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Physics30 questions

Q1JEE Main 2021MCQ4MElectrostatic Potential and Capacitance

If $C$ and $V$ represent capacity and voltage respectively, then what are the dimensions of $\lambda$ , where $\;\frac{C}{V}=\lambda ?$

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📖 Explanation

Given, $C$ and $V$ represent capacity and voltage, respectively. Dimensional formula of $[C]=[M^{-1} L^{-2} T^{4} I^{2}]$ Dimensional formula of $[ {V}]=[ {ML}^{2} {T}^{-3} {I}^{-1}]$ Therefore, dimensional formula of $[C / V]=\;\frac{[M^{-1} L^{-2} T^{4} I^{2}]\$ }{[M L^{2} T^{-3}|^{-1}]$}=[M^{-2} L^{-4} T^{7} I^{3}]$

Q2JEE Main 2021MCQ4MProperties of Matter

The length of metallic wire is $I_{1}$ when tension in it is $T_{1}$ . It is $I_{2}$ when the tension is $T_{2}$ . The original length of the wire will be

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📖 Explanation

Let $I_{0}$ be the original length, $A$ be the area of cross-section, $\alpha$ be the coefficient of linear expansion, $\Delta l$ be the change in length and $Y$ be the Young's modulus of elasticity. As, $\;\; I_{1}=I_{0}(1+\alpha \Delta T)$ $\Rightarrow \;\; I_{1}-I_{0}=I_{0} \alpha \Delta T \Rightarrow \Delta I=I_{0} \alpha \Delta T$ Initially, $Y=\;\frac{\;\text{ Stress }\;}{\;\text{ Strain }\;}=\;\frac{T / A}{\Delta l / I_{0}}$ $\Rightarrow \;\; Y=\;\frac{T_{1} / A}{(I_{1}-I_{0}) / I_{0}}$ . . . (i) Finally, $Y=\;\frac{T_{2} / A}{(I_{2}-I_{0}) / I_{0}}$ . . . (ii) Now, from Eqs. (i) and (ii), we get $\;\;\frac{T_{1} / A}{(I_{1}-I_{0}) / I_{0}} \;\; =\;\frac{T_{2} / A}{(I_{2}-I_{0}) / I_{0}}$ $\Rightarrow \;\;\frac{T_{1}}{I_{1}-I_{0}}= \;\; \;\frac{T_{2}}{I_{2}-I_{0}}$ $\Rightarrow \;T_{1} I_{2}-T_{1} I_{0} \;\; =T_{2} I_{1}-T_{2} I_{0}$ $\Rightarrow \;T_{1} I_{2}-T_{2} I_{1} \;\; =(T_{1}-T_{2}) I_{0}$ $\Rightarrow \;I_{0}=\;\frac{T_{1} I_{2}-T_{2} I_{1}}{T_{1}-T_{2}}$ $\;\text{ or }\;\;I_{0}=\;\frac{T_{2} I_{1}-T_{1} I_{2}}{T_{2}-T_{1}}$

Q3JEE Main 2021MCQ4MElectromagnetic Induction

An aeroplane with its wings spread $10 {m}$ , is flying at a speed of $180 {km} / {h}$ in a horizontal direction. The total intensity of Earth's field at that part is $2.5 \times 10^{-4} {Wb} / {m}^{2}$ and the angle of dip is $60^{\circ}$ . The emf induced between the tips of the plane wings will be

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📖 Explanation

Given, length of aeroplane wing, $I=10 {m}$ Speed of aeroplane, $v=180 {km} / {h}=50 {ms}^{-1}$ Magnetic flux density, $B=2.5 \times 10^{-4} {Wb} / {m}^{2}$ Angle of $dip, \theta=60^{\circ}$ The emf induced between the tips of plane wings, $\epsilon \;\; =B / v \sin \theta$ $\;\; =2.5 \times 10^{-4} \times 10 \times 50 \sin 60^{\circ}$ $\;\; =2.5 \times 5 \times 10^{-2} \times \sqrt{3} / 2$ $\;\; =6.25 \times 1.732 \times 10^{-2}$ $\;\; =108.25 \times 10^{-3}=108.25 {mV}$

Q4JEE Main 2021MCQ4MWaves

A tuning fork $A$ of unknown frequency produces 5 beats/s with a fork of known frequency $340 {Hz}$ . When fork $A$ is filled, the beat frequency decreases to 2 beats/s. What is the frequency of fork $A$ ?

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📖 Explanation

Given, $f=340 {Hz}$ Let beat frequency of tuning fork $A$ is $f_{A}$ and $f_{A}^{'}$ be the frequency of tuning fork $A$ after filling. In first case, beat frequency $=5$ $\Rightarrow \;f_{A}=340 \pm 5$ $\Rightarrow \;f_{A}=345 {Hz} \;\text{ or }\; 335 {Hz}$ After filling tuning fork $A, f_{A}^{'}=340 \pm 2$ $\Rightarrow \;\; f_{A}^{'}=342 \;\text{ or }\; 338 {Hz}$ As on filling, frequency of $A$ increases. Hence, original frequency of tuning fork $A$ is $335 {Hz}$ .

Q5JEE Main 2021MCQ4MOscillations

A particle executes SHM, the graph of velocity as a function of displacement is

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📖 Explanation

Velocity of particle in ${SHM}$ in terms of displacement $x$ is given as $v \;\; =\omega \sqrt{A^{2}-x^{2}}$ $\Rightarrow$ $v \;\; =A \omega \sqrt{1-x^{2} / A^{2}}$ $\Rightarrow$ $\;\frac{v}{A \omega} \;\; =\sqrt{1-x^{2} / A^{2}}$ $\Rightarrow$ $\;\frac{v^{2}}{A^{2} \omega^{2}} \;\; =1-\;\frac{x^{2}}{A^{2}}$ $\Rightarrow$ $\;\frac{v^{2}}{A^{2} \omega^{2}}+$ $\;\frac{x^{2}}{A^{2}} \;\; =1$ . . . (i) As equation of ellipse is $\;\frac{x^{2}}{a^{2}}+\;\frac{y^{2}}{b^{2}}=1$ Eq. (i) is similar to the equation of ellipse. So, the graph between velocity and displacement is an ellipse.

Q6JEE Main 2021MCQ4MMotion in a Plane

The trajectory of a projectile in a vertical plane is $y=\alpha x-\beta x^{2}$ , where $\alpha$ and $\beta$ are constants and $x$ and $y$ are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $\theta$ and the maximum height attained $H$ are respectively given by

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📖 Explanation

Given, equation of trajectory of projectile, $y=\alpha x-\beta x^{2}$ . . . (i) The standard equation of trajectory of projectile, $y=x \tan \theta-\;\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}$ . . . (ii) Now, from Eqs. (i) and (ii), we get $\alpha=\tan \theta \;\text{ and }\; \beta=\;\frac{g}{2 u^{2} \cos ^{2} \theta}$ $\Rightarrow \;\; \alpha^{2}=\;\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$ $\therefore \;\; \;\frac{\beta}{\alpha^{2}}=\;{g}/{2 u^{2} \cos ^{2} \theta \times \;\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}=\;\frac{g}{2 u^{2} \sin ^{2} \theta} \times \;\frac{2}{2}$ $\;\frac{4 \beta}{\alpha^{2}}=\;\frac{2 g}{u^{2} \sin ^{2} \theta}=\;\frac{1}{H} \Rightarrow H=\alpha^{2} / 4 \beta$ $\;\;\; \theta=\tan ^{-1} \alpha \;\text{ and }\; H=\alpha^{2} / 4 \beta$

Q7JEE Main 2021MCQ4MRotational Motion

A cord is wound round the circumference of wheel of radius $r$ . The axis of the wheel is horizontal and the moment of inertia about it is $I$ . A weight $m g$ is attached to the cord at the end. The weight falls from rest. After falling through a distance $h$ , the square of angular velocity of wheel will be

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📖 Explanation

Moment of inertia of wheel, $I=m R^{2}$ Let $v$ be the velocity and $v=r w=c$ Let $v$ be the velocity and $v=r \omega \Rightarrow \omega=v / r$ By using law of conservation of energy, $1 / 2(I+m r^{2}) \omega^{2} \;\; =m g h$ $\omega^{2} \;\; =\;\frac{2 m g h}{I+m r^{2}}$

Q8JEE Main 2021MCQ4MThermodynamics

The internal energy $(U)$ , pressure $(p)$ and volume $(V)$ of an ideal gas are related as $U=3 p V+4$ . The gas is

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📖 Explanation

Given, ideal gas equation, $U=3 p V+4$ Here, $U=$ internal energy, $p=$ pressure and $V=$ volume. By ideal gas equation, $p V=n R \Delta T$ $=R \Delta T$ $( \because n=1) . . .( {i})$ $\Delta U \;\; =n C_{v} \Delta T=3 p V+4$ $\;\; =n \;\frac{f}{2} R \Delta T=3 p V+4$ $\Rightarrow \;f R \Delta T \;\; =6 p V+8$ $\Rightarrow \;f \cdot p V \;\; =6 p V+8$ [From Eq. (i)] $\Rightarrow \;f \;\; =6+\;\frac{8}{p V}$ [From Eq. (i)] As, degree of freedom is more than 6 . $\therefore$ Gas is polyatomic.

Q9JEE Main 2021MCQ4MRay Optics and Optical Instruments

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) For a simple microscope, the angular size of the object equals the angular size of the image. Reason (R) Magnification is achieved as the small object can be kept much closer to the eye than $25 {cm}$ and hence, it subtends a large angle. In the light of the above statements, choose the most appropriate answer from the options given below.

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📖 Explanation

The formation of image with simple microscope is shown below. Here, $\theta^{'}=\;\frac{h}{u_{0}}=\;\frac{h^{'}}{D}=\;\frac{h^{'}}{25}$ where, $D=25 {cm}$ (least distance of distinct vision) Here, $\theta^{'}$ is same for both object and image, hence Assertion is true. Magnification, $m=\;\frac{\theta^{'}}{\theta}=\;\frac{D}{u_{0}}$ Hence, if $u_{0} < D(25 {cm})$ , hence the value of $\theta^{'}$ will obtain large. So, option (c) is the correct.

Q10JEE Main 2021MCQ4MElectrostatics

Given below are two statements: Statement I An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere. Statement II If $R$ is the radius of a solid metallic sphere and $Q$ be the total charge on it. The electric field at any point on the spherical surface of radius $r( < R)$ is zero but the electric flux passing through this closed spherical surface of radius $r$ is not zero. In the light of the above statements, choose the correct answer from the options given below.

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📖 Explanation

Net charge on electyric dipole $=+q-q=0$ Hence, according to Gauss's law, Electric flux, $\varphi =\;{q_{\;\text{net }\;}}/{\epsilon_{0}}=\;\frac{0}{\epsilon_{0}}=0$ Electric field due to electric dipole is non-zero and varies at point to point. Hence, statement ${I}$ is true. Electric field due to charged solid sphere at a distance $r$ from centre. $E=\;\frac{1}{4 \pi \epsilon_{0}} \cdot \;\frac{Q r}{R^{3}} \;\; \begin{bmatrix}\\ \\ \text{ when }\\ \\ r < R \\ R \to \\ \\ \text{ radius }\\ \\ \end{bmatrix}$ which is non-zero. Hence, statement II is false. Hence, option (b) is the correct.

Q11JEE Main 2021MCQ4MAtoms and Nuclei

The recoil speed of a hydrogen atom after it emits a photon in going from $n=5$ state to $n=1$ state will be

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📖 Explanation

As we know, mass of photon, $m=1.6 \times 10^{-27} {kg}$ Using de-Broglie wavelength, $\lambda=\;\frac{h}{p}=\;\frac{h}{m v}$ . . . (i) where, $h$ is Planck's constant $=6.63 \times 10^{-34} {J}- {s}$ and $p$ is momentum of photon. By using Rydberg wavelength equation, $\;\frac{1}{\lambda} \;\; =R[\;\frac{1}{n_{2}^{2}}-\;\frac{1}{n_{1}^{2}}]$ $\;\frac{1}{\lambda} \;\; =1.09 \times 10^{7}[\;\frac{1}{1^{2}}-\;\frac{1}{25}]$ $\Rightarrow \;\; \;\frac{1}{\lambda} \;\; =1.09 \times 10^{7}[\;\frac{24}{25}]$ $\Rightarrow \;\; \lambda \;\; =\;\frac{25}{1.09 \times 10^{7} \times 24}$ Substituting values in Eq. (i), we get $v \;\; =\;\frac{h}{m \lambda}=\;\frac{6.63 \times 10^{-34} \times 1.09 \times 10^{7} \times 24}{1.6 \times 10^{-27} \times 25}$ $\;\; =\;\frac{6.63 \times 1.09 \times 24}{1.6 \times 2.5}=4.17 {ms}^{-1}$

Q12JEE Main 2021MCQ4MAlternating Current

Find the peak current and resonant frequency of the following circuit (as shown in figure).

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📖 Explanation

Given, inductance, $L=100 {mH}$ $=100 \times 10^{-3} {H}$ Capacitance, $C=100 \mu {F}$ $=100 \times 10^{-6} {F}$ Resistance, $R=120 Ω$ Maximum voltage, $V_{0}=30 {V}$ Angular frequency, $\omega=100 {rads}^{-1}$ Impedance, $Z=\sqrt{(X_{L}-X_{C})^{2}+R^{2}}$ Here, $X_{L}=\omega L=100 \times 100 \times 10^{-3}=10 Ω$ $\therefore \;\; X_{C} \;\; =\;\frac{1}{\omega C}=\;\frac{1}{100 \times 100 \times 10^{-6}}=100 Ω$ $\Rightarrow \;\; Z \;\; =\sqrt{X_{C}-X_{L})^{2}+R^{2}}$ $\;\; =\sqrt{(100-10)^{2}+(120)^{2}}=\sqrt{(90^{2}+120^{2}}=153.3 Ω$ As we know that, peak current, $I_{0}=\;\frac{V_{0}}{Z}=\;\frac{30}{153.3}=0.195 {A} ∼eq 0.2 {A}$ Resonance frequency, $\therefore \;\; f=\;\frac{1}{2 \pi \sqrt{L C}}=\;{1}/{2 \pi \sqrt{100 \times 10^{-3} \times 100 \times 10^{-6}}}$ $\Rightarrow \;\; f=\;{1}/{2 \pi \sqrt{10^{-5}}}=\;{1}/{2 \pi \times 10^{-2} \sqrt{10^{-1}}}=\;\frac{100 \pi}{2 \pi}=50 {Hz}$

Q13JEE Main 2021MCQ4MLaws of Motion

An inclined plane making an angle of $30^{\circ}$ with the horizontal is placed in a uniform horizontal electric field $200 {N} / {C}$ as shown in the figure. A body of mass $1 {kg}$ and charge $5 {mC}$ is allowed to slide down from rest at a height of $1 {m}$ . If the coefficient of friction is $0.2$ , find the time taken by the body to reach the bottom. $[$ Take, $g=9.8 {m} / {s}^{2}, \sin 30^{\circ}=\;\frac{1}{2}, \cos 30^{\circ}=\;\frac{\sqrt{3}}{2}]$

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📖 Explanation

Given, mass of block, $m=1 {kg}$ Acceleration due to gravity, $g=9.8 {ms}^{-2}$ Inclination, $\theta=30^{\circ}$ Electric field, $E=200 {N} / {C}$ Coefficient of friction, $\mu =0.2$ Charge, $q=5 {mC}=5 \times 10^{-3} {C}$ Let friction force, $f=\mu N$ where, $N$ be the normal reaction. Since, net force is zero along the perpendicular direction of incline. Therefore, force along $Y$ -axis will be zero. $\Rightarrow \;\; N \;\; =m g \cos 30^{\circ}+q E \sin 30^{\circ}$ $\Rightarrow \;\; N \;\; =1 \times 9.8 \times \sqrt{3} / 2+5 \times 10^{-3} \times 200 \times 1 / 2$ $\;\; =8.49+0.5$ $\;\; =8.99 {N} ∼eq 9 {N}$ $\therefore \;\; f=\mu N=\;\frac{2}{10} \times 9=\;\frac{18}{10}=1.8 {N}$ Now, total force along the plane of incline, $\;m g \sin 30^{\circ}-f-q E \cos 30^{\circ}=m a\;$ $\Rightarrow \;1 \times 9.8 \times 1 / 2-1.8-5 \times 10^{-3} \times 200 \sqrt{3} / 2\;=a$ $\Rightarrow \;5-1.8-1.732 / 2\;=a$ $\Rightarrow \;a=2.34 {ms}^{-2}$ Since, initial velocity of body, $u=0 {ms}^{-1}$ and distance along incline, $s=h / \sin 30^{\circ}=1 / \sin 30^{\circ}=2$ By using second equation of motion, $\;s \;\; =u t+1 / 2 {at}^{2}$ $\Rightarrow \;2 \;\; =0+1 / 2 \times 2.34 \times t^{2}$ $\Rightarrow \;t^{2} \;\; =\;\frac{4}{2.34}$ $\Rightarrow \;t=\;\frac{2}{\sqrt{2.34}}$

Q14JEE Main 2021MCQ4MLaws of Motion

Two masses $A$ and $B$ , each of mass $M$ are fixed together by a massless spring. A force acts on the mass $B$ as shown in figure. If the mass $A$ starts moving away from mass $B$ with acceleration $a$ , then the acceleration of mass $B$ will be

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📖 Explanation

Let ${a}_{ {CM}}$ be the acceleration of centre of mass of system and $F$ be the applied force on spring. $\therefore \;a_{C M}=\;\frac{F}{m_{A}+m_{B}}=\;\frac{m_{A} a_{A}+m_{B} a_{B}}{m_{A}+m_{B}}$ $\Rightarrow \;a_{C M}=\;\frac{F}{2 M}=\;\frac{M a_{A}+M a_{B}}{2 M} \;\; ( \because m_{A}=m_{B}=M)$ $\Rightarrow \;a_{B}=\;\frac{F-M a_{A}}{M}=\;\frac{F-M a}{M}$

Q15JEE Main 2021MCQ4MSemiconductor Electronics

Draw the output signal $Y$ in the given combination of gates

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📖 Explanation

According to the given figure signals are The output of given circuit diagram is By using concept of de-Morgan's law, $Y=\overline{\overline{A}+B}=A \cdot \overline{B}$ Truth table of given signals is Time interval $A$ $B$ $\overline{B}$ $A \overline{B}$ $(0-1) s$ 1 0 1 1 $(1-2) s$ 1 1 0 0 $(2-3) s$ 0 0 1 0 $(3-4) s$ 1 1 0 0 $(4-5) s$ 1 0 1 1Hence, output signal is,

Q16JEE Main 2021MCQ4MAtoms and Nuclei

A radioactive sample is undergoing $\alpha$ -decay. At any time $t_{1}$ , its activity is $A$ and another time $t_{2}$ , the activity is $\;\frac{A}{5}$ . What is the average life time for the sample?

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📖 Explanation

Given, activity pf radioactive element at time $t_{1}$ , $A_{1}=A$ and at time $t_{2}, A_{2}=A / 5$ As we know that, activity at any time $(A)=A_{0} {e}^{-\lambda t}$ where $A_{0}$ is activity at time $t=0$ . $\therefore \;A\;=A_{0} {e}^{-\lambda t_{1}}$ . . . (i) $\;\text{ and }\;\;A / 5=A_{0} {e}^{-\lambda t_{2}}$ . . . (ii) $\Rightarrow \;\; 5= {e}^{-\lambda t_{1}+\lambda t_{2}}$ Taking log on both sides, we get $\ln 5=\lambda(t_{2}-t_{1}) \ln {e}$ $\Rightarrow$ $\ln 5=\lambda(t_{2}-t_{1})$ $[ \because \ln e=1]$ $\Rightarrow$ $\lambda=\ln 5 /(t_{2}-t_{1})$ $\Rightarrow$ Also, average life, $\tau =\;\frac{1}{\lambda}$ $\Rightarrow \;\; \tau =\;\frac{t_{2}-t_{1}}{\ln 5}$

Q17JEE Main 2021MCQ4MMotion in a Straight Line

A scooter accelerates from rest for time $t_{1}$ at constant rate $a_{1}$ and then retards at constant rate $a_{2}$ for time $t_{2}$ and comes to rest. The correct value of $\;\frac{t_{1}}{t_{2}}$ will be

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📖 Explanation

According to given data; Initial speed of scooter, $u_{1}=0 {ms}^{-1}$ and final speed, $v_{2}=0 {ms}^{-1}$ By using first equation of motion, $\;v_{1} \;\; =u+a_{1} t_{1}$ $\;\Rightarrow \;v_{1} \;\; =a_{1} t_{1}$ . . . (i) and for second case, $v_{2} \;\; =v_{1}-a_{2} t_{2}$ $0 \;\; =v_{1}-a_{2} t_{2}$ $v_{1} \;\; =a_{2} t_{2}$ . . . (ii) On dividing Eq. (i) by Eq. (ii), we get $\;v_{1} / v_{1} \;\; =1=\;\frac{a_{1} t_{1}}{a_{2} t_{2}}$ $\Rightarrow \;a_{2} t_{2} \;\; =a_{1} t_{1}$ $\Rightarrow \;t_{1} / t_{2} \;\; =a_{2} / a_{1}$

Q18JEE Main 2021MCQ4MOscillations

Given below are two statements: Statement I A second's pendulum has a time period of $1 {s}$ . Statement II It takes precisely one second to move between the two extreme positions. In the light of the above statements, choosethe correct answer from the options given below.

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📖 Explanation

Statement $I$ is false because time period of second's pendulum is always $2 {s}$ . Therefore, time taken to move between two extreme positions will be $T / 2=2 / 2=1 {s}$ Hence, option (b) is the correct.

Q19JEE Main 2021MCQ4MCurrent Electricity

A wire of $1 Ω$ has a length of $1 {m}$ . It is stretched till its length increases by $25 \%$ . The percentage change in resistance to the nearest integer is

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📖 Explanation

Given, resistance of wire, $R=1 Ω$ Increase in length, $\Delta l=25 \%$ Let $I_{1}, I_{2} A_{1}, A_{2}, V_{1}$ and $V_{2}$ be the length, area and volume, respectively. As, volume remains same and $V=A l$ $\therefore \;\; A_{1} l_{1}=A_{2} l_{2}$ $\Rightarrow \;\; A_{2}=\;\frac{A_{1} I_{1}}{I_{2}}$ a nd resistance, $R=\;\frac{\rho l}{A}$ $\therefore \;\; \;\frac{R_{2}}{R_{1}} \;\; =\;\frac{\rho I_{2}}{A_{2}} \times \;\frac{A_{1}}{\rho I_{1}}$ $\Rightarrow \;\; \;\frac{R_{2}}{R_{1}} \;\; =\;\frac{I_{2} A_{1}}{A_{2} l_{1}}=\;\frac{1.25 I_{1} \cdot A_{1} \times I_{2}}{A_{1} l_{1} \times I_{1}}( \because I_{2}=1.25 I_{1})$ $\Rightarrow \;\; \;\frac{R_{2}}{R_{1}} \;\; =\;\frac{(1.25 I_{1})^{2} A_{1}}{I_{1}^{2} A_{1}} \Rightarrow R_{2}=(1.25)^{2} R_{1}$ $\;\text{ Now, }\; \;\frac{\Delta R}{R} \times 100 \;\; =\;\frac{R_{2}-R_{1}}{R_{1}} \times 100$ $\;\; =\;\frac{(1.25)^{2} R_{1}-R_{1}}{R_{1}} \times 100$ $\therefore \;\; \;\frac{\Delta R}{R} \times 100 \;\; =\;\frac{(1.25)^{2}-1}{1} \times 100$ $\;\; =56.25 \% ∼eq 56 \%$

Q20JEE Main 2021MCQ4MRay Optics and Optical Instruments

The incident ray, reflected ray and the outward drawn normal are denoted by the unit vectors $a$ , $b$ and $c$ , respectively. Then, choose the correct relation for these vectors.

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📖 Explanation

Taking component of $a, b$ and $\text{c} \text{along}\hat{i}, \hat{j}$ and $\hat{k}$ . $\;a=\sin \theta \hat{i}-\cos \theta \hat{j}$ $\;b=\sin \theta \hat{i}+\cos \theta \hat{j}$ $\;c=\hat{j}$ Now, on solving option (c), we get $a-2(a \cdot c) c \;\; =(\sin \theta \hat{i}-\cos \theta \hat{j})-2[(\sin \theta \hat{i}-\cos \theta \hat{j}) \cdot \hat{j}] \hat{j}$ $\;\; =\sin \theta \hat{i}+\cos \theta \hat{j}$ $\;\; =b$

Q21JEE Main 2021NAT4MThermodynamics

The volume $V$ of a given mass of monoatomic gas changes with temperature $T$ according to the relation $V=k T^{2 / 3}$ . The work done when temperature changes by $90 {K}$ will be $x R$ . The value of $x$ is.......... $[R=$ universal gas constant $]$

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📖 Explanation

Given, $\;\; V=k T^{2 / 3} \;\; . . .$ (i) where, $T$ is temperature. Change in temperature, $\Delta T=90 {K}$ Let $p$ be the pressure, $d V$ be the change in volume and work done be $W$ . As we know that, $W=\int p d V \;\; . . . \;\text{ (ii) }\;$ As, $\;\; p V=n R T$ $\therefore \;\; p=\;\frac{n R T}{V}$ Substituting this value in Eq. (ii), we get $\;W=\int n R T \;\frac{d V}{V}$ $\;W=\int n R T \;\frac{d V}{k T^{2 / 3}} \;\; \;\text{ [using Eq. (i)] ... (iii) }\;$ $\;\text{ On differentiating Eq. (i) w.r.t. temperature on both sides, we get }\;$ $\;\frac{d V}{d T} \;\; =k \cdot \;\frac{2}{3} T^{1 / 3}$ $\;\; =\;\frac{2}{3} k T^{-1 / 3}=\;\frac{2}{3} \;\frac{k}{T^{1 / 3}}$ $\therefore \;\; d V \;\; =2 / 3 k T^{-1 / 3} d T$ Substituting this in Eq. (iii), we get $W \;\; =\int n R T \;\frac{2 / 3 k T^{-1 / 3} d T}{k T^{2 / 3}}$ $\Rightarrow \;\; W \;\; =\;\frac{2}{3} n R \int T d T=2 / 3 n R[T]_{T_{1}^{T}}$ $\;\; =2 / 3 n R[T_{2}-T_{1}]=2 / 3 n R \Delta T$ $\;\; =2 / 3 n R 90=60 n R$ Here, $n=1$ So, work done will be $60 R$ . Hence, $x=60$ .

Q22JEE Main 2021NAT4MElectromagnetic Waves

If the highest frequency modulating a carrier is $5 {kHz}$ , then the number of AM broadcast stations accommodated in a $90 {kHz}$ bandwidth are ......... .

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📖 Explanation

Given, highest modulating frequency, $f=5 {kHz}$ Bandwidth $=90 {kHz}$ Since, number of station $=\;\frac{\;\text{ bandwidth }\;}{\;\text{ band width / station (2f) }\;}$ $=\;\frac{\;\text{ Bandwidth }\;}{2 f}=\;\frac{90}{2 \times 5}=9$

Q23JEE Main 2021NAT4MDual Nature of Matter and Radiation

Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is $x: y$ . The value of $x$ is .............. .

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📖 Explanation

Given, first incident energy, $E_{i_{1}}=2$ (work function $\varphi _{0}$ ) Second incident energy, $E_{i_{2}}=10 \varphi _{0}$ If $v_{1}$ and $v_{2}$ be the respective maximum velocities in two cases, then by using the concept of photoelectric effect, ${KE}=E_{i}-\varphi _{0}$ ${KE}=E_{i}-\varphi _{0}$ $\therefore \;\; 1 / 2 m v_{1}^{2}=2 \varphi _{0}-\varphi _{0}=\varphi _{0}$ . . . (i) and $\;\; 1 / 2 m v_{2}^{2}=10 \varphi _{0}-\varphi _{0}=9 \varphi _{0}$ . . . (ii) On dividing Eq. (i) by Eq. (ii), we get $(\;\frac{v_{1}}{v_{2}})^{2}=\;\frac{1}{9}$ ${t} \;\frac{v_{1}}{v_{2}}=\;\frac{1}{3}=\;\frac{x}{y}$ $\therefore \;\; x=1$

Q24JEE Main 2021NAT4MRay Optics and Optical Instruments

A point source of light $S$ , placed at a distance $60 {cm}$ in front of the centre of a plane mirror of width $50 {cm}$ , hangs vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance $1.2 {m}$ from it (see in the figure). The distance between the extreme points, where he can see the image of the light source in the mirror is ......... ${cm}$ .

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📖 Explanation

Given, length of mirror, $m=50 {cm}=50 \times 10^{-2} {m}$ Distance of source from mirror, $d=60 {cm}$ $=60 \times 10^{-2} {m}$ Distance of man from mirror, $d_{m}=1.2 {m}$ By using the concept of ray diagram of plane mirror shown belowNow, using the concept of similar triangle, $\Delta H A l ∼ \triangle G A E \;\text{ and }\; \Delta B A l\;∼ \Delta C A E$ $\therefore \;\;\frac{A I}{A E}=\;\frac{H I}{E G}$ $\Rightarrow \;\;\frac{0.60}{1.8}=\;\frac{0.25}{E G}$ $\Rightarrow \;E G=0.25 \times \;\frac{1.8}{0.6}=0.25 \times 3=0.75 {m}$ $\;\text{ As, }\;\;C G=2 E G$ $\Rightarrow \;C G=0.75 \times 2=1.50 {m}$ $\;\;\frac{A I}{A E} \;\; =\;\frac{H I}{E G}$ $\Rightarrow \;\;\frac{0.60}{1.8} \;\; =\;\frac{0.25}{E G}$ $\Rightarrow \;E G \;\; =0.25 \times \;\frac{1.8}{0.6}=0.25 \times 3=0.75 {m}$ $\;\text{ As, }\;\;C G \;\; =2 E G$ $\Rightarrow \;C G=0.75 \times 2=1.50 {m}$ Hence, distance between the extreme points, where he can see image of light source in mirror is $150 {cm}$ .

Q25JEE Main 2021NAT4MOscillations

A particle executes SHM with amplitude a and time period $T$ . The displacement of the particle when its speed is half of maximum speed is $\;\frac{\sqrt{x} a}{2}$ . The value of $x$ is ........... .

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📖 Explanation

Given, amplitude of particle in ${SHM}= {a}$ Maximum velocity, $v=A \omega$ Let displacement be $y$ when speed, $v_{2}=v / 2=\;\frac{A \omega}{2}$ By using equation of wave velocity, $v_{2}^{2}=\omega^{2}(A^{2}-y^{2})$ $\; \because \;\;(\;\frac{\omega A}{2})^{2} \;\; =\omega^{2}(A^{2}-y^{2})$ $\Rightarrow \;\;\frac{\omega^{2} A^{2}}{4} \;\; =\omega^{2}(A^{2}-y^{2})$ $\Rightarrow \;\;\frac{A^{2}}{4} \;\; =A^{2}-y^{2}$ $\Rightarrow \;y \;\; =\;\frac{\sqrt{3}}{2} A$ Hence, $x=3$

Q26JEE Main 2021NAT4MElectrostatic Potential and Capacitance

27 similar drops of mercury are maintained at $10 {V}$ each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ........... that of a smaller drop.

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📖 Explanation

Given, $n=27$ $V_{1}=10 {V}$ Let $q_{1}$ be the charge of one drop, $r_{1}$ be its radius, $r$ be the radius of bigger drop and $q$ be its charge. As, volume remains constant. $\;\text{ Electric potential }\;(V)=\;\frac{3 k q_{1}^{2}}{5 r}$ $\therefore \;4 / 3 \pi r^{3}\;=4 / 3 \pi r_{1}^{3} \times 27$ $\Rightarrow \;r^{3}\;=27 r_{1}^{3}$ $\Rightarrow \;r\;=3 r_{1}$ where, $k$ is Coulomb constant. Therefore, potential energy of small drop, $U_{s}=\;\frac{3}{5} \;\frac{k q_{1}^{2}}{r_{1}}$ . . . (i) and potential energy of bigger drop, $U_{B}=\;\frac{3}{5} \;\frac{k(q_{1}^{2} \times 27)^{2}}{r}$ $U_{B}=\;\frac{3}{5} \;\frac{k(27)^{2} q_{1}^{2}}{3 r_{1}}$ . . . (ii) On dividing Eq. (ii) by Eq. (i), we get $\;\frac{U_{B}}{U_{s}}=\;\frac{27 \times 27}{3}=243$ $\;U_{B}=243 U_{s}$ Hence, potential energy of big drop is 243 times of small drop.

Q27JEE Main 2021NAT4MOscillations

Time period of a simple pendulum is $T$ . The time taken to complete $5 / 8$ oscillations starting from mean position is $\;\frac{\alpha}{\beta} T$ . The value of $\alpha$ is ......... .

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📖 Explanation

Given, angular displacement to complete $\;\frac{5}{8}(=\;\frac{1}{2}+\;\frac{1}{8})$ rev $\;\; =(\pi+\;\frac{\pi}{6}) {rad}$ $\;\; =(\;\frac{7 \pi}{6}) {rad}$ Since, $\;\; \omega=\;\frac{2 \pi}{T}=\;\frac{\theta}{t}$ $\Rightarrow \;\; \;\; \theta\;=\;\frac{2 \pi}{T} \cdot t \Rightarrow \;\frac{7 \pi}{6}=\;\frac{2 \pi}{T} t$ $\Rightarrow \;\; t\;=\;\frac{7 T}{12}=\;\frac{\alpha}{\beta} T$ Hence, $\alpha=7$

Q28JEE Main 2021NAT4MGravitation

In the reported figure of Earth, the value of acceleration due to gravity is same at point $A$ and $C$ but it is smaller than that of its value at point $B$ (surface of the Earth). The value of $O A: A B$ will be $x: y$ . The value of $x$ is .......... .

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📖 Explanation

Given, $g_{A}=g_{C}$ Radius of Earth, $R=6400 {km}$ Height, $h=3200 {km}=\;\frac{R}{2}$ Since, $g=\;\frac{G M}{R^{2}}$ where, $G$ is gravitational constant. $\therefore \;\; g_{A}=\;\frac{G M(O A)}{R^{3}}$ . . . (i) and $g_{C} \;\; =\;\frac{G M}{(O C)^{2}}=\;\frac{G M}{R+h}$ $\;\; =\;\frac{G M}{(R+R / 2)^{2}}=\;\frac{G M}{(3 R / 2)^{2}}$ . . . (ii) According to given information from Eqs. (i) and (ii), we get $\Rightarrow \;\; O A \;\; =\;\frac{4}{9} R=\;\frac{4}{9} \times 6400$ $A B \;\; =O B-O A$ $\;\text{ and }\;\;A B=6400-\;\frac{4}{9} \times 6400$ $\Rightarrow \;\; =6400(1-4 / 9)=6400 \times 5 / 9$ $\;\text{ Now, }\; O A: A B=\;\frac{6400 \times 4 / 9}{5 / 9 \times 6400}=\;\frac{4}{5}$ $\;\text{ Hence, }\; x=4$

Q29JEE Main 2021NAT4MThermodynamics

1 mole of rigid diatomic gas performs a work of $Q / 5$ when heat $Q$ is supplied to it. The molar heat capacity of the gas during this transformation is $\;\frac{x R}{8}$ , The value of $x$ is .......... . $[R=$ universal gas constant $]$

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📖 Explanation

Given, number of mole, $n=1$ Heat supplied $=Q$ Work done $=Q / 5$ By using first law of thermodynamics, $\Delta Q \;\; =\Delta U+\Delta W$ $Q \;\; =\Delta U+Q / 5$ $\Delta U \;\; =4 Q / 5=n C_{v} \Delta T$ . . . (i) where, $C_{V}$ is heat coefficient at constant volume and $\Delta T$ is change in temperature. Degree of freedom of diatomic gas, $f=5$ $\therefore \;\; Q=\;\frac{5}{4} \;\frac{f}{2} R \Delta T$ [from Eq. (i)] $\Rightarrow \;\; \;\frac{Q}{\Delta T}=C=\;\frac{5}{4} \times \;\frac{5}{2} R=\;\frac{25}{8} R$ Hence, $x=25$

Q30JEE Main 2021NAT4MSemiconductor Electronics

The Zener diode has a $V_{2}=30 {V}$ . The current passing through the diode for the following circuit is ......... mA.

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📖 Explanation

Given, Zener diode voltage, $V_{z}=30 {V}$ Supply voltage, $V=90 {V}$ According to figure, Let $I$ be supply current, then $I_{2}$ be Zener current $(=I-I_{1})$ and $I_{1}$ be current through $5 {k} Ω$ . $\therefore \;\; V_{1}=V_{2}=30 {V}$ By Ohm's law, $\;\; V_{1}=V_{2}=I_{1} R$ $\Rightarrow \;\; I_{1}=V_{2} / R=\;\frac{30}{5000}=6 {mA}$ If $V^{'}$ be the voltage across $4 {k} Ω$ resistor. $\therefore \;\; V^{'}=90-30=60 {V}$ and $I=\;\frac{60}{4000}=15 {mA}$ Therefore, $I_{2}=I-I_{1}$ $=15-6=9 {mA}$

Chemistry30 questions

Q31JEE Main 2021MCQ4MStructure of Atom

Which of the following forms of hydrogen emits low energy $\beta^{-}$ particles ?

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📖 Explanation

Let us calculate $\;\frac{n}{p}$ value (number of neutrons to number of protons) of the isotopes of hydrogen $({ }_{1}^{1} {H},{ }_{1}^{2} {H},{ }_{1}^{3} {H})$ and proton ${H}^{+}$ . Isotopes\Properties Protium $[{ }_{1}^{1} {H}]$ Deuterium $[{ }_{1}^{2} {H}]$ Tritium $[{ }_{1}^{3} {H}]$ Proton $[ {H}^{+}]$ Massnumber $(A)$ 1 2 3 1 Atomicnumber $(Z)$ 1 1 1 1 Number of $n$ 0 1 2 0 Number of $p$ 1 1 2 0 $\;\frac{n}{p}$ value 0 1 2 0So, tritium is radioactive because of its higher $\;\frac{n}{p}$ value $(2 > 1.5)$ which makes its nucleus unstable. Tritium emits low energy $\beta^{-}$ particles ${ }_{-1}^{0}$ e and thus reduces its $\;\frac{n}{p}$ value to attain stability. $\;{ }_{1}^{3} {H} \longrightarrow { }_{2}^{3} {He}+{ }_{-1}^{0} {e}(\beta^{-})$ $\;\frac{n}{p}=2 \;\; \;\frac{n}{p}=0.5$

Q32JEE Main 2021MCQ4Mp Block Elements

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) In ${TII}_{3}$ , isomorphous to ${CSl}_{3}$ , the metal is present in $+1$ oxidation state. Reason (R) Tl metal has fourteen $f$ -electrons in the electronic configuration. In the light of the above statements, choose the most appropriate answer from the options given below.

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📖 Explanation

Assertion (A) is correct. ${TII}_{3}$ and ${CsI}_{3}$ are the triiodide, ${I}_{3}^{-}$ (polyhalide) compounds of ${TI}^{+}$ and ${Cs}^{+}$ ions. Due to inert pair effect of group $-13$ elements, ${Tl}^{+}$ is more stable than ${TH}^{3+}$ . Cs and Tl belong to the same period (6th). Sizes of ${Cs}^{+}$ and ${Tl}^{+}$ are nearly same. So, geometrical network and lattice pattern of both ${CsI}_{3}$ and ${TII}_{3}$ are same (bcc lattice). Csl_3 and ${TII}_{3}$ are also able to form mixed crystals when crystallisation is carried out from a mixture of saturated solutions of ${CsI}_{3}$ and ${TII}_{3}$ . These properties enable ${TII}_{3}$ and ${Csl}_{3}$ to be isomorphous crystals. Reason $(R)$ is also correct. Electronic configuration of $C$ s $(Z=55)$ and $T I(Z=81)$ as follows ${ }_{55} {Cs}:[{ }_{54} {Xe}] 6 s^{1}$ ${ }_{81} {Tl}:[{ }_{54} {Xe}] 4 f^{14}{ }_{5 d}{ }^{10} 6 s^{2} 6 p^{1}$ So, Tl has fourteen $f$ -electrons in the anti-penultimate (4th) shell. ${A}$ and ${R}$ are individually correct and ${R}$ is the correct explanation of ${A}$ .

Q33JEE Main 2021MCQ4MBiomolecules

Match List-I with List-II. List-I List-II A. Sucrose (i) $\;\; \beta$ -D-galactose and $\beta-D-$ glucose B. Lactose (ii) $\alpha$ -D-glucose and $\beta$ -D-fructose C. Maltose (iii) $\alpha$ -D-glucose and $\alpha$ -D-glucoseChoose the correct answer from the options given below.

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📖 Explanation

Sucrose (cane sugar), lactose (milk sugar) and maltose are disaccharides in which two monosaccharides are hold together by a glycosidic linkage. (A) In sucrose, $C_{1}$ of $\alpha-D-$ glucose and $C_{2}$ of $\beta-D-$ fructose are linked together (ii). (B) In lactose, ${C}_{1}$ of $\beta$ - D-galactose and ${C}_{4}$ of $\beta$ -D-glucose are linked together (i). (C) In maltose, ${C}_{1}$ of $\alpha-$ D-glucose is linked to ${C}_{4}$ of another $\alpha- {D}-$ glucose unit (iii).So, option (c) is the correct answer.

Q34JEE Main 2021MCQ4MAmines

A. phenyl methanamine B. N,N-dimethylaniline C. N-methyl aniline D. Benzenamine Choose the correct order of basic nature of the above amines.

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📖 Explanation

In phenyl methanamine, the lone pair on nitrogen of $- {NH}_{2}$ group is localised and does not undergoes resonance as attached to $s p^{3}$ -hybridised C-atom. Phenyl methanamine or benzyl amine $(B),(C)$ and $(D)$ are aromatic amines in which lone pair of electrons of N-atoms goes in resonance (+ $R$ effect) with the benzene ring. So, Lewis basicity or donation of lone of electrons of these amines will be decreased in comparison to (A). $+I$ effects of two $- {CH}_{3}$ groups are in compensation with $-R$ effect of the ring. $+$ l effect of one $- {CH}_{3}$ group is trying to compensate $-R$ effect of the ring.It has no $+$ leffect on $N$ -atom to outweigh $-R$ -effect of the ring. So, $A$ is purely aliphatic $1^{\circ}$ -amine. $B$ is aromatic $3^{\circ}$ -amine with more aliphatic nature (for two $- {CH}_{3}$ groups). ${C}$ is aromatic $2^{\circ}$ -amine with less aliphatic nature (for one $- {CH}_{3}$ group). $D$ is purely aromatic $1^{\circ}$ -amine. Hence basicity order is $A > B > C > D$ .

Q35JEE Main 2021MCQ4MClassification of Elements

The correct order of electron gain enthalpy is

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📖 Explanation

Oxygen is the second most electronegative element in comparison to fluorine. In group - 16 family $( {O}, {S}, {Se}, {Te}), {O}$ -atom is smallest in size. So, electron density on O-atom is very high in group $-16$ . During addition of a free electron to gaseous O-atom, ${O}(g)+ {e}^{-} \longrightarrow {O}^{-}(g)$ We have to supply a significant amount of energy (endothermic) to overcome the electrostatic repulsion between the approaching electron and O-atom of very high electron density. So, the net value of electron affinity (EA) or (negative) electron gain enthalpy $[\Delta _{ {eg}} H$ or $|\Delta _{ {eg}} H|]$ of oxygen decreases to a higher extent in comparison to other elements of group - 16 who have larger size and lower electronegativity. So, the correct order of EA or $|\Delta _{ {eg}} H|$ of group $-16$ elements will be $S$ $> {Se} > {Te} > {O}$ .

Q36JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

$\ln {C}^{1} {H}_{2}={ {C}^{2}}={ {C}^{3}} {H}- {C}^{4} {H}_{3}$ molecule, the hybridisation of carbon $1,2,3$ and 4 respectively are

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📖 Explanation

${C}^{1} {H}_{2}={ {C}^{2}}={ {C}^{3}} {H}- {C}^{4} {H}_{3}$ It is buta $-1,2$ - diene,Here, ${C}_{1} \to s p^{2}, {C}_{2} \to s p, {C}_{3} \to s p^{2}$ and ${C}_{4} \to s p^{3}$ hybridised. So, option (c) is the correct answer.

Q37JEE Main 2021MCQ4MBiomolecules

Seliwanoff test and xanthoproteic test are used for the identification of and respectively.

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📖 Explanation

Seliwanoff and xanthoproteic tests are used for the identification of ketoses and proteins respectively. Seliwanoff test It is a chemical test which distingusihes between aldose sugars (e.g. glucose and maltose) and ketose sugars (e.g., fructose and sucrose). The test is based on the principle that, when heated with acid, ketoses are more readily dehydrated than aldoses. The reagents used in the test consist of conc. ${HCl}$ and resorcinol. Ketoses ${\to }^{\;\text{ Conc. }\; {HCl}}$ Deep cherry red colour Aldoses Resorcinol ${\to }^{\;\text{ Conc. }\; {\text{HCl},\text{Resorcinol}}}$ Faint pink colour For ketoses, a deep cherry red colour is formed rapidly indicating a positive test. Xanthoproteic test It is qualitative test to detect the presence of protein soluble in a solution using ${HNO}_{3}$ . In this test, solution of protein is first heated with conc. ${HNO}_{3}$ and then the mixture is neutralised by $40 \% {NaOH}$ solution. If the colour changes from yellow to orange, this confirms presence of a protein. Protein solution ${\to }_{(\text{ii}) 40 \% {\text{NaOH}}}^{\;\text{ (i) Conc. }\; {\text{HNO}}_{3} / \Delta }$ Yellow colouration which finally changes into orange colour.

Q38JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

2,4 -DNP test can be used to identify

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Q39JEE Main 2021MCQ4MPractical Organic Chemistry

Ceric ammonium nitrate and ${CHCl}_{3}$ /alc. ${KOH}$ are used for the identification of functional groups present in and respectively.

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📖 Explanation

Ceric ammonium nitrate and ${CHCl}_{3} / {alc} . {KOH}$ are used for the identification of functional groups present in alcohol and amine respectively. Ceric ammonium nitrate (CAN) test When an alcohol reacts with few drops of CAN, a red complex ammonium nitrate is developed which confirms presence of alcoholic $- {OH}$ group. ${3 R O H}_{\text{Alcohol} }+{(N H_{4})_{2}[C e(N O_{3})_{6}] }_{\text{Ceric} \text{ammonium}}{{\to }_{\text{Red}\text{colour} }^{-NH_4N O_3 }{ [ {Ce}(N O_{3})_{4}(R O H)_{3}]}_{\text{Red} \text{complex} }$ Test with ${CHCl}_{3}$ and alcoholic ${KOH}$ [Carbylamine test or Saytzeff's isocyanide test] When a primary amine (aliphatic or aromatic) is heated with alcoholic ${KOH}$ and chloroform $( {CHCl}_{3})$ , the isocyanide (carbylamine) is formed as indicated by a foul order. ${R- {NH}_{2}}_{1^{\circ}-\text{amine}}+ {CHCl}_{3}+{3 {KOH}}_{(\text{alc}.)} {\to }_{-3 {KCl}-3 {H}_{2} {O}}^{\Delta } {R- {NC}}_{\text{Carbylamine}}$ Secondary and tertiary amines do not respond to carbylamine test.

Q40JEE Main 2021MCQ4MClassification of Elements

Which pair of oxides is acidic in nature?

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📖 Explanation

(a) ${B}_{2} {O}_{3} \longrightarrow$ oxide of boron (non-metal) :acidic ${CaO} \longrightarrow$ oxide of calcium (metal) : basic (b) ${B}_{2} {O}_{3} \longrightarrow$ oxide of boron (non-metal) : acidic ${SiO}_{2} \longrightarrow$ oxide of silicon (non-metal) : acidic (c) ${N}_{2} {O} \longrightarrow$ oxide of nitrogen (non-metal) : neutral ${BaO} \longrightarrow$ oxide of barium (metal) : basic (d) ${CaO} \longrightarrow$ oxide of calcium (metal): basic ${SiO}_{2} \longrightarrow$ oxide of silicon (non-metal): acidic So, option (b) is the correct answer. ${B}_{2} {O}_{3}$ and ${SiO}_{2}$ are being acidic oxide they react with base to give salt and water. ${B}_{2} {O}_{3}+2 {NaOH} \longrightarrow {2 {NaBO}_{2}}_{\text{Sodium}\text{metaborate}}+ {H}_{2} {O}$ ${SiO}_{2}+2 {NaOH} \longrightarrow {{Na}_{2} {SiO}_{3}}_{\text{Sodium}\text{silicate}}+ {H}_{2} {O}$

Q41JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Identify $A$ in the given chemical reaction,

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📖 Explanation

It is an example of intramolecular aldol condensation. $\alpha$ - hydrogen atom of one - CHO group gets abstracted by ${NaOH}$ form enolate ion which then attacks $= {C}= {O}$ another $- {CHO}$ group to form aldol or $\beta$ - hydroxy aldehyde. The aldol on heating ${C}_{2} {H}_{5} {OH} / {H}_{2} {O}$ forms the final product ( $\alpha, \beta$ -unsaturated aldehyde) as the major product. Mechanism

Q42JEE Main 2021MCQ4MAlcohols, Phenols and Ethers

Identify $A$ in the following chemical reaction.

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📖 Explanation

Here, 4-methoxybenzaldehyde in series of reaction finally forms 4-(iodomethyl) phenol $(A)$ . Let us compute the reaction step by step with mechanism as follows (i) In first step, oxidising agent ( $p$ -methoxybenzaldehyde) reacts with formaldehyde in presence of strong base ${NaOH}$ to give $p$ -methoxybenzyl alcohol along with sodium salt of methanoic acid. It is known as Cannizzaro's reaction. (ii) In second step, deprotonation of $p$ -methoxybenzyl alcohol in presence of sodium hydride $( {NaH})$ to form alkoxide which further react with bromoethane in presence of DMF to give 1-ethoxymethyl-4-methoxybenzene.(iii) In last step, 1-ethoxymethyl-4-methoxybenzene undergoes solvolysis reaction followed by ${S}_{ {N}} 2$ attack of two iodide ion (I $^{-}$ ) to give 4-iodomethyl phenol, iodoform and iodoethane.

Q43JEE Main 2021MCQ4MSolutions

Calgon is used for water treatment. Which of the following statement is not true about Calgon?

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📖 Explanation

Calgon is used to remove permanent hardness of water by removing cation ${Mg}^{2+}$ . Calgon is not an element, it is a compound. The 2 nd most abundant element by weight in the Earth's crust is silicon (Si). So, the statement in option (a) is not true. Calgon is sodium hexametaphosphate $( {NaPO}_{3})_{6}$ or ${Na}_{2}[ {Na}_{4}( {PO}_{3})_{6}]$ . It is a polymeric water soluble compound. It is also named as Graham's salt. ${Mg}^{2+}$ of hard water is removed as ${{Na}_{2}[ {Na}_{4}( {PO}_{4})_{3}]}_{\text{Calgon}}+2 {Mg}^{2+}( {aq}) \longrightarrow {Na}_{2}[ {Mg}_{2}( {PO}_{3})_{6}]( {s})$ $+4 {Na}^{+}( {aq})$ ${Ca}^{2+}$ ion cannot be removed by calgon as ${Ca}^{2+}$ is larger in size than ${Mg}^{2+}$ . So, the statement of option (b), (c) and (d) are true.

Q44JEE Main 2021MCQ4MHaloalkanes and Haloarenes

Match List-I with List-II. List-I A. B. C. $2 {CH}_{3} {CH}_{2} {Cl}+2 {Na} {\longrightarrow }^{\;\text{ Ether }\;} {C}_{2} {H}_{5}- {C}_{2} {H}_{5}+2 {NaCl}$ D. $2 {C}_{6} {H}_{5} {Cl}+2 {Na} {\longrightarrow }^{\;\text{ Ether }\;} {C}_{6} {H}_{5}- {C}_{6} {H}_{5}+2 {NaCl}$ List-II (i) Wurtz reaction (ii) Sandmeyer reaction (iii) Fittig reaction (iv) Gattermann reaction Choose the correct answer from the options given below.

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📖 Explanation

Benzene diazonium chloride undergoes Sandmeyer reaction in presence of ${Cu}_{2} {Cl}_{2}$ while undergoes Gattermann reaction in presence of ${Cu} / {HCl}$ to give chlorobenzene along with nitrogen gas $( {N}_{2})$ . So, (A) - (ii) Sandmeyer reaction (B) - (iv) Gattermann reaction (C) Ethyl chloride reacts with sodium in presence of dry ether to give $n$ -butane. This reaction is known as Wurtz reaction. ${2 {CH}_{3} {CH}_{2} {Cl}}_{\text{Ethyl} \text{chloride} \$ [\text{Aliphatic} 1^{\circ} \text{halide}]$}+2 {Na} {\longrightarrow }^{;\text{ Ether };} {{CH}{3} {CH}{2}- {CH}{2} {CH}{3}}_{n-\text{butane}}+2 {NaCl} $It is Wurtz reaction in which an aliphatic$ 1^{\circ} $or$ 2^{\circ} $halides produces a symmetrical alkane like$ n $-butane. (D) Chlorobenzene reacts with sodium in presence of dry ether to give biphenyl. This is known as Fittig reaction.It is Fittig reaction in which a halobenzene produces biphenyl. So,$ C$ - (i) Wurtz reaction D - (iii) Fittig reaction So, correct match is A-(ii), B-(iv), C-(i) and D-(iii).

Q45JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

considering the above reaction, the major product among the following is

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Q46JEE Main 2021MCQ4Mp Block Elements

Match List-I with List-II List-I(Molecule) List-II(Bond order) A. ${Ne}_{2}$ (i) 1 B. ${N}_{2}$ (ii) 2 C. ${F}_{2}$ (iii) 0 D. ${O}_{2}$ (iv) 3Choose the correct answer from the options given below.

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📖 Explanation

(A) ${Ne}_{2}(20 {e}^{-})$ $\Rightarrow \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \pi 2 p_{x}^{2} \pi 2 p_{y}^{2} \pi^{*} 2 p_{x}^{2} \pi^{*} 2 p_{y}^{2} \sigma^{*} 2 p_{z}^{2}$ Bond order $=\;\frac{10-10}{2}=0 \Rightarrow$ (iii) of List-II. (B) ${N}_{2}(14 {e}^{-})$ $\Rightarrow \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}^{2}, \pi 2 p_{y}^{2} \sigma 2 p$ Bond order $=\;\frac{10-4}{2}=3 \Rightarrow$ (iv) of List-II. (c) ${F}_{2}(18 {e}^{-})$ $\Rightarrow \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \pi 2 p_{x}^{2} \pi 2 p_{y}^{2} \pi^{*} 2 p_{y}^{2} \pi^{*} 2$ Bond order $=\;\frac{10-8}{2}=1 \Rightarrow$ (i) of List-II. (d) ${O}_{2}(16 {e}^{-})$ $\Rightarrow \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}{ }^{2} \pi 2 p_{x}{ }^{2} \pi 2 p_{y}{ }^{2} \pi^{*} 2 p_{x}{ }^{1} \pi^{*} 2 p_{y}{ }^{1}$ Bond order $=\;\frac{10-6}{2}=2 \Rightarrow$ (ii) of List-II. The correct matching is option (a). No. of electrons in antibonding MO (Molecular Orbital) Note : Bond order $=\;\frac{-\;\text{ No. of electrons \in bonding MO }\;}{2}$

Q47JEE Main 2021MCQ4MAlcohols, Phenols and Ethers

Identify $A$ in the given reaction.

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📖 Explanation

The given reaction is an application of Darzen's reaction in which an aliphatic alcohol (preferably $1^{\circ}$ or $2^{\circ}$ ) gets converted into the respective alkyl chloride (in pure form) through ${S}_{ {N}} 2$ pathway when it usesNote Phenolic - OH group does not respond to this reaction as $+R$ effect of phenolic $- {OH}$ group gives double bond character to the ${C}- {O}$ bond.

Q48JEE Main 2021MCQ4Md and f Block Elements

Match List-I with List-II. List-I List-II A. Siderite (i) Cu B. Calamine (ii) Ca C. Malachite (iii) Fe D. Cryolite (iv) Al (v) ZnChose the correct answer from the options given below.

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📖 Explanation

(A) Siderite ({FeCO}{3} $) is the ore of$ {Fe} \Rightarrow $(iii) of List-II. (B) Calamine$ ( {ZnCO}{3}) $is the ore of$ {Zn} \Rightarrow $(v) of List - II. (C) Malachite$ [ {CuCO}{3} ⋅ {Cu}( {OH}){2}] $is the ore of$ {Cu} \Rightarrow $(i) of List II. (D) Cryolite$ ( {Na}{3} {AIF}{6}) $is the ore of$ {Al} \Rightarrow $(iv) of List II. (as well as$ {F}_{2}$ ) So, option -(c) is the correct matching. Note Siderite and calamine are carbonate ores whereas malachite is basic carbonate ore. Cryolite is the thioxide ore.

Q49JEE Main 2021MCQ4MSolutions

The nature of charge on resulting colloidal particles when ${FeCl}_{3}$ is added to excess of hot water is

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📖 Explanation

When ${FeCl}_{3}$ is added to excess of hot water, a positively charged hydrated ferric oxide $[ {Fe}_{2} {O}_{3} \cdot {xH}_{2} {O} / {Fe}^{3+}]$ is formed due to adsorption ${Fe}^{3+}$ ion (adsorbate) on the surface of hydrated ferric oxide (adsorbent). ${FeCl}_{3} {\to }^{ {H}_{2} {O} \;\text{ (Excess) }\;}{[ {Fe}_{2} {O}_{3} \cdot x {H}_{2} {O}] \cdot {Fe}^{3+}}_{(\text{Positively} \text{charged} \text{sol} \text{or} \text{positive} \text{colloid})}+3 {Cl}^{-}(a q)$

Q50JEE Main 2021MCQ4Mp Block Elements

Match List-I with List-II List-I List-II A. Sodium carbonate (i) Deacon B. Titanium (ii) Castner-Kellner C. Chlorine (iii) van-Arkel D. Sodium hydroxide (iv) SolvayChoose the correct answer form the options given below.

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📖 Explanation

(A) Sodium carbonate is manufactured by Solvay process $\Rightarrow$ (iv) of List-ll. Solvay process's reactions are as follows In Solvay's process, first ammonia $( {NH}_{3})$ reacts with ${CO}_{2}$ and ${H}_{2} {O}$ to give ammonium bicarbonate $( {NH}_{4} {HCO}_{3})$ . This bicarbonate react with ${NaCl}$ to give sodium bicarbonate $( {NaHCO}_{3})$ along with ammonium chloride $( {NH}_{4} {Cl})$ . At last, ${NaHCO}_{3}$ is heated to form sodium carbonate as major product. ${NH}_{3}+ {CO}_{2}+ {H}_{2} {O} \longrightarrow {NH}_{4} {HCO}_{3}$ ${NH}_{4} {HCO}_{3}+ {NaCl}_{(\text{Brine})} {\longrightarrow }{NaHCO}_{3} ↓+ {NH}_{4} {Cl}$ $\;2 {NaHCO}_{3} \longrightarrow {\Delta }_{2} \longrightarrow { {Na}_{2} {CO}_{3} }_{\text{Sodium}\text{carbonate}}+ {CO}_{2}+ {H}_{2} {O}$ (B) Titanium is purified by van-Arkel method $\Rightarrow$ (iii) of List-II van-Arkel method Crude titanium ${\longrightarrow }_{I_2}^{780 {K}}$ Til $_{4}(g) {\longrightarrow }^{1800 {K}} {Ti}_{(\text{Pure})}+2 {I}_{2}$ (C) Chlorine is manufactured by Deacon process $\Rightarrow$ (i) of List-II. Deacon process $4 {HCl}(g)+{ {O}_{2}(g) }_{[\text{Air}]}{\to }_{723 \text{K}}^{ {CuCl}_{2} \;\text{ (catalyst) }\;} {2 {Cl}_{2}}_{\text{Chlorine}\text{gas}}+2 {H}_{2} {O}(g)$ (D) Sodium hydroxide is manufactured by Castner-Kellner cell $\Rightarrow$ (ii) of List-II. Castner-Kellner cell reaction Electrolyte used is brine, saturated ${NaCl}$ solution. Cathode $( {Hg})$ : ${Na}^{+}+ {e}^{-} \longrightarrow {Na} {\longrightarrow }^{ {Hg}} {Na} / {Hg}$ Anode (carbon) : ${Cl}^{-}+ {e}^{-} \longrightarrow \;\frac{1}{2} {Cl}_{2} ↑$ $2 {Na} / {Hg}+2 {H}_{2} {O} \longrightarrow {2 {NaOH}}_{\text{Sodium}\text{hydroxide}}+2 {Hg}+ {H}_{2} ↑$

Q51JEE Main 2021NAT4MSolutions

The ${NaNO}_{3}$ weighed out to make $50 {mL}$ of an aqueous solution containing $70.0 {mg} {Na}^{+}$ per ${mL}$ is ......... g. (Rounded off to the nearest integer) [Given : Atomic weight in ${g} {mol}^{-1},- {Na}: 23$ ; $N: 14 ; 0: 16]$ .

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📖 Explanation

$70 {mg} {Na}^{+}$ is present in $1 {mL}$ of ${NaNO}_{3}$ solution. $\therefore 50 {mL}$ of ${NaNO}_{3}$ will contain $=70 \times 50 {mg}$ $=\;\frac{70 \times 50}{1000}=3.5 {gm}$ Moles of ${Na}^{+}$ in solution $=$ Moles of ${NaNO}_{3}$ in solution ${[ \because {NaNO}_{3} \longrightarrow {Na}^{+}+ {NO}_{3}^{-}]\$ }=;\frac{3.5}{23} {mol} ;; [ \because ;\text{ Molar mass of }; {Na}^{+}=23 {gmol}^{-1}] $Mass of$ {NaNO}{3}= $mole$ × $molar mass$ {[ \because ;\text{ molar mass of }; {NaNO}{3}=85 {gmol}^{-1}]$}=;\frac{3.5}{23} \times 85=12.934 {g} ∼eq 13 {g}$

Q52JEE Main 2021NAT4MElectrochemistry

Emf of the following cell at $298 {K}$ in ${V}$ is $x \times 10^{-2}, {Zn}| {Zn}^{2+}(0.1 {M}) \| {Ag}^{+}(0.01 {M})| {Ag}$ The value of $x$ is .......... . (Rounded off to the nearest integer). $[Given, $E_{ {Zn}^{2+} / {Zn}}^{\circ}=-0.76 {V}, E_{ {Ag}^{+} / {Ag}}^{\circ}=+0.80 {V}$ , $\;\frac{2.303 R T}{F}=0.059$ ]$

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📖 Explanation

${Zn}| {Zn}^{2+}(0.1 {M}) \| {Ag}^{+}(0.01 {M})| {Ag}$ Cell reaction : ${Zn}(s)+2 {Ag}^{+} ⇌ {Zn}^{2+}+2 {Ag}(s)$ $\Rightarrow \;\; Q=\;\frac{[ {Zn}^{2+}]\$ [ {Ag}]^{2}}{[ {Zn}]$[ {Ag}^{+}]^{2}}=;\frac{0.1 {M}}{1 \times (0.01)^{2}}=10^{3}n=2 $and$ E_{ {cell}}^{\circ}=E_{ {Ag}^{+} / {Ag}}^{\circ}-E_{ {Zn}^{\circ}}^{2+} / {Zn}=0.8-(-0.76)=1.56 {V}\Rightarrow ;; E_{;\text{cell };} ;; =E_{;\text{cell };}^{\circ}-;\frac{0.059}{n} \log Q=1.56-;\frac{0.059}{2} \log 10^{3};; =1.4715 {V};; =147.15 \times 10^{-2} {V}=147 \times 10^{-2} {V};; =x \times 10^{-2} {V}x ;; =147$

Q53JEE Main 2021NAT4MSolutions

When $12.2 {g}$ of benzoic acid is dissolved in $100 {g}$ of water, the freezing point of solution was found to be $-0.93^{\circ} {C}( {K}_{f}( {H}_{2} {O})=1.86 {Kkg}$ ${mol}^{-1}$ ). The number $(n)$ of benzoic acid molecules associated (assuming $100 \%$ association) is ......... .

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📖 Explanation

${n {PhCOOH}}_{\text{Benzoic} \text{acid}} {\to }^{\;\text{ Association }\;}( {PhCOOH})_{n}$ Assuming $100 \%$ association $(\alpha=1)$ , $\Rightarrow \;\; i=1-\alpha(1-\;\frac{1}{n})=\;\frac{1}{n}[ \because \alpha+1]$ Now, $\Delta T_{f}=K_{f} \times m \times i$ $0-(0.93)=1.86 \times \;\frac{w_{B} \times 1000}{w_{A} \times M_{B}} \times \;\frac{1}{n}$ $[ \because w_{B}=$ mass of ${PhCOOH}=12.2 {g}$ $w_{A}=$ mass of ${H}_{2} {O}=100 {g}$ ${M}_{ {B}}=$ molar mass of ${Ph} {COOH}]$ $=122 {g} {mol}^{-1}$ $=186 \times \;\frac{12.2 \times 1000}{100 \times 122} \times \;\frac{1}{n}$ $\Rightarrow \;\; n=\;\frac{1.86 \times 12.2 \times 1000}{0.93 \times 100 \times 122}=2$ $\therefore$ Number of benzoic acid molecules associated, $n=2$

Q54JEE Main 2021NAT4MChemical Thermodynamics

The average $S-F$ bond energy in $k J{ {mol}}^{-1}$ of ${SF}_{6}$ is ........... (Rounded off to the nearest integer) $[Given, the values of standard enthalpy of formation of ${SF}_{6}(g), {S}(g)$ and ${F}(g)$ are $-1100$ , 275 and $80 {kJ} {mol}^{-1}$ respectively. ]$

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📖 Explanation

So, $\Delta _{f} H^{\circ}[ {S}, g]+6 \times \Delta _{f} H^{\circ}[ {F}, g]=\Delta _{f} H^{\circ}[ {SF}_{6}, g]$ $+6 \times E_{S-F}$ $[ \therefore E_{S-F}=$ Average $S-F$ bond energy in $S F_{6}]$ $275+6 \times \;80=-1100+6 \times E_{S-F}$ $\Rightarrow E_{S-F} \;\; =\;\frac{275+6 \times 80+1100}{6}$ $\;\; =309.16 {kJ} {mol}^{-1}=309 {kJ} {mol}^{-1}$

Q55JEE Main 2021NAT4MStructure of Atom

A ball weighing $10 {g}$ is moving with a velocity of $90 {ms}^{-1}$ . If the uncertainity in its velocity is $5 \%$ , then the uncertainty in its position is $. . . . . . . . . \times 10^{-33} {m}$ (Rounded off to the nearest integer). $[Given, $h=6.63 × 10^{-34} {J}- {s}$ ]$

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📖 Explanation

According to Heisenberg's uncertainty equation, $\Delta x \times m \Delta v \ge \;\frac{h}{4 \pi} \;\; ( \therefore m=10 {g}=10 \times 10^{-3} {kg}, v=90 {ms}^{-1})$ Uncertainty in position, $\Delta x \;\; =\;\frac{h}{4 \pi} \times \;\frac{1}{m \times \Delta v(\;\text{ with uncertainty }\;)}$ $\;\; =\;\frac{6.63 \times 10^{-34}}{4 \times 3.14} \times \;\frac{1}{(10 \times 10^{-3}) \times (90 \times \;\frac{5}{100})} {m}$ $\;\; =1.17 \times 10^{-33} {m}$ $\;\; =1 \times 10^{-33} {m}$

Q56JEE Main 2021NAT4MSolid State

The number of octahedral voids per lattice site in a lattice is ......... . (Rounded off to the nearest integer)

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📖 Explanation

Let us assume, the crystal has fcc or ccp lattice which has octahedral voids. Number of lattice sites occupied $=8$ corner $+6$ face centres $=14$ Number of octahedral voids $=12$ edge centres $+1$ body centre $=13$ Number of octahderal void(s) per lattice site $=\;\frac{13}{14}=0.928 ∼eq 1$

Q57JEE Main 2021NAT4MRedox Reactions

In mild alkaline medium, thiosulphate ion is oxidised by ${MnO}_{4}^{-}$ to " $A^{' '}$ . The oxidation state of sulphur in " $A^{' '}$ is ......... .

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📖 Explanation

${{S}_{2} {O}_{3}^{2-}}_{{\text{Thiosulphate} \text{ion}}_{ [\text{Reducing} \text{agent}]}}+{ {MnO}_{4}^{-} }_{{\text{Permanganate} \text{ion}} _{ [\text{Oxidising} \text{agent}]}}{\to }_{\text{Alkaline}_{\text{medium}}}^{\;\text{ Mild }\;} {{SO}_{4}^{2-}}_{\text{Sulphate} \text{ion}(A)}+ {{MnO}_{4}^{2-}}_{\text{Manganate} \text{ion}}$ Oxidationnumber of sulphur in ${SO}_{4}^{2-}(A)$ is $+6$ $\begin{bmatrix}\because x+4(-2)=-2 \\ \Rightarrow x=+6\end{bmatrix}$

Q58JEE Main 2021NAT4MCoordination Compounds

The number of stereoisomers possible for $[ {Co}( {OX})_{2}( {Br})( {NH}_{3})]^{2-}$ is ......... .[ox $= {Oxalate}] .$

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📖 Explanation

The coordination compound $[{Co}^{\text{III}}( {ox}){2}( {Br})( {NH}{3})]^{2-} $of general formula$ [M(A-A)_{2} B C] $can show both geometrical and optical isomerism (stereoisomerism). The cis-form produces non-superimposable mirror images, i.e. enantiomeric pairs (optically active)The trans-form is optically inactive. So, total number of stereoisomers possible$ =cis(\pm )+trans=3$

Q59JEE Main 2021NAT4MChemical Kinetics

If the activation energy of a reaction is $80.9 {kJ} {mol}^{-1}$ , the fraction of molecules at $700 {K}$ , having enough energy to react to form products is $e^{-x}$ . The value of $x$ is ........... . (Rounded off to the nearest integer) $[$ Use, $R=8.31 {JK}^{-1} {mol}^{-1}] .$

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📖 Explanation

According to Arrhenius equation, the factor ${e}^{-E_{a} / R T}$ corresponds to the fraction of molecules that have kinetic energy greater than $E_{a}$ (activation energy). $\Rightarrow \;\; e^{E_{a} / R T} \;\; =e^{\;\frac{-80.9 \times 1000}{8.31 \times 700}}=e^{-13.90} ∼eq {e}^{-14}= {e}^{-x}$ $x \;\; =14$ ${[ \because E_{a}} \;=80.9 {kJ} {mol}^{-1}=80.9 \times 1000 {J} {mol}^{-1}, R=8.31 {mol}^{-1} {K}^{-1}]$

Q60JEE Main 2021NAT4MIonic Equilibrium

The ${pH}$ of ammonium phosphate solution, if ${p} K_{a}$ of phosphoric acid and ${p} K_{b}$ of ammonium hydroxide are $5.23$ and $4.75$ respectively, is .......... .

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📖 Explanation

Phosphoric acid is a weak tribasic acid $( {H}_{3} {PO}_{4})$ and ${NH}_{4} {OH}$ is a weak monoacidic base. So, hydrolysis of ammonium phosphate $( {NH}_{4})_{3} {PO}_{4}$ can be shown as, $( {NH}_{4})_{3} {PO}_{4}+3 {H}_{2} {O} ⇌ {H}_{3} {PO}_{4}+3{NH}_{4} {OH} \;\; [3: 1$ type salt] So, $\;\; \$ [ {H}^{+}]$-K_{ {a}} \times (;\frac{K_{w}}{K_{a} \times K_{b}})^{12}{pH}= ;; {p} K_{ {a}}+;\frac{1}{2}[ {p} K_{w}- {p} K_{ {a}}- {p} K_{b}]= ;; 5.23+;\frac{1}{2}(14-5.23-4.75){[ \because {p} K_{w}=14{H}{2} {O}}{p} K{ {a}}=5.23( {H}{3} {PO}{4}){p} K_{b}=4.75( {NH}_{4} {OH})]=7.24 ∼eq 7$

Mathematics29 questions

Q61JEE Main 2021MCQ4MVector Algebra

If vectors $a_{1}=x \hat{i}-\hat{j}+\hat{k}$ and $a_{2}=\hat{i}+{y}^{∧} j+z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $x \hat{i}+\hat{j}+z \hat{k}$ is

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📖 Explanation

Given, $a_{1}=x \hat{i}-\hat{j}+\hat{k}$ and $a_{2}=\hat{i}+y \hat{j}+z \hat{k}$ are collinear, then $\;\frac{x}{1}=\;\frac{-1}{y}=\;\frac{1}{z}=\lambda$ (Say) This gives $x=\lambda, y=-\;\frac{1}{\lambda}, z=\;\frac{1}{\lambda}$ Then, unit vector parallel to vector $x \hat{i}+y \hat{j}+$ zk will be $\;\; =\;{\pm ((\lambda) \hat{i}-(\frac{1}{\lambda}) \hat{j}+(\frac{1}{\lambda}) \hat{k})}/{\sqrt{(\lambda)^{2}+(\;\frac{-1}{\lambda})^{2}+(\;\frac{1}{\lambda})^{2}}}$ $\;\; =\pm \;{(\lambda^{2} \hat{i}-\hat{j}+\hat{k}) \lambda}/{\lambda \sqrt{\lambda^{4}+2}}=\pm \;{(\lambda^{2} \hat{i}-\hat{j}+\hat{k})}/{\sqrt{\lambda^{4}+2}}$ Take, $\lambda=1=\;{\pm (\hat{i}-\hat{j}+\frac\hat{k})}{\sqrt{3}}$

Q62JEE Main 2021MCQ4MFunctions

Let $A=\{1,2,3, . . ., 10\}$ and $f: A \to A$ be defined as $f(k)=\begin{cases}k+1 & \\ \\ \text{ if }\\ \\ k \\ \\ \text{ is odd }\\ \\ \\ k & \\ \\ \text{ if }\\ \\ k \\ \\ \text{ is even }\\ \\ \end{cases}$ Then, the number of possible functions $g: A \to A$ , such that $g \circ f=f$ is

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📖 Explanation

$f(x)=\begin{cases}x+1 & x \\ \\ \text{ is odd\end{cases} }\; ; x, , x \;\text{ is even. }\;.$ Given, $g: A \to A$ such that, $g(f(x))=f(x)$ When $x$ is even, then $g(x)=x$ . . . (i) When $x$ is odd, then $g(x+1)=x+1$ . . . (ii) This implies, $g(x)=x, x$ is even. $\Rightarrow$ If $x$ is odd, then $g(x)$ can take any value in set $A$ . So, number of $g(x)=10^{5}$

Q63JEE Main 2021MCQ4MLimits, Continuity and Differentiability

Let $f: R \to R$ be defined as If $f(x)$ is continuous on $R$ , then $a+b$ equals

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📖 Explanation

Given, $f(x)$ is continuous on $R$ . If $f(x)$ is continuous, then $f$ is continuous at $x=1$ $\Rightarrow \;\; \lim _{x^{-}} f(x)=f(1)=\lim f(x)$ $\Rightarrow \;\; |a+{1}^{1^{-}}+b|=\sin \pi {=}^{x} \vec{0}^{+}$ $\Rightarrow \;\; a+b=-1$ . . . (i) Also, $f$ is continuous at $x=-1$ $\Rightarrow \;\; \lim _{x \to -1^{-}} f(x)=f(-1)=\lim _{x \to -1^{+}} f(x)$ $\Rightarrow \;\; 2 \sin (\;\frac{-\pi}{2}(-1)) \;\; =|a-1+b|$ $\Rightarrow \;\; 2 \;\; =|a+b-1|$ . . . (iii) Eq. (ii) is satisfied $\therefore \;\; a+b=-1$

Q64JEE Main 2021MCQ4MDefinite Integrals

For $x > 0$ , if $f(x)=int_{1}^{x} \;\frac{\log _{e} t}{(1+t)} d t$ , then $f(e)+f(\;\frac{1}{e})$ is equal to

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📖 Explanation

$f(x)=int_{1}^{x} \;\frac{\log _{e} t}{(1+t)} d t$ Then, $f(e)=int_{1}^{e \log _{e} t} \;\frac{d t}{1+t}$ . . . (i) and $f(\;\frac{1}{e})=int_{1}^{\;\frac{1}{e}} \;\frac{\log _{e} t}{1+t} d t$ . . .(ii) Let $t=\;\frac{1}{u}, d t=\;\frac{-1}{u^{2}} d u$ and put in Eq. (ii), we get $f(\;\frac{1}{e})=int_{1}^{e} \;\frac{\log (\frac{1}{u})}{1+\;\frac{1}{u}} \cdot \;\frac{-1}{u^{2}} d u=int_{1}^{e} \;\frac{\log u}{u(u+1)} d u$ Using change of variable $f(\;\frac{1}{e})=int_{1}^{e} \;\frac{\log t}{t(t+1)} d t$ . . . (iii) From Eqs. (i) and (iii), we get $f(e)+f(\;\frac{1}{e})=int_{1}^{e} \;\frac{\log t}{1+t} d t+int_{1}^{e} \;\frac{\log t}{t(1+t)} d t=int_{1}^{e \log t} \;\frac{g t}{t}$ Take $\log t=v$ , then $\;\frac{1}{t} d t=d v$ $f(e)+f(\;\frac{1}{e})=int_{0}^{1} v d v=\$ [;\frac{v^{2}}{2}]_{0}^{1}=;\frac{1}{2}∴ ;; f(e)+f(;\frac{1}{e})=;\frac{1}{2}$

Q65JEE Main 2021MCQ4MPermutations and Combinations

A natural number has prime factorisation given by $n=2^{x} 3^{y} 5^{z}$ , where $y$ and $z$ are such that $y+z=5$ and $y^{-1}+z^{-1}=\;\frac{5}{6}, y > z$ . Then, the number of odd divisors of $n$ , including 1 , is

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📖 Explanation

Given, $n=2^{x} 3^{y} 5^{z}$ and $y+z=5$ $\;\frac{1}{y}+\;\frac{1}{z}=\;\frac{5}{6}$ or $\;\frac{y+z}{y z}=\;\frac{5}{6}$ This implies, $y+z=5 \;\text{ and }\; y z=6$ Put $y=\;\frac{6}{z}$ in $y+z=5$ $\;\frac{6}{z}+z \;\; =5 \;\text{ or }\; z^{2}-5 z+6=0$ $\Rightarrow \;\; z^{2}-3 z-2 z+6 \;\; =0$ $\;(z-3)(z-2)=0 \Rightarrow z=3 \;\text{ or }\; 2$ $\;\text{ Using }\; y= \;\; \;\frac{6}{z}, \;\text{ we get }\;$ $y \;\; =2 \;\text{ or }\; 3$ For calculating the odd divisor $x$ must be 0 i.e. $x=0$ $\therefore \;\; n=2^{0} 3^{3} 5^{2} \;\text{ or }\; n=2^{0} 3^{2} 5^{3}$ Total number of odd divisor of $n$ $\;\; =(3+1)(2+1)$ $\;\; =(4)(3)=12$

Q66JEE Main 2021MCQ4MLimits, Continuity and Differentiability

Let $f(x)=\sin ^{-1} x$ and $g(x)=\;\frac{x^{2}-x-2}{2 x^{2}-x-6}$ . If $g(2)=\lim _{x \to 2} g(x)$ , then the domain of the function $x \to g$ is

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📖 Explanation

Given, $g(x)=\;\frac{x^{2}-x-2}{2 x^{2}-x-6}, f(x)=\sin ^{-1} x$ $f(g(x))=\sin ^{-1}(g(x))$ $\Rightarrow \;\; f \circ g(x)=\sin ^{-1}(\;\frac{x^{2}-x-2}{2 x^{2}-x-6})$ For the domain of $f \circ g(x)$ , $|g(x)| \le 1$ $\;\text{ [: Domain of }\; f(x) \;\text{ is }\;[-1,1] \;\text{ ] }\;$ $\Rightarrow \;\; |\;\frac{x^{2}-x-2}{2 x^{2}-x-6}| \le 1 \Rightarrow |\;\frac{(x+1)(x-2)}{(2 x+3)(x-2)}| \le 1$ $\Rightarrow \;\; |\;\frac{x+1}{2 x+3}| \le 1 \Rightarrow -1 \le \;\frac{x+1}{2 x+3} \le 1$ $\Rightarrow \;\; (\;\frac{x+1}{2 x+3})^{2} \le 1 \Rightarrow (x+1)^{2} \le (2 x+3)^{2}$ $\Rightarrow \;\; 3 x^{2}+10 x+8 \ge 0 \Rightarrow (3 x+4)(x+2) \ge 0$ This implies, $x \in(-\infty ,-2] \cup[-\;\frac{4}{3}, \infty )$

Q67JEE Main 2021MCQ4MApplication of Derivatives

The triangle of maximum area that can be inscribed in a given circle of radius $r$ is

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📖 Explanation

Let a $\triangle A B C$ inscribed in a circle with centre $O$ and radius $r$ .Let $∠ O B C=\theta$ Now, area of $\triangle A B C=\;\frac{1}{2} \times$ Base $\times$ Height $A=\;\frac{1}{2} \times (B C) \times (A P)$ . . . (i) Now, $B C=2 B P$ Consider $\Delta O B P$ , where $O B=r$ Then, $B P=r \cos \theta$ Hence, $B C=2 r \cos \theta$ Again, $A P=A O+O P$ where, $A O=r$ Consider $\Delta O B P$ , where $O B=r$ Then, $O P=r \sin \theta$ $\Rightarrow \;\; A P=r+r \sin \theta$ From Eq. (i), we get Area $=\;\frac{1}{2} \times (2 r \cos \theta) \times (r+r \sin \theta)$ $A=r^{2} \cos \theta(1+\sin \theta)$ Now, $\;\frac{d A}{d \theta}=r^{2}(-\sin \theta)(1+\sin \theta)+r^{2} \cos ^{2} \theta$ $\;\; =r^{2}(\cos ^{2} \theta-\sin \theta-\sin ^{2} \theta)$ $\;\; =r^{2}(1-2 \sin ^{2} \theta-\sin \theta)$ $\;\; =r^{2}(1+\sin \theta)(1-2 \sin \theta)$ Equate $\;\frac{d A}{d \theta}=0$ $\Rightarrow \;\; r^{2}(1+\sin \theta)(1-2 \sin \theta)=0$ $r=r^{2}(1+\sin \theta)(1-2 \sin \theta)=0$ $\Rightarrow \;\; \sin \theta=\;\frac{1}{2} \Rightarrow \theta=\;\frac{\pi}{6}$ Now, $\;\frac{d^{2} A}{d \theta^{2}} < 0$ , when $\theta=\;\frac{\pi}{6}$ $\Rightarrow A$ is maximum, when $\theta=\;\frac{\pi}{6}$ $\therefore$ Maximum area $=r^{2} \cos (\;\frac{\pi}{6})(1+\sin \;\frac{\pi}{6})=\;\frac{3 \sqrt{3}}{4} r^{2}$ Height $=A P=\;\frac{3}{2} r$ Consider $\triangle A B P$ , $(A B)^{2}\;=(A P)^{2}+(B P)^{2}\;$ $\;=(\;\frac{3}{2} r)^{2}+(\;\frac{\sqrt{3}}{2} r)^{2}\;{[ \because B P=r \cos \theta]}$ $\;=\;\frac{9}{4} r^{2}+\;\frac{3}{4} r^{2}=3 r^{2}\;$ $\Rightarrow \;\; A B\;=\sqrt{3} r\;$ Hence, the $\triangle A B C$ is an equilateral triangle with side $\sqrt{3}$ .

Q68JEE Main 2021MCQ4MThree Dimensional Geometry

Let $L$ be a line obtained from the intersection of two planes $x+2 y+z=6$ and $y+2 z=4$ . If point $P(\alpha, \beta, \gamma)$ is the foot of perpendicular from $(3,2,1)$ on $L$ , then the value of $21(\alpha+\beta+\gamma)$ equals

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📖 Explanation

Given, $x+2 y+z=6$ . . . (i) and $\;\; y+2 z=4$ . . . (ii) Put $y=4-2 z$ from Eq. (ii) Eq. in (i), we get $\;x+8-4 z+z \;\; =6$ $\Rightarrow \;x \;\; =-2+3 z$ $\Rightarrow \;\;\frac{x+2}{3} \;\; =z$ . . . (iii) $\;y \;\; =4-2 z$ $\Rightarrow \;\;\frac{y-4}{-2} \;\; =z$ . . . (iv) From Eqs. (iii) and (iv), line of intersection of two planes is $\;\frac{x+2}{3}=\;\frac{y-4}{-2}=\;\frac{z}{1}=\lambda$ Then, Direction ratios of $P Q$ is $x-3 \lambda-4-2 \lambda, z=\lambda$ $(3 \lambda-5,-2 \lambda+2, \lambda-1)$ $(3 \lambda-5,-2 \lambda+2, \lambda-1)$ Since, $P Q$ is perpendicular to the line, then $3(3 \lambda-5)-2(-2 \lambda+2)+1(\lambda-1)=0$ $\therefore \;\; \lambda=\;\frac{10}{7}\;$ $\therefore \;\; P(\;\frac{16}{7}, \;\frac{8}{7}, \;\frac{10}{7})\;$ $\;\text{ Then, }\; 21(\alpha+\beta+\gamma) \;\; =21(\;\frac{16}{7}+\;\frac{8}{7}+\;\frac{10}{7})$ $\;\; =21(\;\frac{34}{7})=3 \times 34=102$

Q69JEE Main 2021MCQ4MSets and Relations

Let $F_{1}(A, B, C)=(A ∧ ∼ B) ∨[∼ C ∧(A ∨ B)] ∨ ∼ A$ and $F_{2}(A, B)=(A ∨ B) ∨(B \to ∼ A)$ be two logical expressions. Then,

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📖 Explanation

$F_{1}(A, B, C)=(A ∧ ∼ B) ∨[∼ C ∧(A ∨ B)] ∨ ∼ A$ $\equiv \{(A ∧ ∼ B) ∨[∼ C ∧(A ∨ B)] \;\;$ (Associative law) $\equiv \{(A ∨ ∼ A) ∧(∼ B ∨ ∼ A)\} ∨[(A ∨ B) ∧ C] \;\;$ (distributive law) $\equiv (∼ B ∨ ∼ A) ∨[(A ∨ B) ∧ ∼ C] \;\; ( \because A ∨ ∼ A \equiv T, T ∧ p \equiv p)$ $\equiv [(∼ A ∼ B) ∨(A ∨ B)] ∧[(∼ A ∨ ∼ B) ∨ ∼ C]$ $\equiv T ∧(∼ A ∨ ∼ B) ∨ ∼ C$ $\equiv (∼ A ∨ ∼ B) ∨ ∼ C$ Which is not a tautology. Now, $F_{2}(A, B) \;\; =(A ∨ B) ∨(B \to ∼ A)$ $\;\equiv (A ∨ B) ∨(∼ B ∨ ∼ A)$ Hence, $F_{2}$ is a tautology.

Q70JEE Main 2021MCQ4MDifferential Equations

Let slope of the tangent line to a curve at any point $P(x, y)$ be given by $\;\frac{x y^{2}+y}{x}$ . If the curve intersects the line $x+2 y=4$ at $x=-2$ , then the value of $y$ , for which the point $(3, y)$ lies on the curve, is

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📖 Explanation

Given, slope of tangent line to curve at $(x, y)$ is $\;\frac{x y^{2}+y}{x}$ i.e., $\;\; \;\frac{d y}{d x}=\;\frac{x y^{2}+y}{x}$ $\Rightarrow \;\;\frac{d y}{d x}=y^{2}+\;\frac{y}{x} \Rightarrow x d y=x y^{2} d x+y d x$ $\Rightarrow \;x d y-y d x=x y^{2} d x \Rightarrow \;\;\frac{x d y-y d x}{y^{2}}=x d x$ Integrating both sides, we get $\;\frac{-x}{y}=\;\frac{x^{2}}{2}+C$ . . . (i) The curve intersect line at $x=-2$ Then, $x=-2$ , is satisfied by $x+2 y=4$ Hence, $(-2)+2 y=4$ Gives, $y=3$ $\therefore$ Curve passes through $(2,-3)$ . Use $(2,-3)$ is Eq. (i), we get $\;\frac{-2}{-3}=\;\frac{(-2)^{2}}{2}+C \Rightarrow C=\;\frac{-4}{3}$ $\therefore$ The curve is $\;\frac{-x}{y}=\;\frac{x^{2}}{2}=\;\frac{-4}{3}$ . . . (ii) It also passes through $(3, y)$ . Put $(3,4)$ in Eq. (ii), we get $\Rightarrow \;\; \;\frac{-3}{y}=\;\frac{(3)^{2}}{2}-\;\frac{4}{3} \Rightarrow y=-\;\frac{18}{19}$

Q71JEE Main 2021MCQ4MCircles

If the locus of the mid-point of the line segment from the point $(3,2)$ to a point on the circle, $x^{2}+y^{2}=1$ is a circle of radius $r$ , then $r$ is equal to

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📖 Explanation

Given, circle $x^{2}+y^{2}=1$ Let $(h, k)$ be the mid-point of line segment from point $(3,2)$ to a point on circle. Let point on circle be $(a, b)$ . Then, $\;h=\;\frac{a+3}{2} \Rightarrow a=2 h-3$ $\;k=\;\frac{b+2}{2} \Rightarrow b=2 k-2$ $\because (a, b)$ is point on circle, then $(a)^{2}+(b)^{2} \;\; =1$ $\Rightarrow \;\; (2 h-3)^{2}+(2 k-2)^{2} \;\; =1$ $\Rightarrow \;\; (h-\;\frac{3}{2})^{2}+(k-1)^{2}=\;\frac{1}{4} \;\; =(\;\frac{1}{2})^{2}$ which is an equation of circle with radius $1 / 2$ . $\therefore$ Radius $=1 / 2$ .

Q72JEE Main 2021MCQ4MMatrices and Determinants

Consider the following system of equations $\;x+2 y-3 z=a$ $\;2 x+6 y-11 z=b$ $\;x-2 y+7 z=c$ where, $a, b$ and $c$ are real constants. Then, the system of equations

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📖 Explanation

Given, system of equation can be written as $A X=B$ , where $A=\begin{bmatrix}1 & 2 & -3 \\ 2 & 6 & -11 \\ 1 & -2 & 7\end{bmatrix}, X=\begin{bmatrix}x \\ y \\ z\end{bmatrix}, B=\begin{bmatrix}a \\ b \\ c\end{bmatrix}$ Then, $|A| \;\; =\begin{vmatrix}1 & 2 & -3 \\ 2 & 6 & -1 \\ 1 & -2 & 7\end{vmatrix}$ $\;\; =1(42-22)-2(14+11)-3(-4-6)$ $\;\; =20-50+30=0$ $|A_{1}| \;\; =\begin{vmatrix}a & 2 & -3 \\ b & 6 & -11 \\ c & -2 & 7\end{vmatrix}$ $\;\; =a(42-22)-2(7 b+11 c)$ $\;\; -3(-2 b-6 c)$ $\;\; =20 a-14 b-22 c+6 b+18 c$ $\;\; =20 a-8 b-4 c=4(5 a-2 b-c)$ $|A_{2}| \;\; =\begin{vmatrix}1 & a & -3 \\ 2 & b & -11 \\ 1 & c & 7\end{vmatrix}$ $\;\; =1(7 b+11 c)-a(14+11)-3(2 c-b)$ $\;\; =-25 a+10 b+5 c=-5(5 a-2 b-c)$ $|A_{3}| \;\; =\begin{vmatrix}1 & 2 & a \\ 2 & 6 & b \\ 1 & -2 & c\end{vmatrix}$ $\;\; =1(42-22)-2(14+11)-3(-4-6)$ $\;\; =20-50+30=0$ $|A_{1}| \;\; =\begin{vmatrix}a & 2 & -3 \\ b & 6 & -11 \\ c & -2 & 7\end{vmatrix}$ $\;\; =a(42-22)-2(7 b+11 c)$ $\;\; -3(-2 b-6 c)$ $\;\; =20 a-14 b-22 c+6 b+18 c$ $\;\; =20 a-8 b-4 c=4(5 a-2 b-c)$ $|A_{2}| \;\; =\begin{vmatrix}1 & a & -3 \\ 2 & b & -11 \\ 1 & c & 7\end{vmatrix}$ $\;\; =1(7 b+11 c)-a(14+11)-3(2 c-b)$ $\;\; =-25 a+10 b+5 c=-5(5 a-2 b-c)$ $|A_{3}| \;\; =\begin{vmatrix}1 & 2 & a \\ 2 & 6 & b \\ 1 & -2 & c\end{vmatrix}$ $\;\; =-10 a+4 b+2 c$ $\;\; =-2(5 a-2 b-c)$ For infinite solution, $\;|A|=|A_{1}|=|A_{2}|=|A_{3}|=0$ $\Rightarrow \;\; 5 a-2 b-c=0 \Rightarrow 5 a=2 b+c$

Q73JEE Main 2021MCQ4MInverse Trigonometric Functions

If $0 < a, b < 1$ and $\tan ^{-1} a+\tan ^{-1} b=\;\frac{\pi}{4}$ , then the value of $(a+b)-(\;\frac{a^{2}+b^{2}}{2})+(\;\frac{a^{3}+b^{3}}{3})-(\;\frac{a^{4}+b^{4}}{4})+. . .$ is

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Given, $\tan ^{-1} a+\tan ^{-1} b=\;\frac{\pi}{4}$ $\Rightarrow \;\; \tan ^{-1}[\;\frac{a+b}{1-a b}]=\;\frac{\pi}{4}$ $\Rightarrow \;\; \;\frac{a+b}{1-a b}=\tan \;\frac{\pi}{4}=1$ $\Rightarrow \;\; a+b=1-a b$ $\Rightarrow \;\; (1+a)(1+b)=2$ . . . (i) Now, $(a+b)-(\;\frac{a^{2}+b^{2}}{2})+(\;\frac{a^{3}+b^{3}}{3})-(\;\frac{a^{4}+b^{4}}{4})+. . .$ $=\$ [a-;\frac{a^{2}}{2}+;\frac{a^{3}}{3}-;\frac{a^{4}}{4}+. . .]$+[b-;\frac{b^{2}}{2}+;\frac{b^{3}}{3}-;\frac{b^{4}}{4}+. . .]=\log _{e}(1+a)+\log _{e}(1+b) ;; [\text{Using} $expansion of$ \log _{e}(1+x)]=\log _{e}(1+a)(1+b) ;; [ \therefore \log a+\log b=\log a b]=\log _{e} 2;;$[use Eq. (i)]

Q74JEE Main 2021MCQ4MSequences and Series

The sum of the series $sum_{n=1}^{\infty } \;\frac{n^{2}+6 n+10}{(2 n+1) !}$ is equal to

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📖 Explanation

Let $sum_{n=1}^{\infty } \;\frac{n^{2}+6 n+10}{(2 n+1) !}=S$ $\;\; =sum_{n=1}^{\infty } \;\frac{4 n^{2}+24 n+40}{4(2 n+1) !}=sum_{n=1}^{\infty } \;\frac{(2 n+1)^{2}+(2 n+1) \cdot 10+29}{4(2 n+1) !}$ $\;\; =\;\frac{1}{4}[sum_{n=1}^{\infty } \;\frac{(2 n+1)^{2}}{(2 n+1)(2 n) !}+sum_{n=1}^{\infty } \;\frac{(2 n+1) \cdot 10}{(2 n+1)(2 n) !}]$ $\;\; =\;\frac{1}{4}[sum_{n=1}^{\infty } \;\frac{29}{(2 n+1) !} \;\frac{(2 n+1)}{(2 n) !}+sum_{n=1}^{\infty } \;\frac{10}{(2 n) !}+sum_{n=1}^{\infty } \;\frac{29}{(2 n+1) !}]$ . . . (i) Now, $\;\; =sum_{n=1}^{\infty } \;\frac{(2 n+1)}{(2 n) !}=sum_{n=1}^{\infty } \;\frac{2 n}{(2 n) !}+sum_{n=1}^{\infty } \;\frac{1}{(2 n) !}$ $\;\; =sum_{n=1}^{\infty } \;\frac{1}{(2 n-1) !}+sum_{n=1}^{\infty } \;\frac{1}{(2 n) !}$ $\;\text{ Now, }\; \;\; =sum_{n=1}^{\infty } \;\frac{1}{(2 n-1) !}=\;\frac{1}{1 !}+\;\frac{1}{3 !}+\;\frac{1}{5 !}+. . .=\;{ {e}-\frac{1}{ {e}}}/{2}$ . . . (ii) $\;\text{ and }\; \;\; sum_{n=1}^{\infty } \;\frac{1}{(2 n) !}=\;\frac{1}{2 !}+\;\frac{1}{4 !}+\;\frac{1}{6 !}+. . .=\;\frac{e+\frac{1}{e}-2}{2}$ . . . (iii) and $sum_{n=1}^{\infty } \;\frac{1}{(2 n+1) !}=\;\frac{1}{3 !}+\;\frac{1}{5 !}+\;\frac{1}{7 !}+. . .=\;\frac{e-\frac{1}{e}-2}{2} \;\; . . .$ (iv) Using Eqs. (ii), (iii), (iv) Eq. (i), $S \;\; =\;\frac{1}{4}[\;\frac{e-\frac{1}{e}}{2}+11(\;\frac{e+\frac{1}{e}-2}{2})+29(\;\frac{e-\frac{1}{e}-2}{2})]$ $\;\; =\;\frac{1}{4}[\;\frac{e}{2}-\;\frac{1}{2 e}+\;\frac{11 e}{2}+\;\frac{11}{2 e}+\;\frac{29 e}{2}-\;\frac{29}{2 e}-4]$ $\;\; =\;\frac{41 e}{8}-\;\frac{19}{8 e}-10$

Q75JEE Main 2021MCQ4MLimits, Continuity and Differentiability

Let $f(x)$ be a differentiable function at $x=a$ with $f^{'}(a)=2$ and $f(a)=4$ . Then, $\lim _{x \to a} \;\frac{x f(a)-a f(x)}{x-a}$ equals

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$\lim _{x \to a}\;\;\frac{x f(a)-a f(x)}{x-a}$ $= \;\; \lim _{x \to a} \;\frac{x f(a)-a f(x)+a f(a)-a f(a)}{x-a}$ $= \;\; \lim _{x \to a} \;\frac{(x-a) f(a)-a[f(x)-f(a)]}{x-a}$ $= \;\; \lim _{x \to a} \;\frac{(x-a) f(a)}{x-a}-a \lim _{x \to a} \;\frac{f(x)-f(a)}{x-a}=f(a)-a f^{'}(a)$ $= \;\; 4-a(2)[\;\text{ Given, }\; f(a)=4, f^{'}(a)=2]$ $= \;\; 4-2 a$

Q76JEE Main 2021MCQ4MCircles

Let $A(1,4)$ and $B(1,-5)$ be two points. Let $P$ be a point on the circle $(x-1)^{2}+(y-1)^{2}=1$ , such that $(P A)^{2}+(P B)^{2}$ have maximum value, then the points $P, A$ and $B$ lie on

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📖 Explanation

$P$ be a point on circle $(x-1)^{2}+(y-1)^{2}=1$ So, coordinate of $P$ will be of form $P(1+\cos \theta, 1+\sin \theta)$ Given, points $A(1,4)$ and $B(1,-5)$ Then, $\;(P A)^{2}+(P B)^{2}$ $\;\; =\cos ^{2} \theta+(\sin \theta-3)^{2}+\cos ^{2} \theta+(\sin \theta+6)^{2}$ $\;\; =2 \cos ^{2} \theta+2 \sin ^{2} \theta-6 \sin \theta+12 \sin \theta+45$ $\;\; =47+6 \sin \theta$ Now, $(P A)^{2}+(P B)^{2}=47+6 \sin \theta \;\text{ is }\;$ Maximum, when $\sin \theta=1$ $\theta\;\theta\;=\pi / 2$ $\Rightarrow \;\cos \theta\;=0$ $\;P(1,2), A(1,4), B(1,-5)$ $\therefore P, A, B$ are collinear points on the line $x=1$

Q77JEE Main 2021MCQ4MThree Dimensional Geometry

If the mirror image of the point $(1,3,5)$ with respect to the plane $4 x-5 y+2 z=8$ is $(\alpha, \beta, \gamma)$ , then $5(\alpha+\beta+\gamma)$ equals

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Given, point $P(1,3,5)$ has the mirror image $Q(\alpha, \beta, \gamma)$ with respect to plane $4 x-5 y+2 z=8$ Then, $M$ be the mid-point of line joining point $P$ and $Q$ . and $M$ lies on given plane. Coordinates of $M(a, b, c)$ are as follows $a=\;\frac{\alpha+1}{2}, b=\;\frac{\beta+3}{2}, c=\;\frac{\gamma+5}{2}$ $\because$ M lies on plane, then $\;4 a-5 b+2 c=8$ $\;4(\;\frac{\alpha+1}{2})-5(\;\frac{\beta+3}{2})+2(\;\frac{\gamma+5}{2})=8$ . . . (i) Also, $P Q$ is perpendicular to plane $\Rightarrow \;\; \;\frac{\alpha-1}{4}=\;\frac{\beta-3}{-5}=\;\frac{\gamma-5}{2}=\lambda \;\text{ (say) }\;$ $\Rightarrow \;\; \alpha=4 \lambda+1, \beta=3-5 \lambda, \gamma=2 \lambda+5$ . . . (ii) Use Eq. (ii) in Eq. (i), we get $2(4 \lambda+2)-5(\;\frac{6-5 \lambda}{2})+2 \lambda+10=8$ $\Rightarrow \;\; 8 \lambda+4-15+\;\frac{25 \lambda}{2}+2 \lambda+10=8 \Rightarrow \lambda=\;\frac{2}{5}$ $\therefore \;\; \alpha=4 \lambda+1=4(\;\frac{2}{5})+1=\;\frac{13}{5}$ $\;\; \beta=3-5(\;\frac{2}{5})=\;\frac{5}{5}=1$ and $\;\; \gamma=5+2(\;\frac{2}{5})=\;\frac{29}{5}$ $\therefore 5(\alpha+\beta+\gamma)=5(\;\frac{13}{5}+1+\;\frac{29}{5})=47$

Q78JEE Main 2021MCQ4MDifferential Equations

Let $f(x)=int_{0}^{x} e^{t} f(t) d t+e^{x}$ be a differentiable function for all $x \in R$ . Then, $f(x)$ equals

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Given, $f(x)=int_{0}^{x} e^{t} f(t) d t+e^{x}$ . . . (i) Since, $f(x)$ is differentiable function, differentiate Eq. (i) $f^{'}(x)=e^{x} f(x)+e^{x}$ [Using Newton Leibnitz theorem $]$ $f^{'}(x)= {e}^{x}(f(x)+1) \Rightarrow \;\frac{f^{'}(x)}{f(x)+1}= {e}^{x}$ Integrating it, $\int \;\frac{f^{'}(x)}{f(x)+1} d x=\int e^{x} d x+C$ Let $f(x)+1=u$ , then $f^{'}(x) d x=d u$ $\int \;\frac{d u}{u}=e^{x}+C \Rightarrow \log u=e^{x}+C$ $\log (f(x)+1)=e^{x}+C\;\;$ $\;\;[ \because u=f(x)+1] . . .$ (ii) Now $f(x)=int_{0}^{x} e^{t} f(t) d t+e^{x} \Rightarrow f(0)=e^{0}=1$ Put $x=0$ , in Eq. (ii), we get $\log (2)= {e}^{\circ}+C \Rightarrow C=\log (2)-1$ From Eq. (ii), we get $\log (f(x)+1) \;\; =e^{x}+\log 2-1$ $f(x)+1 \;\; =e^{e^{x}+\log 2-1}=e^{e^{x}} \cdot e^{\log 2} \cdot e^{-1}$ $f(x)+1 \;\; =2 e^{e^{x}} \cdot e^{-1}=2 e^{e^{x}-1}$ $\therefore \;\; f(x) \;\; =2 e^{e^{x}-1}-1$

Q79JEE Main 2021MCQ4MArea Under The Curves

Let $A_{1}$ be the area of the region bounded by the curves $y=\sin x, y=\cos x$ and $y$ -axis in the first quadrant. Also, let $A_{2}$ be the area of the region bounded by the curves $y=\sin x$ , $y=\cos x, x$ -axis and $x=\;\frac{\pi}{2}$ in the first quadrant. Then,

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📖 Explanation

$A_{1}$ is the area of region $O A B$ . $A_{2}$ is the area of region $O B C$ . Coordinate of $B$ is $(\;\frac{\pi}{4}, \;\frac{1}{\sqrt{2}})$ Now, $A_{1} \;\; =int_{0}^{\;\frac{\pi}{4}}(\cos x-\sin x) d x$ $\;\; =[\sin x+\cos x]_{0}^{\pi / 4}$ $\;\; =(\sin \;\frac{\pi}{4}+\cos \;\frac{\pi}{4})-(0+1)=\sqrt{2}-1$ $A_{2} \;\; =int_{0}^{\;\frac{\pi}{4}} \sin x d x+int_{\;\frac{\pi}{4}}^{\;\frac{\pi}{2}} \cos x d x$ $\;\; =[\sin x+\cos x]_{0}^{\pi / 4}$ $\;\; =(\sin \;\frac{\pi}{4}+\cos \;\frac{\pi}{4})-(0+1)=\sqrt{2}-1$ $A_{2} \;\; =int_{0}^{\;\frac{\pi}{4}} \sin x d x+int_{\;\frac{\pi}{4}}^{\;\frac{\pi}{2}} \cos x d x$ $\;A_{2}=[-\cos x]_{0}^{\;\frac{\pi}{4}}+[\sin x]_{\;\frac{\pi}{4}}^{\;\frac{\pi}{2}}=\sqrt{2}(\sqrt{2}-1)$ $\;\text{ Now, }\; A_{1}: A_{2}=(\sqrt{2}-1): \sqrt{2}(\sqrt{2}-1)$ $\;\;\; A_{1}: A_{2}=1: \sqrt{2}$ $\;\text{ and }\; \;\; A_{1}+A_{2}=(\sqrt{2}-1)+\sqrt{2}(\sqrt{2}-1)=(\sqrt{2}-1)(\sqrt{2}+1)$ $\;A_{1}+A_{2}=2-1=1$ Therefore, $\;\text{ and }\; \;\; A_{1}: A_{2} \;\; =1: \sqrt{2},$ $A_{1}+A_{2} \;\; =1$

Q80JEE Main 2021MCQ4MProbability

A seven digit number is formed using digits $3,3,4,4,4,5,5$ . The probability, that number so formed is divisible by 2 , is

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📖 Explanation

Given, digits 3, 3, 4, 4, 4, 5, 5 Total 7 digit number from given digits will be $\;\frac{7 !}{3 ! 2 ! 2 !} \;\; =\;\frac{7 \times 6 \times 5 \times 4}{2 \times 2}$ $\;\; =210$ Now, if the 7 digit number is divisible by 2 , then last digit must be 4 . Now, 7 digit number divisible by 2 will be $=\;\frac{6 !}{2 ! 2 ! 2 !}=\;\frac{6 \times 5 \times 4 \times 3}{2 \times 2}=90$ Required probability $=\;\frac{90}{210}=\;\frac{9}{21}=\;\frac{3}{7}$

Q81JEE Main 2021NAT4MComplex Numbers

Let $z$ be those complex numbers which satisfy $|z+5| \le 4$ and $z(1+i)+\overline{z}(1-i) \ge -10$ , $i=\sqrt{-1}$ . If the maximum value of $|z+1|^{2}$ is $\alpha+\beta \sqrt{2}$ , then the value of $(\alpha+\beta)$ is ......... .

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📖 Explanation

Given, $|z+5| \le 4$ , which is equation of circle. $|z+5| \le 4\;$ $\Rightarrow \;(x+5)^{2}+y^{2} \le 16$ . . . (i) $\;\text{ and }\;\;z(1+i)+\overline{z}(1-i) \ge -10$ $\Rightarrow \;(z+\overline{z})+i(z-\overline{z}) \ge -10$ $\Rightarrow \;x-y+5 \ge 0$ . . . (ii) From Eqs. (i) and (ii), region bounded by inequalities areNow, $|z+1|^{2}=|z-(-1)|^{2}$ Maximum value of $|z+1|^{2}$ will be equal to $(A C)^{2}$ . Now, $(x+5)^{2}+y^{2}=16$ and $x-y+5=0$ Given, $y=\pm 2 \sqrt{2}$ and $x=\pm 2 \sqrt{2}-5$ $\therefore$ Coordinates are $A(-2 \sqrt{2}-5,-2 \sqrt{2})$ $B(2 \sqrt{2}-5,2 \sqrt{2})$ and $\;\; C(-1,$ , $A C^{2} \;\; =(2 \sqrt{2}+4)^{2}+(2 \sqrt{2})^{2}$ $\;\; =32+16 \sqrt{2}$ Given, that maximum value of $|z+1|^{2} \;\text{ is }\; \alpha+\beta \sqrt{2}$ $\Rightarrow \;\alpha+\beta \sqrt{2}\;=32+16 \sqrt{2}$ $\Rightarrow \;\alpha\;=32, \beta=16$ $\therefore \;\alpha+\beta\;=32+16=48$

Q82JEE Main 2021NAT4MCircles

Let the normals at all the points on a given curve pass through a fixed point $(a, b)$ . If the curve passes through $(3,-3)$ and $(4,-2 \sqrt{2})$ and given that $a-2 \sqrt{2} b=3$ , then $(a^{2}+b^{2}+a b)$ is equal to ______.

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📖 Explanation

Let the equation of normal be $(x-x_{1})$ $y-y_{1}=(\;\frac{-1}{m})(x-x_{1})$ . . . (i) where, $m=\;\frac{d y_{1}}{d x_{1}}$ Given that, it passes through point $(a, b)$ , then from Eq. (i), we get $b-y_{1} \;\; =(\;\frac{-1}{m})(a-x_{1})$ $\;\; =\;\frac{-d x_{1}}{d y_{1}}(a-x_{1})$ $(b-y_{1}) d y_{1} \;\; =-(a-x_{1}) d x_{1}$ On integrating, we get $b y_{1}-\;\frac{y_{1}^{2}}{2}=-a x_{1}+\;\frac{x_{1}^{2}}{2}+c$ . . . (ii) Eq. (ii), passes through $(3,-3)$ . and $(4,-2 \sqrt{2)}$ , then from Eq. (ii) $-3 b-\;\frac{9}{2} \;\; =\;\frac{9}{2}-3 a+c$ $\Rightarrow \;3 a-3 b-c=9$ . . . (iii) $\;\text{ and }\; \;\; -2 \sqrt{2} b-4=8-4 a+c$ $\Rightarrow \;4 a-2 \sqrt{2} b-c=12$ . . . (iv) Also, given that $a-2 \sqrt{2 b}=3$ . . . (v) Use Eqs. (iii), (iv), (v) to solve for $a, b, c$ , Subtract Eq. (iv) from Eq. (iii), $-a+(2 \sqrt{2}-3) b=-3$ . . . (vi) Now, add Eqs. (v) and (vi) $b=0$ Put $b=0$ in Eq. (vi) $a=3$ $\therefore a=3, b=0$ $\therefore \;\; a=3, b=0$ Put $a=3, b=0$ in $a^{2}+b^{2}+a b$ $a^{2}+b^{2}+a b=(3)^{2}+0+0=9$

Q83JEE Main 2021NAT4MSequences and Series

Let $\alpha$ and $\beta$ be two real numbers, such that $\alpha+\beta=1$ and $\alpha \beta=-1$ . Let $p_{n}=(\alpha)^{n}+(\beta)^{n}$ , $p_{n-1}=11$ and $p_{n+1}=29$ , for some integer $n \ge 1$ . Then, the value of $p_{n}^{2}$ is _______.

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📖 Explanation

Given that, $\alpha+\beta=1, \alpha \beta=-1$ Let $\alpha, \beta$ be roots of quadratic equation, then the quadratic equation be $x^{2}-x-1=0$ $\;\text{ Now, }\;\;\alpha^{2}-\alpha-1=0$ $\Rightarrow \;\alpha^{2}=\alpha+1$ . . . (i) Similarly, $\beta^{2}=\beta+1$ . . . (ii) Multiply $\alpha^{n-1}$ in Eq. (i), we get $\alpha^{n+1}=\alpha^{n}+\alpha^{n-1}$ . . . (iii) Multiply $\beta^{n-1}$ in Eq. (ii), we get $\beta^{n+1}=\beta^{n}+\beta^{n-1}$ . . . (iv) Add Eqs. (iii) and (iv), we get $\alpha^{n+1}+\beta^{n+1} \;\; =(\alpha^{n}+\beta^{n})+(\alpha^{n-1}+\beta^{n-1})$ $p_{n+1} \;\; =p_{n}+p_{n-1}$ $29 \;\; =p_{n}+11 \Rightarrow P_{n}=18$ $p_{n}^{2} \;\; =(18)^{2}=324$

Q84JEE Main 2021NAT4MDefinite Integrals

If $I_{m \cdot n}=int_{0}^{1} x^{m-1}(1-x)^{n-1} d x$ , for $m, n \ge 1$ and $int_{0}^{1} \;\frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}} d x=\alpha l_{m \cdot n}, \alpha \in R$ , then $\alpha$ equals ______.

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📖 Explanation

Given, $I_{m n}=int_{0}^{1} x^{m-1}(1-x)^{n-1} d x$ Using substitution put $x=\;\frac{1}{t+1}$ Then, $d x=\;\frac{-1}{(t+1)^{2}} d t$ $I_{m n}=int_{\infty }^{0}(-1) \;\frac{1}{(t+1)^{m-1}} \cdot \;\frac{t^{n-1}}{(t+1)^{n-1}} \cdot \;\frac{1}{(t+1)^{2}} d t$ $=-int_{\infty }^{0} \;\frac{t^{n-1}}{(t+1)^{m+n}} d t$ . . . (i) Similarly, $I_{m n} \;\; =int_{0}^{1} x^{n-1}(1-x)^{m-1} d x$ . . . (ii) $\Rightarrow \;\; I_{m n} \;\; =int_{0}^{\infty } \;\frac{t^{m-1}}{(t+1)^{m+n}}$ From Eqs. (i) and (ii), we get $2 I_{m n}=int_{0}^{\infty } \;\frac{t^{n-1}+t^{m-1}}{(t+1)^{m+n}} d t$ $2 I_{m n}=int_{0}^{1 t^{n-1}+t^{m-1}} \;\frac{d t}{(t+1)^{m+n}}+int_{1}^{\infty } \;\frac{t^{n-1}+t^{m-1}}{(t+1)^{m+n}} d t$ $\;\text{ Let }\; I_{1}=int_{1}^{\infty } \frac{t^{n-1}+t^{m-1}}{(t+1)^{m+n}} d t$ $\;\text{ Let }\; t=\frac{1}{z}, \;\text{ then }\; d t=\frac{-1}{z^{2}} d z$ $I_{1}=int_{1}^{0}(-1) \frac{(\frac{1}{z})^{n-1}+(\frac{1}{z})^{m-1}}{(\frac{1}{z}+1)^{m+n}} \cdot \frac{1}{z^{2}} d z=-int_{1}^{0} \frac{z^{n-1}+z^{m-1}}{(z+1)^{m+n}} d z$ $2 l_{m n}=int_{0}^{1 t^{n-1}+t^{m-1}} \frac{d t}{(t+1)^{m+n}} d$ $\;\; -int_{1}^{0} \frac{z^{n-1}+z^{m-1}}{(z+1)^{m+n}} d z$ $\;\; =2 int_{0}^{1 t^{m-1}+t^{n-1}} \frac{(t+1)}{(t+1)^{m+n}} d t$ $\Rightarrow \alpha=1$

Q85JEE Main 2021NAT4MSequences and Series

If the arithmetic mean and geometric mean of the $p$ th and $q$ th terms of the sequence $-16,8,-4,2, . . .$ satisfy the equation $4 x^{2}-9 x+5=0$ , then $p+q$ is equal to _______ .

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📖 Explanation

If ${AM}$ and ${GM}$ satisfy the equation $4 x^{2}-9 x+5=0$ , then ${AM}$ and ${GM}$ are nothing but roots of this quadratic equation, $4 x^{2}-9 x+5=0$ $\Rightarrow \;\; 4 x^{2}-4 x-5 x+5 \;\; =0$ $\Rightarrow \;\; 4 x(x-1)-5(x-1) \;\; =0$ $\Rightarrow \;\; (x-1)(4 x-5) \;\; =0 \Rightarrow x=1, \;\frac{5}{4}$ $M=1$ $[ \because A M \ge G M]$ Then, ${AM}=\;\frac{5}{4}$ and ${GM}=$ Again, the given series is $-16,8,-4,2 . . . . . .$ which is a geometric progression series with common ratio $\;\frac{-1}{2}$ , then $p$ th term $=-16(\;\frac{-1}{2})^{p-1}=t_{p}$ $q$ th term $=-16(\;\frac{-1}{2})^{q-1}=t_{q}$ Arithmetic mean $=\;\frac{5}{4}$ $\Rightarrow \;\; \;\frac{t_{p}+t_{q}}{2}=\;\frac{5}{4}$ and Geometric mean $=1$ $\;\Rightarrow \;\sqrt{t_{p} t_{q}}\;=1$ $\; \because \;t_{p} t_{q}\;=1$ $\Rightarrow \;(-16)(\;\frac{-1}{2})^{p-1}(-16)(\;\frac{-1}{2})^{q-1}\;=1$ $\Rightarrow \;(-16)^{2}(\;\frac{-1}{2})^{p+q-2}\;=1$ $\Rightarrow \;(-2^{4})^{2}(\;\frac{-1}{2})^{p+q-2}\;=1$ $\Rightarrow \;(-2)^{8} \;\frac{(1)^{p+q-2}}{(-2)^{p+q-2}}\;=1$ $\Rightarrow \;(-2)^{8}\;=(-2)^{p+q-2}$ $\;(-2)^{8}\;=(-2)^{p+q-2}$ $\Rightarrow \;p+q-2\;=8$ $\Rightarrow \;p+q\;=10$

Q86JEE Main 2021NAT4MPermutations and Combinations

The total number of 4 -digit numbers whose greatest common divisor with 18 is 3 , is ______.

🚀 Solve in Practice Mode

📖 Explanation

Let $x$ be four digit number, then $g_{C d}(x, 18)=3$ This implies $x$ is divisible by 3 but not divisible by 9 . The 4-digit numbers which is an odd multiple of 3 are 1005,1011 , $1017, . . . . .9999$ These are 1499 in counting i.e. total number of 4-digit numbers which is odd multiple of 3 are $1499 .$ Now, The 4-digit numbers which is an odd multiple of 9 are, 1017,1035 , ...9999 These, are total $499 .$ Then, required 4-digit numbers $=1499-499=1000$

Q87JEE Main 2021NAT4MEllipse

Let $L$ be a common tangent line to the curves $4 x^{2}+9 y^{2}=36$ and $(2 x)^{2}+(2 y)^{2}=31$ . Then, the square of the slope of the line $L$ is _____.

🚀 Solve in Practice Mode

📖 Explanation

Given, curves are $A: 4 x^{2}+9 y^{2}=36$ $B:(2 x)^{2}+(2 y)^{2}=31$ $L$ is common tangent line to curve $A$ and $B$ . Let slope of $L$ be $m$ . Given, curve $A$ is equation of ellipse. $\;\frac{x^{2}}{9}+\;\frac{y^{2}}{4}=1$ Then, equation of tangent to ellipse will be $y=m x \pm \sqrt{9 m^{2}+4}$ . . . (i) Curve $B$ is circle i.e $x^{2}+y^{2}=\;\frac{31}{4}$ $\;\text{ Then, equation of tangent to circle will be }\;$ $\;\;\;\; y=m x \pm \sqrt{(\;\frac{31}{4}) m^{2}+(\;\frac{31}{4})}$ . . . (ii) On comparing Eqs. (i) and (ii), we get On comparing Eqs. (i) and (ii), we get $\;9 m^{2}+4=(\;\frac{31}{4}) m^{2}+(\;\frac{31}{4})$ $\Rightarrow \;36 m^{2}+16=31 m^{2}+31 \Rightarrow 5 m^{2}=15$ $\Rightarrow \;m^{2}=3$ $\therefore$ Square of slope of line $L$ is 3 .

Q88JEE Main 2021NAT4MDifferentiation

Let $a$ be an integer, such that all the real roots of the polynomial $2 x^{5}+5 x^{4}+10 x^{3}+10 x^{2}+10 x+10$ lie in the interval $(a, a+1)$ . Then, $|a|$ is equal to ______.

🚀 Solve in Practice Mode

📖 Explanation

Let $f(x)=2 x^{5}+5 x^{4}+10 x^{3}+10 x^{2}+10 x+10$ Using hit and trial method, $f(-2)=-34 < 0$ and $f(-1)=3 > 0$ Hence, $f(x)$ has a root in $(-2,-1)$ . Again, $f^{'}(x) \;\; =10 x^{4}+20 x^{3}+30 x^{2}+20 x+10$ $\;\; =10 x^{2}(x^{2}+2 x+3+\;\frac{2}{x}+\;\frac{1}{x^{2}})$ $\;\; =10 x^{2}[(x^{2}+\;\frac{1}{x^{2}})+2(x+\;\frac{1}{x})+3]$ $\;\; =10 x^{2}[(x+\;\frac{1}{x})^{2}+1+2(x+\;\frac{1}{x})]$ $\;\; =10 x^{2}\$ [(x+;\frac{1}{x}+1)^{2}]$>0 ∀ x\Rightarrow f(x) $is strictly increasing function, since degree of$ f(x) $is odd.$ \therefore $It has exactly one real root. Therefore,$ f(x) $has atleast one root in$ \Rightarrow ;; |a|=|-2|=2$

Q89JEE Main 2021NAT4MStatistics

Let $X_{1}, X_{2}, . . ., X_{18}$ be eighteen observations, such that $sum_{i=1}^{18}(X_{i}-\alpha)=36$ and $sum_{i=1}^{18}(X_{i}-\beta)^{2}=90$ , where $\alpha$ and $\beta$ are distinct real numbers. If the standard deviation of these observations is 1, then the value of $|\alpha-\beta|$ is ______.

🚀 Solve in Practice Mode

📖 Explanation

Given, $sum_{i=1}^{18}[x_{i}-\alpha]=36$ and $\;\; sum_{i=1}^{18}|x_{i}-\beta|^{2}=90$ Now, $\;\; sum_{i=1}^{18}(x_{i}-\alpha)=36 \Rightarrow sum_{i=1}^{18} x_{i}-18 \alpha=36$ $\Rightarrow \;\; sum_{i=1}^{18} x_{i}=36+18 \alpha$ . . . (i) Again, $sum_{i=1}^{18} x_{i}^{2}+18 \beta^{2}-2 \beta sum_{i=1}^{18} x_{i}=90$ $\Rightarrow sum_{i=1}^{18} x_{i}^{2}+18 \beta^{2}-2 \beta(36+18 \alpha)=90$ [Using Eq. (i)] $\Rightarrow sum_{i=1}^{18} x_{i}^{2}=90-18 \beta^{2}+2 \beta(36+18 \alpha)$ . . . (ii) Given, standard deviation $=1$ i.e $\sigma^{2}=1$ $\Rightarrow \;\; \;\frac{sum_{i=1}^{18} x_{i}^{2}}{18}-(\;\frac{sum_{i}^{18} x_{i}}{18})^{2}=1$ $\Rightarrow \;\; \;\frac{1}{18}(90-18 \beta^{2}+36 \alpha \beta+72 \beta)$ $\Rightarrow \;\; -(\;\frac{36+18 \alpha}{18})^{2}=1$ $\Rightarrow \;\; 5-\beta^{2}+2 \alpha \beta+4 \beta-4-\alpha^{2}+(-4 \alpha)=1$ $\Rightarrow \;-\beta^{2}+2 \alpha \beta+4 \beta-\alpha^{2}-4 \alpha=0$ $\Rightarrow \;\alpha^{2}+\beta^{2}+(-2 \alpha \beta)-4 \beta+4 \alpha=0$ $\Rightarrow \;(\alpha-\beta)^{2}+4(\alpha-\beta)=0$ $\Rightarrow \;(\alpha-\beta)(\alpha-\beta+4)=0$ $\Rightarrow \;\alpha-\beta=-4$ $\;|\alpha-\beta|=|-4|=4$

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