Previous Year Paper

16 Mar 2021 Shift 2

89 Questions356 Marks180 MinTake as Timed Mock →

🎯 Practice smarter, not harder

Practice 16 Mar 2021 Shift 2 questions with full analytics — free

✓ Attempt tracking & analytics
✓ Timer & exam-mode simulation
✓ Adaptive difficulty (Easy → Hard)
✓ Bookmarks, notes & mistake review

Just sign in to unlock everything. Free for all students.

🚀 Start Practice Mode Sign in free →

Physics30 questions

Q1JEE Main 2021MCQ4MSemiconductor Electronics

The following logic gate is equivalent to

🚀 Solve in Practice Mode

📖 Explanation

Let us consider the given logic circuit and solve it using Boolean algebra.Here, $Y_{1}=\overline{\overline{A} \cdot \overline{B}}=\overline{\overline{A}}+\overline{\overline{B}}=A+B$ $\therefore \;\; Y=\overline{Y}_{1}=\overline{A+B}$ which is an expression for NOR Gate.

Q2JEE Main 2021MCQ4MCentre of Mass and Collision

A large block of wood of mass $M=5.99 {kg}$ is hanging from two long massless cords. $A$ bullet of mass $m=10 {g}$ is fired into the block and gets embedded in it. The system (block + bullet) then swing upwards, their centre of mass rising a vertical distance $h=9.8 {cm}$ before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is (Take $g=9.8 {ms}^{-2}$ )

🚀 Solve in Practice Mode

📖 Explanation

Given Mass of large block of wood, $M=5.99 {kg}$ Mass of bullet, $m=10 {g}$ Height at which their centre of mass rise, $h=9.8 {cm}$ From the law of conservation of energy, Energy of the system when bullet gets embedded = Energy of the system till it momentarily comes to rest. $\Rightarrow \;\; \;\frac{1}{2}(M+m) v_{1}^{2}=(M+m) g h$ where, $v_{1}=$ velocity of bullet $+$ block system $\Rightarrow \;\; v_{1}=\sqrt{2 g h}$ .....(i) According to law of conservation of momentum, Momentum before collision $=$ Momentum after collision. $\Rightarrow \;\; m v=(M+m) v_{1}$ [where, $v=$ velocity of bullet before collision] $\Rightarrow \;\; m v=(M+m) \sqrt{2 g h}$ [using Eq. (i)] $\Rightarrow \;\; v=(\;\frac{M+m}{m}) \sqrt{2 g h}$ $\Rightarrow \;\; v=\;\frac{(5.99+0.01)}{10 \times 10^{-3}} \times \sqrt{2 \times 9.8 \times 9.8 \times 10^{-2}}$ $\Rightarrow \;\; v=831.55 {ms}^{-1}$

Q3JEE Main 2021MCQ4MMagnetic Effects of Current and Magnetism

A charge $Q$ is moving $d l$ distance in the magnetic field B. Find the value of work done by $B$ .

🚀 Solve in Practice Mode

📖 Explanation

According to question, a charge $Q$ is moving $d$ distance in magnetic field $B$ . As we know that force acting on a point charge by magnetic field is always perpendicular to the velocity of particle $v$ . $\therefore F=Q(v \times B)$ .....(i) The above, statement means that ${F}$ is perpendicular to $v$ . $\because$ Work done, $W=F \cdot d s$ $=Q(v \times B) \cdot d l$ [using Eq. (i)] $=Q(v \times B) d l \cos \theta$ $=Q(v \times B) d l \cos 90^{\circ}$ $[ \because \theta=90^{\circ} {F} ⟂ v$ or ${dl}]$ $=0$ $\therefore$ The work done by the magnetic field will be zero.

Q4JEE Main 2021MCQ4MProperties of Matter

What will be the nature of flow of water from a circular tap, when its flow rate increased from $0.18 \frac{L}{min}$ to $0.48 \frac{L}{min}$ ? The radius of the tap and viscosity of water are $0.5 {cm}$ and $10^{-3} {Pa}- {s}$ , respectively. (Density of water $=10^{3} \frac{kg}{m}^{3}$ )

🚀 Solve in Practice Mode

📖 Explanation

As we know that the nature of flow of water is determined by Reynold's number, $R_{ {e}}=\;\frac{\rho v D}{η}$ Here, $\rho =10^{3} {kgm}^{-3}, r=0.5 {cm}=0.5 \times 10^{-2} {m}$ $D=2 r=1 \times 10^{-2} {m}$ $η=10^{-3} {Pa}- {s}$ $Q_{1}=0.18 \frac{L}{min}=\;\frac{0.18 \times 10^{-3}}{60} {m}^\frac{3}{s}$ $Q_{2}=0.48 {L} {/} \min =\;\frac{0.48 \times 10^{-3}}{60} {m}^\frac{3}{s}$ Also, $v=\;\frac{Q}{\pi r^{2}} \therefore v_{1}=\;\frac{Q_{1}}{\pi r^{2}}$ Also, $v=\;\frac{Q}{\pi r^{2}} \therefore v_{1}=\;\frac{Q_{1}}{\pi r^{2}}$ $\therefore R_{ {e}_{\;\text{initial }\;}}=\;\frac{\rho v_{1} D}{η}=\;\frac{\rho Q_{1} D}{\pi r^{2} η}$ $=\;\frac{10^{3} \times 0.18 \times 10^{-3}}{\pi \times (0.5 \times 10^{-2})^{2} \times 60} \times \;\frac{1 \times 10^{-2}}{10^{-3}}$ $=381.97 ∼eq 382$ and $R_{ {e}_{\;\text{final }\;}}=\;\frac{\rho v_{2} D}{η}=\;\frac{\rho Q_{2} D}{\pi r^{2} η}$ $=\;\frac{10^{3} \times 0.48 \times 10^{-3} \times 1 \times 10^{-2}}{\pi \times (0.5 \times 10^{-2})^{2} \times 60 \times 10^{-3}}$ $=1019.09$ We know that, When $R_{ {e}}< 1000$ , the flow of water is considered to be steady. When $1000< R_{ {e}}< 2000$ , the flow of water becomes unsteady. When $R_{ {e}}>2000$ , the flow of water becomes turbulent. As the $R_{e_{\;\text{final }\;}}$ value lies in between 1000 and 2000 and $R_{ {e}_{\;\text{initial }\;}}$ value is less than 1000 , so the nature of flow of water will be steady flow to the unsteady one.

Q5JEE Main 2021MCQ4MMotion in a Plane

A mosquito is moving with a velocity $v=(0.5 t^{2} \hat{i}+3 t \hat{j}+9 \hat{k}) {m} / {s}$ and accelerating in uniform conditions. What will be the direction of mosquito after 2s?

🚀 Solve in Practice Mode

📖 Explanation

Given, velocity of mosquito, $v=(0.5 t^{2} \hat{i}+3 t \hat{j}+9 \hat{k}) {m} / {s}$ At $t=2 {s}, v=0.5(2)^{2} \hat{i}+3(2) \hat{j}+9 \hat{k}$ $=2 \hat{i}+6 \hat{j}+9 \hat{k}$ $\therefore$ Angle along $X$ -axis, $\cos \alpha=\;{2}/{\sqrt{2^{2}+6^{2}+9^{2}}}=\;\frac{2}{\sqrt{121}}=\;\frac{2}{11}$ or $\alpha=\cos ^{-1}(\;\frac{2}{11})$ Similarly, angle along $Y$ -axis, $\cos \beta=\;\frac{6}{11}$ $\Rightarrow \;\; \beta=\cos ^{-1}(\;\frac{6}{11})$ Angle along $Z$ -axis, $\cos \gamma=\;\frac{9}{11}$ $\Rightarrow \;\; \gamma=\cos ^{-1}(\;\frac{9}{11})$ $\therefore$ Angle made by the direction of motion of mosquito will be $\cos ^{-1}(\;\frac{2}{11})$ from $X$ -axis $=\tan ^{-1} \;\frac{\sqrt{117}}{2}$ $\cos ^{-1}(\;\frac{6}{11})$ from $Y$ -axis $=\tan ^{-1} \;\frac{\sqrt{85}}{2}$ $\cos ^{-1}(\;\frac{9}{11})$ from $Z$ -axis $=\tan ^{-1}(\;\frac{\sqrt{40}}{9})$

Q6JEE Main 2021MCQ4MElectrostatics

Find out the surface charge density at the intersection of point $X=3 {m}$ plane and $X$ -axis, in the region of uniform line charge of $8 {nC} / {m}$ lying along the $Z$ -axis in free space

🚀 Solve in Practice Mode

📖 Explanation

Given, Linear charge density, $\lambda=8 \frac{nC}{m}=8 \times 10^{-9} \frac{C}{m}$ The relation between surface charge density and linear charge density can be given as $\;\frac{2 k \lambda}{r}=\;\frac{\sigma}{\epsilon_{0}}$ .....(i) where, $k=$ Coulomb's constant, $\lambda=$ linear charge density, $\sigma=$ surface charge density, $\epsilon_{0}=$ absolute electrical permittivity of free space and $r=$ distance. Substituting the values in Eq. (i), we get $\sigma=\;\frac{2 k \lambda \epsilon_{0}}{r}$ $=\;\frac{2 \times 9 \times 10^{9} \times 8 \times 10^{-9} \times 8.85 \times 10^{-12}}{3}$ $=0.424 \times 10^{-9} {Cm}^{-2}=0.424 {nCm}^{-2}$

Q7JEE Main 2021MCQ4MDual Nature of Matter and Radiation

The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of $100 {V}$ . What should nearly be the ratio of their wavelengths? $(m_{p}=1.00727 {u}$ , $m_{e}=0.00055 {u}$ )

🚀 Solve in Practice Mode

📖 Explanation

Given, mass of proton, $m_{p}=1.00727 {u}$ Mass of electron, $m_{ {e}}=0.00055 {u}$ Potential, $V=100 {V}$ As, de-Broglie wavelength can be given as $\lambda=\;\frac{h}{\sqrt{2 m q V}} \Rightarrow \lambda \propto \;\frac{1}{\sqrt{m}}$ $\Rightarrow \;\; \;\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\;\frac{m_{2}}{m_{1}}}$ $\Rightarrow \;\; \;{\lambda_{ {e}}}/{\lambda_{p}}=\sqrt{\;{m_{p}}/{m_{ {e}}}}=\sqrt{\;\frac{1.00727 u}{0.00055 u}}$ $\;\frac{\lambda_{e}}{\lambda_{p}}=\sqrt{1831.4}=42.79$ or $\;\frac{\lambda_{e}}{\lambda_{p}} \approx \;\frac{43}{1} \;\text{ or }\; \lambda_{e}: \lambda_{p}=43: 1$

Q8JEE Main 2021MCQ4MElectromagnetic Induction

For the given circuit, comment on the type of transformer used.

🚀 Solve in Practice Mode

📖 Explanation

Voltage across secondary coil, $=\;\frac{\;\text{ Power across load }\;}{\;\text{ Current passing through load }\;}$ $\Rightarrow \;\; V_{2}=\;\frac{P}{I_{L}}=\;\frac{60}{0.11} \Rightarrow V_{2}=545.45 {V}$ Voltage across primary coil, $V_{1}=220 {V}$ $\Rightarrow \;\; V_{2}>V_{1}$ It means that step-up transformer is used.

Q9JEE Main 2021MCQ4MAtoms and Nuclei

The half-life of ${Au}^{198}$ is $2.7$ days. The activity of $1.50 {mg}$ of ${Au}^{198}$ if its atomic weight is $198 {g} {mol}^{-1}$ is $(N_{A}=6 \times 10^{23} / {mol})$

🚀 Solve in Practice Mode

📖 Explanation

Given, Half-life of ${Au}^{198}, T_{1 {/} 2}=2.7$ days Amount of ${Au}^{198}, m=1.5 {mg}=1.5 \times 10^{-3} {g}$ Atomic weight of ${Au}^{198}, A=198 {g} {mol}^{-1}$ We know that activity, $A_{0}=\lambda N_{0}$ where, $\lambda=$ decay constant $=\;{\ln 2}/{T_{1 {/} 2}}$ $N_{0}=$ initial number of atoms $=\;\frac{m}{A} N_{A}$ $\Rightarrow =A_{0} \lambda \cdot \;\frac{m}{A} N_{A}=\;\frac{\ln 2}{T_{1 / 2}} \times \;\frac{1.5 \times 10^{-3}}{198} \times 6.023 \times 10^{23}$ $\Rightarrow A_{0}=\;\frac{\ln 2 \times 1.5 \times 10^{-3} \times 6 \times 10^{23}}{2.7 \times 3600 \times 24 \times 198 \times 3.7 \times 10^{10}} {Ci}$ $[ \because 1$ curie $( {Ci})=3.7 \times 10^{10} {Bq}]$ $\Rightarrow A_{0}=3.65 \times 10^{2} {Ci}=365 {Ci}$ The answer is close to $357 {Ci}$ .

Q10JEE Main 2021MCQ4MThermodynamics

Calculate the value of mean free path $(\lambda)$ for oxygen molecules at temperature $27^{\circ} {C}$ and pressure $1.01 \times 10^{5} {Pa}$ . Assume the molecular diameter $0.3 {nm}$ and the gas is ideal. $(k=1.38 \times 10^{-23} {JK}^{-1})$

🚀 Solve in Practice Mode

📖 Explanation

Given, Pressure of the gas, $p=1.01 \times 10^{5} {Pa}$ Absolute temperature of gas, $T=27^{\circ} {C}$ $=27+273=300 {K}$ Molecular diameter, $d=0.3 {nm}=0.3 \times 10^{-9} {m}$ Boltzmann constant, $k_{B}=1.38 \times 10^{-23} {J} / {K}$ Mean free path, $\lambda=\;\frac{k_{B} T}{\sqrt{2} \pi d^{2} p}$ ....(i) Substituting the given values in Eq. (i), we get $\lambda=\;\frac{1.38 \times 10^{-23} \times 300}{\sqrt{2} \times 3.14 \times (0.3 \times 10^{-9})^{2} \times 1.01 \times 10^{5}}$ $=102 {nm}$

Q11JEE Main 2021MCQ4MRay Optics and Optical Instruments

The refractive index of a converging lens is 1.4. What will be the focal length of this lens if it is placed in a medium of same refractive index? (Assume the radii of curvature of the faces of lens are $R_{1}$ and $R_{2}$ respectively)

🚀 Solve in Practice Mode

📖 Explanation

Consider a convex lens of refractive index $\mu _{2}$ and $\mu _{1}$ is the refractive index of the medium in which it is placed. $\Rightarrow \;\; \mu _{2}=\mu _{1}$ .....(i) According to the lens Maker's formula, $\;\frac{1}{f}=[\;\frac{\mu _{1}}{\mu _{2}}-1][\;\frac{1}{R_{1}}-\;\frac{1}{R_{2}}]$ ....(ii) where, $f$ is the focal length of the lens, $R_{1}$ and $R_{2}$ are the radii of curvature of respective faces of lens. From Eqs. (i) and (ii), we can write $\;\frac{1}{f}=[1-1][\;\frac{1}{R_{1}}-\;\frac{1}{R_{2}}]=0$ $\Rightarrow \;\; \;\frac{1}{f}=0 \Rightarrow f=$ Infinite

Q12JEE Main 2021MCQ4MUnits and Measurements

In order to determine the Young's modulus of a wire of radius $0.2 {cm}$ (measured using a scale of least count $=0.001 {cm}$ ) and length $1 {m}$ (measured using a scale of least count $=1 {mm}$ ), a weight of mass $1 {kg}$ (measured using a scale of least count $=1 {g}$ ) was hanged to get the elongation of $0.5 {cm}$ (measured using a scale of least count $0.001 {cm}$ ). What will be the fractional error in the value of Young's modulus determined by this experiment?

🚀 Solve in Practice Mode

📖 Explanation

Young's modulus, $Y=\;\frac{\;\text{ Stress }\;}{\;\text{ Strain }\;}=\;\frac{F L}{A l}$ $\Rightarrow Y=\;\frac{m g L}{\pi R^{2} I}$ .....(i) $[ \because F=m g$ and $A=\pi R^{2}]$ To determine the fractional errors, we can write Eq. (i) as follows $\;\frac{\Delta Y}{Y}=\;\frac{\Delta m}{m}+\;\frac{\Delta L}{I}+\;\frac{2 \cdot \Delta R}{R}+\;\frac{\Delta l}{I}$ $\Rightarrow \;\; \;\frac{\Delta Y}{Y} \times 100=\;\frac{\Delta m}{m} \times 100+\;\frac{\Delta L}{L} \times 100+\;\frac{2 \Delta R}{R} \times 100+\;\frac{\Delta l}{l} \times 100$ $\Rightarrow \;\; \;\frac{\Delta Y}{Y} \times 100=100[\;\frac{\Delta m}{m}+\;\frac{\Delta L}{L}+\;\frac{2 \Delta R}{R}+\;\frac{\Delta l}{I}]$ $=100[\;\frac{1}{1000}+\;\frac{1}{1000}+2(\;\frac{0.001}{0.2})+\;\frac{0.001}{0.5}]$ $=\;\frac{1}{10}+\;\frac{1}{10}+1+\;\frac{1}{5}=\;\frac{14}{10}=1.4 \%$

Q13JEE Main 2021MCQ4MProperties of Matter

A bimetallic strip consists of metals $A$ and $B$ . It is mounted rigidly as shown. The metal $A$ has higher coefficient of expansion compared to that of metal $B$ . When the bimetallic strip is placed in a cold bath, it will

🚀 Solve in Practice Mode

📖 Explanation

As per the question, coefficient of expansion of metal $A$ is higher than that of $B$ . It means, $\alpha_{A}>\alpha_{B}$ If the bimetallic strip is placed in a cold bath, then the length of both strips will decrease. i.e., $I_{A}>I_{B}$ $\therefore$ The bimetallic strip will bend towards left. It can be shown as below.

Q14JEE Main 2021MCQ4MCurrent Electricity

A resistor develops $500 {J}$ of thermal energy in $20 {s}$ , when a current of $1.5 {A}$ is passed through it. If the current is increased from $1.5 {A}$ to $3 {A}$ , what will be the energy developed in 20 s?

🚀 Solve in Practice Mode

📖 Explanation

Given, Heat energy, $H_{1}=500 {J}$ Initial current, $I_{1}=1.5 {A}$ , final current, $I_{2}=3 {A}$ and time, $t=20 {s}$ According to Joule's law of heating, $H=I^{2} R t$ $\Rightarrow \;\; H_{1}=l_{1}^{2} R t$ .....(i) and $H_{2}=I_{2}^{2} R t$ ....(ii) [ Resistance and time being the same in both cases] Dividing Eq. (i) by Eq. (ii), we get $\Rightarrow \;\; \;\frac{H_{1}}{H_{2}}=\;\frac{I_{1}^{2} R t}{I_{2}^{2} R t} \Rightarrow \;\frac{H_{1}}{H_{2}}=(\;\frac{I_{1}}{I_{2}})^{2}$ $\Rightarrow \;\; \;\frac{H_{1}}{H_{2}}=(\;\frac{1.5}{3})^{2}=(\;\frac{15}{30})^{2} \Rightarrow \;\frac{H_{1}}{H_{2}}=(\;\frac{1}{2})^{2}$ $\Rightarrow \;\; \;\frac{H_{1}}{H_{2}}=\;\frac{1}{4} \Rightarrow \;\frac{500}{H_{2}}=\;\frac{1}{4} \Rightarrow H_{2}=500 \times 4$ $\Rightarrow \;\; H_{2}=2000 {J}$

Q15JEE Main 2021MCQ4MCircular Motion

Statement I A cyclist is moving on an unbanked road with a speed of $7 {kmh}^{-1}$ and takes a sharp circular turn along a path of radius of $2 {m}$ without reducing the speed. The static friction coefficient is $0.2$ . The cyclist will not slip and pass the curve $(g=9.8 \frac{m}{s}^{2})$ Statement II If the road is banked at an angle of $45^{\circ}$ , cyclist can cross the curve of $2 {m}$ radius with the speed of $18.5 {kmh}^{-1}$ without slipping. In the light of the above statements, choose the correct answer from the options given below.

🚀 Solve in Practice Mode

📖 Explanation

The maximum speed of cyclist on turn of unbanked road without slipping is given as $v_{\max }=\sqrt{\mu g R}=\sqrt{0.2 \times 10 \times 2}=2 {ms}^{-1} \;\; [ \because \mu =0.2$ (given) $]$ Given, speed $=7 \frac{km}{h}$ $=\;\frac{7000}{3600} {ms}^{-1}=\;\frac{70}{36}=1.94 {ms}^{-1}$ As given speed is lesser than $v_{\max }$ , so the cyclist will not slip. Therefore, Statement I is true. As per Statement II, angle of banking, $\theta=45^{\circ}$ We know that, for banked road, $V_{\max }=\sqrt{\;\frac{g R(\mu +\tan \theta)}{(1-\mu \tan \theta)}}$ and $\;\; v_{\min }=\sqrt{\;\frac{g R(\tan \theta-\mu )}{1+\mu \tan \theta}}$ $\Rightarrow \;\; v_{\max }=\sqrt{\;\frac{10 \times 2(0.2+\tan 45^{\circ})}{1-0.2 \tan 45^{\circ}}}$ and $\;\; v_{\min }=\sqrt{\;\frac{10 \times 2(1-0.2)}{1+0.2}}$ and $\;\; v_{\min }=\sqrt{\;\frac{10 \times 2(1-0.2)}{1+0.2}}$ $\Rightarrow \;\; v_{\max }=5.47 {ms}^{-1}$ and $v_{\min }=3.65 {ms}^{-1}$ $\because \;\; v=18.5 {km} / {h}=\;\frac{18.5 \times 1000}{5600}=5.13 {ms}^{-1}$ $[ \because$ given speed $=18.5 {kmh}^{-1}]$ As, $v_{\min } < v < v_{\max }$ , so the cyclist will not slip. $\therefore$ Statement II is also true. Hence, option (d) is the correct.

Q16JEE Main 2021MCQ4MElectromagnetic Waves

Two identical antennas mounted on identical towers are separated from each other by a distance of $45 {km}$ . What should nearly be the minimum height of receiving antenna to receive the signals in line of sight? (Assume, radius of earth is $6400 {km}$ .)

🚀 Solve in Practice Mode

📖 Explanation

Given, distance between two antennas, $D=45 {km}$ Radius of Earth, $R=6400 {km}$ Height of the receiving antenna, $h=$ ? We know that, $D=2 \sqrt{2 R h}$ $\Rightarrow \;\; h=\;\frac{D^{2}}{8 R}$ .....(i) Substituting the given values in Eq. (i), we get $h=\;\frac{(45)^{2}}{8 \times 6400}=\;\frac{(45)^{2}}{8 \times 6400}$ $=39.55 {m}$

Q17JEE Main 2021MCQ4MElectromagnetic Induction

The magnetic field in a region is given by $B=B_{0}(\;\frac{x}{a}) \hat{k}$ . A square loop of side $d$ is placed with its edges along the $X$ and $Y$ -axes. The loop is moved with a constant velocity $v=v_{0} \hat{i}$ . The emf induced in the loop is

🚀 Solve in Practice Mode

📖 Explanation

The given situation can be shown as From the above figure, it can be seen that emf induced in side $D C$ and $A B$ will be zero. It is because $v$ is parallel to length of conductor along $X$ -axis. $\therefore$ Net emf induced $=E_{B C}-E_{A D}$ $=\;\frac{B_{0}(x+d) v_{0} d}{a}-\;\frac{B_{0} x v_{0} d}{a}$ $=\;\frac{B_{0} v_{0} d^{2}}{a}$

Q18JEE Main 2021MCQ4MOscillations

Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass $=500 {g}$ , decay constant $=20 \frac{g}{s}$ , then how much time is required for the amplitude of the system to drop to half of its initial value? $(\ln 2=0.693)$

🚀 Solve in Practice Mode

📖 Explanation

Given, Mass, $m=500 {g}$ Decay constant, $\lambda=20 {g} / {s}$ For damped oscillation, $A=A_{0} {e}^{-b {bt} / 2 m}$ ....(i) Here, $b=$ damped constant $=$ decay constant As per question, the amplitude of the system is dropped to half of its initial value. So, Eq. (i) becomes $\;\frac{A_{0}}{2}=A_{0} {e}^{-b t / 2 m} \Rightarrow \;\frac{1}{2}= {e}^{-b t / 2 m}$ $\Rightarrow \;\; 2= {e}^{b t / 2 m}$ $\Rightarrow \;\; 2= {e}^{b t / 2 m}$ Taking log on both sides of above equation, we get $\log 2=\log (e^{b t / 2 m})$ $\Rightarrow \;\; \log 2=\;\frac{b t}{2 m} \Rightarrow t=\;\frac{2 m}{b} \log 2$ $=\;\frac{2 \times 500 \times 0.693}{20}$ $[ \because \log 2=0.693]$ $\Rightarrow t=34.65 {s}$

Q19JEE Main 2021MCQ4MAtoms and Nuclei

Calculate the time interval between $33 \%$ decay and $67 \%$ decay if half-life of a substance is $20 {min}$ .

🚀 Solve in Practice Mode

📖 Explanation

As we know that, $N=N_{0} {e}^{-\lambda t}$ where, $N=$ number of radioactive atoms left undecayed, $N_{0}=$ original number of radioactive atoms, $t=$ elapsed time and $\lambda=$ decay constant. $\Rightarrow \;\; N_{1}=N_{0} e^{-\lambda \cdot t_{1}}$ ....(i) and $N_{2}=N_{0} e^{-\lambda t_{2}}$ ....(ii) Consider Eq. (i), $N_{1}=N_{0} {e}^{-\lambda t_{1}}$ $\Rightarrow$ $\;\frac{N_{1}}{N_{0}}= {e}^{-\lambda t_{1}} \Rightarrow 0.67= {e}^{-\lambda t_{1}}$ $\Rightarrow \;\; \log (0.67)=\log (e^{-\lambda t_{1}})$ $\Rightarrow \;\; \log 0.67=-\lambda t_{1}$ .....(iii) Consider Eq. (ii), $N_{2}=N_{0} {e}^{-\lambda t_{2}}$ $\Rightarrow \;\; \;\frac{N_{2}}{N_{0}}= {e}^{-\lambda t_{2}} \Rightarrow 0.33= {e}^{-\lambda t_{2}}$ $\Rightarrow \;\; \log (0.33)=\log ( {e}^{-\lambda t_{2}})$ $\Rightarrow \;\; \log (0.33)=-\lambda t_{2}$ .....(iv) Subtracting Eq. (iv) from Eq. (iii), we get $\log 0.67-\log 0.33=-\lambda t_{1}+\lambda t_{2}$ $\Rightarrow \;\; \log (\;\frac{0.67}{0.33})=\lambda(t_{2}-t_{1}) \Rightarrow \lambda(t_{2}-t_{1})=\log 2$ $\Rightarrow \;\; t_{2}-t_{1}=\;\frac{\log 2}{\lambda}=t_{1 {/} 2}$ $\therefore$ Half-life $=t_{1 {/} 2}=20 {min}$

Q20JEE Main 2021MCQ4MElectromagnetic Waves

Red light differs from blue light as they have

🚀 Solve in Practice Mode

📖 Explanation

Since, $\lambda v=c=$ constant where, $\lambda=$ wavelength of light and $\;\; v=$ frequency of light. Red light and blue light have different wavelengths and different frequencies but same speed.

Q21JEE Main 2021NAT4MCurrent Electricity

The energy dissipated by a resistor is $10 {mJ}$ in $1 {s}$ when an electric current of $2 {mA}$ flows through it. The resistance is ........ $Ω$ . (Round off to the nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

Since, $\lambda v=c=$ constant where, $\lambda=$ wavelength of light and $v=$ frequency of light. Red light and blue light have different wavelengths and different frequencies but same speed. $\Rightarrow \;\; R=\;\frac{H}{I^{2} T}$ .....(i) Substituting the given values in Eq. (i), we get $R=\;\frac{10 \times 10^{-3}}{(2 \times 10^{-3})^{2} \times 1}$ $\Rightarrow \;\; R=\;\frac{10^{-2}}{4 \times 10^{-6}} \Rightarrow R=0.25 \times 10^{4}$ $\Rightarrow \;\; R=2500 Ω$

Q22JEE Main 2021NAT4MElectrostatic Potential and Capacitance

In a parallel plate capacitor set up, the plate area of capacitor is $2 {m}^{2}$ and the plates are separated by $1 {m}$ . If the space between the plates are filled with a dielectric material of thickness $0.5 {m}$ and area $2 {m}^{2}$ (see figure) the capacitance of the set-up will be ...... $\epsilon_{0}$ . (Dielectric constant of the material $=3.2$ ) (Round off to the nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

The equivalent capacitance of the capacitor when dielectric material is partially filled, is given as $C=\;\frac{\epsilon_{0} A}{(d-t)+\;\frac{t}{k}}=\;{\epsilon_{0} A}/{(d-\;\frac{d}{2})+\;\frac{d {/} 2}{k}}=\;\frac{\epsilon_{0} A}{\frac{d}{2 k}+\;\frac{d}{2}}$ where, $\epsilon_{0}=$ absolute electrical permittivity of free space, $A=$ area of the plates of capacitor $=2 {m}^{2}$ $K=$ dielectric constant $=3.2$ $t=$ thickness of dielectric material $=\;\frac{d}{2}$ and $d=$ distance between the plates $=2 {m}$ . $C=\;\frac{2 \epsilon_{0} A}{\frac{d}{k}+d}=\;\frac{2 \times 2 \epsilon_{0}}{\frac{1}{3.2}+1}=\;\frac{4 \times 3.2}{4.2} \epsilon_{0}$ $\Rightarrow \;\; C=3.04 \epsilon_{0}$ The required value after rounding off to the nearest integer is 3 .

Q23JEE Main 2021NAT4MRotational Motion

A force $F=4 \hat{i}+3 \hat{j}+4 \hat{k}$ is applied on an intersection point of $x=2$ plane and $X$ -axis. The magnitude of torque of this force about a point $(2,3,4)$ is ....... (Round off to the nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

Given, Force, $F=4 \hat{i}+3 \hat{j}+4 \hat{k}$ We know that Torque, $\tau =r \times F$ where, $r$ is the perpendicular distance. $r=(2 \hat{i})-(2 \hat{i}+3 \hat{j}+4 \hat{k})=-3 \hat{j}-4 \hat{k}$ $\Rightarrow \;\; \tau =r \times F=\begin{vmatrix}i^{∧} & j^{∧} & k^{∧} \\ 0 & -3 & -4 \\ 4 & 3 & 4\end{vmatrix}$ $\Rightarrow \;\; \tau =\hat{i}(-12+12)-\hat{j}(0+16)+\hat{k}(0+12)$ $\Rightarrow \;\; \tau =-16 \hat{j}+12 \hat{k}$ $\therefore$ Magnitude of torque $|\tau |=\sqrt{(16)^{2}+(12)^{2}}$ $=\sqrt{256+144}=\sqrt{400}$ $\Rightarrow \;\; |\tau |=20$

Q24JEE Main 2021NAT4MGravitation

If one wants to remove all the mass of the earth to infinity in order to break it up completely. The amount of energy that needs to be supplied will be $\;\frac{x}{5} \;\frac{G M^{2}}{R}$ , where $x$ is (Round off to the nearest integer) ( $M$ is the mass of earth, $R$ is the radius of earth and $G$ is the gravitational constant.)

🚀 Solve in Practice Mode

📖 Explanation

We know that binding energy of earth, ${BE}=-\;\frac{3}{5} \;\frac{G M^{2}}{R}$ $\therefore$ Energy required to break the earth into pieces $=- {BE}=\;\frac{3}{5} \;\frac{G M^{2}}{R}$ .......(i) According to question, the amount of energy that needs to be supplied is $\;\frac{x}{5} \;\frac{G M^{2}}{R}$ . Comparing it with value in Eq. (i), we get, $x=3$

Q25JEE Main 2021NAT4MRay Optics and Optical Instruments

A deviation of $2^{\circ}$ is produced in the yellow ray when prism of crown and flint glass are achromatically combined. Taking dispersive powers of crown and flint glass are $0.02$ and $0.03$ respectively and refractive index for yellow light for these glasses are $1.5$ and 1.6, respectively. The refracting angles for crown glass prism will be ...... (in degree). (Round off to the nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

Given Dispersive power of crown glass, $\omega_{1}=0.02$ Dispersive power of flint glass, $\omega_{2}=0.03$ Refractive index of yellow light, for crown glass, $\mu _{1}=1.5$ and for flint glass, $\mu _{2}=1.6$ This is a case of achromatic combination. $\therefore \;\; \theta_{\;\text{net }\;}=0$ $\Rightarrow \;\; \theta_{1}-\theta_{2}=0 \Rightarrow \theta_{1}=\theta_{2}$ $\Rightarrow \;\; \omega_{1} \delta _{1}=\omega_{2} \delta _{2} \;\; [ \because \theta=\omega \delta ] . . .$ (i) and $\delta _{\;\text{net }\;}=\delta _{1}-\delta _{2}=2^{\circ}$ ......(ii) From Eqs. (i) and (ii), we get $\delta _{1}-\;\frac{\omega_{1} \delta _{1}}{\omega_{2}}=2^{\circ}$ $\Rightarrow \;\; \delta _{1}(1-\;\frac{\omega_{1}}{\omega_{2}})=2^{\circ} \Rightarrow \delta _{1}(1-\;\frac{2}{3})=2^{\circ}$ $\Rightarrow \;\; \delta _{1}=6^{\circ}$ Also, $\delta _{1}=(\mu _{1}-1) A_{1}$ $\Rightarrow \;\; 6^{\circ}=(1.5-1) A_{1} \Rightarrow A_{1}=12^{\circ}$ $\therefore$ The refracting angle for crown glass prism will be $12^{\circ}$ .

Q26JEE Main 2021NAT4MLaws of Motion

A body of mass $2 {kg}$ moves under a force of $(2 \hat{i}+3 \hat{j}+5 \hat{k}) {N}$ . It starts from rest and was at the origin initially. After $4 {s}$ , its new coordinates are $(8, b, 20)$ . The value of $b$ is (Round off to the nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

Given Force, $F=(2 \hat{i}+3 \hat{j}+5 \hat{k}) {N}$ Mass, $m=2 {kg}$ Time, $t=4 {s}$ We know that Force $=$ mass $\times$ acceleration $\Rightarrow \;\; F=m \times a \Rightarrow a=\;\frac{F}{m}$ $\Rightarrow \;\; a=\;{2 \hat{i}+3 \hat{j}+5 \frac\hat{k}}{2}$ .....(i) From second equation of motion, $s=u t+\;\frac{1}{2} a t^{2}$ ...(ii) From Eqs. (i) and (ii), we get $s=u t+\;\frac{1}{2} \;{(2 \hat{i}+3 \hat{j}+5 \frac\hat{k})}{2} \cdot t^{2}$ $=0+\;\frac{1}{4}(2 \hat{i}+3 \hat{j}+5 \hat{k}) \cdot (4)^{2} \;\; [ \because u=0$ and $t=4 {s}]$ $=0+8 \hat{i}+12 \hat{j}+20 \hat{k}=8 \hat{i}+12 \hat{j}+20 \hat{k}$ Let $s=x \hat{i}+y \hat{j}+z \hat{k}$ $\Rightarrow x \hat{i}+y \hat{j}+z \hat{k}=8 \hat{i}+12 \hat{j}+20 \hat{k}$ .....(iii) According to question,new coordinates are $(8, b, 20)$ , it means $x \hat{i}+y \hat{j}+z \hat{k}=8 \hat{i}+b \hat{j}+20 \hat{k}$ ......(iv) Comparing Eqs. (iii) and (iv), we get $b=12$

Q27JEE Main 2021NAT4MMotion in a Plane

A swimmer can swim with velocity of $12 \frac{km}{h}$ in still water. Water flowing in a river has velocity $6 \frac{km}{h}$ . The direction with respect to the direction of flow of river water he should swim in order to reach the point on the other bank just opposite to his starting point is ........ (in degree). (Round off to the nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

The situation is depicted in the following figure.where, $V_{M R}=$ velocity of $man=12 \frac{km}{h}$ and $v_{R}=$ velocity of water flow in river $=6 \frac{km}{h}$ As, $v_{M R}$ should be along $C D$ . $\Rightarrow \;\; v_{R}-v_{M R} \sin \theta=0$ $\Rightarrow \;\; 6-12 \sin \theta=0 \Rightarrow \sin \theta=\;\frac{6}{12}$ $\Rightarrow \;\; \sin \theta=\;\frac{1}{2}$ $\Rightarrow \;\; \theta=\sin ^{-1}(\;\frac{1}{2})=\sin ^{-1}(\sin 30^{\circ}) \;\; [ \because \sin 30^{\circ}=\;\frac{1}{2}]$ $\Rightarrow \;\; \theta=30^{\circ}$ $\therefore \;\; \alpha=90^{\circ}+\theta=90^{\circ}+30^{\circ}=120^{\circ}$ $\Rightarrow \;\; \alpha=120^{\circ}$

Q28JEE Main 2021NAT4MWaves

A closed organ pipe of length $L$ and an open organ pipe contain gases of densities $\rho _{1}$ and $\rho _{2}$ respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open pipe is $\;\frac{x}{3} L \sqrt{\;\frac{\rho _{1}}{\rho _{2}}}$ where $x$ is ...... (Round off to the nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

First overtone of open pipe $=\;\frac{V_{2}}{L_{2}}$ First overtone of closed pipe at one end $=\;\frac{3 {v}}{4 L}$ As per question, $\;\frac{3 V}{4 L}=\;\frac{V_{2}}{L}$ $\Rightarrow \;\; \sqrt{\;\frac{B}{\rho _{1}}} \cdot \;\frac{3}{4 L}=\sqrt{\;\frac{B}{\rho _{2}}} \cdot \;\frac{1}{L_{2}}$ $( \because V=\sqrt{\;\frac{B}{\rho }})$ $\Rightarrow \;\; L_{2}=\;\frac{4 L}{3} \sqrt{\;\frac{\rho _{1}}{\rho _{2}}}$ ....(i) According to question, the length of the open pipe is $\;\frac{x}{3} L \sqrt{\;\frac{\rho _{1}}{\rho _{2}}}$ ....(ii) Comparing Eqs. (i) and (ii), we get $x=4$

Q29JEE Main 2021NAT4MRotational Motion

A solid disc of radius $a$ and mass $m$ rolls down without slipping on an inclined plane making an angle $\theta$ with the horizontal. The acceleration of the disc will be $\;\frac{2}{b} g \sin \theta$ , where $b$ is (Round off to the nearest integer) ( $g=$ acceleration due to gravity) $(\theta=$ angle as shown in figure)

🚀 Solve in Practice Mode

📖 Explanation

We know that, on an inclined plane, Acceleration, $a=\;{g \sin \theta}/{1+\;\frac{I}{m R^{2}}}$ $\Rightarrow \;\; a=\;\frac{g \sin \theta}{1+\;\frac{1}{2}}$ $[ \because For disc, $I=\;\frac{m R^{2}}{2}$ ]$ $\Rightarrow \;\; a=\;\frac{2}{3} g \sin \theta$ .....(i) As per question, acceleration of the disc will be $\;\frac{2}{b} g \sin \theta$ . Comparing it with Eq. (i), we get $b=3$

Q30JEE Main 2021NAT4MThermodynamics

For an ideal heat engine, the temperature of the source is $127^{\circ} {C}$ . In order to have $60 \%$ efficiency the temperature of the sink should be ...... ${ }^{\circ} {C}$ . (Round off to the nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

Given, temperature of source, $T_{H}=127^{\circ} {C}$ $=273+127=400 {K}$ Efficience, $η=60 \%=0.6$ The efficiency of Carnot (ideal) heat engine is given by $η=(1-\;\frac{T_{L}}{T_{H}})$ where, $T_{L}=$ temperature of sink, $0.6=(1-\;\frac{T_{L}}{T_{H}}) \Rightarrow \;\frac{T_{L}}{T_{H}}=1-0.6$ $\;\frac{T_{L}}{T_{H}}=0.4 \Rightarrow T_{L}=0.4 \times T_{H}$ $=0.4 \times 400=160 {K}$ $=160-273=-113^{\circ} {C}$

Chemistry29 questions

Q31JEE Main 2021MCQ4MSome Basic Concepts of Chemistry

The green house gas/es is (are) (A) carbon dioxide (B) oxygen (C) water vapour (D) methane Choose the most appropriate answer from the options given below.

🚀 Solve in Practice Mode

Q32JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

In the above reaction, the reagent ' $A'$ is

🚀 Solve in Practice Mode

📖 Explanation

In the given reaction, ${A}$ is alkaline ${KMnO}_{4}, {H}^{+}$ . Alkaline ${KMnO}_\frac{4}{H}^{+}$ is a strong oxidising agent and oxidises alkyl benzene to benzoic acid. While ${NaBH}_{4}, {LiAlH}_{4}$ and ${Zn}- \frac{Hg}{HCl}$ are reducing agent.

Q33JEE Main 2021MCQ4MChemical Thermodynamics

Which of the following reduction reaction cannot be carried out with coke?

🚀 Solve in Practice Mode

📖 Explanation

Reduction of alumina $( {Al}_{2} {O}_{3})$ to ${Al}$ cannot be achieved by carbon reduction. As aluminium has greater reactivity with oxygen than carbon and hence cannot be reduced by carbon. So, electrometallurgy (Hall-Heroult process) is used to get Al. $2 {Al}_{2} {O}_{3}(I)+3 {C} \longrightarrow 4 {Al}+3 {CO}_{2}$ Rest oxides can be reduced by coke as follows ${Fe}_{2} {O}_{3}+3 {C} \longrightarrow 2 {Fe}+3 {CO}$ ${Cu}_{2} {O}+ {C} \longrightarrow 2 {Cu}+ {CO}$ ${ZnO}+ {C} {\to }^{\;\text{ Coke, }\; 1673 {K}} {Zn}+ {CO}$

Q34JEE Main 2021MCQ4MClassification of Elements

Identify the elements $X$ and $Y$ using the ionisation energy values given below.Ionisation energy (Ist) $\frac{kJ}{mol}\;(\text{IInd})$ X4954563Y711450

🚀 Solve in Practice Mode

📖 Explanation

Given data, $\text{IE}_1$ $\text{IE}_2$ X4954563 kJ/molY711450 kJ/mol $X$ is ${Na}(3 s^{1})$ as it has very high second ionisation energy due to removal of electron from highly stable noble gas configuration. ${{Na} {\longrightarrow }^{ {IE}_{1}}}_{1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}} \;\;\;\; {{Na}^{+}+ {e}^{-} ; \;\; {IE}_{1}}_{1 s^{2} 2 s^{2} 2 p^{6} \text{or} [ {Ne}]}=495 \frac{kJ}{mol}$ ${Na}^{+}_{1 s^{2} 2 s^{2} 2 p^{6}} {\longrightarrow }^{ {IE}_{2}} {{Na}^{2+}+ {e}^{-} ; }_{1 s^{2} 2 s^{2} 2 p^{5}}{IE}_{2}=4563 \frac{kJ}{mol}$ (Stable noble gas configuration). In case of ${Mg}$ , noble gas configuration is achieved after removal of second electron. So, ${IE}_{2}$ is comparatively less. ${Mg} _{[{Ne}] 3 s^{2}}\longrightarrow {Mg}^{+}_{[Ne]3s^1}+ {e}^{-} ; {IE}_{1}=731 \frac{kJ}{mol}$ ${Mg}^{+}_{[Ne]3s^1} \longrightarrow {Mg}^{2+}_{[Ne]}+ {e}^{-} ; {IE}_{2}=1450 \frac{kJ}{mol}$

Q35JEE Main 2021MCQ4MHaloalkanes and Haloarenes

Identify the reagent(s) ' $A$ ' and condition(s) for the reaction:

🚀 Solve in Practice Mode

📖 Explanation

The reaction can be achieved by ${Cl}_{2}$ {/}UV light $(A)$ . Given, reaction follows free radical substitution mechanism.Mechanism(I) Initiation : This involves formation of radicals. ${Cl}_{2} \;{{\longrightarrow }^{U {V}}_{\;\text{ light }\;}} 2 {Cl}^{*}$ (II) Propagation : Reactive free radicals react with stable molecules to form new free radicals.Similarly second $\mathrm{Cl}$ atom is also attached at allylic position.

Q36JEE Main 2021MCQ4MBiomolecules

The secondary structure of protein is stabilised by

🚀 Solve in Practice Mode

📖 Explanation

In secondary structure of protein, hydrogen bond is formed between $- {\text{C}}^{||^{\text{O}}}-$ and $- {\text{NH}}-$ groups of the peptide bond. Peptide linkage is an amide formed between $- \text{COOH}$ group and $- {\text{NH}}_{2}$ group while connecting two amino acids. The two monosaccharides are joined together by an oxide linkage formed by the loss of a water molecule is called glycosidic linkage.

Q37JEE Main 2021MCQ4Md and f Block Elements

Fex $_{2}$ and ${Lt}_{3}$ are known when $x$ and $y$ are

🚀 Solve in Practice Mode

📖 Explanation

${Fel}_{3}$ does not exist because of ${I}^{-}$ . Being a very good reducing agent, ${I}^{-}$ reduces ${Fe}^{3+}$ to ${Fe}^{2+}$ . $2 {Fe}^{3+}+2 {I}^{-} \longrightarrow 2 {Fe}^{2+}+ {I}_{2}$ Reducing nature order ${I}> {Br}> {Cl}> {F}$ ${Fe}^{3+}+ {e}^{-} \longrightarrow {Fe}^{2+}\;E^{\circ} \;\; =0.77 {V}$ ${I}_{2}+2 {e}^{-} \longrightarrow 2 {I}^{-}\;E^{\circ} \;\; =-0.54 {V}$ As standard reduction potential value of ${Fe}^{3+}$ is more than ${I}_{2}$ , so it is reduced to ${Fe}^{2+}$ . Hence, except ${Fel}_{3}$ , all other halogens can form iron halide compounds.

Q38JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Which of the following polymer is used in the manufacture of wood laminates ?

🚀 Solve in Practice Mode

Q39JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

Statement I Sodium hydride can be used as an oxidising agent. Statement II The lone pair of electrons on nitrogen in pyridine makes it basic. Choose the correct answer from the options given below.

🚀 Solve in Practice Mode

📖 Explanation

${NaH}$ is not an oxidising agent as ${Na}^{+}$ and ${H}^{-}$ ions are formed. ${Na}$ exists only in $+1$ and hydrogen is present in minimum $-1$ state so, cannot be reduced further. ${NaH}$ is used as reducing agent. Lone pair on nitrogen in pyridine is localised and does not take part in resonance. So, it can accept ${H}^{+}$ acting as a base.So, statement I is true but statement II is false.

Q40JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

The incorrect statement regarding the structure of $C_{60}$ is

🚀 Solve in Practice Mode

📖 Explanation

${C}_{60}$ molecule has a shape like soccer ball and called Buckminster fullerene. It is an allotrope of carbon. It contains twenty-six membered rings and twelve-five membered rings. A six membered ring is fused with six or five membered rings but a five membered ring can only fuse with six membered rings. All the carbon atoms are equal and they undergo $s p^{2}$ -hybridisation. Each carbon atom forms three sigma bonds with other three carbon atoms. So, statement (d) is wrong.

Q41JEE Main 2021MCQ4Mp Block Elements

The correct statements about ${H}_{2} {O}_{2}$ are a. used in the treatment of effluents. b. used as both oxidising and reducing agents. c. the two hydroxyl groups lie in the same plane. d. miscible with water. Choose the correct answer from the options given below.

🚀 Solve in Practice Mode

📖 Explanation

${H}_{2} {O}_{2}$ act as both oxidising and reducing agent. Oxygen is present in $-1$ oxidation state, so it can be oxidised to ${O}_{2}$ as well as reduced to water. ${{H}_{2} {O}_{2}^{-1}}_{\text{Oxidising agent}} \longrightarrow {H}_{2} {O}^{-2} \;\text{ (Reduction) }\;$ ${{H}_{2} {O}_{2}^{-1}}_{\text{Reducing agent}} \longrightarrow {O}_{2}^{0}$ (Oxidation) ${H}_{2} {O}_{2}$ is miscible with water due to intermolecular ${H}$ -bonding. Due to open book like structure both $-OH$ groups are not in same plane and hence, ${H}_{2} {O}_{2}$ is non-planar.(a) ${H}_{2} {O}_{2}$ structure in gas phase, dihedral angle is $111.5^{\circ}$ . (b) ${H}_{2} {O}_{2}$ structure in solid phase at $110 {K}$ , dihedral angle is $90.2^{\circ}$ . ${H}_{2} {O}_{2}$ used in treatment of effluents.

Q42JEE Main 2021MCQ4MAmines

Ammonolysis of alkyl halides followed by the treatment with ${NaOH}$ solution can be used to prepare primary, secondary and tertiary amines. The purpose of ${NaOH}$ in the reaction is

🚀 Solve in Practice Mode

📖 Explanation

Ammonolysis of alkyl halides $R-X \;{──>}^{{NH_{3}}}_{- {HX}} R- {NH}_{2} \;{──>}^{R-X}_{- HX} R_{2} {NH} \;{──>}^{R-X}_ {-HX}R_{3} {N} {──>}^{R X} R_{4} {N^{⊕}X^{⊖}}$ During the reaction ${HX}$ (acid) is formed. Hence, we use ${NaOH}$ to remove this acidic impurities.

Q43JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

An unsaturated hydrocarbon $X$ on ozonolysis gives $A$ . Compound $A$ when warmed with ammonical silver nitrate forms a bright silver mirror along the sides of the test tube. The unsaturated hydrocarbon $X$ is

🚀 Solve in Practice Mode

📖 Explanation

On ozonolysis, unsaturated hydrocarbon X will form carbonyl compound, which gives positive Tollen's test. So, it should be aldehyde. (a) 2,3 -dimethylbut-2-ene undergoes ozonolysis reaction to give 2 molecules of acetone. ${{H}_{3} {C}_{|_{\;\;\;\;\;\text{CH}_3}}- {C}= {C}_{|_{\;\;\;\;\;\text{CH}_3}}=- {CH}_{3}}_{\text{2-3-dimethylbut-2-ene}}{\to }^{\;\text{ (i) }\; {O}_{3}}_{\text{ (ii) }\; \frac{Zn}{H}_{2} {O}}{ 2 {CH}_{3}-\;{ {C}}_{|_{\;\;\;\;\;\text{CH}_3}}= {O}}_{\text{Ketone (acetone)} }$ (b) Isopropylidiene undergoes ozonolysis reaction to give acetone along with cyclopropanone. ${CH_3-C^{|^{\;\;\;\;\;\text{CH}_3}}=◁}_{\text{Isopropylidiene}}{\to }^{\;\text{ (i) }\; {O}_{3}}_{\text{ (ii) }\; \frac{Zn}{H}_{2} {O}}{H_3C-C^{|^{\;\;\;\;\;\text{CH}_3}}=O}_{\text{Acetone}}{+O=◁}_{\text{Cyclopropanone}}$ (c) Butyne reacts with ${O}_{3}$ and $\frac{Zn}{H}_{2} {O}$ to give 2-oxobutyraldehyde. The aldehyde will give Tollen's test to give silver mirror. Note Tollen's reagent contain silver nitrate $( {AgNO}_{3})$ , ammonia and sodium hydroxide $( {NaOH})$ which react with aldehydes to give ${Ag}$ (silver mirror). ${{CH} \equiv {C}- {CH}_{2}}_{\text{Butyne}}- {CH}_{3} {\to }^{ {O}_{3}}_{ {Zn} {/ } {H}_{2} {O}} {{HC^{||^{\text{O}}}}- {C^{||^{\text{O}}}}- {CH}_{2}- {CH}_{3}}_{2-\text{oxobutyraldehyde}}{\to }^{\text{Tollen}'s}_{reagent}\text{Silver mirror}$ (d) But-2-yne undergoes ozonolysis to give butan-2, 3-dione. ${CH_3-C \equiv C-CH_3}_{\text{But}-2-\text{yne}}{\to }^{ {O}_{3}}_{ {Zn} {/ } {H}_{2} {O}}{H_3C-C^{||^{\text{O}}}-C^{||^{\text{O}}}-CH_3}_{\text{Butan}-2, 3-\text{dione}}$ Only product of option (c) contains aldehyde group.

Q44JEE Main 2021MCQ4MAmines

Which of the following is least basic?

🚀 Solve in Practice Mode

📖 Explanation

Basic nature is directly proportional to electron releasing tendency. If lone pair is delocalised, electron density decreases and basic strength decreases.So, there is delocalisation of electron in both options (a) and (c). But it option (c) there is delocalisation of electron with both carbonyl group. Hence, it is weakest base. There is a subtle interplay of the inductive effect, solvation effect and steric hinderance of the alkyl group, which decides the basic strength of alkyl amines in the aqueous state. The order of basic strength in case of ethyl substituted amines in aqueous solution is as follows $( {C}_{2} {H}_{5})_{2} {NH}>( {C}_{2} {H}_{5})_{3} {N}> {C}_{2} {H}_{5} {NH}_{2}> {NH}_{3}$ $\therefore ( {C}_{2} {H}_{5})_{2}{ {N^{..}H}}$ is strong base.

Q45JEE Main 2021MCQ4MClassification of Elements

The characteristics of elements $X, Y$ and $Z$ with atomic numbers, respectively, 33, 53 and 83 are

🚀 Solve in Practice Mode

📖 Explanation

Atomic number 33 is for arsenic (As), 53 is for iodine (I) and 83 is for bismuth (Bi). $X=$ As is metalloid, $Y=I$ is non-metal and $Z=B i$ is metal.

Q46JEE Main 2021MCQ4MPractical Organic Chemistry

Match List-I with List-II.List-I (Test / Reagents / Observation(s))List-II (Species detected)A. Lassaigne's test(i) CarbonB. Cu(II) oxide (ii) SulphurC. Silver nitrate(iii) N,S,P, and halogenD. The sodium fusion extract gives black precipitate with acetic acid and lead acetate(iv) Halogen specificallyThe correct match is

🚀 Solve in Practice Mode

📖 Explanation

(A) Lassaigne test is used to detect ${N}, {S}, {P}, X$ elements. The elements present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. Following reactions take place. ${Na}+ {C}+ {N} {\longrightarrow }^{\Delta } {NaCN}$ $2 {Na}+ {S} {\longrightarrow }^{\Delta } {Na}_{2} {S}$ ${Na}+X {\longrightarrow }^{\Delta } {NaX} \;\; (X= {Cl}, {Br}$ or ${l})$ (B) Carbon and hydrogen are detected by heating the compound with copper (II) oxide. ${C}+2 {CuO} \longrightarrow 2 {Cu}+ {CO}_{2}$ $2 {H}+ {CuO} \longrightarrow {Cu}+ {H}_{2} {O}$ (C) Halides are detected by silver nitrate. They form precipitate of $Ag X$ except fluorine. ${Ag}^{+}( {aq})+X^{-}( {aq}) \longrightarrow {Ag} X(s)$ (D) Sodium fusion extract gives black precipitate of PbS with acetic acid and lead acetate to confirm the presence of sulphur. ${Na}_{2} {S}+( {CH}_{3} {COO}){ }_{2} {Pb} \longrightarrow \;{ {PbS}}_{\;\text{ Black ppt. }\;}+2 {CH}_{3} {COONa}^{2}$

Q47JEE Main 2021MCQ4MSolutions

The incorrect statements below regarding colloidal solutions is

🚀 Solve in Practice Mode

📖 Explanation

Statement (b) is incorrect whereas all other statements are correct. (a) Colloidal solution exhibits colligative properties. (b) An ordinary filter paper cannot stop the flow of colloidal particles as the pores are bigger as compared to size of colloidal particles. However, the pores of filter paper can be reduced in size by impregnating with collodion solution to stop the flow of colloidal particles. The usual collodion is a $4 \%$ solution of nitrocellulose in a mixture of alcohol and ether. An ultra-filter paper may be prepared by soaking the filter paper in a collodion solution, hardening by formaldehyde and then finally drying it. Thus, by using ultra-filter paper, the colloidal particles are separated from rest of the materials. (c) The flocculating power of ${Al}^{+3}$ is more than that of ${Na}^{+}$ . The greater the valence of the flocculating ion added, the greater is its power to cause precipitation. This is known as Hardy schulze rule. In the coagulation of a negative sol, the flocculating power is in the order ${Al}^{+3}> {Ba}^{+2}> {Na}^{+}$ . (d) When colloidal solutions are viewed under a powerful ultramicroscope, the colloidal particles appear to be in a state of continuous zig-zag motion all over the field of view. This is known as Brownian movement.

Q48JEE Main 2021MCQ4MCoordination Compounds

Arrange the following metal complex/compounds in the increasing order of spin only magnetic moment. Presume all the three, high spin system. (Atomic numbers ${\text{Ce}}=58, {\text{Gd}}=64$ and ${\text{Eu}}=63$ .) A. $( {\text{NH}}_{4})_{2}[ {\text{Ce}}( {\text{NO}}_{3})_{6}]$ B. ${\text{Gd}}( {\text{NO}}_{3})_{3}$ and C. ${\text{Eu}}( {\text{NO}}_{3})_{3}$

🚀 Solve in Practice Mode

📖 Explanation

(A) ${ }_{58} {Ce} \longrightarrow [ {Xe}] 4 f^{2} 5 d^{0} 6 s^{2}$ In complex ${Ce}^{+4} \longrightarrow [ {Xe}] 4 f^{0} 5 d^{0} 6 s^{0}$ There is no unpaired electron, so $\mu _{m}=0 {BM}$ (B) ${ }_{64} {Gd} \longrightarrow [ {Xe}] 4 f^{7} 5 {d}^{1} 6 s^{2}$ In complex, ${ }_{64} {Gd}^{2+} \longrightarrow [ {Xe}] 4 f^{7} 5 d^{1}$ There are 8 unpaired electrons. $\therefore \;\; \mu _{m}=\sqrt{8(8+2)}=\sqrt{80}=8.94 {BM}$ (C) ${ }_{63} {Eu} \longrightarrow [ {Xe}] 4 f^{9} 5 d^{0} 6 s^{0}$ In complex ${ }_{63} {Eu}^{+3} \longrightarrow [ {Xe}] 4 f^{6} 5 {d}^{0} 6 {s}^{0}$ contain six unpaired electrons. So, $\mu _{m}=\sqrt{6(6+2)}=\sqrt{48} {BM}=6.93 {BM}$ Hence, order of spin only magnetic moment $B>C>A$

Q49JEE Main 2021MCQ4MIonic Equilibrium

The exact volumes of $1{\text{M}} \; {\text{NaOH}}$ solution required to neutralise $50 {mL}$ of $1{M}\; {\text{H}}_{3}{\text{PO}}_{3}$ solution and $100 {mL}$ of $2 {M}\; {\text{H}}_{3} {\text{PO}}_{2}$ solution, respectively, are

🚀 Solve in Practice Mode

📖 Explanation

${{H}_{3} {PO}_{3}}_{{50 \text{ml}}_{1\text{M}}}+{2 {NaOH}}_{{1M}_{\text{V}=?}} \longrightarrow {Na}_{2} {HPO}_{3}+2 {H}_{2} {O}$ Millimoles of ${H}_{3} {PO}_{3}=M \times V=1 \times 50=50$ For 1 millimole of ${H}_{3} {PO}_{3}$ , we require 2 millimoles of ${NaOH}$ . For 50 millimole of ${H}_{3} {PO}_{3}$ , we require $(2 \times 50)=100$ millimoles of ${NaOH}$ . Millimoles of ${NaOH}=M \times V=100$ $1 \times V \;\; =100$ $V \;\; =100 {mL}$ ${{H}_{3} {PO}_{2}}_{{100\text{mL}}_{2\text{M}}}+ {NaOH} _{{1M}_{\text{V}=?}}\to {NaH}_{2} {PO}_{2}+ {H}_{2} {O}$ Millimoles of ${H}_{3} {PO}_{2}=M \times V=2 \times 100=200$ For 1 millimole of ${H}_{3} {PO}_{2}$ , we require 1 millimoles of ${NaOH}$ . For 200 millimole of ${H}_{3} {PO}_{2}$ , we require 200 millimoles of ${NaOH}$ . So, volume of ${NaOH}=200 {mL}$

Q50JEE Main 2021NAT4MSolid State

Ga (atomic mass $70 {u}$ ) crystallises in a hexagonal close packed structure. The total number of voids in $0.581 {g}$ of ${Ga}$ is ...... $\times 10^{21}$ . (Round off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

For HCP structure, Effective number of atoms per unit cell $(Z)=6$ Number of tetrahedral void $=2 Z=12$ Number of octahedral void $=Z=6$ Total number of voids per unit cell $=18$ Moles of Ga $=\;\frac{\;\text{ Given mass }\;}{\;\text{ Molar mass }\;}=\;\frac{0.581}{70}$ No. of atoms $=\;\frac{0.581}{70} \times N_{A}$ No. of unit cells $=\;\frac{0.581}{70} \times \;\frac{N_{A}}{6}$ No. of voids $=(\;\frac{0.581}{70} \times \;\frac{6.02 \times 10^{23}}{6}) \times 18=14.99 \times 10^{21} .$ Nearest integer $=15 \times 10^{21}$ So, correct answer is 15 .

Q51JEE Main 2021NAT4MElectrochemistry

A $5.0 {m} {mol} {dm}^{-3}$ aqueous solution of ${KCl}$ has a conductance of $0.55 {mS}$ when measured in a cell constant $1.3 {cm}^{-1}$ . The molar conductivity of this solution is ${mSm}^{2} {mol}^{-1}$ . (Round off to the nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

$Lambda_{m}=\;\frac{K}{1000 \times C}$ where, $Lambda_{ {m}}=$ molar conductivity $κ=\;\text{ conductivity, }\; C=\;\text{ molarity }\;$ Conductivity $(κ)=G \;\frac{I}{A}=0.55 \times (1.3 \times 100)$ $=55 \times 1.3$ $Lambda_{ {m}}=\;\frac{55 \times 1.3}{1000 \times 5 \times 10^{-3}}$ $Lambda_{ {m}}=14.3$ Closest integer, $Lambda_{ {m}}=14$

Q52JEE Main 2021NAT4MChemical Kinetics

$A$ and $B$ decompose via first order kinetics with half-lives $54.0$ min and $18.0 {min}$ respectively. Starting from an equimolar non-reactive mixture of $A$ and $B$ , the time taken for the concentration of $A$ to become 16 times that of $B$ is ......... min. (Round off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

Given, $(t_{1 / 2})_{1}=54 {min}$ $(t_{1 {/} 2})_{2}=18 {min}$ $\text{table} A,B;t=0\text{,} 'x' {M}, t=0\text{,} 'x' {M}$ To calculate, $\;\; [A_{t}]=16 \times [B_{t}]$ ...(i) time $=$ ? For Ist order reaction, $[A_{t}]=\;\frac{A_{0}}{(2)^{n}}$ $n=$ number of half-lives From Eq. (i), $[A_{t}]=16 \times [B_{t}]$ $\;{x}/{(2)^{n_{1}}}=\;{x}/{(2)^{n_{2}}} \times 16$ $(2)^{n_{2}}=(2)^{n_{1}} \times (2)^{4}$ $\Rightarrow \;\; n_{2}=n_{1}+4$ $\;\frac{t}{(t_{1 / 2})_{2}}=\;\frac{t}{(t_{1 / 2})_{1}}+4 \Rightarrow t(\;\frac{1}{18}-\;\frac{1}{54})=4$ $\Rightarrow \;\; t=\;\frac{4 \times 18 \times 54}{36}=108 {min}$

Q53JEE Main 2021NAT4MPractical Organic Chemistry

In Duma's method of estimation of nitrogen, $0.1840 {g}$ of an organic compound gave $30 {mL}$ of nitrogen collected at $287 {K}$ and $758 {mm}$ of ${Hg}$ pressure. The percentage composition of nitrogen in the compound is (Round off to the nearest integer). [Given: Aqueous tension at $287 {K}=14 {mm}$ of ${Hg}$ ]

🚀 Solve in Practice Mode

📖 Explanation

Given, mass of organic compound $0.1840 {g}$ Total pressure $=758 {mm}$ of ${Hg}$ Aqueous tension $=14 {mm}$ of ${Hg}$ Partial pressure of dry ${N}_{2}=758-14$ $=744 {mm}$ of ${Hg}$ For ${N}_{2}$ $V=30 {mL}=0.03 {L}$ $T=287 {K}$ $p=744 {mm}$ of ${Hg}=\;\frac{744}{760} {atm}$ So, moles of ${N}_{2}(n)=\;\frac{p V}{R T}=\;\frac{744}{760} \times \;\frac{0.03}{0.0821 \times 287}$ $n=1.25 \times 10^{-3}$ moles of ${N}_{2}$ Weight of ${N}_{2}=$ Moles $\times$ Molar mass $=1.25 \times 10^{-3} \times 28$ $=35 \times 10^{-3} {g}$ $\%$ composition of ${N}_{2}$ $=\;\frac{\;\text{ Mass of }\; {N}_{2}}{\;\text{ Mass of organic compound }\;} \times 100$ $=\;\frac{35 \times 10^{-3}}{0.1840} \times 100$ $=19.02 \%$

Q54JEE Main 2021NAT4MStructure of Atom

The number of orbitals with $n=5, m_{1}=+2$ is (Round off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

Given, $n=5, m_{l}=+2$ For $n=5$ , possible value of $I=0,1,2,3,4$ For $\;\; l=0, m_{1}=0$ $l=1, m_{l}=-1,0,1$ $l=2, m_{l}=-2,-1,0,1,2$ $l=3, m_{1}=-3,-2,-1,0,1,2,3$ $l=4, m_{l}=-4,-3,-2,-1,0,1,2,3,4$ Possible value of $m_{l}$ for a given value of $l$ $=0, \pm 1, \pm 2, \pm 3 . . . \pm 1$ So, number of orbitals having $n=5$ and $m_{l}=\pm 2$ are 3 .

Q55JEE Main 2021NAT4MSolutions

At $363 {K}$ , the vapour pressure of $A$ is $21 {kPa}$ and that of $B$ is $18 {kPa}$ . One mole of $A$ and 2 moles of $B$ are mixed. Assuming that this solution is ideal, the vapour pressure of the mixture is ......... kPa. (Round off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

From Raoult's law, $p_{T}=p_{A}^{\circ} χ_{A}+p_{B}^{\circ} χ_{B}$ ..........(i) where, $p_{T}=$ total pressure of solution $p_{B}^{\circ}, p_{A}^{\circ}=$ vapour pressure of pure liquid $B$ and $A$ respectively. $χ_{A}, χ_{B}=$ mole fraction of $A$ and $B$ in solution phase. Given, $p_{A}^{\circ}=21 {kPa}, p_{B}^{\circ}=18 {kPa}$ $n_{A}=1, n_{B}=2$ $χ_{A}=\;\frac{n_{A}}{n_{A}+n_{B}}=\;\frac{1}{3}$ $χ_{B}=1-χ_{A}=\;\frac{2}{3}$ Putting the values in equation (i), $p_{T}=(21 \times \;\frac{1}{3})+(18 \times \;\frac{2}{3})=7+12=19 {kPa}$

Q56JEE Main 2021NAT4MIonic Equilibrium

Sulphurous acid $( {H}_{2} {SO}_{3})$ has $K_{a_{1}}=1.7 \times 10^{-2}$ and $K_{a_{2}}=6.4 \times 10^{-8}$ . The ${pH}$ of $0.588 {M}$ is (Round off to the nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

For ${H}_{2} {SO}_{3}, K_{ {a}_{1}}>>K_{ {a}_{2}}$ So, we mainly consider 1st ionisation $\;\;\;\;\;\;\;\;\;\;\;\;\;\;{H}_{2} {SO}_{3} ⇌ {H}^{+}+ {HSO}_{3}^{-} ; K_{ {a}_{1}}=1.7 \times 10^{-2}$ $\text{table} t=0,0.558,0, 0;t=t_{ {eq}},0.558-0.558 \alpha ,0.558 \alpha ,0.558 \alpha$ $K_{ {a}_{1}}=\;\frac{[ {H}^{+}]\$ $[ {HSO}{3}^{-}]$}{[ {H}{2} {SO}{3}]}=;\frac{(0.558 alpha)(0.558 alpha)}{0.558(1-alpha)}=;\frac{0.558 alpha^{2}}{1-alpha}alpha< < 1 $for weak acid$ (1-alpha) ≈ 1alpha=\sqrt{;{K{a_{1}}}/{0.558}}=\sqrt{;\frac{1.7 × 10^{-2}}{0.558}}alpha=1.7 × 10^{-1}=0.17[ {H}]^{+}=0.558 alpha=9.9 × 10^{-2}{pH}=-\log $[ {H}^{+}]$=-\log (9.9 × 10^{-2}) ;; [ \because \log 9.9 ≈ 1]=2-\log 9.9=2-1{pH}=1$

Q57JEE Main 2021NAT4MIonic Equilibrium

When $35 {mL}$ of $0.15 {M}$ lead nitrate solution is mixed with $20 {mL}$ of $0.12 {M}$ chromic sulphate solution, ........ $\times 10^{-5}$ moles of lead sulphate precipitate out. (Round off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

$\text{table}3 {Pb}( {NO}_{3})_{2},+ ,{Cr}_{2}( {SO}_{4})_{3} ,\longrightarrow , 3{PbSO}_{4},+,2{Cr}( {NO}_{3})_{3};35\text{mL},,20\text{mL};0.15\text{ML},,0.12\text{ML};\text{M}\times V,,20\times 0.12;35\times 0.15,,=2.4,,0,,0;=5.25$ For 3 moles of ${Pb}( {NO}_{3})_{2}$ , we require 1 mole of ${Cr}_{2}( {SO}_{4})_{3}$ For $5.25$ moles of ${Pb}( {NO}_{3})_{2}$ , we require $\;\frac{1}{3} \times 5.25$ mole of ${Cr}_{2}( {SO}_{4})_{3}=1.75$ moles But we have $2.4$ moles. So, ${Cr}_{2}( {SO}_{4})_{3}$ is excess reagent and ${Pb}( {NO}_{3})_{2}$ is limiting reagent, $( {LR})$ $\text{table}3{Pb}( {NO}_{3})_{2},+ ,{Cr}_{2}( {SO}_{4})_{3} ,\longrightarrow ,3 {PbSO}_{4},+,2 {Cr}( {NO}_{3})_{3};\text{t} = 0 5.25 \text{m} \text{mol},,2.4,,0,,0;t=t\;5.25 525,,2.4-1.7=0.7,,5.25;=0$ Moles of ${PbSO}_{4}$ formed $=$ moles of ${Pb}( {NO}_{3})_{2}$ consumed $=5.25 {m} {mol}$ $=525 \times 10^{-5}$

Q58JEE Main 2021NAT4MChemical Thermodynamics

At $25^{\circ} {C}, 50 {g}$ of iron reacts with ${HCl}$ to form ${FeCl}_{2}$ . The evolved hydrogen gas expands against a constant pressure of 1 bar. The work done by the gas during this expansion is ......J. (Round off to the nearest integer) $[Given, $R=8.314 {J} {mol}^{-1} {K}^{-1}$ . Assume, hydrogen is an ideal gas]$ [Atomic mass off Fe is $55.85 {u}$ ]

🚀 Solve in Practice Mode

📖 Explanation

${{Fe}(s)}_{50\text{g}}+ {HCl}( {aq}) \longrightarrow {FeCl}_{2}(a q)+ {H}_{2}(g)$ $($ Moles of ${Fe}=$ Moles of ${H}_{2})$ moles $=\;\frac{50}{55.85} \;\; 0.8952 {mol} \;\; 0.8952 {mol}$ $=0.8952 {mol}$ Work done can be given by $W=-p_{\;\text{ext }\;} \Delta V$ $W=-\Delta n_{g} R T$ where, $\Delta n_{g}=$ { gaseous moles on product side } - {gaseous mole on reactant side} $=0.8952-0$ $W=-0.8952 \times 8.314 \times 298$ $=-2217.92 {J}$ Closest integer $=2218$

Q59JEE Main 2021NAT4MCoordination Compounds

$[ {Ti}( {H}_{2} {O})_{6}]^{3+}$ absorbs light of wavelength $498 {nm}$ during a $d-d$ transition. The octahedral splitting energy for the above complex is ......... $\times 10^{-9} {J}$ (Round off to the nearest intger). $[h=6.626 \times 10^{-34} {Js}$ , $c=3 \times 10^{8} {ms}^{-1}$ ]

🚀 Solve in Practice Mode

📖 Explanation

$[ {Ti}( {H}_{2} {O})_{6}]^{3+}$ is an octahedral complexFor transition of electron from $t_{2 g}$ to ${e}_{g}$ orbital, $\Delta _{o}$ amount of energy is required. Given, $\lambda=498 {nm}=498 \times 10^{-9} {m}$ $\Delta _{o}=h c {/} \lambda$ where, $h=$ Planck's constant $c=$ speed of light $\lambda=$ wavelength of radiation So, $\Delta _{o}=\;\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{498 \times 10^{-9}}$ $=3.98 \times 10^{-19} {J}=4 \times 10^{-19} {J}$

Mathematics30 questions

Q60JEE Main 2021MCQ4MApplication of Derivatives

The maximum value of $f(x)={\begin{vmatrix}\sin ^{2} x & 1+\cos ^{2} x & \cos 2 x \\ 1+\sin ^{2} x & \cos ^{2} x & \cos 2 x \\ \sin ^{2} x & \cos ^{2} x & \sin 2 x\end{vmatrix}}, x \in R$ is

🚀 Solve in Practice Mode

📖 Explanation

Given, $f(x)={\begin{vmatrix}\sin ^{2} x & 1+\cos ^{2} x & \cos 2 x \\ 1+\sin ^{2} x & \cos ^{2} x & \cos 2 x \\ \sin ^{2} x & \cos ^{2} x & \sin 2 x\end{vmatrix}}$ On applying $C_{1} \to C_{1}+C_{2}$ , we get $f(x)={\begin{vmatrix}\sin ^{2} x+1+\cos ^{2} x & 1+\cos ^{2} x & \cos 2 x \\ 1+\sin ^{2} x+\cos ^{2} x & \cos ^{2} x & \cos 2 x \\ \sin ^{2} x+\cos ^{2} x & \cos ^{2} x & \sin 2 x\end{vmatrix}}$ $f(x)={\begin{vmatrix}2 & 1+\cos ^{2} x & \cos 2 x \\ 2 & \cos ^{2} x & \cos 2 x \\ 1 & \cos ^{2} x & \sin 2 x\end{vmatrix}}$ On applying $R_{1} \to R_{1}-R_{2}$ $f(x)={\begin{vmatrix}0 & 1 & 0 \\ 2 & \cos ^{2} x & \cos 2 x \\ 1 & \cos ^{2} x & \sin 2 x\end{vmatrix}}$ $f(x)=-1(2 \sin 2 x-\cos 2 x)$ As, we know that, if $f(\theta)=A \sin \theta+B \cos \theta$ Then, $-\sqrt{A^{2}+B^{2}} \le f(\theta) \le \sqrt{A^{2}+B^{2}}$ Here, we have, $f(x)=\cos 2 x-2 \sin 2 x$ $-\sqrt{2^{2}+1^{2}} \le f(x) \le \sqrt{2^{2}+1^{2}}$ $-\sqrt{5} \le f(x) \le \sqrt{5}$ So, maximum value of $f(x)$ is $\sqrt{5}$ .

Q61JEE Main 2021MCQ4MProbability

Let $A$ denote the event that a 6-digit integer formed by $0,1,2,3,4,5,6$ without repetitions, be divisible by 3 . Then, probability of event $A$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Given, digits $=\{0,1,2,3,4,5,6\}$ Number of ways in which 6 digit number can be formed using these 7 digits $=6 \cdot 6 !$ Number of ways \Rightarrow $\text{table}\ov↓\;\ov↓\;\ov↓\;\ov↓\;\ov↓\;\ov↓; 6\;6\;5\;4\;3\;2$ ( 0 can't be filled here) If the required number is divisible by 3 , then the sum of the digits must be divisible by 3 . Sum of all 7 digits $=0+1+2+3+4+5+6$ $=\;\frac{6 \cdot 7}{2}=21$ Now, this further implies we have to remove anyone digit from the given digits. We can remove only the multiple of 3 as the sum of all 7 digits is already 21 . So, removing any other digit will lead to a number of 6 digit which will not be a multiple of 3 . Possible digits $\Rightarrow \{1,2,3,4,5,6\}$ $\{0,1,2,4,5,6\}$ $\{0,1,2,3,4,5\}$ Case I $\{1,2,3,4,5,6\}$ Number of 6 digit numbers $=6 !$ Case II $\{0,1,2,4,5,6\}$ Case III $\{0,1,2,3,4,5\}$ Number of 6 digit numbers $=5 \cdot 5 !$ $\therefore$ Probability $=(\;\frac{6 !+5 \cdot 5 !+5 \cdot 5 !}{6 \cdot 6 !})$ $=(\;\frac{1}{6}+\;\frac{5}{36}+\;\frac{5}{36})=\;\frac{16}{36}=\;\frac{4}{9}$

Q62JEE Main 2021MCQ4MLimits, Continuity and Differentiability

Let $\alpha \in R$ be such that the function $f(x)=\begin{cases}\\ \\ \frac{\cos ^{-1}(1-\{x\}^{2}) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^{3}} & x \ne 0 \\ \alpha & x=0\end{cases}$ is continuous at $x=0$ , where $\{x\}=x-[x],[x]$ is the greatest integer less than or equal to $x$ . Then,

🚀 Solve in Practice Mode

📖 Explanation

Given, $f(x)=\begin{cases}\\ \\ \frac{\cos ^{-1}(1-\{x\}^{2}) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^{3}} & x \ne 0 \\ \alpha & x=0\end{cases}$ $\{x\}=x-[x]$ ${So}$ , when $x \to 0^{+}$ $\Rightarrow \;\; \{x\}=x-0=x$ And, when $x \to 0^{-}$ $\Rightarrow \;\; \{x\}=x+1$ $L H L=\lim _{x \to 0^{-}} f(x)$ $=\lim _{x \to 0} \;\frac{\cos ^{-1}\$ [1-(1+x)^{2}]$ \sin ^{-1}[1-(1+x)]}{(1+x)-(1+x)^{3}}=\lim _{x \to 0} ;\frac{\cos ^{-1}(-x^{2}-2 x) \sin ^{-1}(-x)}{(1+x)(1+1+x)(1-1-x)}=\lim _{x \to 0} ;\frac{\cos ^{-1}(-x^{2}-2 x)}{(1+x)(x+2)}=;\frac{\cos ^{-1}(0)}{1.2}=;\frac{\pi}{4}R H L=\lim _{x \to 0^{+}} f(x)=\lim _{x \to 0} ;\frac{\cos ^{-1}(1-x^{2}) \sin ^{-1}(1-x)}{x(1-x)(1+x)}=;\frac{\pi}{2} \lim _{x \to 0} ;\frac{\cos ^{-1}(1-x^{2})}{x} $Applying L-Hospital Rule,$ =;\frac{\pi}{2} \lim _{x \to 0} ;{(-1)(-2 x)}/{\sqrt{1-(1-x^{2})^{2}}}=;\frac{\pi}{2} \cdot 2 \lim _{x \to 0} ;{x}/{\sqrt{2 x^{2}-x^{4}}}=\pi \lim _{x \to 0} ;{1}/{\sqrt{2-x^{2}}}=;\frac{\pi}{\sqrt{2}}L H L=;\frac{\pi}{4} $and RHL$ =;\frac{\pi}{\sqrt{2}} $Hence,$ L H L \ne R H L $So, the function will be discontinuous for every value of$ \alpha \in R $.$ \therefore $No such$ \alpha$ exist.

Q63JEE Main 2021MCQ4MThree Dimensional Geometry

If $(x, y, z)$ be an arbitrary point lying on a plane $P$ , which passes through the points $(42,0,0),(0,42,0)$ and $(0,0,42)$ , then the value of expression $3+\;\frac{x-11}{(y-19)^{2}(z-12)^{2}}+\;\frac{y-19}{(x-11)^{2}(z-12)^{2}}$ $+\;\frac{z-12}{(x-11)^{2}(y-19)^{2}}-\;\frac{x+y+z}{14(x-11)(y-19)(z-12)}$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Equation of plane passing through $A(42,0,0), B(0,42,0)$ and $C(0,0,42)$ will be $\;\frac{x}{42}+\;\frac{y}{42}+\;\frac{z}{42}=1$ $\Rightarrow \;\; x+y+z=42$ $(x-11)+(y-19)+(z-12)=0$ Now, $3+\;\frac{x-11}{(y-19)^{2}(z-12)^{2}}+\;\frac{z-12}{(x-11)^{2}(y-19)^{2}}$ $+\;\frac{y-19}{(x-11)^{2}(z-12)^{2}}-\;\frac{x+y+z}{14(x-11)(y-19)(z-12)}$ $\Rightarrow \;\frac{3(x-11)^{2}(y-19)^{2}(z-12)^{2}+(x-11)^{3}+(y-19)^{3}+(z-12)^{3}}{(x-11)^{2}(y-19)^{2}(z-12)^{2}}-\;\frac{42}{14(x-11)(y-19)(z-12)}$ $\Rightarrow \;\frac{(x-11)^{3}+(y-19)^{3}+(z-12)^{3}-3(x-11)(y-19)(z-12)+3(x-11)^{2}(y-19)^{2}(z-12)^{2}}{(x-11)^{2}(y-19)^{2}(z-12)^{2}}$ ⇒ If $A+B+C=0$ Then, $A^{3}+B^{3}+C^{3}=3 A B C$ $\Rightarrow (x-11)^{3}+(y-19)^{3}+(z-12)^{3}=3(x-11)(y-19)(z-12)$ $\Rightarrow \;\frac{3(x-11)(y-19)(z-12)-3(x-11)(y-19)(z-12)+3(x-11)^{2}(y-19)^{2}(z-12)^{2}}{(x-11)^{2}(y-19)^{2}(z-12)^{2}}$ $\Rightarrow \;\; \;\frac{3(x-11)^{2}(y-19)^{2}(z-12)^{2}}{(x-11)^{2}(y-19)^{2}(z-12)}=3$

Q64JEE Main 2021MCQ4MDefinite Integrals

Consider the integral $I=int_{0}^{10} \;\frac{[x] e^{[x]}}{e^{x-1}} d x$ where $[x]$ denotes the greatest integer less than or equal to $x$ . Then, the value of $/$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

We have, $int_{0}^{10} \;\frac{[x] e^{[x]}}{e^{x-1}} d x$ $=e int_{0}^{10} \;\frac{[x] e^{[x]}}{e^{x}} d x$ $=e int_{0}^{1} \;\frac{0}{e^{x}} d x+e int_{1}^{2} \;\frac{e}{e^{x}} d x+e int_{2}^{3} \;\frac{2 e^{2}}{e^{x}} d x+. . .$ $\Rightarrow int_{a}^{b} e^{-x} d x=\;\frac{e^{-x}}{-1}|_{a} ^{b} \Rightarrow (e^{-a}-e^{-b})$ $\Rightarrow \;\; {e}^{2}(\;\frac{1}{ {e}}-\;\frac{1}{ {e}^{2}})+2 {e}^{3}(\;\frac{1}{ {e}^{2}}-\;\frac{1}{ {e}^{3}})+3 {e}^{4}(\;\frac{1}{ {e}^{3}}-\;\frac{1}{ {e}^{4}})+. . .+9 e^{10}(\;\frac{1}{e^{9}}-\;\frac{1}{e^{10}})$ $=(e-1)+2(e-1)+3(e-1)+. . .+9(e-1)$ $=(1+2+3+. . .+9)(e-1)=(\;\frac{9 \times 10}{2})(e-1)$ $=45(e-1)$

Q65JEE Main 2021MCQ4MParabola

Let $C$ be the locus of the mirror image of a point on the parabola $y^{2}=4 x$ with respect to the line $y=x$ . Then, the equation of tangent to $C$ at $P(2,1)$ is

🚀 Solve in Practice Mode

📖 Explanation

The mirror image of any point $(\alpha, \beta)$ with respect to line $y=x$ is simply $(\beta, \alpha)$ . Let $(h, k)$ be the mirror image of a point on parabola $y^{2}=4 a x$ Then, $(k, h)$ will be the mirror image of $(h, k)$ and it will lie on parabola. ${SO}$ $y^{2}=4 x$ $h^{2}=4 k$ $\Rightarrow \;\; x^{2}=4 y$ Hence, Locus is $x^{2}=4 y$ ....(i) For finding equation of tangent differentiate Eq. (i) w.r.t. $x$ $2 x=4 \;\frac{d y}{d x}$ $\Rightarrow \;\; \;\frac{d y}{d x}=\;\frac{2 x}{4}=(\;\frac{x}{2}) \Rightarrow (\;\frac{d y}{d x})_{2,1}=(\;\frac{2}{2})=1$ $\Rightarrow \;\; \;\frac{y-1}{x-2}=1 \Rightarrow y-1=x-2$ $y=x-1$ $\therefore$ Equation of tangent $\Rightarrow y=x-1$ $\Rightarrow \;\; x-y=1$

Q66JEE Main 2021MCQ4MDifferential Equations

If $y=y(x)$ is the solution of the differential equation $\;\frac{d y}{d x}+(\tan x) y=\sin x, 0 \le x \le \;\frac{\pi}{3}$ , with $y(0)=0$ , then $y(\;\frac{\pi}{4})$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Given, $\;\frac{d y}{d x}+(\tan x) y=\sin x, x \in[0, \pi / 3]$ which is a linear differential equation of the form of $\;\frac{d y}{d x}+P y=Q$ Here, $P=\tan x$ $\therefore \;\; \;\; {IF}= {e}^{\int P d x}$ $\Rightarrow {e}^{\int \tan x d x}= {e}^{\log (\sec x)}=\sec x$ Multiplying by $\sec x$ on both sides $\;\frac{d y}{d x}+(\tan x) y=\sin x$ $\sec x \;\frac{d y}{d x}+(\tan x \sec x) y=\sin x \sec x$ $\Rightarrow \;\; \;\frac{d}{d x}(y \sec x)=\tan x \Rightarrow y \sec x=\int \tan x d x$ $\Rightarrow \;\; \;\; y \sec x=\log (\sec x)+c$ $y=\cos x \log (\sec x)+c \cdot \cos x$ $y(0)=0$ $\Rightarrow \;\; 0=1 \cdot 0+c \cdot 1 \Rightarrow c=0$ $\therefore \;\; y=\cos x \cdot \log (\sec x)$ $\Rightarrow \;\; y(\;\frac{\pi}{4})=\cos (\;\frac{\pi}{4}) \cdot \log (\sec \;\frac{\pi}{4})$ $=\;\frac{1}{\sqrt{2}} \log (\sqrt{2})$ $=\;\frac{1}{2 \sqrt{2}} \log 2$

Q67JEE Main 2021MCQ4MSets and Relations

Let $A=\{2,3,4,5, . . . ., 30\}$ and ' $=$ ' be an equivalence relation on $A \times A$ , defined by $(a, b)=(c, d)$ , if and only if $a d=b c$ . Then, the number of ordered pairs, which satisfy this equivalence relation with ordered pair $(4,3)$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

$A=\{2,3,4,5, . . ., 30\}$ $a d=b c$ $(a, b) R(4,3)$ $\Rightarrow \;\; 3 a=4 b \Rightarrow a=(\;\frac{4}{3}) b$ $b$ must be a multiple of $3, b$ can be $(3,6,9, . . . 30)$ . Also, a must be less than or equal to $30 .$ $(a, b)=(4,3),(8,6),(12,9),(16,12),(20,15)$ $(24,18),(28,21)$ $\Rightarrow 7$ ordered pairs

Q68JEE Main 2021MCQ4MCircles

Let the lengths of intercepts on $x$ -axis and $y$ -axis made by the circle $x^{2}+y^{2}+a x+2 a y+c=0,(a< 0)$ be $2 \sqrt{2}$ and $2 \sqrt{5}$ , respectively. Then, the shortest distance from origin to a tangent to this circle which is perpendicular to the line $x+2 y=0$ , is equal to

🚀 Solve in Practice Mode

📖 Explanation

$x^{2}+y^{2}+a x+2 a y+c=0(a< 0)$ For a circle in standard form $x^{2}+y^{2}+2 g x+2 f y+c=0$ $x$ -intercept $=2 \sqrt{g^{2}-c}$ Here, $2 g=a \Rightarrow g=a / 2$ $2 f=2 a \Rightarrow f=a$ $x$ -intercept $\Rightarrow 2 \sqrt{\;\frac{a^{2}}{4}-c}=2 \sqrt{2}$ $\Rightarrow \;\; \;\frac{a^{2}}{4}-c=2$ .....(i) $y$ -intercept $\Rightarrow 2 \sqrt{a^{2}-c}=2 \sqrt{5}$ $a^{2}-c=5$ ...(ii) Subtracting Eq. (ii) from Eq. (i), we get $(\;\frac{3}{4}) a^{2}=3$ $\Rightarrow a^{2}=4 \Rightarrow a=\pm 2$ As, $a< 0$ , so, $a=-2, c=-1$ So, $x^{2}+y^{2}-2 x-4 y-1=0$ $(x-1)^{2}+(y-2)^{2}-1=1+4$ $\Rightarrow \;\; (x-1)^{2}+(y-2)^{2}=6$ Centre $(1,2)$ , Radius $=\sqrt{6}$ Equation of tangent is perpendicular to $x+2 y=0$ . So, equation of tangent will be $2 x-y+\lambda=0$ Now, perpendicular distance of this line from $(1,2)$ will be $\sqrt{6}$ units. $\;\frac{|2.1-2+\lambda|}{\sqrt{5}}=\sqrt{6}$ $\lambda=\pm \sqrt{30}$ So, $T \Rightarrow 2 x-y \pm \sqrt{30}=0$ Now, shortest distance from origin $(0,0)$ to tangent $T$ will be $\;{|2.0-0 \pm \sqrt{30}|}/{\sqrt{2^{2}+1^{2}}}=\;\frac{\sqrt{30}}{\sqrt{5}}=\sqrt{6}$

Q69JEE Main 2021MCQ4MComplex Numbers

The least value of $|z|$ , where $z$ is a complex number which satisfies the inequality $\exp (\;\frac{(|z|+3)(|z|-1)}{(|z|+1)} \log _{e} 2) \ge \log _{\sqrt{2}}|5 \sqrt{7}+9 i|$ . $i=\sqrt{-1}$ , is equal to

🚀 Solve in Practice Mode

📖 Explanation

$\exp [\;\frac{(|z|+3)(|z|-1)}{(|z|+1)} \times \log _{e} 2] \ge \log _{\sqrt{2}}|5 \sqrt{7}+9 i|$ $\exp [\;\frac{(|z|+3)(|z|-1)}{(|z|+1)} \times \log _{e} 2] \ge \log _{\sqrt{2}} 16$ $\Rightarrow 2^{[\;\frac{(|z|+3)(|z|-1)}{(|z|+1)}]} \ge 2^{3} \Rightarrow \;\frac{(|z|+3)(|z|-1)}{(|z|+1)} \ge 3$ $\Rightarrow |z|^{2}+2 z|-3 \ge 3 z|+3 \Rightarrow |z|^{2}-|z|-6 \ge 0$ $\Rightarrow (|z|-3)(|z|+2) \ge 0$ $\Rightarrow |z _{\min }=3$

Q70JEE Main 2021MCQ4MPermutations and Combinations

Consider a rectangle $A B C D$ having $5,7,6,9$ points in the interior of the line segments $A B$ , $C D, B C, D A$ , respectively. Let $\alpha$ be the number of triangles having these points from different sides as vertices and $\beta$ be the number of quadrilaterals having these points from different sides as vertices. Then, $(\beta-\alpha)$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Number of triangles that can be formed from the points on 3 of the sides. ${ }^{5} C_{1}^{7} C_{1}^{6} C_{1}+{ }^{5} C_{1}^{7} C_{1}^{9} C_{1}+{ }^{5} C_{1}^{6} C_{1}^{9} C_{1}+{ }^{6} C_{1}^{7} C_{1}^{9} C_{1}$ $=210+315+270+378$ $\Rightarrow \;\; \alpha=1173$ Number of quadrilaterals that can be formed by taking one point from each of the four vertex ${ }^{5} C_{1}^{7} C_{1}^{6} C_{1}^{9} C_{1}=5 \times 6 \times 7 \times 9=1890$ $\Rightarrow \;\; \beta=1890$ $\therefore \;\; \beta-\alpha=1890-1173=717$

Q71JEE Main 2021MCQ4MEllipse

If the point of intersections of the ellipse $\;\frac{x^{2}}{16}+\;\frac{y^{2}}{b^{2}}=1$ and the circle $x^{2}+y^{2}=4 b, b>4$ lie on the curve $y^{2}=3 x^{2}$ , then $b$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Given, equation of ellipse $\;\frac{x^{2}}{16}+\;\frac{y^{2}}{b^{2}}=1$ ....(i) and equation of circle $x^{2}+y^{2}=4 b$ .....(ii) From Eqs. (i) and (ii), $\;\frac{4 b-y^{2}}{16}+\;\frac{y^{2}}{b^{2}}=1$ $\Rightarrow \;\; 4 b^{3}-y^{2} b^{2}+16 y^{2}=16 b^{2}$ $\Rightarrow y^{2}=\;\frac{16 b^{2}-4 b^{3}}{16-b^{2}}$ As, $(x, y)$ lie on $y^{2}=3 x^{2}$ So, $y^{2}=3(4 b-y^{2})$ [from Eq. (ii)] $\Rightarrow \;\; 4 y^{2}=12 b \Rightarrow y^{2}=3 b$ Now, $\;\frac{16 b^{2}-4 b^{3}}{16-b^{2}}=3 b$ $\Rightarrow \;\; 16 b^{2}-4 b^{3}=48 b-3 b^{3}$ $\Rightarrow b^{3}-16 b^{2}+48 b=0 \Rightarrow b(b^{2}-16 b+48)=0$ $\Rightarrow b(b-4)(b-12)=0$ As, $b>4$ So, $b=12$

Q72JEE Main 2021MCQ4MInverse Trigonometric Functions

Given that the inverse trigonometric functions take principal values only. Then, the number of real values of $x$ which satisfy $\sin ^{-1}(\;\frac{3 x}{2})+\sin ^{-1}(\;\frac{4 x}{5})=\sin ^{-1} x$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Given, $\sin ^{-1}(\;\frac{3 x}{5})+\sin ^{-1}(\;\frac{4 x}{5})=\sin ^{-1} x$ $\Rightarrow \;\; \sin ^{-1}(\;\frac{3 x}{5} \sqrt{1-\;\frac{16 x^{2}}{25}}+\;\frac{4 x}{5} \sqrt{1-\;\frac{9 x^{2}}{25}})=\sin ^{-1} x$ $\Rightarrow \;\; 3 x \sqrt{25-16 x^{2}}+4 x \sqrt{25-9 x^{2}}=25 x$ $\Rightarrow \;\; \sqrt{225 x^{2}-114 x^{4}}+\sqrt{400 x^{2}-144 x^{4}}=25 x$ ....(i) $(400 x^{2}-144 x^{4})-(225 x^{2}-144 x^{4})=175 x^{2}$ ...(ii) On dividing Eq. (ii) by Eq. (i), $\sqrt{400 x^{2}-144 x^{4}}-\sqrt{225 x^{2}-144 x^{2}}=7 x$ .....(iii) Now, adding Eqs. (i) and (iii), $\Rightarrow \;\; 400 x^{2}-144 x^{4}=256 x^{2} \Rightarrow 144 x^{2}-144 x^{4}=0$ $144 x^{2}(1-x^{2})=0 \Rightarrow x=0,-1,1$ Hence, 3 real values for $x$ satisfies the equation.

Q73JEE Main 2021MCQ4MStraight Lines and Pair of Straight Lines

Let $A(-1,1), B(3,4)$ and $C(2,0)$ be given three points. A line $y=m x, m>0$ intersects lines $A C$ and $B C$ at point $P$ and $Q$ , respectively. Let $A_{1}$ and $A_{2}$ be the areas of $\triangle A B C$ and $\triangle P Q C$ , respectively, such that $A_{1}=3 A_{2}$ , then the value of $m$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Given, points $A(-1,1), B(3,4), C(2,0)$ Equation of $A C=\;\frac{y-1}{x+1}=\;\frac{0-1}{2+1}=\;\frac{-1}{3}$ $\Rightarrow \;\; 3 y-3=-x-1$ $\Rightarrow \;\; x+3 y=2$ ...(i) On solving Eq. (i) and $y=m x$ , we get $P(\;\frac{2}{3 m+1}, \;\frac{2 m}{3 m+1})$ Equation of $B C=\;\frac{y-0}{x-2}=\;\frac{4-0}{3-2}$ $\Rightarrow \;\; y=4 x-8$ ....(ii) Similarly, on solving Eq. (ii) and $y=m x$ , we get $Q(\;\frac{8}{4-m}, \;\frac{8 m}{4-m})$ Area of $\triangle A B C=3$ Area of $\triangle P Q C$ (given) $\;\frac{1}{2}\begin{vmatrix}-1 & 1 & 1 \\ 2 & 0 & 1 \\ 3 & 4 & 1\end{vmatrix}=3 \times \;\frac{1}{2}\begin{vmatrix}\\ \\ \frac{2}{4-m} & \\ \\ \frac{8 m}{2-m} & 1 \\ \\ \\ \frac{2 m}{3 m+1} & \\ \\ \frac{2 m}{3 m+1} & 1\end{vmatrix}$ $\Rightarrow 13=3(\;\frac{1}{4-m})(\;\frac{1}{3 m+1})\begin{vmatrix}2 & 0 & 1 \\ 8 & 8 m & 4-m \\ 2 & 2 m & 3 m+1\end{vmatrix}$ $\Rightarrow \;\; 13=\;\frac{3}{4+11 m-3 m^{2}} \times (52 m^{2})$ $\Rightarrow 15 m^{2}-11 m-4=0$ $\Rightarrow \;\; m=1, \;\frac{-4}{15}$ [but $m>0$ ] $\Rightarrow \;\; m=1$

Q74JEE Main 2021MCQ4MApplication of Derivatives

Let $f$ be a real valued function, defined on $R-\{-1,1\}$ and given by $f(x)=3 \log _{e}|\;\frac{x-1}{x+1}|-\;\frac{2}{x-1}$ Then, in which of the following intervals, function $f(x)$ is increasing?

🚀 Solve in Practice Mode

📖 Explanation

Given, $f(x)=3 \log _{e}|\;\frac{x-1}{x+1}|-\;\frac{2}{x-1}$ $\Rightarrow \;\; f^{'}(x)=\;\frac{3}{(\;\frac{x-1}{x+1})} \cdot \;\frac{(x+1) \cdot 1-(x-1) \cdot 1}{(x+1)^{2}}+\;\frac{2}{(x-1)^{2}}$ $\Rightarrow \;\; f^{'}(x)=3(\;\frac{x+1}{x-1})\$ [;\frac{2}{(x+1)^{2}}]$+;\frac{2}{(x-1)^{2}}\Rightarrow ;; f^{'}(x)=(;\frac{2}{x-1})(;\frac{3}{x+1}+;\frac{1}{x-1})\Rightarrow ;; f^{'}(x)=(;\frac{2}{x-1})[;\frac{3 x-3+x+1}{(x-1)(x+1)}]\Rightarrow ;; f^{'}(x)=;\frac{2 \cdot 2 \cdot (2 x-1)}{(x-1)^{2}(x+1)} $So,$ f^{'}(x)=;\frac{4(2 x-1)}{(x-1)^{2}(x+1)} $Now,For$ f(x) $to be an increasing function,$ f^{'}(x)>0 $. And$ f^{'}(x)>0 $at$ x \in(-\infty ,-1) \cup[;\frac{1}{2}, \infty ) $But domain of$ f(x) $is$ x \in(-\infty ,-1) \cup(1, \infty ) $So,$ f^{'}(x)>0 $at$ x \in(-\infty ,-1) \cup[;\frac{1}{2}, 1) \cup(1, \infty ) $or$ x \in(-\infty ,-1) \cup([;\frac{1}{2}, \infty )-{1})$

Q75JEE Main 2021MCQ4MDifferentiation

Let $f: S \to S$ , where $S=(0, \infty )$ be a twice differentiable function, such that $f(x+1)=x \;\; f(x)$ . If $g: S \to R$ be defined as $g(x)=\log _{e} f(x)$ , then the value of $g^{' '}(5)-g^{' '}(1)|$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

We have, $f: S \to S, S=(0, \infty )$ $f(x+1)=x \cdot f(x)$ $g: S \to R$ $g(x)=\log _{e} f(x)$ To find $\Rightarrow g^{' '}(5)-g^{' '}(1)|$ $\Rightarrow g(x+1)=\log _{e} f(x+1)$ $\Rightarrow g(x+1)=\log [x \cdot f(x)]$ $\Rightarrow g(x+1)=\log x+\log f(x)$ $\Rightarrow g(x+1)=\log x+g(x)$ $\Rightarrow g(x+1)-g(x)=\log x$ $\Rightarrow g^{'}(x+1)-g^{'}(x)=1 / x$ $\Rightarrow g^{' '}(x+1)-g^{' '}(x)=\;\frac{-1}{x^{2}}$ $x=1, g^{' '}(2)-g^{' '}(1) \;\; =-1$ ....(i) $x=2, g^{' '}(3)-g^{' '}(2) \;\; =-1 / 4$ ....(ii) $x=3, g^{' '}(4)-g^{' '}(3) \;\; =-1/ 9$ ....(iii) $x=4, g^{' '}(5)-g^{' '}(4) \;\; =-1 / 16$ ....(iv) Adding Eqs. (i), (ii), (iii) and (iv), $g^{' '}(5)-g^{' '}(1)=-1-\;\frac{1}{4}-\;\frac{1}{9}-\;\frac{1}{16}$ $=-(\;\frac{144+36+16+9}{144})=\;\frac{-205}{144}$ So, $|g^{' '}(5)-g^{' '}(1)|=\;\frac{205}{144}$

Q76JEE Main 2021MCQ4MDefinite Integrals

Let $P(x)=x^{2}+b x+c$ be a quadratic polynomial with real coefficients, such that $int_{0}^{1} P(x) d x=1$ and $P(x)$ leaves remainder 5 when it is divided by $(x-2)$ . Then, the value of $9(b+c)$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

$P(x)=x^{2}+b x+c$ $int_{0}^{1} P(x) d x=1$ $\Rightarrow \;\; int_{0}^{1}(x^{2}+b x+c) d x=1$ $\Rightarrow \;\; \$ [;\frac{x^{3}}{3}+;\frac{b x^{2}}{2}+c x]_{0}^{1}=1\Rightarrow ;; ;\frac{1}{3}+;\frac{b}{2}+c=1\Rightarrow ;; b+2 c=4 / 3 $....(i) And,$ P(x)=(x-2) ⋅ Q(x)+5 $When,$ x=2P(2)=54+2 b+c=5c=1-2 b $.....(ii) Putting$ c=1-2 b $\in Eq. (i),$ b+2(1-2 b)=4 / 3\Rightarrow ;; -3 b+2=4 / 3 ⇒ b=2 / 9∴ ;; C=1-4 / 9=5 / 99(b+c)=9(;\frac{2}{9}+;\frac{5}{9})=7$

Q77JEE Main 2021MCQ4MThree Dimensional Geometry

If the foot of the perpendicular from point $(4,3,8)$ on the line $L_{1}: \;\frac{x-a}{1}=\;\frac{y-2}{3}=\;\frac{z-b}{4}$ , $I \ne 0$ is $(3,5,7)$ , then the shortest distance between the line $L_{1}$ and line $L_{2}: \;\frac{x-2}{3}=\;\frac{y-4}{4}=\;\frac{z-5}{5}$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

$L_{1} \Rightarrow \;\frac{x-a}{I}=\;\frac{y-2}{3}=\;\frac{z-b}{4}$ Foot of perpendicular from $A(4,3,8)$ to $L_{1}$ is $B(3,5,7)$ . $A B=O B-O A$ $=(3 \hat{i}+5 \hat{j}+7 \hat{k})-(4 \hat{i}+3 \hat{j}+8 \hat{k})$ $=-\hat{i}+2 \hat{j}-\hat{k}$ Now, $A B$ is perpendicular to direction cosine of $L_{1}$ , So, $(-\hat{i}+2 \hat{j}-\hat{k}) \cdot (\hat{i}+3 \hat{j}+4 \hat{k})=0$ $\Rightarrow \;\; -I+6-4=0 \Rightarrow I=2$ As, $(3,5,7)$ lies on $L_{1}$ , $\;\frac{3-a}{2}=\;\frac{5-2}{3}=\;\frac{7-b}{4}$ $3-a=2$ So, $a=1,7-b=4$ So, $b=3$ $L_{1} \Rightarrow \;\frac{x-1}{2}=\;\frac{y-2}{3}=\;\frac{z-3}{4}$ and $L_{2}=\;\frac{x-2}{3}=\;\frac{y-4}{4}=\;\frac{z-5}{5}$ $\therefore$ Shortest distance will be along common normal. So, common normal $=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{vmatrix}$ $\Rightarrow \;\; n=-\hat{i}+2 \hat{j}-\hat{k} \Rightarrow \hat{n}=\;\frac{1}{\sqrt{6}}(-\hat{i}+2 \hat{j}-\hat{k})$ Shortest distance will be the projection of $(2-1) \hat{i}+(4-2) \hat{j}+(5-3) \hat{k}$ or $\hat{i}+2 \hat{j}+2 \hat{k}$ along $\hat{n}$ $\Rightarrow (i^{∧}+2 j^{∧}+2 k^{∧}) \;(-i^{∧}+2 j^{∧}-k^{∧})/{\sqrt{6}}=\;\frac{-1+4-2}{\sqrt{6}}=\;\frac{1}{\sqrt{6}}$

Q78JEE Main 2021MCQ4MDifferential Equations

Let $C_{1}$ be the curve obtained by the solution of differential equation $2 x y \;\frac{d y}{d x}=y^{2}-x^{2}, x>0$ . Let the curve $C_{2}$ be the solution of $\;\frac{2 x y}{x^{2}-y^{2}}=\;\frac{d y}{d x}$ . If both the curves pass through $(1,1)$ , then the area enclosed by the curves $C_{1}$ and $C_{2}$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Given, $2 x y \;\frac{d y}{d x}=y^{2}-x^{2}, x>0$ $\;\frac{d y}{d x}=(\;\frac{y^{2}-x^{2}}{2 x y}), x>0$ Let $y=v x$ $\;\frac{d y}{d x}=v+x \;\frac{d v}{d x}$ $v+x \;\frac{d v}{d x}=\;\frac{v^{2} x^{2}-x^{2}}{2 x \cdot v x}=\;\frac{v^{2}-1}{2 v}$ $\Rightarrow \;\; x \;\frac{d v}{d x}=\;\frac{v^{2}-1}{2 v}-v \Rightarrow x \;\frac{d v}{d x}=\;\frac{v^{2}-1-2 v^{2}}{2 v}$ $\Rightarrow x \;\frac{d v}{d x}=\;\frac{-(v^{2}+1)}{2 v} \Rightarrow \int-(\;\frac{2 v}{v^{2}+1}) d v=\int \;\frac{d x}{x}$ $\Rightarrow \;\; -\log |v^{2}+1|=\log x+c$ $\Rightarrow \;\; \log |x|+\log |v^{2}+1|=c$ $\Rightarrow \;\; \log \mid v^{2}+1) x \mid =c \Rightarrow (\;\frac{y^{2}}{x^{2}}+1) x=c$ $\Rightarrow \;\; \;\frac{(y^{2}+x^{2})}{x}=c \Rightarrow y^{2}+x^{2}=c x$ Similarly, for second curve, $x^{2}+y^{2}=c y$ Both passes through $(1,1)$ , $C_{1} \Rightarrow 1+1=C_{1}$ $\Rightarrow \;\; C_{1}=2$ $C_{1} \Rightarrow x^{2}+y^{2}=2 x$ $C_{2} \Rightarrow x^{2}+y^{2}=C_{2} y$ $\Rightarrow \;\; 1+1=C_{2} \Rightarrow C_{2}=2$ $C_{2} \Rightarrow x^{2}+y^{2}=2 y$ $\therefore$ Required area $=2[\;\frac{\pi \cdot 1^{2}}{4}-\;\frac{1}{2} \cdot 1 \cdot 1]$ $=2[(\;\frac{\pi-2}{4})]=\;\frac{\pi}{2}-1$

Q79JEE Main 2021MCQ4MVector Algebra

Let $a=\hat{i}+2 \hat{j}-3 \hat{k}$ and $b=2 \hat{i}-3 \hat{j}+5 \hat{k}$ . If $r \times a=b \times r, r \cdot (\alpha \hat{i}+2 \hat{j}+\hat{k})=3$ and r. $(2 \hat{i}+5 \hat{j}-a \hat{k})=-1, \alpha \in R$ , then the value of $\alpha+|r|^{2}$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Given, $a=(\hat{i}+2 \hat{j}-3 \hat{k})$ and $b=(2 \hat{i}-3 \hat{j}+5 \hat{k})$ If $r \times a=b \times r, r \cdot (\alpha \hat{i}+2 \hat{j}+\hat{k})=3$ $r \cdot (2 \hat{i}+5 \hat{j}-\alpha \hat{k})=-1$ $\Rightarrow r \times a=b \times r$ $(r \times a)=-(r \times b) \Rightarrow (r \times a)+(r \times b)=0$ $\Rightarrow r \times (a+b)=0 \Rightarrow r=\lambda(a+b)$ $\Rightarrow \;\; r=\lambda[(1+2) \hat{i}+(2-3) \hat{j}+(-3+5) \hat{k}]$ $\Rightarrow r=\lambda(3 \hat{i}-\hat{j}+2 \hat{k})$ $r \cdot (\alpha \hat{i}+2 \hat{j}+\hat{k})=3$ $\Rightarrow \lambda(3 \hat{i}-\hat{j}+2 \hat{k})(\alpha \hat{i}+2 \hat{j}+\hat{k})=3$ $\Rightarrow \;\; \lambda(3 \alpha-2+2)=3$ $\Rightarrow \alpha \lambda=1$ $r \cdot (2 \hat{i}+5 \hat{j}-\alpha \hat{k})=-1$ $\Rightarrow \lambda(3 \hat{i}-\hat{j}+2 \hat{k})(2 \hat{i}+5 \hat{j}-\alpha \hat{k})=-1$ $\Rightarrow \;\; \lambda(6-5-2 \alpha)=-1 \Rightarrow \lambda(1-2 \alpha)=-1$ $\Rightarrow \;\; \lambda-2 \alpha \lambda=-1 \Rightarrow \lambda-2=-1$ $\Rightarrow \;\; \lambda=1$ So, $\alpha=1$ $r=(3 \hat{i}-\hat{j}+2 \hat{k})$ $\Rightarrow \;\; |r|^{2}=9+1+4=14$ $\therefore \;\; \alpha+|r|^{2}=1+14=15$

Q80JEE Main 2021NAT4MThree Dimensional Geometry

If the distance of the point $(1,-2,3)$ from the plane $x+2 y-3 z+10=0$ measured parallel to the line, $\;\frac{x-1}{3}=\;\frac{2-y}{m}=\;\frac{z+3}{1}$ is $\sqrt{\;\frac{7}{2}}$ , then the value of $|m|$ is equal to.........

🚀 Solve in Practice Mode

📖 Explanation

Given, point $A \Rightarrow (1,-2,3)$ Plane $\Rightarrow x+2 y-3 z+10=0$ Distance of point from plane along the vector $(3 \hat{i}-m \hat{j}+\hat{k})$ is $\sqrt{7 {/} 2}$ . Line passing through $(1,-2,3)$ in the direction of $(3 \hat{i}-m \hat{j}+\hat{k})$ is $\;\frac{x-1}{3}=\;\frac{y+2}{-m}=\;\frac{z-3}{1}=\lambda$ Any general point $B$ will be $(3 \lambda+1,-m \lambda-2, \lambda+3)$ Now, this point $B$ lies on plane So, $x+2 y-3 z+10=0$ $(3 \lambda+1)+2(-m \lambda-2)-3(\lambda+3)+10=0$ $=(3-2 m-3) \lambda=2$ $\Rightarrow \;\; \lambda=-1 {/} m$ Now, $A=(1,-2,3)$ $B=(3 \lambda+1,-m \lambda-2, \lambda+3)$ $|A B|^{2}=(3 \lambda+1-1)^{2}+(-m \lambda-2+2)^{2}+(\lambda+3-3)^{2}$ $\Rightarrow 7 {/} 2=9 \lambda^{2}+m^{2} \lambda^{2}+\lambda^{2}$ $\Rightarrow \;\; 7 {/}2=10 \lambda^{2}+1$ $[ \because m \lambda=-1]$ $\Rightarrow \;\; 10 \lambda^{2}=5 {/}2 \Rightarrow \lambda^{2}=1{/}4$ $\Rightarrow =\pm 1 {/} 2 \Rightarrow m=-1{/} \lambda$ $\therefore \;\; m=\pm 2$ and $|m|=2$

Q81JEE Main 2021NAT4MStatistics

Consider the statistics of two sets of observations as followsSizeMeanVarianceObservation I1022Observation IIn31If the variance of the combined set of these two observations is $17 / 9$ , then the value of $n$ is equal to ..........

🚀 Solve in Practice Mode

📖 Explanation

SizeMeanVarianceObservation I1022Observation IIn31Variance of combined set is $17 / 9$ . Let the observations for group I be set $A:\{x_{1}, x_{2}, x_{3}, . . ., x_{10}\}$ $\Rightarrow \;\; \;{sum_{x_{i}}}/{10}=2$ $\Rightarrow \;\; \;\; \sum x_{i}=20$ and $\sum(\;\frac{x_{i}^{2}}{10})-(\;\frac{\sum x_{i}}{10})^{2}=2$ $\Rightarrow \;\; \;\frac{\sum x_{i}^{2}}{10}-\;\frac{400}{100}=2$ $\Rightarrow \;\; \;\frac{\sum x_{i}^{2}}{10}=6$ $\Rightarrow \;\; \;\; \sum x_{i}^{2}=60$ Let the observation for group II be set $B:\{y_{1}, y_{2}, y_{3}, . . ., y_{10}\}$ $\;{\sum{y_{i}}}/{n}=3 \Rightarrow sum_{y_{i}}=3 n$ and $\;\frac{\sum y_{i}^{2}}{n}-(\;\frac{\sum y_{i}}{n})^{2}=1$ $\;\frac{\sum y_{i}^{2}}{n}-9=1 \Rightarrow \sum y_{i}^{2}=10 n$ Combined variance $=17 / 9$ $\sigma^{2}=\;\frac{\sum(x_{i}^{2}+y_{i}^{2})}{n+10}-[\sum(\;\frac{x_{i}+y_{i}}{n+10})]^{2}$ $\Rightarrow \;\; \;\frac{17}{9}=\;\frac{60+10 n}{n+10}-[\;\frac{20+3 n}{n+10}]^{2}$ $\Rightarrow \;\; \;\frac{17}{9}=\;\frac{(10 n+60)(n+10)-(3 n+20)^{2}}{(n+10)^{2}}$ $\Rightarrow \;\; \;\frac{17}{9}=\;\frac{(10 n^{2}+160 n+600)-(9 n^{2}+120 n+400)}{(n+10)^{2}}$ $\Rightarrow \;\; \;\frac{17}{9}=\;\frac{n^{2}+40 n+200}{n^{2}+20 n+100}$ $\Rightarrow \;\; 17 n^{2}+340 n+1700=9 n^{2}+360 n+1800$ $\Rightarrow \;\; 8 n^{2}-20 n-100=0$ $\Rightarrow \;\; 2 n^{2}-5 n-25=0$ $\Rightarrow \;\; 2 n^{2}-10 n+5 n-25=0$ $\Rightarrow \;\; 2 n(n-5)+5(n-5)=0$ $\Rightarrow \;\; (2 n+5)(n-5)=0$ $\therefore \;\; n=5$

Q82JEE Main 2021NAT4MMatrices and Determinants

Let $A=\begin{bmatrix}a_{1} \\ a_{2}\end{bmatrix}$ and $B=\begin{bmatrix}b_{1} \\ b_{2}\end{bmatrix}$ be two $2 \times 1$ matrices with real entries such that $A=X B$ , where $X=\;\frac{1}{\sqrt{3}}\begin{bmatrix}1 & -1 \\ 1 & k\end{bmatrix}$ and $k \in R$ . If $a_{1}^{2}+a_{2}^{2}=\;\frac{2}{3}(b_{1}^{2}+b_{2}^{2})$ and $(k^{2}+1) b_{2}^{2} \ne -2 b_{1} b_{2}$ then the value of $k$ is ..........

🚀 Solve in Practice Mode

📖 Explanation

$A=\begin{bmatrix}a_{1} \\ a_{2}\end{bmatrix} \;\text{ and }\; B=\begin{bmatrix}b_{1} \\ b_{2}\end{bmatrix}$ $A=X B$ $X=\;\frac{1}{\sqrt{3}}\begin{bmatrix}1 & -1 \\ 1 & k\end{bmatrix}$ $X B=\;\frac{1}{\sqrt{3}}\begin{bmatrix}1 & -1 \\ 1 & k\end{bmatrix}\begin{bmatrix}b_{1} \\ b_{2}\end{bmatrix}$ $X B=\;\frac{1}{\sqrt{3}}\begin{bmatrix}b_{1}-b_{2} \\ b_{1}+k b_{2}\end{bmatrix}$ As, $A=X B$ So, $\begin{bmatrix}a_{1} \\ a_{2}\end{bmatrix}=\;\frac{1}{\sqrt{3}}\begin{bmatrix}b_{1}-b_{2} \\ b_{1}+k b_{2}\end{bmatrix}$ $\sqrt{3} a_{1}=b_{1}-b_{2}$ ....(i) $\sqrt{3} a_{2}=b_{1}+k b_{2}$ .......(ii) And as given, $a_{1}^{2}+a_{2}^{2}=\;\frac{2}{3}(b_{1}^{2}+b_{2}^{2})$ Squaring and adding Eqs. (i) and (ii), $3 a_{1}^{2}+3 a_{2}^{2}=(b_{1}-b_{2})^{2}+(b_{1}+k b_{2})^{2}$ $\Rightarrow \;\; 2(b_{1}^{2}+b_{2}^{2})=2 b_{1}^{2}+b_{2}^{2}(k^{2}+1)+2 b_{1} b_{2}(k-1)$ $\Rightarrow \;\; b_{2}^{2}(k^{2}+1-2)+2 b_{1} b_{2}(k-1)=0$ $\Rightarrow \;\; (k-1)\$ [b_{2}^{2}(k+1)+2 b_{1} b_{2}]$=0 $So,$ k=1$

Q83JEE Main 2021NAT4MIndefinite Integrals

For real numbers $\alpha, \beta, \gamma$ and $\delta$ , if $\;\int \;{(x^{2}-1)+\tan ^{-1}(\frac{x^{2}+1}{x})}/{(x^{4}+3 x^{2}+1) \tan ^{-1}(\;\frac{x^{2}+1}{x})} d x$ $= \;\; \alpha \log _{e}\$ [\tan ^{-1}(;\frac{x^{2}+1}{x})]$ +\beta \tan ^{-1}(;\frac{\gamma(x^{2}-1)}{x})+\delta \tan ^{-1}(;\frac{x^{2}+1}{x})+C $where$ C $is an arbitrary constant, then the value of$ 10(\alpha+\beta \gamma+\delta )$ is equal to..........

🚀 Solve in Practice Mode

📖 Explanation

Let $I=\int \;{(x^{2}-1)+\tan ^{-1}(\frac{x^{2}+1}{x})}/{(x^{4}+3 x^{2}+1) \tan ^{-1}(\;\frac{x^{2}+1}{x})} d x$ $\Rightarrow \;\; I=\int \;{x^{2}-1}/{(x^{4}+3 x^{2}+1) \tan ^{-1}(\;\frac{x^{2}+1}{x})} d x+\int \;\frac{1}{x^{4}+3 x^{2}+1} d x$ Again let $I_{1}=\int \;{x^{2}-1}/{(x^{4}+3 x^{2}+1) \tan ^{-1}(\;\frac{x^{2}+1}{x})} d x$ and $\;\; I_{2}=\int \;\frac{d x}{x^{4}+3 x^{2}+1}$ $\therefore \;\; I=I_{1}+I_{2}$ ......(i) Now, $\;\; I_{1}=\int \;{(x^{2}-1)}/{(x^{4}+3 x^{2}+1) \tan ^{-1}(\;\frac{x^{2}+1}{x})} d x$ Let $\tan ^{-1}(\;\frac{x^{2}+1}{x})=t$ $\Rightarrow \;\frac{x^{2}-1}{(x^{4}+3 x^{2}+1)} d x=d t$ $\therefore I_{1}=\int \;\frac{d t}{t}=\log |t|+C_{1}=\log |\tan ^{-1}(\;\frac{x^{2}+1}{x})|+C_{1}$ $I_{2}=\int \;\frac{1}{x^{4}+3 x^{2}+1} d x=\;\frac{1}{2} \int \;\frac{(x^{2}+1)-(x^{2}-1)}{x^{4}+3 x^{2}+1} d x$ $=\;\frac{1}{2} \int \;\frac{x^{2}+1}{x^{4}+3 x^{2}+1} d x-\;\frac{1}{2} \int \;\frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x$ $=\;\frac{1}{2} \int \;\frac{1+1 / x^{2}}{x^{2}+3+1 / x^{2}} d x-\;\frac{1}{2} \int \;\frac{1-1 / x^{2}}{x^{2}+3+1 / x^{2}} d x$ $I_{2}=\;\frac{1}{2} \int \;\frac{1+1 / x^{2}}{(x-\;\frac{1}{x})^{2}+5} d x-\;\frac{1}{2} \int \;\frac{1-1 / x^{2}}{(x+\;\frac{1}{x})^{2}+1} d x$ $=\;\frac{1}{2 \sqrt{5}} \tan ^{-1}(\;\frac{x^{2}-1}{\sqrt{5} x})-\;\frac{1}{2} \tan ^{-1}(\;\frac{x^{2}+1}{x})+C_{2}$ $I=\log |\tan ^{-1}(\;\frac{x^{2}+1}{x})|+\;\frac{1}{2 \sqrt{5}} \tan ^{-1}(\;\frac{x^{2}-1}{\sqrt{5} x})-\;\frac{1}{2} \tan ^{-1}(\;\frac{x^{2}+1}{x})+C$ $=\alpha \log _{e}(\tan ^{-1}(\;\frac{x^{2}+1}{x}))+\beta \tan ^{-1}(\;\frac{\gamma(x^{2}-1)}{x})+\delta \tan ^{-1}(\;\frac{x^{2}+1}{x})+C$ (given) $\therefore \;\; \alpha=1, \beta=\;\frac{1}{2 \sqrt{5}} \gamma=\;\frac{1}{\sqrt{5}}$ and $\delta =-\;\frac{1}{2}$ $\therefore$ Required value of $10(\alpha+\beta \gamma+\delta )$ $=10(1+\;\frac{1}{10}-\;\frac{1}{2})$ $=10(\;\frac{10+1-5}{10})$ $=6$

Q84JEE Main 2021NAT4MLimits, Continuity and Differentiability

Let $f: R \to R$ and $g: R \to R$ be defined as $f(x)= \begin{cases}x+a & x< 0 \\ { |x-1|} & x \ge 0$ and $g(x)=\{\text{table} x+1, , x< 0 \\ (x-1)^{2}+b & x \ge 0\end{cases}$ where $a, b$ are non-negative real numbers. If $(g \circ f)(x)$ is continuous for all $x \in R$ , then $a+b$ is equal to.........

🚀 Solve in Practice Mode

📖 Explanation

Given, $f(x)=\begin{cases}x+a & x< 0 \\ {|x-1|} & x \ge 0\end{cases}$ $g(x)=\begin{cases}x+1 & x< 0 \\ (x-1)^{2}+b & x \ge 0\end{cases}$ $g[f(x)]=\begin{cases}f(x)+1 & f(x)< 0 \\ {[f(x)-1]^{2}+b,} & f(x) \ge 0\end{cases}$ Case I $x+a< 0$ and $x< 0 \Rightarrow x< -a$ Case II $|x-1|< 0$ and $x \ge 0 \Rightarrow$ Not possible $f(x) \ge 0$ Case I $x+a \ge 0$ and $x< 0 \Rightarrow x \in[-a, 0)$ Case II $|x-1| \ge 0$ and $x \ge 0 \Rightarrow x \ge 0$ $g[f(x)]=\begin{cases}x+a+1 & x< 0 \text{ and }f(x)< 0 \\ {|x-1|}+1 & x \ge 0 \\ \\ \text{ and } f(x)< 0 \\ (x+a-1)^{2}+b & x< 0 \\ \\ \text{ and } f(x) \ge 0 \\ ({|x-1|}-1)^{2}+b & x \ge 0 \\ \\ \text{ and } f(x) \ge 0\end{cases}$ $g[f(x)]=\begin{cases}x+a+1 & x< -a \\ (x+a-1)^{2}+b & -a \le x< 0 \\ (|x-1|-1)^{2}+b & x \ge 0\end{cases}$ This is continuous function. Since, $g[f(x)]$ is continuous for all $x \in R$ So, $g(f(x))$ will be continuous at $x=-a$ and $x=0$ Now, at $x=-a$ ${LHL}= {RHL}=$ value of function $\Rightarrow \;\; (a-1)^{2}+b=b$ $\Rightarrow (a-1)^{2}=0$ $\Rightarrow \;\; a=1$ Hence, $a+b=1$

Q85JEE Main 2021NAT4MSequences and Series

Let $\;\frac{1}{16}, a$ and $b$ be in G. P. and $\;\frac{1}{a}, \;\frac{1}{b}, 6$ be in A. P., where $a, b>0$ . Then, $72(a+b)$ is equal to .........

🚀 Solve in Practice Mode

📖 Explanation

Given, $G P=\;\frac{1}{16}, a, b$ $\Rightarrow \;\; a^{2}=\;\frac{b}{16}$ and given, ${AP}=1 {/} {a}, 1 {/} {b}, 6$ $\Rightarrow \;\; \;\frac{2}{b}=\;\frac{1}{a}+6$ $\Rightarrow \;\; \;\frac{2}{16 a^{2}}=\;\frac{1}{a}+6$ $\Rightarrow \;\; \;\frac{1}{8 a^{2}}=\;\frac{1+6 a}{a}$ $\Rightarrow \;\; 1=8 a(1+6 a)$ $\Rightarrow \;\; 1=48 a^{2}+8 a$ $\Rightarrow \;\; 48 a^{2}+8 a-1=0$ $\Rightarrow \;\; (4 a+1)(12 a-1)=0$ $a=-1 {/} 4$ or $1 {/} 12$ As per the question, $a>0$ $\therefore \;\; a=1 / 12$ $b=16 a^{2}=16 \cdot \;\frac{1}{144}=\;\frac{1}{9}$ $\therefore 72(a+b)=72(\;\frac{1}{12}+\;\frac{1}{9})$ $=6+8=14$

Q86JEE Main 2021NAT4MCircles

In $\triangle A B C$ , the lengths of sides $A C$ and $A B$ are $12 {cm}$ and $5 {cm}$ , respectively. If the area of $\triangle A B C$ is $30 {cm}^{2}$ and $R$ and $r$ are respectively the radii of circumcircle and incircle of $\triangle A B C$ , then the value of $2 R+r$ (in ${cm}$ ) is equal to

🚀 Solve in Practice Mode

📖 Explanation

Area of $\triangle A B C=30 {cm}^{2}$ $\;\frac{1}{2} \cdot 5 \cdot 12 \sin A=30 {cm}^{2}$ $\Rightarrow \sin A=1$ $\Rightarrow \;\; A=90^{\circ}$ Here, $A B^{2}+A C^{2}=B C^{2}$ $\Rightarrow \;\; 12^{2}+5^{2}=B C^{2}$ $\Rightarrow \;\; B C=13$ $\therefore \;\; R=\;\frac{a}{2 \sin A}=\;\frac{13}{2 \cdot \sin 90}=\;\frac{13}{2}$ $r=\;\frac{\Delta }{s}=\;\frac{30}{(\;\frac{12+5+13}{2})}=\;\frac{30}{15}=2$ $\therefore \;\; 2 R+r=2 \cdot \;\frac{13}{2}+2=15$

Q87JEE Main 2021NAT4MBinomial Theorem

Let $n$ be a positive integer. Let $A=sum_{k}^{n}(-1)^{k}{ }_{n} C^{k}[ (\;\frac{1}{2})^{k}+(\;\frac{3}{4})^{k}+(\;\frac{7}{8})^{k} +(\;\frac{15}{16})^{k}+(\;\frac{31}{32})^{k} ]$ If $63 A=1-\;\frac{1}{2^{30}}$ , then $n$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Given, $A=sum_{k=0}^{n}(-1)^{k} \cdot C_{k}[(\;\frac{1}{2})^{k}+(\;\frac{3}{4})^{k}+(\;\frac{7}{8})^{k}+(\;\frac{15}{16})^{k}+(\;\frac{31}{32})^{k}]$ $\Rightarrow \;\; A=sum_{k=0}^{n}(-1)^{k}\begin{bmatrix}{ }^{n} C_{k}(\\ \\ \frac{1}{2})^{k}+{ }^{n} C_{k}(\\ \\ \frac{3}{4})^{k}+{ }^{n} C_{k}(\\ \\ \frac{7}{8})^{k} \\ +{ }^{n} C_{k}(\\ \\ \frac{15}{16})^{k}+{ }^{n} C_{k}(\\ \\ \frac{31}{32})^{k}\end{bmatrix}$ $\Rightarrow (1-x)^{n}=sum_{r=0}^{n}(-1)^{t} \cdot C_{t} x^{t}$ $x=\;\frac{1}{2}$ $\Rightarrow (1-\;\frac{1}{2})^{n}=C_{0}-C_{1}(\;\frac{1}{2})+C_{2}(\;\frac{1}{2})^{2} . . .$ $\Rightarrow \;\; x=\;\frac{3}{4}$ $\Rightarrow \;\; x=\;\frac{3}{4}$ $(1-\;\frac{3}{4})^{n}=C_{0}-C_{1}(\;\frac{3}{4})+C_{2}(\;\frac{3}{4})^{2} ⋯$ Similarly, we will get $A=sum_{k=0}^{n}(-1)^{k}\begin{bmatrix}{ }^{n} C_{k}(\\ \\ \frac{1}{2})^{k}+{ }^{n} C_{k}(\\ \\ \frac{3}{4})^{k}+{ }^{n} C_{k}(\\ \\ \frac{7}{8})^{k} \\ +{ }^{n} C_{k}(\\ \\ \frac{15}{16})^{k}+{ }^{n} C_{k}(\\ \\ \frac{31}{32})^{k}\end{bmatrix}$ $\Rightarrow A=(1-\;\frac{1}{2})^{n}+(1-\;\frac{3}{4})^{n}+(1-\;\frac{7}{8})^{n}+(1-\;\frac{15}{16})^{n}+(1-\;\frac{31}{32})^{n}$ $\Rightarrow A=\;\frac{1}{2^{n}}+\;\frac{1}{2^{2 n}}+\;\frac{1}{2^{3 n}}+\;\frac{1}{2^{4 n}}+\;\frac{1}{2^{5 n}}$ $\Rightarrow A=\;\frac{1}{2^{n}} \cdot [\;{1-\frac{1}{(2^{n})^{5}}}/{1-\;\frac{1}{2^{n}}}]$ $\Rightarrow A=(\;\frac{1}{2^{n}})(\;\frac{2^{5 n}-1}{2^{n}-1})(\;\frac{2^{n}}{2^{5 n}})$ $\Rightarrow A=(\;\frac{2^{5 n}-1}{2^{n}-1})(\;\frac{1}{2^{5 n}})$ $\because \;\; 63 A=1-\;\frac{1}{2^{30}}$ $\Rightarrow \;\; 63 \;\frac{(2^{5 n}-1)}{2^{5 n}(2^{n}-1)}=\;\frac{2^{30}-1}{2^{30}}$ $\Rightarrow (\;\frac{63}{2^{n}-1})(1-\;\frac{1}{2^{5 n}})=(1-\;\frac{1}{2^{30}})$ $n=6$ , satisfies the equation.

Q88JEE Main 2021NAT4MVector Algebra

Let $\vec{c}$ be a vector perpendicular to the vectors $\vec{a}={i}^{∧}+{j}^{∧}-{k}^{∧}$ and $\vec{b}={i}^{∧}+2 {j}^{∧}+{k}^{∧}$ If $\vec{c} \cdot ({i}^{∧}+{j}^{∧}+3 {k}^{∧})=8$ then the value of $\vec{c} \cdot (\vec{a} \times \vec{b})$ is equal to ..........

🚀 Solve in Practice Mode

📖 Explanation

Since, $c$ is perpendicular to $a$ and $b$ . ${So}, c=\lambda(a \times b)$ $a=\hat{i}+\hat{j}-\hat{k}$ $b=\hat{i}+2 \hat{j}+\hat{k}$ Now, $a \times b=\begin{vmatrix}i^{∧} & j^{∧} & k^{∧} \\ 1 & 1 & -1 \\ 1 & 2 & 1\end{vmatrix}$ $=(1+2) \hat{i}-(1+1) \hat{j}+(2-1) \hat{k}$ $=3 \hat{i}-2 \hat{j}+\hat{k}$ $\Rightarrow \;\; c=\lambda(3 \hat{i}-2 \hat{j}+\hat{k})$ Now, c $\cdot (\hat{i}+\hat{j}+3 {k}^{∧})=8$ $\lambda(3 \hat{i}-2 \hat{j}+\hat{k})(\hat{i}+\hat{j}+3 \hat{k})=8$ $\Rightarrow \;4 \lambda=8$ $\therefore \;\lambda=2$ So, $c=2(3 \hat{i}-2 \hat{j}+\hat{k})$ So, $c \cdot (a \times b)=2(3 \hat{i}-2 \hat{j}+\hat{k})(3 \hat{i}-2 \hat{j}+\hat{k})$ $=2(9+4+1)$ $=28$

Q89JEE Main 2021NAT4MSequences and Series

Let $S_{n}(x)=\log _{a^{1 {/} 2}} x+\log _{a^{1 {/} 3}} x+\log _{a^{1 {/} 6}} x +\log _{a^{1{/}11}} x+\log _{a^{1 {/} 18}} x+\log _{a^{1 {/}27}} x+. . .$ up to $n$ -terms, where $a>1$ . If $S_{24}(x)=1093$ and $S_{12}(2 x)=265$ , then value of $a$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Given, $S_{n}(x)=\log _{ {a}^{1 {/} 2}} x+\log _{ {a}^{1{/} 3}} x+\log _{ {a}^{1{/}6}} x+\log _{ {a}^{1 {/} 1}} x+\log _{a^{1{/}8}} x+\log _{a^{1 {/}27}} x+. . .$ upto $n$ -terms $S_{n}(x)=2 \log _{a} x+3 \log _{a} x+6 \log _{a} x+11 \log _{a} x+18 \log _{a} x+27 \log _{a} x+. . .$ $S_{n}(x)=(2+3+6+11+18+27+. . .) \log _{a} x$ Let $S_{n}=2+3+6+11+18+. . . T_{n}$ $S_{n}=2+3+6+11+. . .+T_{n}$ $\Rightarrow \;\; 0=2+1+3+5+7+. . .(T_{n}-T_{n-1})-T_{n}$ $\Rightarrow T_{n}=2+1+3+5+. . . . . .+(n-1)$ terms $T_{n}=2+(n-1)^{2}$ $S_{1}=Σ T_{n}=2 n+\;\frac{n(n-1)(2 n-1)}{6}$ $\therefore \;\; S_{n}(x)=(2 n+\;\frac{n(n-1)(2 n-1)}{6}) \log _{a} x$ We have, $S_{24}(x)=1093$ $\log _{a} x(48+\;\frac{24 \cdot 23 \cdot 47}{6})=1093$ $\Rightarrow \;\; \log x=\;\frac{1}{4}$ .....(i) $S_{12}(2 x)=\$ [sum_{n=0}^{11}(2+(n-1)^{2})]$ \log _{a} 2 x=[2 \times 12+;\frac{12 \cdot 11 \cdot 23}{6}] \log _{a} 2 x=265\log _{a} 2 x=;\frac{1}{2} $...(ii) On subtracting Eq. (ii) from Eq. (i), we get$ \log _{a} 2 x-\log _{a} x=;\frac{1}{4}=\log _{a} 2=;\frac{1}{4}\Rightarrow ;; a=16$

← All JEE Main 2021 papers Take as Timed Mock →