side walls of the groove will be equal to the centripetal force acting on it. i.e. $N=\;\frac{m v^{2}}{r}$ .....(i) where, $r \;\; =20 {cm}=0.2 {m}$ $m \;\; =200 {g}=200 \times 10^{-3} {kg}$ and $v=r \omega=\;\frac{2 \pi r}{T}=\;\frac{2 \pi \times 0.2}{40} \frac{m}{s}$ Substituting the given values in Eq. (i), we get $N \;\; =\;\frac{(200 \times 10^{-3}) \times (\frac{2 \pi \times 0.2}{40})^{2}}{0.2}$ $\;\; \approx 9.859 \times 10^{-4} {N}$

Q20JEE Main 2021MCQ4MElectromagnetic Induction

A conducting bar of length $L$ is free to slide on two parallel conducting rails as shown in the figure Two resistors $R_{1}$ and $R_{2}$ are connected across the ends of the rails. There is a uniform magnetic field B pointing into the page. An external agent pulls the bar to the left at a constant speed $v$ . The correct statement about the directions of induced currents $I_{1}$ and $I_{2}$ flowing through $R_{1}$ and $R_{2}$ respectively is

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📖 Explanation

According to Lenz's law, "An induced current flows in a direction in such a way that it always opposes the cause that induced it". Considering this, it can be concluded that $I_{1}$ is in the clockwise direction and $I_{2}$ is in the anti-clockwise direction and the effective circuit with the directions of $I_{1}$ and $I_{2}$ can be shown as follows

Q21JEE Main 2021NAT4MCurrent Electricity

In the figure given, the electric current flowing through the $5 {k} Ω$ resistor is $x {mA}$ . The value of $x$ to the nearest Integer is ...3...

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📖 Explanation

According to the figure given in question, all $3 Ω$ resistances are in parallel combination. So, their equivalent resistance is $\;{1}/{R_{\;\text{parallel }\;}}=\;\frac{1}{3}+\;\frac{1}{3}+\;\frac{1}{3}$ $\;{1}/{R_{\;\text{parallel }\;}}=\;\frac{3}{3}$ $\Rightarrow \;\; R_{\;\text{parallel }\;}=1 {k} Ω$ $5 \mathrm{k} \Omega$ and $1 \mathrm{k} \Omega$ resistance are in series to the equivalent of all $3 \Omega$ resistances. $\therefore \;\; R_{\;\text{net }\;}=5+1+R_{\;\text{parallel }\;}$ ...(i) $\Rightarrow \;\; R_{\;\text{net }\;}=(5+1+1) {k} Ω=7 {k} Ω=7 \times 10^{3} Ω$ $\therefore$ The value of electric current flowing through $5 {k} Ω$ resistor will be $I=\;{V}/{R_{\;\text{net }\;}}=\;\frac{21}{7 \times 10^{3}}=3 \times 10^{-3} {A}$ $=3 {mA}$ Comparing with the given value in the question i.e., $x {mA}$ , the value of $x=3$ .

Q22JEE Main 2021NAT4MWave Optics

A fringe width of $6 {mm}$ was produced for two slits separated by $1 {mm}$ apart. The screen is placed $10 {m}$ away. The wavelength of light used is $x {nm}$ . The value of $x$ to the nearest integer is .........

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📖 Explanation

Given, Slit width, $d=1 {mm}=10^{-3} {m}$ Fringe width, $\beta=6 {mm}=6 \times 10^{-3} {m}$ Distance between the screen and the slit, $D=10 {m}$ Wavelength of light used $\lambda=$ ? As, $\beta=\;\frac{D \lambda}{d}$ $\Rightarrow \;\; \lambda=\;\frac{\beta d}{D}$ $=\;\frac{6 \times 10^{-3} \times 10^{-3}}{10}=6 \times 10^{-6} \times 10^{-1}$ $=6 \times 10^{-7} {m}=600 \times 10^{-9} {m}$ $=600 {nm}$

Q23JEE Main 2021NAT4MRotational Motion

Consider a $20 {kg}$ uniform circular disc of radius $0.2 {m}$ . It is pin supported at its centre and is at rest initially. The disc is acted upon by a constant force $F=20 {N}$ through a massless string wrapped around its periphery as shown in the figure.Suppose the disc makes $n$ number of revolutions to attain an angular speed of $50 {rad} {s}^{-1}$ . The value of $n$ to the nearest integer, is .......... . (Given, in one complete revolution, the disc

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📖 Explanation

Given, $M=20 {kg}, R=0.2 {m}$ , $F=20 {N}, \omega=50 {rad} {} {s}^{-1}$ $\theta=6.28 {rad}$ (for one revolution). We know that, Torque $=$ Moment of inertia $\times$ Angular acceleration $\Rightarrow \;\; \tau =I \times \alpha$ $\Rightarrow \;\; \alpha=\;\frac{\tau }{l}=\;{F \times R}/{\frac{M R^{2}}{2}}=\;\frac{2 F}{M R} \;\; [ \because \tau =F \times R$ and $I_{ {disc}}=\;\frac{1}{2} M R^{2}]$ $\Rightarrow \;\; \alpha=\;\frac{2 \times 20}{20 \times (0.2)}=10 {rad} / {s}^{2}$ In angular terms, third equation of motion, $\omega^{2}=\omega_{0}^{2}+2 \alpha \Delta \theta$ $\Rightarrow \;(50)^{2}=(0)^{2}+2 \times 10 \times \Delta \theta$ $\Rightarrow \;\Delta \theta=\;\frac{2500}{20}$ $\Rightarrow \;\Delta \theta=125 {rad}$ $\therefore$ Number of revolution $=\;\frac{\Delta \theta}{\theta}=\;\frac{125}{6.28} \approx 20$

Q24JEE Main 2021NAT4MAtoms and Nuclei

The first three spectral lines of H-atom in the Balmer series are given $\lambda_{1}, \lambda_{2}, \lambda_{3}$ considering the Bohr atomic model, the wavelengths of first and third spectral lines $(\;\frac{\lambda_{1}}{\lambda_{3}})$ are related by a factor of approximately $x \times 10^{-1}$ . The value of $x$ to the nearest integer, is ..........

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📖 Explanation

Since, we know that, $\;\frac{1}{\lambda}=R z^{2}(\;\frac{1}{n_{1}^{2}}-\;\frac{1}{n_{2}^{2}})$ ......(i) where, $\lambda=$ wavelength of light emitted. $R=$ Rydberg's constant, $Z=$ atomic number, $n_{1}=$ principal quantum number of lower energy level and $\;\; n_{2}=$ principal quantum number of higher energy level. Therefore, for $1 {st}$ spectral line of Balmer series, $n_{1}=2$ and $n_{2}=3$ $\;\frac{1}{\lambda_{1}}=R z^{2}(\;\frac{1}{2^{2}}-\;\frac{1}{3^{2}})$ $\Rightarrow \;\; \;\frac{1}{\lambda_{1}}=R z^{2}(\;\frac{5}{36})$ ...(ii) Similarly, for 3rd spectral line, $n_{1}=2$ and $n_{2}=5$ $\;\frac{1}{\lambda_{3}}=R z^{2}(\;\frac{1}{2^{2}}-\;\frac{1}{5^{2}})$ $\Rightarrow \;\; \;\frac{1}{\lambda_{3}}=R z^{2}(\;\frac{21}{100})$ ......(iii) Now, dividing Eq. (iii) by Eq. (ii), we get $\;{\;\frac{1}{\lambda_{3}}}/{\frac{1}{\lambda_{1}}}=\;\frac{R z^{2}(\frac{21}{100})}{R z^{2}(\;\frac{5}{36})} \Rightarrow \;\frac{\lambda_{1}}{\lambda_{3}}=\;\frac{21}{100} \times \;\frac{36}{5}$ $\Rightarrow \;\; \;\frac{\lambda_{1}}{\lambda_{3}}=1.512=15.12 \times 10^{-1}$ Comparing with the given value in the question i.e., $x \times 10^{-1}$ , the value of $x=15$ .

Q25JEE Main 2021NAT4MSemiconductor Electronics

The value of power dissipated across the Zener diode $(V_{z}=15 {V})$ connected in the circuit as shown in the figure is $x \times 10^{-1} {W}$ . The value of $x$ , to the nearest integer, is .........

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📖 Explanation

Consider the figure with current in the directions shown below. Voltage across $R_{s}, V_{s}=22-15=7 {V}$ Current through $R_{s}, I=\;\frac{V_{s}}{R_{S}}$ $\Rightarrow \;\; I=\;\frac{7}{35}=\;\frac{1}{5} {A}$ Current through $90 Ω$ resistance, $I_{1}=\;\frac{V_{z}}{R_{L}}=\;\frac{15}{90}=\;\frac{1}{6} {A}$ $\therefore$ Current through Zener diode, $I_{2}=\;\frac{1}{5}-\;\frac{1}{6}=\;\frac{1}{30} {A}$ $\therefore$ Power through Zener diode, $P=V_{2} I_{2}$ $=15 \times \;\frac{1}{30}=\;\frac{1}{2}=0.5 {W}=5 \times 10^{-1} {W}$ Comparing with the given value in question i.e., $x \times 10^{-1}$ , the value of $x=5$ .

Q26JEE Main 2021NAT4MAlternating Current

A sinusoidal voltage of peak value $250 {V}$ is applied to a series $L-C-R$ circuit, in which $R=8 Ω, L=24 {mH}$ and $C=60 \mu {F}$ . The value of power dissipated at resonant condition is $x {kW}$ . The value of $x$ to the nearest integer is

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📖 Explanation

Given, $V_{0}=250 {V}, R=8 Ω, L=24 {mH}$ and $C=60 \mu {F}$ We know that, at resonance power, $P=\;{V_{\;\text{rms }\;}^{2}}/{R}$ $\Rightarrow \;\; P=\;\frac{(250 {/} \sqrt{2})^{2}}{8} \;\; [ \because V_{\;\text{rms }\;}=\;\frac{V_{0}}{\sqrt{2}}=\;\frac{250}{\sqrt{2}} {V}]$ $=(250)^{2} / 16$ $=\;\frac{62500}{16}=3906.25 {W} \approx 4 {kW}$ $x=4$ . Comparing with the given value in the question i.e., $x {kW}$ , the value of $x=4$ .

Q27JEE Main 2021NAT4MSemiconductor Electronics

In the logic circuit shown in the figure, if input $A$ and $B$ are 0 to 1 respectively, the output at $Y$ would be $x$ . The value of $x$ is ...... .

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📖 Explanation

Consider the figure given in the question and solve it using Boolean identities.Now, put the value of $A$ and $B$ in the output, we get $Y=(\overline{0+1})(0+1)$ $Y=\overline{1} \cdot 1=0$ Alternate method We can directly put the given values in the logic circuit given in the question and can find its output. Let us consider the given logic circuit,The output is $Y=0.1=0$ $\therefore$ The value of $x$ is 0 .

Q28JEE Main 2021NAT4MUnits and Measurements

The resistance $R=\;\frac{V}{I}$ , where $V=(50 \pm 2) {V}$ and $I=(20 \pm 0.2)$ A. The percentage error in $R$ is $x \%$ . The value of $x$ to the nearest integer is..........

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📖 Explanation

Given, $R=\;\frac{V}{I}$ ....(i) where, $&V=(50 \pm 2) {V}$ $I=(20 \pm 0.2) {A}$ From Eq. (i) $\;\frac{\Delta R}{R} \times 100=\;\frac{\Delta V}{V} \times 100+\;\frac{\Delta l}{l} \times 100$ $\%$ error in $R=[\;\frac{2}{50} \times 100+\;\frac{0.2}{20} \times 100] \%$ $=[2 \times 2+0.2 \times 5] \%$ $=5 \%$

Q29JEE Main 2021NAT4MLaws of Motion

Consider a frame that is made up of two thin massless rods $A B$ and $A C$ as shown in the figure. $A$ vertical force $P$ of magnitude $100 {N}$ is applied at point $A$ of the frame.

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📖 Explanation

If the force $P$ of magnitude $100 {N}$ is resolved parallel to the arms $A B$ and $A C$ of the frame, the above figure will be represented as followsComponent of force along $A C=100 \cos 35^{\circ} {N}$ $=100 \times 0.819 {N}=81.9 {N} \approx 82 {N}$ This is the required magnitude of the resolved component along the arm $A C$ . Compare with given in question, $x=82$

Q30JEE Main 2021NAT4MCentre of Mass and Collision

A ball of mass $10 {kg}$ moving with a velocity $10 \sqrt{3} {ms}^{-1}$ along $X$ -axis, hits another ball of mass $20 {kg}$ , which is at rest. After collision, the first ball comes to rest and the second one disintegrates into two equal pieces. One of the pieces starts moving along $Y$ -axis at a speed of $10 {m} / {s}$ . The second piece starts moving at a speed of $20 {m} / {s}$ at an angle $\theta$ (degree) with respect to the $X$ -axis. The configuration of pieces after collision is shown in the figure. The value of $\theta$ to the nearest integer is

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📖 Explanation

We can represent the given situation in figure asIt means linear momentum is conserved along $X$ -axis. According to the law of conservation of linear momentum, $p_{i}=p_{f}$ m_{A} u_{A}+m_{B} u_{B}=m_{A} v_{A}+(m_{B} v_{B} \cos theta) $=m_{A} v_{A}+m_{B_{1}} v_{B_{1}} \cos 90^{\circ}+m_{B_{2}} v_{B_{2}} \cos \theta$ $\Rightarrow 10 \times 10 \sqrt{3}+20 \times 0=10 \times 0+10 \times 10 \times 0+10 \times 20 \cos \theta$ $\Rightarrow \;\; 10 \times 10 \sqrt{3}=200 \cos \theta$ where, $\cos \theta$ being the horizontal component i.e., along $X$ -axis $\Rightarrow \;\; \cos \theta=\sqrt{3} {/}2$ $\Rightarrow \;\; \theta=30^{\circ} \;\; [ \because \cos 30^{\circ}=\sqrt{3} {/} 2]$

Chemistry30 questions

Q31JEE Main 2021MCQ4Md and f Block Elements

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) Size of ${Bk}^{3+}$ ion is less than ${Np}^{3+}$ ion. Reason (R) The above is a consequence of the lanthanoid contraction. In the light of the above statements, choose the correct answer from the options given below

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📖 Explanation

${ }_{97} {Bk}$ (Berkelium) and ${ }_{93} {Np}$ (Neptunium) are present in actinide series. Due to actinoid contraction, there is gradual decrease in size of ${M}^{3+}$ ions across the period due to poor shielding of $5 {f}$ -electrons. So, ${Bk}^{3+}$ is smaller than that of ${Np}^{3+}$ due to actinoid contraction. So, $A$ is true and $R$ is false.

Q32JEE Main 2021MCQ4MBiomolecules

Which among the following pair of vitamins is stored in our body relatively for longer duration?

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Q33JEE Main 2021MCQ4MClassification of Elements

Given below are two statements. Statement I Both ${CaCl}_{2} \cdot 6 {H}_{2} {O}$ and ${MgCl}_{2} \cdot 8 {H}_{2} {O}$ undergo dehydration on heating. Statement II BeO is amphoteric, whereas the oxides of other elements in the same group are acidic. In the light of the above statements, choose the correct answer from the options given below.

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📖 Explanation

The dehydration of hydrated chlorides, bromides and iodides of ${Ca}, {Sr}, {Ba}$ is possible but hydrated halides of Be and ${Mg}$ on heating suffer hydrolysis due to their small size or high charge density. ${CaCl}_{2} \cdot 6 {H}_{2} {O} {\longrightarrow }^{\Delta } {{CaCl}_{2}}_{\text{Anhydrous}}+6 {H}_{2} {O}$ ${MgCl}_{2} \cdot 8 {H}_{2} {O} {\longrightarrow }^{\Delta } {MgO}+7 {H}_{2} {O}+2 {HCl}$ ${MgCl}_{2}$ can be made anhydrous by heating in presence of dry ${HCl}$ . ${MgCl}_{2} \cdot 6 {H}_{2} {O} {\to }^{\Delta \;\text{ (Calcination) }\;}_{\;\text{ Dry HCl(g) }\;} {MgCl}_{2}+8 {H}_{2} {O}$ Hence, on heating ${CaCl}_{2} \cdot 6 {H}_{2} {O}$ undergoes dehydration but ${MgCl}_{2} \cdot 8 {H}_{2} {O}$ does not. Metal oxides are generally basic. In alkaline earth metal BeO is amphoteric and shows anomalous behaviour while other oxides are basic. So, both statement I and statement II are false.

Q34JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

The product $P$ in the above reaction is

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Q35JEE Main 2021MCQ4Mp Block Elements

Match List -I with List II.List-I (Industrial process)List-II (Application)A. Haber's process(i) ${HNO}_{3}$ synthesisB. Ostwald's process(ii) Aluminium extractionC. Contact process(iii) ${NH}_{3}$ synthesisD. Hall-Heroult process(iv) ${H}_{2} {SO}_{4}$ synthesisChoose the correct answer from the options given below:

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📖 Explanation

A. Haber's process $- {NH}_{3}$ synthesis ${N}_{2}( {g})+3 {H}_{2}( {g}) \longrightarrow 2 {NH}_{3}(g)$ B. Ostwald's process $- {HNO}_{3}$ synthesis $4 {NH}_{3}(g)+5 {O}_{2}(g) \longrightarrow 4 {NO}(g)+6 {H}_{2} {O}(g)$ $2 {NO}( {g})+ {O}_{2}( {g}) \longrightarrow 2 {NO}_{2}( {g})$ $4 {NO}_{2}(g)+2 {H}_{2} {O}(I)+ {O}_{2}(g) \longrightarrow 4 {HNO}_{3}( {aq})$ C. Contact process $- {H}_{2} {SO}_{4}$ synthesis $2 {SO}_{2}(g)+ {O}_{2}(g) \longrightarrow 2 {SO}_{3}(g)$ ${SO}_{3}(g)+ {H}_{2} {SO}_{4}( {aq}) \longrightarrow {{H}_{2} {S}_{2} {O}_{7}(l)}_{\;\text{ Oleum }$ ${H}_{2} {S}_{2} {O}_{7}(I)+ {H}_{2} {O}(I) \longrightarrow 2 {H}_{2} {SO}_{4}( {aq})$ D. Hall-Heroult process- Aluminium extraction $2 {Al}_{2} {O}_{3}+3 {C} \longrightarrow 4 {Al}+3 {CO}_{2}$ Correct match is, ${A} \to$ (iii), ${B} \to ( {i}), {C} \to ( {iv}), {D} \to$ (ii)

Q36JEE Main 2021MCQ4MHydrocarbons

Among the following, the aromatic compounds areChoose the correct answer from the following options.

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📖 Explanation

(b) For aromatic compounds following conditions should be satisfied (i) planarity (ii) complete delocalisation of $\pi$ -electrons in the ring. (iii) presence of $(4 n+2) \pi$ -electrons in the ring, where $n=0,1,2, . . .$ . This is called Huckel rule. $\;\text{ e.g. }\; n \;\; =0,2 \pi {e}^{-}$ $n \;\; =1,6 \pi {e}^{-}$ $n \;\; =2,10 \pi {e}^{-}$

Q37JEE Main 2021MCQ4MAmines

In the above chemical reaction, intermediate $X$ and reagent / condition $A$ are

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📖 Explanation

Aniline reacts with $\mathrm{NaNO}_{2}, \mathrm{HCl}$ at $273-278 \mathrm{~K}$ to give benzene diazonium salt $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \overline{\mathrm{Cl}}=X\right)$ which on hydrolysis $\left(\mathrm{H}_{2} \mathrm{O} / \Delta\right)$ gives phenol. Complete reaction is as follows

Q38JEE Main 2021MCQ4Md and f Block Elements

Given below are two statements. Statement I The $E^{\circ}$ value of $C {C}^\frac{4+}{Ce}^{3+}$ is $+1.74 {V}$ . Statement II Ce is more stable in ${Ce}^{4+}$ state than ${Ce}^{3+}$ state. In the light of the above statements, choose the most appropriate answer from the options given below.

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📖 Explanation

$E_{ {Ce}^{4+} / {Ce}^{+3}}^{\circ}=1.74$ (data given) ${Ce}^{4+}$ can be reduced to ${Ce}^{3+}$ and can oxidise water. $C {e}^{4+}+ {e}^{-} \longrightarrow {Ce}^{3+}$ ${Ce}^{4+}$ is having noble gas configuration [Xe] $4 f^{0}$ but it is strong oxidant and gets converted to ${Ce}^{3+}$ . Statement I is correct but statement II is incorrect.

Q39JEE Main 2021MCQ4MBiomolecules

The functions of antihistamine are

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Q40JEE Main 2021MCQ4MHydrocarbons

Which of the following is Lindlar catalyst?

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📖 Explanation

Lindlar catalyst $( {H}_{2} \/ {Pd}, {CaCO}_{3})$ is poisoned palladium catalyst. It is known as also partially deactivated palladised charcoal. It is composed of powdered calcium carbonate coated with palladium and poisoned with quinoline. It is used for reduction of alkyne to alkene and acid chloride to aldehyde. Stereochemistry involved is syn-addition. ${H}_{3} {C}- {C} \equiv {C}-2 {CH}_{3}{\to }^{ {H}_\frac{2}{Pd}}_{ {CaCO}_{3}}$

Q41JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

The product $A$ and $B$ formed in above reactions are

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📖 Explanation

Alcohols undergo dehydration (removal of water molecule) in the presence of protic acids (conc. ${H}_{2} {SO}_{4}, {H}_{3} {PO}_{4}$ ) at high temperature to give Saytzeff alkene (more alkylated) as major product.Bulky bases like tert-butoxide, due to steric hindrance, a bulky base abstracts a less hindered proton, that leads to formation of the less substituted product as, the major one, called the Hofmann product.

Q42JEE Main 2021MCQ4Mp Block Elements

Given below are two statements. Statement I ${H}_{2} {O}_{2}$ can act as both oxidising and reducing agent in basic medium. Statement II In the hydrogen economy, the energy is transmitted in the form of dihydrogen. In the light of the above statements, choose the correct answer from the options given below:

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📖 Explanation

${H}_{2} {O}_{2}$ can act as both oxidising and reducing agent in basic medium as in hydrogen peroxide oxygen is present in intermediate-1 oxidation state. So, it can get oxidised to ${O}_{2}$ and reduced to water. ${{H}_{2} {O}_{2}}_{-1} \longrightarrow { {H}_{2} {O}}_{-2}$ (reduction) Oxidising agent ${{H}_{2} {{O}_{2}}_{-1} \longrightarrow { {O}}_{0}$ (reduction) Reducing agent In the hydrogen economy, the energy is transmitted in the form of dihydrogen and not as electric power. So, both statement I and statement II are true.

Q43JEE Main 2021MCQ4MRedox Reactions

The type of pollution that gets increased during the day time and in the presence of ${O}_{3}$ is

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📖 Explanation

Oxidising smog (photochemical smog) gets increased during the day time and in the presence of ${O}_{3}$ . It occurs in warm, dry and sunny climate. When fossil fuels are burnt, hydrocarbons (unburnt fuels) and nitric oxide (NO) are produced. Ozone reacts rapidly with the ${NO}$ to generate ${NO}_{2}$ . ${NO}(g)+ {O}_{3}(g) \longrightarrow {NO}_{2}(g)+ {O}_{2}(g)$ Ozone is a toxic gas and both ${NO}_{2}$ and ${O}_{3}$ are strong oxidising agents and can react with the unburnt hydrocarbons in the polluted air to produce chemicals such as formaldehyde, acrolein and peroxyacetyl nitrate (PAN).

Q44JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Assertion (A) Enol form of acetone $[ {CH}_{3} {COCH}_{3}]$ exists in $< 0.1 \%$ quantity. However, the enol form of acetyl acetone $[ {CH}_{3} {COCH}_{2} {OCCH}_{3}]$ exists in approximately $15 \%$ quantity. Reason (R) Enol form of acetyl acetone is stabilised by intramolecular hydrogen bonding, which is not possible in enol form of acetone. Choose the correct statement.

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📖 Explanation

Enol form of acetone $[ {CH}_{3} {COCH}_{3}]$ exists in $< 0.1 \%$ quantity as monocarbonyl are more stable in keto form due to high bond energy. $CH_3-C^{||^{O}}_{\text{Keto}}-CH_3{\to }^{\text{Tautomerisation}}CH_3-C^{|^{\text{OH}}}_{\text{Enol}}=CH_2$ The enol form of acetyl acetone $[ {CH}_{3} {COCH}_{2} {OCCH}_{3}]$ exists in approximately $15 \%$ quantity as it is stabilised by intramolecular hydrogen bonding, which is not possible in enol form of acetone.So, both $A$ and $R$ are true and $R$ is the correct explanation of A.

Q45JEE Main 2021MCQ4MAmines

Which of the following reaction does not involve Hoffmann bromamide degradation?

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📖 Explanation

Methyl ketone gives haloform reaction to form carboxylic acid salt which on heating with ${NH}_{3}$ forms amide. Amide is reduced to amine by ${LiAlH}_{4}$ .(a) Benzamide undergoes Hofmann bromamide degradation when treated with bromine and aqueous sodium hydroxide, to produces benzylamine.(b) Benzene nitrile undergoes alkaline hydrolysis to form acidamide which further give Hofmann bromamide degradation.(d) Acid chloride undergoes nucleophilic substitution with ${NH}_{3}$ to form acid amide. Acid amide undergoes Hoffmann bromamide degradation.

Q46JEE Main 2021MCQ4Md and f Block Elements

The process that involves the removal of sulphur from the ores is

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📖 Explanation

The process that involves the removal of sulphur from the ores is called roasting. In roasting, the ore is heated in a regular supply of air in a furnace at a temperature below the melting point of the metal. Some of the reactions involving sulphide ores are $2 {ZnS}+3 {O}_{2} \longrightarrow 2 {ZnO}+2 {SO}_{2}$ $2 {PbS}+3 {O}_{2} \longrightarrow 2 {PbO}+2 {SO}_{2}$ $2 {Cu}_{2} {S}+3 {O}_{2} \longrightarrow 2 {Cu}_{2} {O}+2 {SO}_{2}$

Q47JEE Main 2021MCQ4Mp Block Elements

Match List-I with List-II. List-I (Name of oxo acid)List-II (Oxidationstate of P)A. Hypophosphorus acid(i) $+5$ B. Orthophosphoric acid(ii) $+4$ C. Hypophosphoric acid(iii) $+3$ D. Orthophosphorus acid(iv) $+2$ (v) $+1$ Choose the correct answer from the options given below:

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📖 Explanation

Following are the structure and oxidation state of given acids. (A) ${H}_{3} {PO}_{2} \Rightarrow 3+x-4=0$ $x=+1$ (B) ${H}_{3} {PO}_{4} \Rightarrow +3+x-8 \;\; =0$ $x \;\; =+5$ (C) ${H}_{4} {P}_{2} {O}_{6} \Rightarrow +4+2 x-12 \;\; =0$ $\Rightarrow \;x=+4$ (D) ${H}_{3} {PO}_{3}=+3+x-6=0$ $\Rightarrow \;\; x=+3$ So, the correct match is A - v, B - i, C - ii, D - iii.

Q48JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) The ${H}- {O}- {H}$ bond angle in water molecule is $104.5^{\circ}$ . Reason (R) The lone pair - lone pair repulsion of electrons is higher than the bond pair - bond pair repulsion.

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📖 Explanation

Hybridisation of ' ${O}^{'}$ in ${H}_{2} {O}$ is $s p^{3}$ as it is having two lone pair and two sigma bond. Bond angle of $s p^{3}$ hybridised compound should be $109.5^{\circ}$ . But in case of water, it is only $104.5^{\circ}$ due to repulsion between the lone pairs. Lone pair-lone pair repulsion is higher than bond pair-bond pair repulsion. (VSEPR Theory)So, both $A$ and $R$ are true and $R$ is the correct explanation of $A$ .

Q49JEE Main 2021MCQ4MPractical Organic Chemistry

In chromatography technique, the purification of compound is independent of

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Q50JEE Main 2021MCQ4Mp Block Elements

A group 15 element, which is a metal and forms a hydride with strongest reducing power among group 15 hydrides. The element is

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📖 Explanation

${BiH}_{3}$ is the strongest reducing agent and have strongest reducing power. The reducing character of the group 15 element hydrides increases down the group. Ammonia is only a mild reducing agent while ${BiH}_{3}$ is the strongest reducing agent amongst all the hydrides. This is due to the decrease in bond energy down the group.So, correct answer is Bi (bismuth).

Q51JEE Main 2021NAT4MChemical Equilibrium

For the reaction, $A(g) ⇌ B(g)$ at $495 {K}$ , $\Delta , G^{\circ}=-9.478 {kJ} {mol}^{-1}$ . If we start the reaction in a closed container at $495 {K}$ with 22 millimoles of $A$ , the amount of $B$ is the equilibrium mixture is ......... millimoles (Round off to the nearest integer). $[R=8.314 {J} {mol}^{-1} {K}^{-1}, \ln 10=2.303]$

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📖 Explanation

$A(g) ⇌ B(g)$ Given, $T=495 {K}, \Delta _{r} C^{\circ}=-9.478 {kJ} / {mol}$ We know, $\Delta G^{\circ}=-2.303 R T \log K$ $\because \;\; K=\;\frac{[B]}{[A]}$ where, $K=$ equilibrium constant Now, $\log K=\;\frac{-\Delta G^{\circ}}{2.303 R T}$ $\log K=\;\frac{9.478 \times 1000}{2.303 \times 8.314 \times 495}$ $\therefore \;\; \log K=1$ $\Rightarrow \;\; K=10$ $\Rightarrow \;\; \;\frac{[B]}{[A]}=\;\frac{n_{B}}{n_{A}}=10$ Now, $\text{table} ,A(g), ⇌ ,B(g);t=0, 22,,0;t=t,22-x,,x$ $K=\;\frac{[B]}{[A]}=\;\frac{x}{22-x}=10$ So, $\;\; x=20$ Milimoles of $B=20$

Q52JEE Main 2021NAT4MOrganic Chemistry - Some Basic Principles

Complete combustion of $750 {g}$ of an organic compound provides $420 {g}$ of ${CO}_{2}$ and $210 {g}$ of ${H}_{2} {O}$ . The percentage composition of carbon and hydrogen in organic compound is $15.3$ and ...... respectively (Round off to the nearest integer).

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📖 Explanation

${C}_{x} {H}_{y}+(x+\;\frac{y}{4}) {O}_{2} \longrightarrow x {CO}_{2}+\;\frac{y}{2} {H}_{2} {O}$ Any hydrocarbon, on combustion gives ${CO}_{2}$ and ${H}_{2} {O}$ . This is Liebig's method for estimation of ' ${C}$ ' and ' ${H}^{'}$ percentage. Mass of water formed $=210 {g}$ $18 {g}$ of ${H}_{2} {O}$ contains $=2 {g}$ of hydrogen $210 {g}$ of ${H}_{2} {O}$ contains $=(\;\frac{2}{18} \times 210=\;\frac{70}{3}) {g}$ of hydrogen. Given, mass of organic compound $=750 {g}$ Percentage of hydrogen $=\;\frac{\;\text{ Mass of hydrogen }\;}{\;\text{ Mass of organic compound }\;} \times 100$ $=\;\frac{70}{3 \times 750} \times 100$ $=3.11 \%$ Nearest integer $=3$

Q53JEE Main 2021NAT4MRedox Reactions

$2 {MnO}_{4}^{-}+b {C}_{2} {O}_{4}^{2-}+ {CH}^{+} \longrightarrow x {Mn}^{2+}+y {CO}_{2}+ {ZH}_{2} {O}$ If the above equation is balanced with integer coefficients, the value of $c$ is

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📖 Explanation

Balancing by half-reaction method $\;{2 {MnO}_{4}^{-}_{+7}+ {bC}_{2} {O}_{4}^{2-}+ {cH}^{+}{\longrightarrow } {xMn}^{+2}_{+2}+ {y}{ {CO}_{2}}_{+4}+ {ZH}_{2} {O}$ Step 1 Separate two half-reactions Oxidation half (O.H.) ${C}_{2} {O}_{4}^{2-} \longrightarrow {CO}_{2}$ Reduction half (R.H.) ${MnO}_{4}^{-} \longrightarrow {Mn}^{2+}$ Step 2 Balance all the atoms except oxygen and 'H'. O.H. ${C}_{2} {O}_{4}^{2-} \longrightarrow 2 {CO}_{2}$ R.H. ${MnO}_{4}^{-} \longrightarrow {Mn}^{2+}$ Step 3 Balance 'O' by adding ${H}_{2} {O}$ and ' ${H}^{'}$ by adding ${H}^{+}$ . O.H. ${C}_{2} {O}_{4}^{2-} \longrightarrow 2 {CO}_{2}$ R.H. $8 {H}^{+}+ {MnO}_{4}^{-} \longrightarrow {Mn}^{2+}+4 {H}_{2} {O}$ Step 4 Balance charge by adding electron O.H ${C}_{2} {O}_{4}^{2-} \longrightarrow 2 {CO}_{2}+2 {e}^{-}$ R.H. $5 {e}^{-}+8 {H}^{+}+ {MnO}_{4}^{-} \longrightarrow {Mn}^{2+}+4 {H}_{2} {O}$ Step 5 Balance electron by multiplying O.H. by '5' and R.H. by ' 2 '. O.H. $5 {C}_{2} {O}_{4}^{2-} \longrightarrow 10 {CO}_{2}+10 {e}^{-}$ R.H. $10 {e}^{-}+16 {H}^{+}+2 {MnO}_{4}^{-} \longrightarrow 2 {Mn}^{2+}+8 {H}_{2} {O}$ Step 6 Add both half-reaction $16 {H}^{+}+5 {C}_{2} {O}_{4}^{2-}+2 {MnO}_{4}^{-} \longrightarrow 10 {CO}_{2}+2 {Mn}^{2+}+8 {H}_{2} {O}$ So, $c=16$

Q54JEE Main 2021NAT4MSolutions

$A B_{2}$ is $10 \%$ dissociated in water to $A^{2+}$ and $B^{-}$ . The boiling point of a $10.0$ molal aqueous solution of $A B_{2}$ is ...... ${ }^{\circ} {C}$ (Round off to the nearest integer). $[Given, molal elevation constant of water $K_{b}=0.5 {K} {kg} {mol}{ }^{-1}$ boiling point of pure water $=100^{\circ} {C}$ ]$

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📖 Explanation

$A B_{2} \longrightarrow A^{2+}+2 B^{-}$ Elevation in boiling point, $\Delta T_{b}=i K_{b} m$ where, $i=$ van't Hoff factor $K_{b}=$ molal elevation constant $m=$ molality $i=1+(n-1) \alpha$ $(n=3, \alpha=$ degree of dissociation) $i=1+(3-1) \cdot 1$ $=1.2$ Putting the values, $\Delta T_{b} \;\; =1.2 \times 0.5 \times 10$ $\Delta T_{b} \;\; =6$ $T_{b} \;\; =100^{\circ} {C}+6^{\circ} {C}$ $T_{b} \;\; =106^{\circ} {C}$

Q55JEE Main 2021NAT4MCoordination Compounds

The equivalents of ethylene diamine required to replace the neutral ligands from the coordination sphere of the trans-complex of ${COCl}_{3} \cdot 4 {NH}_{3}$ is (Round off to the nearest integer).

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📖 Explanation

$[ {Co}( {NH}_{3})_{4} {Cl}_{2}] {Cl}+2 {en} \longrightarrow [ {Co}( {en})_{2} {Cl}_{2}]+4 {NH}_{3}$ ${NH}_{3}$ is the neutral monodentate ligand. Ethylene diamine is a neutral didentate ligand. ${H}_{2} { {N}}^{..}- {CH}_{2}- {CH}_{2}-{ {N^{..}H}_{2}(\;\text{ en }\;)}$ So, two ethylene diamine are equivalent to four ${NH}_{3}{ }^{'}$ ligand.

Q56JEE Main 2021NAT4MSolutions

A $6.50$ molal solution of ${KOH}(a q)$ has a denisity of $1.89 {g} {cm}^{-3}$ . The molarity of the solution is ...... ${mol} {dm}^{-3}$ (Round off to the nearest integer). [Atomic masses: ${K}: 39.0 {u}, {O}: 16.0 {u}, {H}: 1.0 {u}$ ]

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📖 Explanation

Molality $(m)=\;{M \times 1000}/{\{(1000 d)-M \times M_{\;\text{solute }\;}\}}$ where, $M=$ molarity $d=$ density of solution $M_{\;\text{solute }\;}=$ molar mass of solute Putting the values, $6.50=\;\frac{M \times 1000}{(1000 \times 1.89)-(M \times 56)}$ On solving, $M=\;\frac{12285}{1364}$ $M=9$ molar

Q57JEE Main 2021NAT4MStructure of Atom

When light of wavelength $248 {nm}$ falls on a metal of threshold energy $3.0 {eV}$ , the de-Broglie wavelength of emitted electrons is ...... $Å$ . [Round off to the nearest integer] [Use: $\sqrt{3}=1.73, h=6.63 \times 10^{-34} {Js}$ ${m}_{e}=9.1 \times 10^{-31} {kg}, c=3.0 \times 10^{8} {ms}^{-1}$ $1 {eV}=1.6 \times 10^{-19} {J}]$

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📖 Explanation

Given wavelength, if incident light $(\lambda)=248 {nm}=2480 Å$ Threshold energy $=3.0 {eV}$ We know, for electron de-Broglie wavelength is given by $\lambda=\;\frac{h}{m v}$ $\lambda=\;{h}/{\sqrt{2 m( {KE})}}$ $K E=q \times V$ (where, $q=$ charge of particle $)$ $V=$ voltage applied $\lambda=\;\frac{h}{\sqrt{2 m(q V)}}$ Putting mass of electron, $m_{e}=9.1 \times 10^{-31} {kg}$ Charge of electron, $q=1.6 \times 10^{-19} {C}$ $\lambda=\sqrt{\;\frac{150}{V}} Å$ ....(i) We know, $\;\; E_{\;\text{applied }\;}=\varphi +K_{\max }$ ....(ii) where, $\varphi =\;\text{ threshold energy }\;$ $\varphi =3 {eV}$ $E_{\;\text{applied }\;}=\;\frac{12400}{\lambda_{Å}}=\;\frac{12400}{2480}=5 {eV}$ Putting value in Eq. (ii), $5=3+K_{\max }$ $\Rightarrow \;\; {KE}_{\max }=2 {eV}$ $\;\text{ Voltage }\; \;\; =2$ $\lambda_{e}=\sqrt{\;\frac{150}{2}}=8.6 Å$ [Eq. (i)] Nearest integer $=9$

Q58JEE Main 2021NAT4MIonic Equilibrium

Two salts $A_{2} X$ and $M X$ have the same value of solubility product of $4.0 \times 10^{-12}$ . The ratio of their molar solubilities i.e $\;\frac{S(A_{2} X)}{S(M X)}=. . . . . .$ (Round off to the nearest integer)

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📖 Explanation

Let the solubility of $A_{2} X$ be ' 'S'. $A_{2} X(s) ⇌ 2 A^{+}(aq)_{2 s}+X_{s}^{2-}(a q) ; K_{sp1}=4 \times 10^{-12}$ $K_{s p_{1}}=\$ [A^{+}]^{2}[X^{2-}]4 × 10^{-12}=(2 S)^{2}(S)4 × 10^{-12}=4 S^3\Rightarrow ;; S=10^{-3} {M} $Let the solubility of$ M Y $be '$ S_{1} $'.$ M Y(s) ⇌ M^{+}(a q){s_1}+Y^{-}(a q){s_1}, K_{s p_{2}}=4 × 10^{-12}K_{s p_{2}}=$[M^{+}]$[Y^{-}]K_{s p_{2}}=(S_{1})^{2}4 × 10^{-12}=(S_{1})^{2}\Rightarrow ;; S_{1}=2 × 10^{-6} {M};\frac{[A_{2} Y]}{[M Y]}=;\frac{S}{S_{1}}=;\frac{10^{-4}}{2 × 10^{-6}}=50$

Q59JEE Main 2021NAT4MSolid State

A certain element crystallises in a bcc lattice of unit cell edge length $27 Å$ . If the same element under the same conditions crystallises in the fcc lattice, the edge length of the unit cell in A will be...... (Round off to the nearest integer) [Assume each lattice point has a single atom ] [Assume $\sqrt{3}=1.73, \sqrt{2}=1.41$ ]

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📖 Explanation

In bcc structure, spheres touch along body diagonal for bcc structure, $4 r=\sqrt{3} a$ where, $r=$ radius of sphere $a=$ edge length $r=\;\frac{\sqrt{3} \times a_{b c c}}{4}$ ....(i) For fcc structure, $4 r=\sqrt{2} a$ as spheres touch along face diagonals. So, $r=\;{\sqrt{2} a_{ {fcc}}}/{4}$ ....(ii) So, from Eq. (i) and (ii), Given, $\;\; a_{b c c}=27 Å$ Putting the value, $\sqrt{3} \times 27=\sqrt{2} \times a_{ {fcc}}$ $\Rightarrow \;\; a_{f c c}=33 Å$

Q60JEE Main 2021NAT4MChemical Kinetics

The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at $300 {K}$ is $1.0 \times 10^{-3} {s}^{-1}$ and the activation energy, $E_{a}=11.488 {kJ} {mol}^{-1}$ , the rate constant at $200 {K}$ is ...... $\times 10^{-5} {s}^{-1}$ (Round off to the nearest integer). $[Given, $R=8.314 {J} {mol}^{-1} {K}^{-1}$ ]$

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📖 Explanation

Given, $T_{1}=200 {K}, T_{2}=300 {K}$ $E_{a}=$ Activation energy $=11.488 {kJ} / {mol}$ Rate constant $(k_{2})$ at $300 {K}=10^{-3} {s}^{-1}$ To find $=$ rate constant $(k_{1})$ at $200 {K}$ We know, $\log \;\frac{k_{2}}{k_{1}}=\;\frac{E}{2.303 R}(\;\frac{1}{T_{1}}-\;\frac{1}{T_{2}})$ $\log \;\frac{10^{-3}}{k_{1}}=\;\frac{11.488 \times 10^{3}}{2.303 \times 8.314}(\;\frac{1}{200}-\;\frac{1}{300})$ $\log \;\frac{10^{-3}}{k_{1}}=1$ $\log \;\frac{10^{-3}}{k_{1}}=\log 10$ $\Rightarrow \;\; k_{1} \;\; =10^{-4}$ $\;\; =10 \times 10^{-5} {s}^{-1}$

Mathematics30 questions

Q61JEE Main 2021MCQ4MQuadratic Equation and Inequalities

The number of elements in the set $\{x \in R:(|x|-3)|x+4|=6\}$ is equal to

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📖 Explanation

Given, set $=\{x \in R:(|x|-3)|x+4|=6\}$ As, we already know $|x|=\begin{cases}x & x \ge 0 \\ -x & x< 0\end{cases}$ and $|x+4|=\begin{cases}x+4 & x \ge -4 \\ -(x+4) & x< -4\end{cases}$ Case I $x< -4$ $(-x-3)(-x-4)=6 \Rightarrow (x+3)(x+4)=6$ $x^{2}+7 x+12=6 \Rightarrow x^{2}+7 x+6=0$ $(x+6)(x+1)=0$ $x=-6$ or $x=-1$ We will reject $x=-1$ as, $-1>-4$ $\therefore$ When $x< -4, x=-6$ is the solution. Case II $-4 \le x< 0$ $(-x-3)(x+4)=6 \Rightarrow -(x+3)(x+4)=6$ $\Rightarrow \;-(x^{2}+7 x+12)=6$ $\Rightarrow \;x^{2}+7 x+18=0$ As, the discriminant of this quadratic equation is $D=7^{2}-4.18=49-72=-23$ $\because \;\; D=-23$ and $D< 0$ So, no real roots and as per the question, $x \in R$ . No solution when $-4 \le x< 0$ . Case III $x \ge 0$ $(|x|-3)|x+4|=6 \Rightarrow (x-3)(x+4)=6$ $\Rightarrow \;\; x^{2}+x-12=6$ $\Rightarrow \;\; x^{2}+x-18=0$ $x=\;\frac{-1 \pm \sqrt{1+72}}{2}=\;\frac{-1 \pm \sqrt{73}}{2}$ We will reject $x=\;\frac{-1-\sqrt{73}}{2}$ as $\;\frac{-1-\sqrt{73}}{2}< 0$ and here, $x \ge 0$ So, $x=\;\frac{-1+\sqrt{73}}{2}$ , when $x \ge 0$ . $\therefore x=-6$ and $x=\;\frac{-1+\sqrt{73}}{2}$ are the two solutions which belong to the set. Hence, number of solutions $=2$

Q62JEE Main 2021MCQ4MVector Algebra

Let a vector $\alpha \hat{i}+\beta \hat{j}$ be obtained by rotating the vector $\sqrt{3} \hat{i}+\hat{j}$ by an angle $45^{\circ}$ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices $(\alpha, \beta),(0, \beta)$ and $(0,0)$ is equal to

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📖 Explanation

Let OA be $\sqrt{3} \hat{i}+\hat{j}$ and $O B$ be $\alpha \hat{i}+\beta \hat{j}$ .As, we can notice in $O A, \;\frac{1}{\sqrt{3}}=\tan 30^{\circ}$ . So, it makes an angle of $30^{\circ}$ with the $X$ -axis. Now, when $O A$ is rotated further by $45^{\circ}$ anticlockwise, the resultant vector $O B$ makes an angle of $75^{\circ}$ with the $X$ -axis. So, OB $=|O A|(\cos 75^{\circ} \hat{i}+\sin 75^{\circ} \hat{j})$ Let $\triangle O B C$ be the required triangle whose area we have to determine. Area of $\triangle O B C=(1 {/} 2) \times ($ Base $) \times ($ Height $)$ $=1 {/}2 \times \beta \times \alpha$ $=\;\frac{1}{2}(2 \sin 75^{\circ})(2 \cos 75^{\circ})=2 \sin 75^{\circ} \cos 75^{\circ}$ $=\sin 150^{\circ}=\sin 30^{\circ}$ $=1 {/} 2$ Hence, the area is $1{/} 2 {sq}$ . unit.

Q63JEE Main 2021MCQ4MThree Dimensional Geometry

If for $a>0$ , the feet of perpendiculars from the points $A(a,-2 a, 3)$ and $B(0,4,5)$ on the plane $I x+m y+n z=0$ are points $C(0,-a,-1)$ and $D$ respectively, then, the length of line segment $C D$ is equal to

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📖 Explanation

Given, $A \Rightarrow (a,-2 a, 3)$ $B \Rightarrow (0,4,5)$ $C \Rightarrow (0,-a,-1)$ Equation of plane $P \Rightarrow I x+m y+n z=0$ As, C is foot of perpendicular from $A$ to plane $P$ . So, CA $\| N$ , where $N$ is the normal vector to the plane. $C A=(a-0) \hat{i}+(-2 a+a) \hat{j}+(3+1) \hat{k}$ $=a \hat{i}-a \hat{j}+4 \hat{k}$ Now, CA $\| N$ So, $\;\frac{ {a}}{ {l}}=\;\frac{- {a}}{m}=\;\frac{4}{n}=\lambda$ where $\lambda$ is any real number. $P \Rightarrow (a / \lambda) x-(a / \lambda) y+(4 / \lambda) z=0$ $P \Rightarrow a x-a y+4 z=0$ C lies on plane. So, $a \cdot 0-a(-a)+4(-1)=0$ $a^{2}-4=0 \Rightarrow a=\pm 2$ As per the question, $a>0$ , so $a=2$ So, equation of plane $P \Rightarrow 2 x-2 y+4 z=0$ $P \Rightarrow x-y+2 z=0$ Coordinates of $D$ $\;\frac{x-0}{1}=\;\frac{y-4}{-1}=\;\frac{z-5}{2}=\;\frac{-(0-4+10)}{[1^{2}+(-1)^{2}+2^{2}]\$ } $If$ (x, y, z) $be the foot of perpendicular drawn from$ (x_{1}, y_{1}, z_{1}) $to the plane$ a x+b y+c z+d=0 $. Then,$ ;\frac{x-x_{1}}{a}=;\frac{y-y_{1}}{b}=;\frac{z-z_{1}}{c}=;\frac{-(a x_{1}+b y_{1}+c z_{1}+d)}{a^{2}+b^{2}+c^{2}} $Here,$ (x, y, z)=(0,4,5)\Rightarrow ;; x-0=-(y-4)=;\frac{z-5}{2}=;\frac{-6}{6}∴ ;; x=-1, y=5, z=3C=(0,-2,-1) \Rightarrow D=(-1,5,3)∴ ;; C D=\sqrt{(0+1)^{2}+(-2-5)^{2}+(-1-3)^{2}}=\sqrt{1+49+16}C D=\sqrt{66}$

Q64JEE Main 2021MCQ4MStatistics

Consider three observations $a, b$ and $c$ , such that $b=a+c$ . If the standard deviation of $a+2, b+2, c+2$ is $d$ , then which of the following is true?

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📖 Explanation

Given, three observations $=a, b, c$ $b=a+c$ Standard deviation $=\sigma$ $\sigma=\sqrt{\;\frac{\sum(x_{i}-\mu )^{2}}{N}}$ where, $\mu$ is the mean of observation $x_{i}$ and $N$ is the total number of observations. When $x_{1}=a+2, x_{2}=b+2, x_{3}=c+2$ $\mu =\;\frac{a+2+b+2+c+2}{3}=\;\frac{2 b}{3}+2$ $x_{1}-\mu =(a+2)-(2 b {/} 3+2)=a-2 b {/}3$ $x_{2}-\mu =(b+2)-(2 b {/} 3+2)=b-2 b {/}3$ $x_{3}-\mu =(c+2)-(2 b {/} 3+2)=c-2 b {/} 3$ $d=\sqrt{\;\frac{(a-2 b {/}3)^{2}+(b-2 b {/}3)^{2}+(c-2 b{/}3)^{2}}{3}}$ $3 d^{2}=a^{2}+b^{2}+c^{2}+\;\frac{12 b^{2}}{9}-\;\frac{4}{3}(a b+b^{2}+b c)$ $\Rightarrow 3 d^{2}=a^{2}+b^{2}+c^{2}+\;\frac{4}{3} b^{2}-\;\frac{4}{3}[b(a+c)+b^{2}]$ $\Rightarrow 3 d^{2}=a^{2}+b^{2}+c^{2}+\;\frac{4}{3} b^{2}-\;\frac{4}{3}[b \cdot b+b^{2}]$ $\Rightarrow 3 d^{2}=a^{2}+b^{2}+c^{2}+\;\frac{4}{3} b^{2}-\;\frac{4}{3} \cdot 2 b^{2}$ $\Rightarrow 3 d^{2}=a^{2}+c^{2}-\;\frac{b^{2}}{3}$ $\Rightarrow b^{2}=3 a^{2}+3 c^{2}-9 d^{2}$

Q65JEE Main 2021MCQ4MLogarithms

If for $x \in(0, \;\frac{\pi}{2}), \log _{10} \sin x+\log _{10} \cos x=-1$ and $\log _{10}(\sin x+\cos x)=\;\frac{1}{2}(\log _{10} n-1), n>0$ , then the value of $n$ is equal to

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📖 Explanation

$\log _{10} \sin x+\log _{10} \cos x=-1, x \in(0, \pi{/} 2)$ $\log _{10}(\sin x \cos x)=-1$ $\Rightarrow \;\; \sin x \cos x=10^{-1}=1 {/} 10$ $\log _{10}(\sin x+\cos x)=1 {/}2(\log _{10} n-1), n>0$ $2 \log _{10}(\sin x+\cos x)=(\log _{10} n-\log _{10} 10)$ $\Rightarrow \;\; \log _{10}(\sin x+\cos x)^{2}=\log _{10}(n{/} 10)$ $\Rightarrow \;\; (\sin x+\cos x)^{2}=n {/} 10$ $\Rightarrow \sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=\;\frac{n}{10}$ $\Rightarrow \;\; 1+2(1 {/}10)=n {/}10 \Rightarrow 12{/} 10=n{/}10$ $\therefore \;\; n=12$

Q66JEE Main 2021MCQ4MMatrices and Determinants

Let $A=\begin{bmatrix}i & -i \\ -i & i\end{bmatrix}, i=\sqrt{-1}$ . Then, the system of linear equations $A^{8}\begin{bmatrix}x \\ y\end{bmatrix}=\begin{bmatrix}8 \\ 64\end{bmatrix}$ has

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📖 Explanation

Let $A=\begin{bmatrix}i & -i \\ -i & i\end{bmatrix}$ $A^{2}=A \cdot A=\begin{bmatrix}i & -i \\ -i & i\end{bmatrix}\begin{bmatrix}i & -i \\ -i & i\end{bmatrix}$ $=\$ \begin{bmatrix}i^{2}+i^{2} & -i^{2}-i^{2} \ -i^{2}-i^{2} & i^{2}+i^{2}\end{bmatrix}$=\begin{bmatrix}-2 & 2 \ 2 & -2\end{bmatrix}A^{4}=A^{2} \cdot A^{2}=\begin{bmatrix}-2 & 2 \ 2 & -2\end{bmatrix}\begin{bmatrix}-2 & 2 \ 2 & -2\end{bmatrix}=\begin{bmatrix}4+4 & -4-4 \ -4-4 & 4+4\end{bmatrix}=\begin{bmatrix}8 & -8 \ -8 & 8\end{bmatrix} $Similarly,$ A^{8}=A^{4} \cdot A^{4}=\begin{bmatrix}8 & -8 \ -8 & 8\end{bmatrix}\begin{bmatrix}8 & -8 \ -8 & 8\end{bmatrix}=\begin{bmatrix}64+64 & -64-64 \ -64-64 & 64+64\end{bmatrix}=\begin{bmatrix}128 & -128 \ -128 & 128\end{bmatrix}Now, A^{8}\begin{bmatrix}x \ y\end{bmatrix}=\begin{bmatrix}8 \ 64\end{bmatrix}\begin{bmatrix}128 & -128 \ -128 & 128\end{bmatrix}\begin{bmatrix}x \ y\end{bmatrix}=\begin{bmatrix}8 \ 64\end{bmatrix}128 x-128 y=8\Rightarrow ;; x-y=;\frac{1}{16} $.....(i) and$ -128 x+128 y=64\Rightarrow ;; x-y=-;\frac{1}{2}$........(ii) From Eqs. (i) and (ii), we get these two lines are parallel. So, there will be no solution.

Q67JEE Main 2021MCQ4MParabola

If the three normals drawn to the parabola, $y^{2}=2 x$ pass through the point $(a, 0), a \ne 0$ , then $a$ must be greater than

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📖 Explanation

Given, equation of parabola $\Rightarrow y^{2}=2 x$ Equation of normal of parabola, $y^{2}=4 a x$ is $t x+y=2 a t+a t^{3}$ Here, $4 a=2$ So, $a=1 {/ }2$ So, equation of normal $\Rightarrow t x+y=t+t^{3} / 2$ $t^{3}+(2-2 x) t-2 y=0$ As, there are three normals which are passing through $(a, 0)$ , so there must be three roots of this equation. Let's say $t_{1}, t_{2}, t_{3}$ . $t^{3}+(2-2 a) t-2.0=0$ $\Rightarrow \;\; t^{3}+(2-2 a) t=0$ $\therefore \;\; t_{1}+t_{2}+t_{3}=0$ and $t_{1} t_{2}+t_{2} t_{3}+t_{3} t_{1}=2-2 a$ $t_{1}, t_{2}, t_{3} \in R$ $t_{1}^{2}+t_{2}^{2}+t_{3}^{2} > 0$ $(t_{1}+t_{2}+t_{3})^{2}-2(t_{1} t_{2}+t_{2} t_{3}+t_{3} t_{1}) > 0$ $\Rightarrow 0-2(2-2 a)>0 \Rightarrow 4 a>4$ $\Rightarrow a>1$

Q68JEE Main 2021MCQ4MVector Algebra

Let the position vectors of two points $P$ and $Q$ be $3 \hat{i}-\hat{j}+2 \hat{k}$ and $\hat{i}+2 \hat{j}-4 \hat{k}$ , respectively. Let $R$ and $S$ be two points such that the direction ratios of lines $P R$ and $Q S$ are $(4,-1,2)$ and $(-2,1,-2)$ , respectively. Let lines $P R$ and $Q S$ intersect at $T$ . If the vector TA is perpendicular to both PR and QS and the length of vector TA is $\sqrt{5}$ units, then the modulus of a position vector of $A$ is

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📖 Explanation

$P \;\; =(3 \hat{i}-\hat{j}+2 \hat{k}) \;\text{ and }\; Q=(\hat{i}+2 \hat{j}-4 \hat{k})$ $V_{P R} \;\; =(4,-1,2) \;\text{ and }\; V_{Q S}(-2,1,-2)$ Equation of line $P R=(3 \hat{i}-\hat{j}+2 \hat{k})+\lambda(4 \hat{i}-\hat{j}+2 \hat{k})$ Equation of lineQS $=(\hat{i}+2 \hat{j}-4 \hat{k})+\mu (-2 \hat{i}+\hat{j}-2 \hat{k})$ Let $T$ be the point of intersection. $T=(3+4 \lambda,-1-\lambda, 2+2 \lambda)$ $T=(1-2 \mu , 2+\mu ,-4-2 \mu )$ $3+4 \lambda=1-2 \mu$ $\Rightarrow \;\; 2 \lambda+\mu =-1$ .......(i) $-1-\lambda=2+\mu$ $\Rightarrow \lambda+\mu =-3$ .....(ii) From Eqs. (i) and (ii), $\lambda=2$ and $\mu =-5$ $T=[3+4(2)],-1-(2), 2+2(2)=(11,-3,6)$ Now, $D C$ of TA will be $V_{P R} \times V_{Q S}$ $\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & -2 \\ 4 & -1 & 2\end{vmatrix}=0 \hat{i}-4 \hat{j}-2 \hat{k}$ $L_{ {TA}} \Rightarrow (11 \hat{i}-3 \hat{j}+6 \hat{k})+x(-4 \hat{j}-2 \hat{k})$ Let $A=(11,-3-4 x, 6-2 x)$ $T A=\sqrt{5}$ $\Rightarrow \sqrt{(11-11)^{2}+(-3-4 x+3)^{2}+(6-2 x-6)^{2}}=\sqrt{5}$ $\Rightarrow (4 x)^{2}+(2 x)^{2}=5 \Rightarrow 20 x^{2}=5$ $\Rightarrow \;\; x^{2}=\;\frac{1}{4} \Rightarrow x=\pm \;\frac{1}{2}$ $A=[11,-3-4(1 / 2), 6-2(1 / 2)]$ $A=(11,-5,5)$ Or $A=[11,-3+4(1 / 2), 6+2(1 / 2)]$ $A=(11,-1,7)$ $\therefore \;\; |A|=\sqrt{11^{2}+5^{2}+5^{2}}$ or $\Rightarrow \;\; |A|=\sqrt{11^{2}+1^{2}+7^{2}}$ $\Rightarrow \;\; |A|=\sqrt{171}$ or $\sqrt{171}$ $\therefore \;\; |A|=\sqrt{171}$

Q69JEE Main 2021MCQ4MLimits, Continuity and Differentiability

Let the functions $f: R \to R$ and $g: R \to R$ be defined as $f(x)=\begin{cases}x+2 & x< 0 \\ x^{2} & x \ge 0\end{cases}$ and $g(x)=\begin{cases}x^{3} & x< 1 \\ 3 x-2 & x \ge 1\end{cases}$ Then, the number of points in $R$ , where $(f o g)(x)$ is NOT differentiable is equal to

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📖 Explanation

$f(x)=\begin{cases}x+2 & x< 0 \\ x^{2} & x \ge 0\end{cases}$ $g(x)=\begin{cases}x^{3} & x< 1 \\ 3 x-2 & x \ge 1\end{cases}$ $f[g(x)]=\begin{cases}g(x)+2 & g(x)< 0 \\ g^{2}(x) & g(x) \ge 0\end{cases}$ When $g(x)< 0 \Rightarrow g(x)=x^{3}, x< 0$ When $g(x) \ge 0 \Rightarrow g(x)=\begin{cases}x^{3} & 0 \le x< 1 \\ 3 x-2 & x \ge 1\end{cases}$ $f[g(x)]=\begin{cases}x^{3}+2 & x< 0 \\ x^{6} & 0 \le x< 1 \\ (3 x-2)^{2} & x \ge 1\end{cases}$ As, polynomial function is continuous everywhere in its domain. So, $f[g(x)]$ will be continuous everywhere at $x< 0,0< x< 1$ and $x>1$ . We will check the behaviour of $f o g(x)$ only at boundary points which is $x=0$ and $x=1$ . At $x=0, \lim ^{-(}(x^{3}+2)=2$ $\lim _{x \to 0^{+}} x^{6}=0$ Clearly, ${LHL} \ne$ RHL at $x=0$ So, $f \circ g(x)$ is discontinuous at $x=0$ . $\Rightarrow f o g(x)$ is not differentiable at $x=0$ At $x=1, \lim _{x \to 1^{-}} x^{6}=1$ $\lim _{x \to 1^{+}}(3 x-2)^{2}=1$ . Also $f(1)=1$ fog $(x)$ is continuous at $x=1$ Derivative test at $x=1$ , ${LHD}=\lim _{h \to 0} \;\frac{f(1)-f(1-h)}{h}$ $=\lim _{h \to 0} \;\frac{1-(1-h)^{6}}{h}=\lim _{h \to 0} 6(1-h)^{5}=6$ ${RHD}=\lim _{h \to 0} \;\frac{f(1+h)-f(1)}{h}$ $=\lim _{h \to 0} \;\frac{[3(1+h)-2]^{2}-1}{h}=\lim _{h \to 0} 2[3(1+h)-2] \cdot 3=6$ $\therefore f \circ g(x)$ is continuous and differentiable at $x=1$ . $\therefore f o g(x)$ is discontinuous and non-differentiable at $x=0$ . So, number of points of non-differentiability of $f \circ g(x)$ is 1 .

Q70JEE Main 2021MCQ4MSets and Relations

Which of the following Boolean expression is a tautology?

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📖 Explanation

Let $p$ and $q$ are two statements. Let's make the truth table and see $(p ∧ q)$ implies $(p ∨ q)$ or not. pq $p ∧ q$ $p ∨ q$ $p \to q$ $(p ∧ q) \to$ $(p \to q)$ TTTTTTFTFTTTTFFTFTFFFFTTSince, last column of the truth table for $(p ∧ q) \to (p \to q)$ contains $T$ only. So, $(p ∧ q) \to (p \to q)$ is a tautology.

Q71JEE Main 2021MCQ4MLogarithms

Let a complex number $z,|z| \ne 1$ , satisfy $\log _{\;\frac{1}{\sqrt{2}}}(\;\frac{|z|+11}{(|z|-1)^{2}}) \le 2$ . Then, the largest value of $|z|$ is equal to

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📖 Explanation

Given, $\log _{\;\frac{1}{\sqrt{2}}}(\;\frac{|z|+11}{(|z|-1)^{2}}) \le 2,|z| \ne 1$ As, $\log _{a} x$ is a decreasing function when $0< a< 1$ . So, $\log _{a}(x_{1}) \le k$ , $x_{1} \ge a^{k}$ Similarly, $\frac{|z|+11}{(|z|-1)^{2}} \ge (\;\frac{1}{\sqrt{2}})^{2}$ $\Rightarrow \frac{|z|+11}{(|z|-1)^{2}} \ge \;\frac{1}{2}$ $\Rightarrow \;\; 2|z|+22 \ge (|z|-1)^{2}$ $\Rightarrow \;\; |z|^{2}+1-2|z|-2|z|-22 \le 0$ $\Rightarrow \;|z|^{2}-4|z|-21 \le 0$ $\Rightarrow \;(|z|+3)(|z|-7) \le 0$ $0 \le |z| \le 7$ ${SO}_{,}|z|_{\max }=7$

Q72JEE Main 2021MCQ4MBinomial Theorem

If $n$ is the number of irrational terms in the expansion of $(3^{1 {/} 4}+5^{1{/}8})^{60}$ , then $(n-1)$ is divisible by

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📖 Explanation

$(3^{1 {/} 4}+5^{1 {/} 8})^{60}$ By using Binomial expansion, $(r+1)$ th term, $T_{t+1}={ }^{60} C_{r}(3^{1 {/} 4})^{t}(5^{1 {/} 8})^{60-t}={ }^{60} C_{r} 3^{t {/}4} 5^{\;\frac{60-t}{8}}$ For this term to be a rational number, $r$ should be a multiple of 4 and $(60-r)$ should be a multiple of 8 . Let $A$ be a set when $r$ is the multiple of 4 . $A=\{4,8,12, . . ., 56,60\}$ $n(A)=15$ Let $B$ be a set of $r$ , when $(60-r)$ is the multiple of 8 . $B=\{4,12,20,28,36,44,52,60\}$ $n(B)=8$ Now, $n(A \cap B)=8$ So, there are only 8 terms out of 61 terms which will be rational numbers. 53 terms will be irrational. So, $n=53$ and $n-1=52$ which is divisible only by 26 among the given options.

Q73JEE Main 2021MCQ4MThree Dimensional Geometry

Let $P$ be a plane $l x+m y+n z=0$ containing the line, $\;\frac{1-x}{1}=\;\frac{y+4}{2}=\;\frac{z+2}{3}$ . If plane $P$ divides the line segment $A B$ joining points $A(-3,-6,1)$ and $B(2,4,-3)$ in ratio $k: 1$ , then the value of $k$ is equal to

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📖 Explanation

$P \Rightarrow l x+m y+n z=0$ $P$ contains $L_{1}$ $L_{1} \Rightarrow \;\frac{x-1}{-1}=\;\frac{y+4}{2}=\;\frac{z+2}{3}$ So, $(1,-4,-2)$ lies on plane. $1-4 m-2 n=0$ ...(i) And $(-1,2,3)$ will be perpendicular to $(I, m, n)$ . $-1+2 m+3 n=0$ ....(ii) Adding Eqs. (i) and (ii), $-2 m+n=0$ $n=2 m$ $l-4 m-4 m=0$ $l=8 m$ So, $l=8 {m}$ and $n=2 {m}$ Plane $\Rightarrow 8 x+y+2 z=0$ Now, $A(-3,-6,1)$ and $B(2,4,-3)$ Plane $P$ divides $A B$ in the ratio of $k: 1$ . Let plane $P$ intersect the line $A B$ at point $O$ . So, $O=(\;\frac{2 k-3}{k+1}, \;\frac{4 k-6}{k+1}, \;\frac{-3 k+1}{k+1})$ And $O$ lies on plane $P$ , So, $8(2 k-3)+(4 k-6)+2(-3 k+1)=0$ $\Rightarrow \;14 k-28 \;\; =0$ $\therefore \;k \;\; =2$

Q74JEE Main 2021MCQ4MApplication of Derivatives

The range of $a \in R$ for which the function $f(x)=(4 a-3)(x+\log _{e} 5)+2(a-7)$ $\cot (\;\frac{x}{2}) \sin ^{2}(\;\frac{x}{2}), x \ne 2 n \pi, n \in N$ , has critical points, is

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📖 Explanation

Given, $f(x)=(4 a-3)(x+\log _{e} 5)+2(a-7) \cot \;\frac{x}{2} \cdot \sin ^{2} \;\frac{x}{2}$ $\Rightarrow \;\; f(x)=(4 a-3)(x+\log _{e} 5)+2(a-7) \cos \;\frac{x}{2} \sin \;\frac{x}{2}$ $\Rightarrow \;\; f(x)=(4 a-3)(x+\log _{e} 5)+(a-7) \sin x$ $\Rightarrow \;\; f^{'}(x)=(4 a-3)(1+0)+(a-7) \cos x$ $\Rightarrow \;\; f^{'}(x)=(4 a-3)+(a-7) \cos x$ When $f^{'}(x)=0,(4 a-3)+(a-7) \cos x=0$ $\Rightarrow \;\; \cos x=\;\frac{4 a-3}{7-a}$ As, $-1 \le \cos x \le 1$ So, $-1 \le \;\frac{4 a-3}{7-a} \le 1$ $\Rightarrow \;\; \;\frac{4 a-3}{7-a}+1 \ge 0 \Rightarrow \;\frac{4 a-3+7-a}{7-a} \ge 0$ $\Rightarrow \;\; \;\frac{3 a+4}{7-a} \ge 0$ $a \in[-4 {/} 3,7)$ ...(i) Now, $\;\frac{4 a-3}{7-a}-1 \le 0$ $\Rightarrow \;\; \;\frac{4 a-3-7+a}{7-a} \le 0$ $\Rightarrow \;\; \;\frac{5 a-10}{7-a} \le 0$ $a \in(-\infty , 2] \cup(7, \infty )$ ...(ii) From Eqs. (i) and (ii), $a \in[-\;\frac{4}{3}, 2]$

Q75JEE Main 2021MCQ4MProbability

A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is

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📖 Explanation

Let $S$ and $N S$ be the respective events of choosing a spade card and a card which is not a spade. $P(S)=\;\frac{{ }^{13} C_{1}}{{ }^{52} C_{1}}=\;\frac{1}{4}$ $P(N S)=\;\frac{{ }^{39} C_{1}}{{ }^{52} C_{1}}=\;\frac{3}{4}$ Let $A$ be the event when both the chosen cards are spade. Let $\overline{A}$ be the event when both the chosen cards are not spade.We have to find the probability of missing card not being spade when both the chosen cards are spade $=P(\overline{S} \mid A)$ By using Bayes theorem, $P(\overline{S} \mid A)=\;\frac{P(\overline{S} \cap A)}{P(A)}$ $=\;\frac{P(A \mid \overline{S}) P(\overline{S})}{P(A \mid S) P(S)+P(A \mid \overline{S}) P(\overline{S})}$ $=\;{(\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}})(\frac{{ }^{39} C_{1}}{{ }^{52} C_{1}})}/{(\;\frac{{ }^{13} C_{1}}{{ }^{52} C_{1}})(\;\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}})+(\;\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}})(\;\frac{{ }^{39} C_{1}}{{ }^{52} C_{1}})}$ $=\;\frac{(\frac{13 \cdot 12}{51 \cdot 50})(\frac{3}{4})}{(\;\frac{1}{4})(\;\frac{12 \cdot 11}{51 \cdot 50})+(\;\frac{3}{4})(\;\frac{13 \cdot 12}{51 \cdot 50})}$ $=\;\frac{13 \times 12 \times 3}{(12 \times 11)+(3 \times 12 \times 13)}$ $=\;\frac{39}{50}$

Q76JEE Main 2021MCQ4MBinomial Theorem

Let $[x]$ denote greatest integer less than or equal to $x$ . If for $n \in N,(1-x+x^{3})^{n}=sum_{j=0}^{3 n} a_{j} x^{j}$ , then $sum_{j=0}^{[\;\frac{3 n}{2}]} a_{2 j}+4 sum_{j=0}^{[\;\frac{3 n-1}{2}]} a_{2 j}+1$ is equal to

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📖 Explanation

Given, $(1-x+x^{3})^{n}=sum_{i=0}^{3 n} a_{j} \times j$ $(1-x+x^{3})^{n}=a_{0}+a_{1} x+a_{2} x^{2}+. . .+a_{3 n} x^{3 n}$ Putting $x=1$ , $(1-1+1)^{n} \;\; =a_{0}+a_{1}+a_{2}+. . .+a_{3 n}$ $1 \;\; =a_{0}+a_{1}+a_{2}+. . .+a_{3 n}$ .....(i) Putting $x=-1$ , $(1+1-1)^{n}=a_{0}-a_{1}+a_{2}-a_{3}+. . .(-1)^{3 n} a_{n}$ $1=a_{0}-a_{1}+a_{2}-a_{3}+. . .(-1)^{3 n} a_{n}$ ....(ii) Adding Eqs. (i) and (ii), we get $2=2(a_{0}+a_{2}+a_{4}+a_{6} . . .)$ $a_{0}+a_{2}+a_{4}+. . .=1$ On subtracting Eq. (ii) from Eq. (i), we get $0=2(a_{1}+a_{3}+a_{5}+. . .)$ $a_{1}+a_{3}+a_{5}+. . .=0$ $\;\;$ Now, $\sum^{[\;\frac{3 n}{2}] }_{i=0} a_{2 j}+4 \sum^{ [\;\frac{3 n-1}{2}]}_{i=0}a_{2 j}+1$ $=[a_{0}+a_{2}+a_{4}+. . .]+4[a_{1}+a_{3}+a_{5}+. . .]$ $=1+4 \times 0$ $=1$

Q77JEE Main 2021MCQ4MDifferential Equations

If $y=y(x)$ is the solution of the differential equation, $\;\frac{d y}{d x}+2 y \tan x=\sin x, y(\;\frac{\pi}{3})=0$ , then the maximum value of the function $y(x)$ over $R$ is equal to

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📖 Explanation

Given, $\;\frac{d y}{d x}+2 y \tan x=\sin x$ This differential equation is of the form $\;\frac{d y}{d x}+P y=Q$ where $P$ and $Q$ is function of $x$ . which is a linear differential equation. Here, $P=2 \tan x$ and $Q=\sin x$ The integrating factor of linear differential equation is ${e}^{\int P d x}$ . Here, ${e}^{\int 2 \tan x d x}= {e}^{\int \;\frac{2 \sin x}{\cos x} d x}= {e}^{-2 \log (\cos x)}=\sec ^{2} x$ Now, $\;\frac{d y}{d x}+2 y \tan x=\sin x$ On multiplying $\sec ^{2} \times$ both the sides, $\sec ^{2} x \;\frac{d y}{d x}+2 y \sec ^{2} x \tan x=\sin x \sec ^{2} x$ $\Rightarrow \;\; \;\frac{d}{d x}(y \sec ^{2} x)=\sin x \sec ^{2} x$ $\Rightarrow \;\; y \sec ^{2} x=\int \sin x \sec ^{2} x d x$ $\Rightarrow \;\; y \sec ^{2} x=\int \;\frac{\sin x}{\cos ^{2} x} d x$ Let $\cos x=t$ $(-\sin x) d x=d t$ $\int \;\frac{-d t}{t^{2}}=\;\frac{1}{t}+c$ ${So}, y \sec ^{2} x=\;\frac{1}{\cos x}+c$ $y \sec ^{2} x=\sec x+c$ Now, $x=\pi / 3, y=0$ $0=2+c$ $\Rightarrow \;\; c=-2$ So, $y \sec ^{2} x=\sec x-2$ $y=\cos ^{2} x(\;\frac{1}{\cos x}-2)=-2 \cos ^{2} x+\cos x$ $\Rightarrow \;\; y=-2(\cos ^{2} x-\;\frac{\cos x}{2})$ $\Rightarrow \;\; y=-2[(\cos ^{2} x-\;\frac{\cos x}{2}+\;\frac{1}{16})-\;\frac{1}{16}]$ $\Rightarrow \;\; y=-2[(\cos x-1 / 4)^{2}-\;\frac{1}{16}]$ $\Rightarrow \;\; y=\;\frac{1}{8}-2(\cos x-1 / 4)^{2}$ So, $y_{\min }=\;\frac{1}{8}$

Q78JEE Main 2021MCQ4MHyperbola

The locus of the mid-points of the chord of the circle, $x^{2}+y^{2}=25$ which is tangent to the hyperbola, $\;\frac{x^{2}}{9}-\;\frac{y^{2}}{16}=1$ is

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📖 Explanation

Given, equation of circle $\Rightarrow x^{2}+y^{2}=25$ Equation of hyperbola $\Rightarrow \;\frac{x^{2}}{9}-\;\frac{y^{2}}{16}=1$ Let the mid-point of the chord of the circle be $(h, k)$ . The equation of a chord through its mid-point is simply $T=S_{1}$ where $T$ is the equation of tangent and $S_{1}$ is the value of $S$ by putting $(h, k)$ . Here, $\;\; h x+k y=h^{2}+k^{2}$ The equation of tangent to hyperbola $\;\frac{x^{2}}{a^{2}}-\;\frac{y^{2}}{b^{2}}=1$ is $y=m x \pm \sqrt{a^{2} m^{2}-b^{2}}$ Here, $y=m x \pm \sqrt{9 m^{2}-16}$ So, $y=\;\frac{-h x}{k}+\;\frac{h^{2}+k^{2}}{k}$ .....(i) and $y=m x \pm \sqrt{9 m^{2}-16}$ .....(ii) Eqs. (i) and (ii) are identical, so $\;\frac{1}{1}=\;\frac{-h / k}{m}=\;{h^{2}+k^{2}}/{k \sqrt{9 m^{2}-16}} \Rightarrow m=\;\frac{-h}{k}$ $\Rightarrow 9 m^{2}-16=\;\frac{(h^{2}+k^{2})^{2}}{k^{2}} \Rightarrow 9(\;\frac{h^{2}}{k^{2}})-16=\;\frac{(h^{2}+k^{2})^{2}}{k^{2}}$ $\Rightarrow 9 h^{2}-16 k^{2}=(h^{2}+k^{2})^{2}$ $\{\text{table} h \to x ; k \to y$ So, $9 x^{2}-16 y^{2}=(x^{2}+y^{2})^{2}$ or $(x^{2}+y^{2})-9 x^{2}+16 y^{2}=0$

Q79JEE Main 2021MCQ4MTrigonometric Ratios and Identities

The number of roots of the equation, $(81)^{\sin ^{2} x}+(81)^{\cos ^{2} x}=30$ in the interval $[0, \pi]$ is equal to

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📖 Explanation

Given, $81^{\sin ^{2} x}+81^{\cos ^{2} x}=30$ $\Rightarrow \;\; 81^{\sin ^{2} x}+81^{(1-\sin ^{2} x)}=30$ $\Rightarrow \;\; 81^{\sin ^{2} x}+\;{81}/{81^{\sin ^{2} x}}=30$ Let $81^{\sin ^{2} x}=y$ . $\therefore$ $y+\;\frac{81}{y}=30$ $\Rightarrow \;\; y^{2}-30 y+81=0$ $\Rightarrow \;\; (y-27)(y-3)=0$ $\Rightarrow \;\; y=3$ or $y=27$ $81^{\sin ^{2} x}=3$ or $81^{\sin ^{2} x}=27$ $3^{4 \sin ^{2} x}=3$ or $3^{4 \sin ^{2} x}=3^{3}$ $\Rightarrow \;\; 4 \sin ^{2} x=1$ or $4 \sin ^{2} x=3$ $\Rightarrow \;\; \sin ^{2} x=1 {/} 4$ or $\sin ^{2} x=3 / 4$ $\Rightarrow \;\; \sin ^{2} x=\sin ^{2}(\pi {/} 6)$ or $\sin ^{2} x=\sin ^{2}(\pi{/} 3)$ $\Rightarrow \;\; x=n \pi \pm \pi{/} 6$ or $x=n \pi \pm \pi {/}3$ From $[0, \pi]$ , $x=\pi {/} 6,5 \pi{/} 6$ or $x=\pi {/}3,2 \pi{/} 3$ Hence, the total number of solutions $=4$

Q80JEE Main 2021MCQ4MSequences and Series

Let $S_{k}=sum_{r=1}^{k} \tan ^{-1}(\;\frac{6^{r}}{2^{2 r+1}+3^{2 r+1}})$ . Then, $\lim _{k \to \infty } S_{k}$ is equal to

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📖 Explanation

$S_{K}=sum_{r=1}^{k} \tan ^{-1}(\;\frac{6^{r}}{2^{2 r+1}+3^{2 r+1}})$ $=\sum \tan ^{-1} \;\frac{2^{r} 3^{r+1}-3^{r} 2^{r+1}}{2^{2 r+1}\$ [1+(;\frac{3}{2})^{2 r+1}]$}=\sum \tan ^{-1} ;\frac{2^{2 r+1}$[(\frac{3}{2})^{r+1}-(\frac{3}{2})^{r}]$}{2^{2 t+1}$[1+(;\frac{3}{2})^{2 r+1}]$}=sum_{r=1}^{k} \tan ^{-1} ;\frac{(\frac{3}{2})^{r+1}-(\frac{3}{2})^{r}}{1+(;\frac{3}{2})^{2 r+1}}=sum_{r=1}^{k} \tan ^{-1} ;\frac{(\frac{3}{2})^{r+1}-(\frac{3}{2})^{r}}{1+(;\frac{3}{2})^{2 r+1}}=sum_{r=1}^{k} \tan ^{-1}(;\frac{3}{2})^{r+1}-\tan ^{-1}(;\frac{3}{2})^{r}=\tan ^{-1}(;\frac{3}{2})^{2}-\tan ^{-1}(;\frac{3}{2})+\tan ^{-1}(;\frac{3}{2})^{3}-\tan ^{-1}(;\frac{3}{2})^{2}+\tan ^{-1}(;\frac{3}{2})^{k+1}-\tan ^{-1}(;\frac{3}{2})^{k}\Rightarrow ;; S_{k}=\tan ^{-1}(;\frac{3}{2})^{k+1}-\tan ^{-1}(;\frac{3}{2}) $When$ k \to \infty , \tan ^{-1}(;\frac{3}{2})^{k+1} \to \pi / 2\lim {k \to \infty } s{k}=;\frac{\pi}{2}-\tan ^{-1}(;\frac{3}{2})=\cot ^{-1}(;\frac{3}{2})$

Q81JEE Main 2021NAT4MSequences and Series

Consider an arithmetic series and a geometric series having four initial terms from the set $\{11,8,21,16,26,32,4\}$ . If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to.........

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📖 Explanation

Given, set $\{11,8,21,16,26,32,4\}$ By observation, we can say that ${AP}=\{11,16,21,26, . . .\}$ ${GP}=\{4,8,16,32, . . .\}$ Since we are looking for common terms in both the series where $5 m+6$ is $m^{\;\text{th }\;}$ term of AP and $2^{n+1}$ is $n^{\;\text{th }\;}$ term of GP $5 m+6=4 \cdot 2^{n-1}$ $5 m+6=2^{n+1}$ So, $(2^{n+1}-6)$ should be a multiple of 5 . The unit digit of $2^{k}$ is $2,4,6,8$ . So, when 6 is subtracted from $2^{n+1}$ , the possible unit digits will be 6 , $8,0,2$ . Only 0 is divisible by 5 . Hence, $2^{n+1}$ unit digit has to be 6 . $2^{n+1}=2^{4}, 2^{8}, 2^{12}, 2^{16} . . .$ As, $2^{16}$ will not be a 4 digit number, so, common terms $=\{16,256,4096\}$ $\therefore$ Number of common terms $=3$

Q82JEE Main 2021NAT4MDefinite Integrals

Let $f:(0,2) \to R$ be defined as $f(x)=\log _{2}[1+\tan (\;\frac{\pi x}{4})]$ Then, $\lim _{n \to \infty } \;\frac{2}{n}[f(\;\frac{1}{n})+f(\;\frac{2}{n})+. . .+f(1)]$ is equal to..........

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📖 Explanation

$f(x)=\log _{2}[1+\tan (\;\frac{\pi x}{4})]$ Then, $=\lim _{n \to \infty } \;\frac{2}{n}[f(\;\frac{1}{n})+f(\;\frac{2}{n})+. . .+f(1)]=2 \lim _{n \to \infty } sum_{r=1}^{n}(\;\frac{1}{n}) f(\;\frac{r}{n})$ Let $l=\;\frac{2}{\log _{n} 2} int_{0}^{1} \log _{n}[1+\tan (\;\frac{\pi x}{4})] d x$ .....(i) as, $int_{a}^{b} f(x) d x=int_{a}^{b} f(a+b-x) d x$ ${So}, {x} \to 1- {x}$ $l=\;\frac{2}{\log _{n} 2} int_{0}^{1} \log _{n}[1+\tan \;\frac{\pi}{4}(1-x)] d x$ $=\;\frac{2}{\log _{n} 2} int_{0}^{1} \log _{n}[1+\tan (\;\frac{\pi}{4}-\;\frac{\pi x}{4})] d x$ $=\;\frac{2}{\log _{n} 2} int_{0}^{1} \log _{n}[1+(\;\frac{1-\tan \pi x / 4}{1+\tan \pi x / 4})] d x$ $=\;\frac{2}{\log _{n} 2} int_{0}^{1} \log _{n}(\;\frac{2}{1+\tan \;\frac{\pi x}{4}}) d x$ $=\;\frac{2}{\log _{n} 2} int_{0}^{1} \log _{n} 2-\log _{n}(1+\tan \;\frac{\pi x}{4}) d x$ .....(ii) Adding Eqs. (i) and (ii), we get $2 l=\;\frac{2}{\log _{n} 2} int_{0}^{1} \log _{n} 2 d x$ $l=1$

Q83JEE Main 2021NAT4MCircles

Let $A B C D$ be a square of side of unit length. Let a circle $C_{1}$ centred at $A$ with unit radius is drawn. Another circle $C_{2}$ , which touches $C_{1}$ and the lines $A D$ and $A B$ are tangent to it, is also drawn. Let a tangent line from the point $C$ to the circle $C_{2}$ meet the side $A B$ at $E$ . If the length of $E B$ is $\alpha+\sqrt{3} \beta$ , where $\alpha, \beta$ are integers, then $\alpha+\beta$ is equal to ...... .

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📖 Explanation

Given, $E B=\alpha+\sqrt{3} \beta$ Let $O$ be the centre of circle $C_{2}$ . $O Q^{2}+O R^{2}=A O^{2}$ $\Rightarrow \;\; A O+O P=A P=1$ $\Rightarrow \;\; A O+r=1$ $O Q^{2}+O R^{2}=A O^{2}$ $\Rightarrow \;\; r^{2}+r^{2}=(1-r)^{2}$ $\Rightarrow \;\; 2 r^{2}=r^{2}+1-2 r$ $\Rightarrow \;\; r^{2}+2 r+1=1+1$ $\Rightarrow \;\; (r+1)^{2}=2 \Rightarrow r+1=\sqrt{2}$ $\Rightarrow \;\; r=\sqrt{2}-1$ As, $A O=1-r$ $A C=\sqrt{2}$ So, $O C=A C-A O$ $=\sqrt{2}-(1-\sqrt{2}+1)=2(\sqrt{2}-1)$ In $\triangle O C N$ , we have $\;\frac{O N}{O C}=\sin \theta$ $\Rightarrow \;\; \;\frac{\sqrt{2}-1}{2(\sqrt{2}-1)}=\;\frac{1}{2}=\sin \theta \Rightarrow \theta=\pi / 6$ In $\triangle E B C$ , we have $\;\frac{E B}{B C}=\tan 15^{\circ}$ $E B=(\tan 15^{\circ}) B C$ $=(\tan 15^{\circ})$ $[ \because B C=1$ unit $]$ NOW, $\tan 2 \theta=\;\frac{2 \tan \theta}{1-\tan ^{2} \theta}$ $\therefore \;\; \tan 30^{\circ}=\;\frac{2 \tan 15^{\circ}}{1-\tan ^{2} 15^{\circ}}$ $\Rightarrow \;\; 1-\tan ^{2} 15^{\circ}=2 \sqrt{3} \tan 15^{\circ}$ $\Rightarrow \;\; \tan ^{2} 15^{\circ}+2 \sqrt{3} \tan 15^{\circ}-1=0$ $\tan 15^{\circ}=\;\frac{-2 \sqrt{3} \pm \sqrt{12+4}}{2} \Rightarrow \tan 15^{\circ}=-\sqrt{3}+2$ $\therefore \;\; E B=2-\sqrt{3}=\alpha+\sqrt{3} \beta \Rightarrow \alpha=2, \beta=-1$ $\therefore \;\; \alpha+\beta=2+(-1)=1$

Q84JEE Main 2021NAT4MLimits, Continuity and Differentiability

If $\lim _{x \to 0} \;\frac{a e^{x}-b \cos x+c e^{-x}}{x \sin x}=2$ , then $a+b+c$ is equal to .........

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📖 Explanation

We have, $\lim _{x \to 0} \;\frac{a e^{x}-b \cos x+c e^{-x}}{x \sin x}=2$ $\Rightarrow \lim _{x \to 0} \;{a(1+x+\;\frac{x^{2}}{2 !} . . .)-b(1-\;\frac{x^{2}}{2}+\;\frac{x^{4}}{4 !} . . .)+c(1-x+\frac{x^{2}}{2 !} . . .)}/{x(x-\;\frac{x^{3}}{3 !}+. . .)}$ $\Rightarrow \lim _{x \to 0} \;{(a-b+c)+(a-c) x+(\frac{a}{2}+\frac{b}{2}+\frac{c}{2}) x^{2}+. . .}/{x^{2}-\;\frac{x^{4}}{6}+. . .}$ Here, in numerator, all the coefficients of $x^{k}$ , where $k< 2$ has to be zero, then only limit will exist. $a-b+c=0$ $a-c=0$ $\Rightarrow \;\; a=c \Rightarrow b=2 a$ After solving limit, $\;\frac{a+b+c}{2}=2$ So, $a+2 a+a=4$ $\Rightarrow \;\; a=1$ $\therefore$ $a=1, b=2 \;\text{ and }\; c=1$ $a+b+c=1+2+1=4$

Q85JEE Main 2021NAT4MMatrices and Determinants

The total number of $3 \times 3$ matrices $A$ having entries from the set $(0,1,2,3)$ , such that the sum of all the diagonal entries of $A A^{T}$ is 9 , is equal to .........

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📖 Explanation

Set $S:\{0,1,2,3\}$ Let $A=\begin{bmatrix}a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3}\end{bmatrix}$ $A^{T}=\begin{bmatrix}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{bmatrix}$ $A A^{T}=\begin{bmatrix}a \cdot a & a \cdot b & a \cdot c \\ b \cdot a & b \cdot b & b \cdot c \\ c \cdot a & c \cdot b & c \cdot c\end{bmatrix}$ where $a \;\; =a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$ $b \;\; =b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$ $c \;\; =C_{1} \hat{i}+C_{2} \hat{j}+c_{3} \hat{k}$ Now, as per question $a \cdot a+b \cdot b+c \cdot c=9$ $\Rightarrow \;\; (a_{1}^{2}+a_{2}^{2}+a_{3}^{2})+(b_{1}^{2}+b_{2}^{2}+b_{3}^{2})+(c_{1}^{2}+c_{2}^{2}+c_{3}^{2})=9$ $[a_{i}, b_{i}, c_{i} \in S]$ $9=(1+1+1+1+1+1+1+1+1)$ or $\;\; 9=(1+4+4+0+0+0+0+0+0)$ or $9=(9+0+0+0+0+0+0+0+0)$ or $\;\; 9=(4+1+1+1+1+1+0+0+0)$ Total permutations in case $1=1$ Total permutations in case $2=\;\frac{9 !}{6 ! 2 !}=252$ In case $3=\;\frac{9 !}{8 !}=9$ In case $4=\;\frac{9 !}{5 ! 3 !}=504$ Total permutations $=1+252+9+504$ $=766$

Q86JEE Main 2021NAT4MMatrices and Determinants

Let $P=\begin{bmatrix}-30 & 20 & 56 \\ 90 & 140 & 112 \\ 120 & 60 & 14\end{bmatrix}$ and $A=\begin{bmatrix}2 & 7 & \omega^{2} \\ -1 & -\omega & 1 \\ 0 & -\omega & -\omega+1\end{bmatrix}$ where, $\omega=\;\frac{-1+i \sqrt{3}}{2}$ , and $I_{3}$ be the identity matrix of order 3 . If the determinant of the matrix $(P^{-1} A P-I_{3})^{2}$ is $\alpha \omega^{2}$ , then the value of $\alpha$ is equal to.........

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📖 Explanation

Given, $P=\begin{bmatrix}-30 & 20 & 56 \\ 90 & 140 & 112 \\ 120 & 60 & 14\end{bmatrix}$ $A=\begin{bmatrix}2 & 7 & \omega^{2} \\ -1 & -\omega & 1 \\ 0 & -\omega & -\omega+1\end{bmatrix}$ $|(P^{-1} A P-I_{3})|^{2}=\alpha \omega^{2}$ $\Rightarrow \;\; |(P^{-1} A P-I_{3})(P^{-1} A P-I_{3})|=\alpha \omega^{2}$ $\Rightarrow P^{-1} A P P^{-1} A P-P^{-1} A P I_{3}-I_{3} P^{-1} A P+I_{3} \cdot I_{3}|=\alpha \omega^{2}$ $\Rightarrow \;\; P^{-1} A^{2} P-P^{-1} A P-P^{-1} A P+I_{3}| \;\; =\alpha \omega^{2}$ $\;{[ \because P P^{-1}=I \;\text{ and }\; I A=A]\$ }\Rightarrow ;; |P^{-1} A^{2} P-2 P^{-1} A P+P P^{-1}|=alpha omega^{2}\Rightarrow ;; |P^{-1}(A^{2}-2 A+I_{3}) P|=alpha omega omega^{2}\Rightarrow ;; |P^{-1}||A-I_{3}|^{2}|P|=alpha omega^{2}\Rightarrow ;; ;; |P^{-1} P||A-I_{3}|^{2}=alpha omega^{2}\Rightarrow |A-I_{3}|^{2}=alpha omega^{2} $Consider,$ ;; A-I_{3}=$\begin{bmatrix}2 & 7 & \omega^{2} \ -1 & -\omega & 1 \ 0 & -\omega & -\omega+1\end{bmatrix}$-\begin{bmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix}|A-I_{3}|=\begin{vmatrix}1 & 7 & omega^{2} \ -1 & -omega-1 & 1 \ 0 & -omega & -omega\end{vmatrix} $On applying$ C_{2} → C_{2}-C_{3} $, we get$ |A-I_{3}|=\begin{vmatrix}1 & 7-omega^{2} & omega^{2} \ -1 & -omega-2 & 1 \ 0 & 0 & -omega\end{vmatrix} $On applying$ C_{2} → C_{2}-C_{3} $, we get$ |A-I_{3}|=\begin{vmatrix}1 & 7-omega^{2} & omega^{2} \ -1 & -omega-2 & 1 \ 0 & 0 & -omega\end{vmatrix}=-omega[(-omega-2)-(-7+omega^{2})]=-omega(-omega-2+7-omega^{2})=-omega(1-2+7)=-6 omega|A-I_{3}|=-6 omega|A-I_{3}|^{2}=36 omega^{2}=alpha omega^{2}∴ ;; alpha=36$

Q87JEE Main 2021NAT4MApplication of Derivatives

If the normal to the curve $y(x)=int_{0}^{x}(2 t^{2}-15 t+10) d t$ at a point $(a, b)$ is parallel to the line $x+3 y=-5, a>1$ , then the value of $|a+6 b|$ is equal to.........

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📖 Explanation

Given, $y(x)=int_{0}^{x}(2 t^{2}-15 t+10) d t$ $\Rightarrow \;\; y^{'}(x)=2 x^{2}-15 x+10$ Since equation of normal is parallel to $x+3 y=-5$ $\therefore$ Slope of normal to $y(x)=$ Slope of lime $\Rightarrow \;\; \;\frac{-1}{[y^{'}(x)]_{a, b}}=\;\frac{-1}{3}$ or $[y^{'}(x)]_{a, b}=3$ $2 a^{2}-15 a+10=3$ $\Rightarrow \;\; 2 a^{2}-15 a+7=0$ $\Rightarrow \;\; (2 a-1)(a-7)=0$ $a=\;\frac{1}{2}$ or 7 As, $a>1$ , so, $a=7$ Now, $(7, b)$ lies on $y(x)$ , $\therefore \;\; b=int_{0}^{a}(2 t^{2}-5 t+10) d t$ $\Rightarrow \;\; b=\;\frac{2}{3} a^{3}-\;\frac{15}{2} a^{2}+10 a$ $\Rightarrow \;\; b=\;\frac{2}{3}(7)^{3}-\;\frac{15}{2}(7)^{2}+10(7)$ $\Rightarrow \;\; b=\;\frac{-413}{6}$ So, $a+6 b=7-6(\;\frac{413}{6})=-406$ $\therefore \;\; |a+6 b|=406$

Q88JEE Main 2021NAT4MDifferential Equations

Let the curve $y=y(x)$ be the solution of the differential equation $\;\frac{d y}{d x}=2(x+1)$ . If the numerical value of area bounded by the curve $y=y(x)$ and $x$ -axis is $\;\frac{4 \sqrt{8}}{3}$ , then the value of $y(1)$ is equal to.........

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📖 Explanation

We have, $\;\frac{d y}{d x}=2(x+1)$ $y=\int 2(x+1) d x=2(\;\frac{x^{2}}{2}+x)+C$ $y=x^{2}+2 x+C$ ...(i)Now, $int_{\alpha}^{\beta} y d x=\;\frac{4 \sqrt{8}}{3}$ $int_{\alpha}^{\beta}(x^{2}+2 x+c) d x=\;\frac{4 \sqrt{8}}{3}$ $\Rightarrow \;\; \$ [;\frac{x^{3}}{3}+x^{2}+c x]_{alpha}^{beta}=;\frac{4 \sqrt{8}}{3}\Rightarrow ;; (;\frac{beta^{3}-alpha^{3}}{3})+(beta^{2}-alpha^{2})+c(beta-alpha)=;\frac{4 \sqrt{8}}{3} $....(ii) From Eq. (i),$ alpha+beta=-2alpha beta=c∴ ;; beta-alpha=\sqrt{(alpha+beta)^{2}-4 alpha beta}=\sqrt{4-4 c}=2 \sqrt{1-c} $From Eq. (ii),$ ;; ;\frac{1}{3}(beta-alpha)(alpha^{2}+beta^{2}+alpha beta)+(beta-alpha)(beta+alpha)+c(beta-alpha)=;\frac{4 \sqrt{8}}{3}\Rightarrow ;\frac{1}{3}(2 \sqrt{1-c})(4-c)+(2 \sqrt{1-c})(-2+c)=;\frac{4 \sqrt{8}}{3}\Rightarrow ;; (2 \sqrt{1-c})[4-c+(-6+3 c)]=4 \sqrt{8}\Rightarrow (2 \sqrt{1-c})[2(c-1)]=4 \sqrt{8}\Rightarrow ;; (1-c)^{3 / 2}=-\sqrt{8}\Rightarrow ;; (1-c)^{3}=8\Rightarrow ;; 1-c=2∴ ;; c=-1 $Now,$ y=x^{2}+2 x-1∴ ;; y(1)=1^{2}+2 ⋅ 1-1=2$

Q89JEE Main 2021NAT4MDefinite Integrals

Let $f: R \to R$ be a continuous function such that $f(x)+f(x+1)=2$ , for all $x \in R$ . If $I_{1}=int_{0}^{8} f(x) d x$ and $I_{2}=int_{-1}^{3} f(x) d x$ , then the value of $I_{1}+2 I_{2}$ is equal to..........

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📖 Explanation

Given, $f(x)+f(x+1)=2$ ....(i) $I_{1}=int_{0}^{8} f(x)$ and $I_{2}=int_{-1}^{3} f(x) d x$ Let $f(0)=a$ Put $x=0$ in Eq. (i) $f(0)+f(1) \;\; =2$ $f(1) \;\; =2-a$ Put $x=1$ in Eq. (i) $f(1)+f(2)=2$ $f(2)=a$ and so on ${So}, \;\; f(0)=f(2)=f(4) . . .=a$ $f(1)=f(3)=f(5) . . .=2-a$ Clearly, $f(x)$ is periodic with its period 2 units. So, $\;\; I_{1}=int_{0}^{24} f(x) d x$ $\Rightarrow \;\; I_{1}=4 int_{0}^{2} f(x) d x$ Now, $I_{2}=int_{-1}^{3} f(x) d x$ $x \to x+1$ $I_{2}=int_{0}^{4} f(x+1) d x=int_{0}^{4}[2-f(x)] d x$ $\Rightarrow \;\; I_{2}=8-2 int_{0}^{2} f(x) d x$ $\Rightarrow \;\; 2 I_{2}=16-4 int_{0}^{2} f(x) d x$ $\Rightarrow \;2 I_{2}\;=16-I_{1}$ $\therefore \;\; I_{1}+2 I_{2}\;=16$

Q90JEE Main 2021NAT4MComplex Numbers

Let $z$ and $w$ be two complex numbers, such that $w=z \overline{z}-2 z+2,|\;\frac{z+i}{z-3 i}|=1$ and $Re(w)$ has minimum value. Then, the minimum value of $n \in N$ for which $w^{n}$ is real, is equal to

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📖 Explanation

Given, $w=z \overline{z}-2 z+2$ $|\;\frac{z+i}{z-3 i}|=1$ $\Rightarrow \;\; |z+i|=|z-3 i|$ Let $z=x+i y$ $\Rightarrow \;\; |x+i(y+1)|=|x+i(y-3)|$ $\Rightarrow \;\; x^{2}+(y+1)^{2}=x^{2}+(y-3)^{2}$ $\Rightarrow \;\; 2 y+1=-6 y+9$ $\therefore \;\; y=1$ Now, $w=z \overline{z}-2 z+2$ $w=|z|^{2}-2 z+2$ $\Rightarrow \;w=x^{2}+y^{2}-2(x+i y)+2$ $\Rightarrow \;w=(x^{2}+y^{2}-2 x+2)+i(-2 y)$ $\Rightarrow \;w=(x^{2}+1-2 x+2)+i(-2)$ $w=(x-1)^{2}+2-2 i$ $Re(w)$ has minimum value. So, $(x-1)^{2}+2$ is minimum when $x=1$ $\therefore \;\; W=2-2 i$ $=2(1-i)$ $=2 \sqrt{2}(\;\frac{1}{\sqrt{2}}-\;\frac{i}{\sqrt{2}})$ $w=2 \sqrt{2} {e}^{-i \pi / 4}$ Now, $\;\; w^{n}=(2 \sqrt{2})^{n} e^{\;\frac{-i n \pi}{4}}$ $=(2 \sqrt{2})^{n}[\cos (\;\frac{n \pi}{4})-i {i \sin (\;\frac{n \pi}{4})}]$ This has to be zero for $w^{n}$ to be real. So, $\sin (\;\frac{n \pi}{4})=0$ $\Rightarrow \;\; \;\frac{n \pi}{4}=0, \pi, 2 \pi, 3 \pi . . .$ $\Rightarrow \;\; n=0,4,8,12 . . .$ The minimum value of $n$ is $4(n \in N)$ .

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