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Physics30 questions

Q1JEE Main 2021MCQ4MProperties of Matter

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion (A) When a rod lying freely is heated, no thermal stress is developed in it. Reason (R) On heating, the length of the rod increases. In the light of the above statements, choose the correct answer from the options given below

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Q2JEE Main 2021MCQ4MWaves

A student is performing the experiment of resonance column. The diameter of the column tube is $6 {cm}$ . The frequency of the tuning fork is $504 {Hz}$ . Speed of the sound at the given temperature is $336 {m} / {s}$ . The zero of the meter scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is

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📖 Explanation

Given, diameter of the column tube, $d=6 {cm}=6 \times 10^{-2} {m}$ Frequency of tuning fork, $f=504 {Hz}$ Speed of sound at given temperature, $v=336 {ms}^{-1}$ As, this is a closed organ pipe. Let $L$ be the length of tube and $\lambda$ be the wavelength, thenand $L+ {e} \;\; =\lambda / 4=v / 4 f \;\; ( \because \lambda=\;\frac{v}{f})$ ${e} \;\; =0.6 \times d / 2$ $\;\; =\;\frac{6}{10} \times \;\frac{6 \times 10^{-2}}{2}=0.018$ $\Rightarrow \;\; L+ {e} \;\; =\;\frac{v}{4 f}=\;\frac{336}{4 \times 504}$ $\Rightarrow \;\; L \;\; =\;\frac{336}{2016}-0.018=0.1667-0.018$ $\;\; =0.1487 {m}=14.87 {cm}$ $\;∼eq 14.8 {cm}$

Q3JEE Main 2021MCQ4MGravitation

Two satellites $A$ and $B$ of masses $200 {kg}$ and $400 {kg}$ are revolving around the Earth at height of $600 {km}$ and $1600 {km}$ , respectively. If $T_{A}$ and $T_{B}$ are the time periods of $A$ and $B$ respectively, then the value of $T_{B}-T_{A}$ is(Given, radius of Earth $=6400 {km}$ , mass of Earth $=6 \times 10^{24} {kg}$ )

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📖 Explanation

Given, $M_{A}=200 {kg}, M_{B}=400 {kg}, H_{A}=600 {km}, H_{B}=1600 {km}$ and $\;\; R_{A}=R_{E}+H_{A}=6400+600=7000 {km}$ $R_{B}=R_{E}+H_{B}=6400+1600=8000 {km}$ Let $T_{A}, T_{B}, \omega_{A}, \omega_{B}, R_{A}$ and $R_{B}$ be the time period, angular frequencies and radii of satellites $A$ and $B$ , respectively. Force on satellite $A, F_{A}=m_{A} \omega_{A}^{2} R_{A}=\;\frac{G M m_{A}}{R_{A}^{2}}$ $\Rightarrow \;\; \omega_{A}^{2}=\;\frac{G M}{R_{A}^{3}}$ but, $\;\; \omega_{A}=\;\frac{2 \pi}{T_{A}}$ $\therefore \;\; (\;\frac{2 \pi}{T_{A}})^{2}=\;\frac{G M}{R_{A}^{3}} \Rightarrow T_{A}=\sqrt{\;\frac{4 \pi^{2} R_{A}^{3}}{G M}}$ Similarly, $\;\; T_{B}=\sqrt{\;\frac{4 \pi^{2} R_{B}^{3}}{G M}}$ $\therefore \;\; T_{B}-T_{A} \;\; =\sqrt{\;\frac{4 \pi^{2}}{G M}}(\sqrt{R_{B}^{3}}-\sqrt{R_{A}^{3}})$ $\;\; =\;\frac{2 \pi}{\sqrt{G M}}[\sqrt{(8 \times 10^{3})^{3}}-\sqrt{(7 \times 10^{3})^{3}}]$ $=\;\frac{2 \pi \times 10^{9}}{\sqrt{G M}}(8 \sqrt{8}-7 \sqrt{7})$ $=\;{2 \pi \times 10^{9}}/{\sqrt{6.67 \times 10^{-11} \times 6 \times 10^{24}}}(4.107)$ $=\;{2 \pi}/{\sqrt{4 \times 10^{14}}} \times 10^{9}(4.107)$ $=\pi \times 10^{2} \times 4.107=12.9 \times 10^{2} {s}$ $=1.33 \times 10^{3} {s}$

Q4JEE Main 2021MCQ4MAlternating Current

The angular frequency of alternating current in an L-C-R circuit is $100 {rad} / {s}$ . The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.

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📖 Explanation

Given, angular frequency, $\omega=100 {rads}^{-1}$ $R \;\; =60 Ω, V_{R}=15 {V},$ $R^{'} \;\; =40 Ω, V_{R^{'}}=V_{L}=20 {V}$ $V_{C}=10 {V} \;\; V_{R^{'}}=V_{L}=20 {V}$ By using Ohm's law, $V \;\; =I R \Rightarrow I=V / R$ $I \;\; =15 / 60=1 / 4 {A}$ . . . (i) $\;\text{ and }\; \;\; I_{1} \;\; =\;{V_{R^{'}}}/{R^{'}}=20 / 40=1 / 2 {A}$ . . . (ii) $\;\text{ As, }\; \;\; X_{C} \;\; =\;\frac{V_{C}}{I}=\;\frac{10}{1 / 4}=40 Ω$ $\;\text{ and }\; \;\; X_{C} \;\; =\;\frac{1}{\omega C}$ $\Rightarrow \;\; C \;\; =\;\frac{1}{X_{C}(1)}=\;\frac{1}{40 \times 100}$ $\;\; =0.25 \times 10^{-3} {F}=0.25 {mF}$ $\;\; =250 \mu {F}$ $\;\text{ By using }\; {KCL} \;\text{ \in }\; loop 2,\;\;\text{ (ii) }\;$ $\;I_{2}=I-I_{1}$ $\;\; =1 / 4-1 / 2=-1 / 4 {A}$ $\;X_{L}=\;\frac{V_{L}}{|I_{2}|}=\;\frac{20}{1 / 4}=80 Ω \Rightarrow \;\; \omega L=80$ $\;L=\;\frac{80}{\omega}=\;\frac{80}{100}=0.8 {H}$

Q5JEE Main 2021MCQ4MMagnetic Effects of Current and Magnetism

A proton, a deuteron and an $\alpha$ -particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic forces acting on them is ......... and their speed is ......... in the ratio.

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📖 Explanation

Let $F_{p}, F_{d}$ and $F_{\alpha}$ be the forces and $v_{p}, v_{d}$ and $v_{\alpha}$ be the velocities of proton, deuteron and $\alpha-$ particle, respectively. Since, $F=B q v$ On dividing and multiplying $F$ by $m$ , we get $F=B q v \;\frac{m}{m}$ $\Rightarrow \;\; F=B \;\frac{q}{m} p \;\; ( \because p=m v)$ $\Rightarrow \;\; F \propto q / m \;\; ( \because p$ and $B$ are same) $\therefore \;\; F_{p}: F_{d}: F_{\alpha} \;\; =\;\frac{+q}{m}: \;\frac{+q}{2 m}: \;\frac{+2 q}{4 m}\;$ $\; \;\; =1: 1 / 2: 1 / 2=2: 1: 1$ $\;\;\text{ and }\;\; \;\; =B q v$ $\Rightarrow \;\; \propto F / q$ $\therefore \;\; v_{p}: v_{d}: v_{\alpha} \;\; =\;\frac{F_{p}}{q_{p}}: \;\frac{F_{d}}{q_{d}}: \;\frac{F_{\alpha}}{q_{\alpha}}=\;\frac{2 x}{q}: \;\frac{x}{q}: \;\frac{x}{2 q}$ $\;\; =2: 1: 1 / 2=4: 2: 1$

Q6JEE Main 2021MCQ4MSemiconductor Electronics

Given, below are two statements Statement I A speech signal of $2 {kHz}$ is used to modulate a carrier signal of $1 {MHz}$ . The bandwidth requirement for the signal is $4 {kHz}$ . Statement II The side band frequencies are $1002 {kHz}$ and $998 {kHz}$ . In the light of the above statements, choose the correct answer from the options given below

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📖 Explanation

Given, frequency of modulated signal, $f_{m}=2 {kHz}=2 \times 10^{3} {Hz}$ Frequency of carrier signal, $f_{c}=1 {MHz}$ $\;\; =1 \times 10^{6} {Hz}$ $\;\; =1000 {kHz}$ Then, bandwidth $=2 f_{m}=4 {kHz}$ and side band frequency $=f_{c} \pm f_{m}$ $\;\; =(1000 \pm 2) {kHz}$ $\;\; =1002 {kHz} \;\text{ and }\; 998 {kHz}$ Hence, option (c) is the correct.

Q7JEE Main 2021MCQ4MOscillations

If the time period of a $2 {m}$ long simple pendulum is $2 {s}$ , the acceleration due to gravity at the place, where pendulum is executing SHM is

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📖 Explanation

Given, length of simple pendulum, $I=2 {m}$ Time period, $T=2 {s}$ Let $g_{\;\text{eff }\;}$ be the acceleration due to gravity. $\therefore \;\text{ Time period, }\; T \;\; =2 \pi \sqrt{\;{1}/{g_{\;\text{eff }\;}}}$ $\Rightarrow \;\; g_{\;\text{eff }\;} \;\; =4 \pi^{2} \;\frac{1}{T^{2}}$ $\;\; =4 \pi^{2} \cdot \;\frac{2}{4}=2 \pi^{2} {ms}^{-2}$

Q8JEE Main 2021MCQ4MUnits and Measurements

The pitch of the screw gauge is $1 {mm}$ and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72 nd division on circular scale coincides with the reference line. The radius of the wire is

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📖 Explanation

Given, pitch of screw gauge, $P=1 {mm}$ Number of division, $n=100$ $\therefore$ Least count $( {LC})=\;\frac{P}{n}=1 / 100=0.01 {mm}$ As, zero of circular division lies 8 divisions below. $\therefore$ Zero error $=8 \times {LC}=8 \times 0.01=0.08 {mm}$ Since, 1st linear scale division coinside with 72nd circular scale division. $\therefore Radius(r) \;\; =\;\frac{[P+(72 \times {LQ}-\;\text{ Zero error }\;]}{2}$ $\;\; =[1+(72 \times 0.01)-0.08] / 2$ $\;\; =(1.72-0.08) / 2=1.64 / 2$ $\;\; =0.82 {mm}$

Q9JEE Main 2021MCQ4MSemiconductor Electronics

A $5 {V}$ battery is connected across the points $X$ and $Y$ . Assume $D_{1}$ and $D_{2}$ to be normal silicon diodes. Find the current supplied by the battery, if the positive terminal of the battery is connected to point $X$ .

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📖 Explanation

Given, supply voltage, $V=5 {V}$ The circuit diagram, when positive terminal of the battery is connected to $X$ is as shown belowLet $I$ current is coming from battery. $\therefore D_{1}$ will act as closed circuit as forward biased and $D_{2}$ will act as open circuit as reverse biased. Now, by using Kirchhoff's voltage law, $\;5-V_{D_{1}}-10 I \;\; =0$ $\Rightarrow \;\; 5-0.7-10 I \;\; =0\;$ $\;\; ( \because V_{D_{1}}=0.7 {V})$ $\Rightarrow \;\; 4.3 \;\; =101\;$ $\Rightarrow \;\; I \;\; =0.43 {A}$

Q10JEE Main 2021MCQ4MDual Nature of Matter and Radiation

An $\alpha$ -particle and a proton are accelerated from rest by a potential difference of $200 {V}$ . After this, their de-Broglie wavelengths are $\lambda_{\alpha}$ and $\lambda_{p}$ , respectively. The ratio $\;\frac{\lambda_{p}}{\lambda_{\alpha}}$ is

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📖 Explanation

de-Broglie wavelength, $\lambda \;\; =\;\frac{h}{\sqrt{2 m q V}}$ $\Rightarrow \;\; \;\frac{\lambda_{p}}{\lambda_{\propto }} \;\; =\sqrt{\;\frac{m_{\alpha} q_{\alpha}}{m_{p} q_{p}}}=\sqrt{\;\frac{4 m_{p} \cdot 2 e}{m_{p} \cdot {e}}}$ $\;\; =\;\frac{2 \sqrt{2}}{1}=2 \times 1.414=2.8$

Q11JEE Main 2021MCQ4MThermodynamics

A diatomic gas having $C_{p}=\;\frac{7}{2} R$ and $C_{V}=\;\frac{5}{2} R$ , is heated at constant pressure. The ratio $d U: d Q: d W$ is

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📖 Explanation

Given, $C_{p}=7 / 2 R, C_{V}=5 / 2 R$ Since, change in internal energy $(d U)=n C_{V} d T$ Heat change $(d Q)=n C_{p} d T$ Work done $(d W)=n R d T$ $\therefore \;\; d U: d Q: d W \;\; =n C_{V} d T: n C_{p} d T: n R d T$ $\;\; =C_{V}: C_{p}: R=\;\frac{5}{2} R: \;\frac{7}{2} R: R$ $\;\; =5 R: 7 R: 2 R$ $\;\; =5: 7: 2$

Q12JEE Main 2021MCQ4MMotion in a Straight Line

An engine of a train moving with uniform acceleration, passes the signal-post with velocity $u$ and the last compartment with velocity $v$ . The velocity with which middle point of the train passes the signal post is

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📖 Explanation

Given, initial speed of engine $=u$ Speed of engine last compartment $=v$ Let the length of train be $I$ and the speed of mid-point of train be $v^{'}$ . Using third equation of motion, $v^{2}=u^{2}+2 a s$ $\Rightarrow \;\; v^{' 2}=u^{2}+2 a l / 2$ $\Rightarrow \;\; v^{' 2}-u^{2}=a l$ . . . (i) $\;\text{ Also, }\; \;\; v^{2}-u^{2}=2 a l \Rightarrow a=\;\frac{v^{2}-u^{2}}{2 l}$ Substituting the value of $a$ in Eq. (i), we get $v^{' 2}-u^{2} \;\; =\;\frac{v^{2}-u^{2}}{2 l} \cdot l=\;\frac{v^{2}-u^{2}}{2}$ $\Rightarrow \;\; v^{'} \;\; =\sqrt{\;\frac{v^{2}-u^{2}}{2}+u^{2}}=\sqrt{\;\frac{v^{2}+u^{2}}{2}}$

Q13JEE Main 2021MCQ4MUnits and Measurements

Match List-I with List-II List-I List-II A. $h$ (Planck's constant) 1. $[ {M} {L} {T}^{-1}]$ B. $E$ (kinetic energy) 2 $[ {M} {L}^{2} {T}^{-1}]$ C. $V$ (electric potential) 3. $[ {ML}^{2} {T}^{-2}]$ D. $P$ (linear momentum) 4. $[ {M} {L}^{2} {I}^{-1} {T}^{-3}]$ Choose the correct answer from the options given below.

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📖 Explanation

The dimensional formulae of given terms are Planck's constant $(h)=[ {ML}^{2} {T}^{-1}]$ Kinetic energy $(E)=[ {ML}^{2} {T}^{-2}]$ Electric potential $(V)=[M L^{2}{ }^{-1}{ }^{1} {T}^{-3}]$ Linear momentum $(p)=[ {MLT}^{-1}]$ So, the correct match is ${A} \to 2, {B} \to 3, {C} \to 4, {D} \to 1 .$

Q14JEE Main 2021MCQ4MMagnetic Effects of Current and Magnetism

Magnetic fields at two points on the axis of a circular coil at a distance of $0.05 {m}$ and $0.2 {m}$ from the centre are in the ratio $8: 1$ . The radius of coil is

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📖 Explanation

Given, $d_{1}=0.05 {m}, d_{2}=0.2 {m}$ and $B_{1}: B_{2}=8: 1$ Let radius of coil be $r$ . As we know that, magnetic field at a point on axis of a coil, $B=\;\frac{\mu _{0}}{2} \cdot \;\frac{I r^{2}}{(d_{1}^{2}+r^{2})^{3 / 2}}$ $\; \therefore \;\;\frac{B_{1}}{B_{2}} \;\; =\;\frac{r^{2} \cdot (d_{2}^{2}+r^{2})^{3 / 2}}{(d_{1}^{2}+r^{2})^{3 / 2} \cdot r^{2}}=\;\frac{(d_{2}^{2}+r^{2})^{3 / 2}}{(d_{1}^{2}+r^{2})^{3 / 2}}$ $\;\;\frac{8}{1}=(\;\frac{d_{2}^{2}+r^{2}}{d_{1}^{2}+r^{2}})^{3 / 2} \Rightarrow \;\frac{2^{3}}{1}=(\;\frac{d_{2}^{2}+r^{2}}{d_{1}^{2}+r^{2}})^{3 / 2}$ $\;\;\frac{4}{1}=\;\frac{d_{2}^{2}+r^{2}}{d_{1}^{2}+r^{2}}$ $\Rightarrow \;4 d_{1}^{2}+4 r^{2} \;\; =d_{2}^{2}+r^{2}$ $\Rightarrow \;4(0.05)^{2}+4 r^{2} \;\; =(0.2)^{2}+r^{2} \Rightarrow 3 r^{2}=0.03$ $\Rightarrow \;r=0.1 {m}$

Q15JEE Main 2021MCQ4MGravitation

A solid sphere of radius $R$ gravitationally attracts a particle placed at $3 R$ from its centre with a force $F_{1}$ . Now, a spherical cavity of radius $(\;\frac{R}{2})$ is made in the sphere (as shown in figure) and the force becomes $F_{2}$ . The value of $F_{1}: F_{2}$ is

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📖 Explanation

Given, radius of sphere $=R$ Distance of particle from centre of Earth $=3 R$ Force between sphere and particle $=F_{1}$ Radius of cavity $=R / 2$ Let mass of sphere $=M$ Mass of sphere with cavity $=M^{'}$ Mass of particle $=m$ Now, $\;\; M^{'}=\rho \cdot \;\frac{4}{3} \pi(\;\frac{R}{2})^{3}=\;\frac{M}{8}$ By using concept of gravitational force, $F_{1} \;\; =\;\frac{G M m}{(3 R)^{2}}=\;\frac{G M m}{9 R^{2}}$ . . . (i) and $F_{\;\text{cavity }\;} \;\; =\;\frac{G M^{'} m}{(A B)^{2}}=\;\frac{(2}{25 \times 8 R^{2}}$ $\;\; =\;\frac{4 G M m}{25}$ . . . (ii) $\therefore \;\; F_{2} \;\; =F_{1}-F_{\;\text{cavity }\;}=\;\frac{G M m}{9 R^{2}}-\;\frac{4}{25 \times 8} \;\frac{G M m}{R^{2}}$ $\;\; =\;\frac{G M m}{R^{2}}(\;\frac{1}{9}-\;\frac{1}{50})=\;\frac{41}{50 \times 9} \;\frac{G M m}{R^{2}}$ $\because \;\; \;\frac{F_{1}}{F_{2}} \;\; =\;\frac{50}{41}$

Q16JEE Main 2021MCQ4MAtoms and Nuclei

Two radioactive substances $X$ and $Y$ originally have $N_{1}$ and $N_{2}$ nuclei, respectively. Half-life of $X$ is half of the half-life of $Y$ . After three half-lives of $Y$ , number of nuclei of both are equal. The ratio $\;\frac{N_{1}}{N_{2}}$ will be equal to

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📖 Explanation

Given, initial amount of $X$ and $Y$ be $N_{1}$ and $N_{2}$ . Let half-life of $X$ be $t_{x}$ and $y$ be $t_{y}$ . According to question, $t_{x}=\;\frac{t_{y}}{2}=t$ $\Rightarrow \;\; t_{x}=t$ and $t_{y}=2 t$ After 3 half-lives of $Y$ , $3 t_{y}=6 t$ As we know that, $N=N_{0} {e}^{-\lambda t}$ where, $N$ is the number of nuclei left undecayed. and $t_{1 / 2} \;\; =\;\frac{0.693}{\lambda}$ $\lambda \;\; =\;\frac{0.693}{t_{1 / 2}}$ $\lambda_{1} \;\; =\;\frac{0.693}{t} \;\text{ and }\; \lambda_{2}=\;\frac{0.693}{2 t}$ Since, after 3 half-lives of $Y$ number of nuclei of both become equal. $\therefore$ $\therefore \;N_{1} {e}^{-\lambda_{1} 6 t}=N_{2} {e}^{-\lambda_{2} 6 t}$ $\Rightarrow \;N_{1} / N_{2}= {e}^{6 t(-\lambda_{2}+\lambda_{1})}$ $\Rightarrow \;N_{1} / N_{2}= {e}^{6 t(\;\frac{0.693}{t}-\;\frac{0.693}{2 t})}$ $\;\; =e^{0.693(\;\frac{6 t}{t}-\;\frac{6 t}{2 t})}=e^{0.693 \times 3}=7.9 ∼eq 8$ $\;\frac{N_{1}}{N_{2}} \;\; =\;\frac{8}{1}$

Q17JEE Main 2021MCQ4MMotion in a Plane

In an octagon $A B C D E F G H$ of equal side, what is the sum of ${AB}+ {AC}+ {AD}+ {AE}+ {AF}+ {AG}+ {AH}$ $\;\text{ if, }\; \overline{A O}=2 \hat{i}+3 \hat{j}-4 \hat{k} ?$

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📖 Explanation

Given, $A O=2 \hat{i}+3 \hat{j}-4 \hat{k}$ By using triangle law, $\;\text{ Similarly, }\;\;A C=A O+O B$ $A D \;\; =A O+O C$ $A E \;\; =A O+O E$ $A F \;\; =A O+O F$ $A G \;\; =A O+O G$ $A H \;\; =A O+O H$ Now, adding all vectors $\;A B+A C+A D+A E+A F+A G+A H$ $\;\; =7 A O+(O B+O C+O D+O E+O F+O G+O H)$ . . . (i) By using cyclic vector, ${OA}+ {OB}+ {OC}+ {OD}+ {OE}+ {OF}+ {OG}+ {OH}=0$ $\Rightarrow {OB}+ {OC}+ {OD}+ {OE}+ {OF}+ {OG}+ {OH}$ $=0- {OA}=0+ {AO}$ Substituting in Eq. (i), we get $A B+A C+A D \;+A E+A F+A G+A H=7 A O+A O=8 A O$ $\;\; =8(2 \hat{i}+3 \hat{j}-4 \hat{k})$ $\;\; =16 \hat{i}+24 \hat{j}-32 \hat{k}$

Q18JEE Main 2021MCQ4MGravitation

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion $A$ The escape velocities of planet $A$ and $B$ are same. But $A$ and $B$ are of unequal mass. Reason $R$ The product of their mass and radius must be same, $M_{1} R_{1}=M_{2} R_{2}$ In the light of the above statements, choose the most appropriate answer from the options given below.

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📖 Explanation

Let $v_{A}, v_{B}, M_{A}, M_{B}$ and $R_{A}, R_{B}$ be the escape velocities, masses and radii of planet $A$ and $B$ , respectively. As we know that, $v=\sqrt{2 g R}=\sqrt{\;\frac{2 G M}{R^{2}} \cdot R}=\sqrt{\;\frac{2 G M}{R}}$ where, $G$ is the gravitational constant. $\Rightarrow \;\; v \propto \sqrt{\;\frac{M}{R}}$ Since, $v_{A}=v_{B}$ $\because \;\; \;\frac{M_{A}}{R_{A}}=\;\frac{M_{B}}{R_{B}} \Rightarrow M_{A} R_{B}=M_{B} R_{A}$ or $M_{1} R_{2}=M_{2} R_{1}$ Hence, option (b) is the correct.

Q19JEE Main 2021MCQ4MElectromagnetic Induction

The current $(i)$ at time $t=0$ and $t=\infty$ respectively for the given circuit is

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📖 Explanation

As we know that at time $t=0$ , inductor acts as open circuit. Then, the circuit becomes Therefore, $R_{ {eq}}=\;\frac{(5+1)(5+4)}{(5+1)+(5+4)}=\;\frac{6 \times 9}{6+9}=\;\frac{54}{15}=\;\frac{18}{5}$ By using Ohm's law, $V\;=I R_{ {eq}}$ $I\;=\;\frac{E \times 5}{18}=\;\frac{5 E}{18}$ $\;\; [ \because V=E]$ At $t=\infty$ , inductor will act as short circuit. It is shown belowTherefore, $R_{\;\text{eq }\;}=\;\frac{5 \times 5}{5+5}+\;\frac{1 \times 4}{1+4}=\;\frac{25}{10}+\;\frac{4}{5}=\;\frac{5}{2}+\;\frac{4}{5}=\;\frac{33}{10} Ω$ and $I=\;\frac{E}{\frac{33}{10}}=\;\frac{10 E}{33}$

Q20JEE Main 2021MCQ4MWave Optics

Two coherent light sources having intensity in the ratio $2 x$ produce an interference pattern. The ratio $\;\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}$ will be

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📖 Explanation

Given, $\;\frac{I_{1}}{I_{2}}=2 x$ $\Rightarrow \;\; I_{1}=2 I_{2} x$ As we know, and $\;\; I_{\min }=(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}$ $\therefore \;\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\;{2 \sqrt{I_{1} I_{2}}}/{l_{1}+I_{2}} \;\;$ (using dividendo rule) $=\;{2 \sqrt{2 I_{2} x \cdot I_{2}}}/{2 I_{2} x+I_{2}}=\;\frac{2 \sqrt{2 x}}{2 x+1}$

Q21JEE Main 2021NAT4MAlternating Current

A transmitting station releases waves of wavelength $960 {m}$ . A capacitor of $2.56 \mu {F}$ is used in the resonant circuit. The self-inductance of coil necessary for resonance is ......... $\times 10^{-8} {H}$ .

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📖 Explanation

Given, wavelength of transmission signal, $\lambda \;\; =960 {m}$ $\;\text{ Capacitance, }\; C \;\; =2.56 \mu {F}=2.56 \times 10^{-6} {F}$ As we know resonance frequency, $f=\;\frac{1}{2 \pi \sqrt{L C}}$ Also, frequency $(f)=\;\frac{\;\text{ speed }\;(v)}{\;\text{ wavelength }\;(\lambda)}$ $\; \therefore \;\; \;\frac{v}{\lambda}\;=\;\frac{1}{2 \pi \sqrt{L C}}$ $\Rightarrow \;\; \sqrt{L C}=\;\frac{\lambda}{v \times 2 \pi}$ On squaring both sides, we get $\Rightarrow \;\; L C=\;\frac{\lambda^{2}}{v^{2} \times 4 \pi^{2}} \Rightarrow L=\;\frac{\lambda^{2}}{v^{2} \times 4 \pi^{2} \times C}$ $\Rightarrow \;\; L=\;\frac{(960)^{2}}{(3 \times 10^{8})^{2} \times 4 \pi^{2} \times 2.56 \times 10^{-6}}$ $\therefore \;\; L=10 \times 10^{-8} {H}$

Q22JEE Main 2021NAT4MElectrostatics

The electric field in a region is given $E=(\;\frac{3}{5} E_{0} \hat{i}+\;\frac{4}{5} E_{0} \hat{j}) \;\frac{N}{C}$ . The ratio of flux of reported field through the rectangular surface of area $0.2 {m}^{2}$ (parallel to YZ-plane) to that of the surface of area $0.3 {m}^{2}$ (parallel to $X Z$ - plane) is $a: b$ , where $a=. . . . . . . . .$ . [Here $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along $X, Y$ and $Z$ -axes, respectively]

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📖 Explanation

Given, ${E}=\;\frac{3 E_{0}}{5} { {i}^{∧}+\;\frac{4}{5} E_{0} { {j}^{∧}$ , $A_{1}=0.2 {m}^{2} \hat{i}$ and $A_{2}=0.3 {m}^{2} \hat{j}$ Let $\varphi _{1}$ and $\varphi _{2}$ be the flux linked with area $A_{1}$ and $A_{2}$ , respectively. As we know that, $\varphi \;\; =∮ E \cdot d S=E \cdot A$ $\Rightarrow \;\; \varphi _{1} \;\; =(3 / 5 E_{0} \hat{i}+4 / 5 E_{0} \hat{j}) \cdot 0.2 \hat{i}=3 / 5 E_{0} \times 0.2$ and similarly, $\;\; \varphi _{2}=4 / 5 E_{0} \times 0.3$ Now, $\;\; \;\frac{\varphi _{1}}{\varphi _{2}}=\;\frac{3 / 5 E_{0} \times 0.2}{4 / 5 E_{0} \times 0.3}=\;\frac{0.6}{1.2}=\;\frac{1}{2}$ $\therefore \;\; a=1$

Q23JEE Main 2021NAT4MThermodynamics

In a certain thermodynamical process, the pressure of a gas depends on its volume as $k V^{3}$ . The work done when the temperature changes from $100^{\circ} {C}$ to $300^{\circ} {C}$ will be ........... $n R$ , where $n$ denotes number of moles of a gas.

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📖 Explanation

Given, pressure $(p) \propto k V^{3}$ $T_{1} \;\; =100^{\circ} {C}, T_{2}=300^{\circ} {C}$ $\Delta T \;\; =T_{2}-T_{1}=300-100=200^{\circ} {C}$ $\Rightarrow$ By using ideal gas equation, $p V=n R T$ $k V^{3} \cdot V=n R T \Rightarrow k V^{4}=n R T$ On differentiating both sides w.r.t temperature, we get $4 k V^{3} \;\frac{d V}{d T} \;\; =n R$ $\Rightarrow \;\; 4 k V^{3} d V \;\; =n R d T \Rightarrow k V^{3} d V=n R d T / 4$ $\Rightarrow \;\; p d V \;\; =n R d T / 4$ $\;\text{ As, work done }\;(W) \;\; =p d V=n R d T / 4$ $\;\; =\;\frac{n R}{4} \Delta T=\;\frac{n R}{4} \times 200=50 n R$

Q24JEE Main 2021NAT4MCircular Motion

A small bob tied at one end of a thin string of length $1 {m}$ is describing a vertical circle, so that the maximum and minimum tension in the string are in the ratio $5: 1$ . The velocity of the bob at the highest position is ......... ${m} / {s}$ . (Take, $g=10 {m} / {s}^{2}$ )

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📖 Explanation

Given, length of string, $l=1 {m}$ $T_{\max }$ and $T_{\min }$ be the tension in string and $v_{1}$ and $v_{2}$ be the velocities of bob at bottom and top in vertical circle. $\;T_{\max } \;\; =m g+m v_{1}^{2} / l$ $\;\;\text{ and }\;\;T_{\min } \;\; =m v_{2}^{2} / l-m g$ $\; \because \;\;\frac{T_{\max }}{T_{\min }} \;\; =\;\frac{m g+m v_{1}^{2} / l}{m v_{2}^{2} / l-m g}=\;\frac{5}{1}$ $\Rightarrow \;m g+m v_{1}^{2} / l \;\; =5 m v_{2}^{2} / l-5 m g$ $\;\;\text{ Here, }\;\;v_{1} \;\; =\sqrt{v_{2}^{2}+4 g l}$ $\Rightarrow \;m g+\;\frac{m}{l}(v_{2}^{2}+4 g l) \;\; =\;\frac{5 m v_{2}^{2}}{l}-5 m g$ $\Rightarrow \;g+\;\frac{v_{2}^{2}+4 g l}{l} \;\; =\;\frac{5 v_{2}^{2}}{l}-5 g$ $\Rightarrow \;6 g l \;\; =5 v_{2}^{2}-v_{2}^{2}-4 g l \Rightarrow 10 g l=4 v_{2}^{2}$ $\Rightarrow \;v_{2} \;\; =\sqrt{\;\frac{10 g l}{4}}=\sqrt{\;\frac{10 \times 10 \times 1}{4}}$ $\; \;\; =\sqrt{25}=5 {m} / {s}$

Q25JEE Main 2021NAT4MCurrent Electricity

In the given circuit of potentiometer, the potential difference $E$ across $A B$ (10 m length) is larger than $E_{1}$ and $E_{2}$ as well. For key $K_{1}$ (closed), the jockey is adjusted to touch the wire at point $J_{1}$ , so that there is no deflection in the galvanometer. Now, the first battery $(E_{1})$ is replaced by second battery $(E_{2})$ for working by making $K_{1}$ open and $K_{2}$ closed. The galvanometer gives then null deflection at $J_{2}$ . The value of $\;\frac{E_{1}}{E_{2}}$ is $\;\frac{a}{b^{'}}$ , where $a=$ .......... .

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📖 Explanation

Given, length of $A B=10 {m}=1000 {cm}$ Length of one ${arm}=\;\frac{1000}{10}=100 {cm}$ For no deflection, In first case, $l_{1}=3 \times 100+80=380 {cm}$ In 2 nd case, $I_{2}=7 \times 100+60=760 {cm}$ As we know that in balanced potentiometer, $\Rightarrow \;\; \;\frac{a}{b}=\;\frac{380}{760}=\;\frac{1}{2}$ $\therefore \;\; a=1$

Q26JEE Main 2021NAT4MRay Optics and Optical Instruments

The same size images are formed by a convex lens when the object is placed at $20 {cm}$ or at $10 {cm}$ from the lens. The focal length of convex lens is ........... ${cm}$ .

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📖 Explanation

Let $v$ be the position of image, $h$ be the height of image and $h_{o}$ be the height of object. Given, $h_{i}=h_{o}$ Since, magnification, $m=h_{i} / h_{o}=h_{i} / h$ . . . (i) By using lens formula, $1 / f=1 / v-1 / u \Rightarrow 1 / v=1 / f+1 / u$ $\;\; v=\;\frac{f u}{u+f}$ $\;\; m=\;\frac{v}{u}=\;\frac{f}{f+u}$ Now, from Eq. (i), m can be $\pm 1$ . For, $m=+1=\;\frac{f}{-10+f}$ . . . (ii) For $m=-1=\;\frac{f}{-20+f}$ . . . (iii) On dividing Eq. (iii) by Eq. (ii), we get $-1=\;\frac{-10+f}{-20+f}$ $\Rightarrow \;\; 20-f=-10+f \Rightarrow 30=2 f$ $\Rightarrow \;\; f=15 {cm}$

Q27JEE Main 2021NAT4MElectrostatic Potential and Capacitance

512 identical drops of mercury are charged to a potential of $2 {V}$ each. The drops are joined to form a single drop. The potential of this drop is.

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📖 Explanation

Given, number of mercury drops, $n=512$ Voltage of each drop, $V=2 V$ Let $r, R$ be the radius of drop small and combined spherical drop, respectively. Now, when all drops are joined into single drop, volume remains constant, i.e. $\;\; 512 \times \;\frac{4}{3} \pi r^{3}=\;\frac{4}{3} \pi R^{3}$ $\because \;\; V=\;\frac{k q}{r}=2$ . . . (i) $\therefore \;\; V_{\;\text{net }\;}=\;\frac{k Q}{R}$ where, $Q$ be the charge of bigger sphere. $\Rightarrow \;\; V_{\;\text{net }\;}=\;\frac{k \times 512 q}{8 r}$ . . . (ii) On dividing Eq. (ii) by Eq. (i), we get $\;{V_{\;\text{net }\;}}/{V}=\;\frac{k \times 512 q \times r}{8 r \times k q}=\;\frac{512}{8}$ $\;V_{\;\text{net }\;}=2 \times \;\frac{512}{8}=128 {V}$

Q28JEE Main 2021NAT4MElectromagnetic Induction

A coil of inductance $2 {H}$ having negligible resistance is connected to a source of supply whose voltage is given by $V=3 t V$ (where, $t$ is in second). If the voltage is applied when $t=0$ , then the energy stored in the coil after $4 {s}$ is ........... J.

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📖 Explanation

Given, inductance of coil, $L=2 {H}$ Supply voltage, $V=3 t {V}$ Let $E$ be the energy stored in the coil. Since, $\;\;$ emf $V=L \cdot \;\frac{d l}{d t}$ $3 t=L \;\frac{d l}{d t} \Rightarrow 3 t d t=L d l$ $3 t \;\; =L \;\frac{d l}{d t} \Rightarrow 3 t d t=L d l$ $\;\text{ On integrating both sides, we get }\;\;$ $3 \;\frac{t^{2}}{2} \;\; =L I$ At $t=4 {s}$ , $\;\text{ At }\; t=4 {s},\;$ $\;\frac{3}{2} \times 4^{2} \;\; =L I \Rightarrow \;\frac{3}{2} \times \;\frac{16}{L}=I$ $\;\frac{24}{L} \;\; =I$ . . . (i) As, $E \;\; =1 / 2 L I^{2}$ $E \;\; =1 / 2 L \;\frac{24^{2}}{L^{2}} \;\;$ [From Eq. (i)] $=\;\frac{1}{2} \;\frac{24^{2}}{L}=\;\frac{24^{2}}{2 \times 2}$ $\therefore \;\; =144 {J}$

Q29JEE Main 2021NAT4MThermodynamics

A monoatomic gas of mass $4.0 {u}$ is kept in an insulated container. Container is moving with velocity $30 {m} / {s}$ . If container is suddenly stopped, then change in temperature of the gas $(R=$ gas constant $)$ is $\;\frac{x}{3 R} .$ Value of $x$ is ........ .

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📖 Explanation

Given, atomic mass of monoatomic gas, $m=4 u$ Velocity of container, $u=30 {ms}^{-1}$ Final velocity of container, $v=0 {ms}^{-1}$ Let $\Delta T$ be the change in temperature and $\Delta U$ be the internal energy change. Therefore, $\Delta {KE}=\Delta U$ $\;\; =(1 / 2) m(v^{2}-u^{2})=n C_{v} \Delta T$ $\Delta T=\;\frac{1}{2} \;\frac{m(v^{2}-u^{2})}{n \cdot 3 / 2 R}$ $\Rightarrow \;\; \;\; \;\; \;\; \;\; \;\; ( \because C_{v}=\;\frac{3}{2} R)$ $\;\; \;\frac{1}{2} \times m\$ [v^{2}-u^{2}]$=n \cdot ;\frac{3}{2} \cdot R \Delta T;\text{ Now, }; ;\frac{1}{2} \times 4|0^{2}-30^{2}|=1 \times ;\frac{3}{2} R \Delta T\Delta T=;\frac{4 \times 900}{3 R}=;\frac{x}{3 R};; x=3600$

Q30JEE Main 2021NAT4MProperties of Matter

The potential energy (U) of a diatomic molecule is a function dependent on $r$ (interatomic distance) as $U=\;\frac{\alpha}{r^{10}}-\;\frac{\beta}{r^{5}}-3$ where, $\alpha$ and $\beta$ are positive constants. The equilibrium distance between two atoms will be $(\;\frac{2 \alpha}{\beta})^{\;\frac{a}{b}}$ , where $a=. . . . . . . . .$

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📖 Explanation

Given, potential energy, $U=\;\frac{\alpha}{r^{10}}-\;\frac{\beta}{r^{5}}-3$ As we know for equilibrium, differentiation of potential energy with respect to distance $(\;\frac{d u}{d r})=0$ $\Rightarrow \;\; \;\frac{d}{d r}(\;\frac{\alpha}{r^{10}}-\;\frac{\beta}{r^{5}}-3)=0$ $\Rightarrow \;\;$ $\;\frac{d}{d r}(\alpha r^{-10}-betar^{-5}-3) \;\; =0$ $-10 \alpha r^{-11}+5 \beta r^{-6}-0 \;\; =0$ $\Rightarrow \;\; 10 \alpha r^{-11} \;\; =5 \beta r^{-6}$ $\Rightarrow \;\; 2 \alpha r^{-11} \;\; =\beta r^{-6}$ $\Rightarrow \;\; \;\frac{2 \alpha}{\beta}=\;\frac{r^{-6}}{r^{-11}}=r^{5} \Rightarrow r \;\; =(\;\frac{2 \alpha}{\beta})^{1 / 5}$ $\Rightarrow \;\;\;a=1$ $\;\Rightarrow \;\;\frac{d}{d r}(\;\frac{\alpha}{r^{10}}-\;\frac{\beta}{r^{5}}-3) \;\; =0$ $\Rightarrow \;\;\frac{d}{d r}(\alpha r^{-10}-\beta r^{-5}-3) \;\; =0$ $\Rightarrow \; \;\; -10 \alpha r^{-11}+5 \beta r^{-6}-0 \;\; =0$ $\Rightarrow \;10 \alpha r^{-11} \;\; =5 \beta r^{-6}$ $\Rightarrow \;2 \alpha r^{-11} \;\; =betar^{-6}$ $\Rightarrow \;\;\frac{2 \alpha}{\beta}=\;\frac{r^{-6}}{r^{-11}}=r^{5} \Rightarrow r \;\; =(\;\frac{2 \alpha}{\beta})^{1 / 5}$ $\;\Rightarrow \;a \;\; =1$

Chemistry30 questions

Q31JEE Main 2021MCQ4MStructure of Atom

The plots of radial distribution functions for various orbitals of hydrogen atom against ' $r$ ' are given below.The correct plot for $3 s$ -orbital is

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📖 Explanation

The correct plot for $3 s$ -orbital isFor $3 s$ , value of $I=0$ value of $n=3$ Number of peak $=n-I=3-0=3$ In graph $D$ , three peaks are present, so this is the correct plot for 3s-orbital.

Q32JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

According to molecular orbital theory, the species among the following that does not exist is

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📖 Explanation

According to Molecular Orbital Theory (MOT), electronic configuration and their bond order of given options are as follows - ${O}_{2}^{2-}=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2}, \pi 2 p_{y}^{2}, \pi^{*} 2 p_{x}^{2}, \pi^{*} 2 p_{y}^{2}$ [No. of bonding electrons Bond order $=$ $\;r=\;\frac{-\;\text{ No. of antibonding electrons] }\;}{2}$ $\;\; =\;\frac{10-8}{2}=\;\frac{2}{2}=1$ - ${He}_{2}^{-}=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{1}$ Bond order $=\;\frac{3-2}{2}=\;\frac{1}{2}=0.5$ - ${Be}_{2}=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}$ Bond order $=\;\frac{4-4}{2}=0$ - ${He}_{2}^{+}=\sigma 1 s^{2}, \sigma^{*} 1 s^{1}$ Bond order $=\;\frac{2-1}{2}=\;\frac{1}{2}=0.5$ If bond order of chemical species is zero then that chemical species does not exist. Therefore, ${Be}_{2}$ does not exist.

Q33JEE Main 2021MCQ4MIonic Equilibrium

The solubility of AgCN in a buffer solution of ${pH}=3$ is $x$ . The value of $x$ is...... . [Assume : No cyano complex is formed; $K_{ {sp}}( {AgCN})=2.2 \times 10^{-16}$ and $K_{a}( {HCN})=6.2 \times 10^{-10}]$

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📖 Explanation

${pH}$ of ${AgCN}$ buffer solution $=3$ ${[ {H}^{+}]\$ =10^{-3}}K_{ {sp}}( {AgCN})=2.2 × 10^{-16}K_{ {a}}[ {HCN}]=6.2 × 10^{-10}{AgCN} ⇌ {Ag}^{+}+ {CN}^{-} . . . . . . . K_{ {sp}}{CN}^{-}+ {H}^{+} ⇌ {HCN} . . . . . . ;{1}/{K_{ {a}}}{AgCN}+ {H}^{+} ⇌ {HCN}+ {Ag}^{+}K_{ {sp}} × ;{1}/{K_{ {a}}}=;\frac{[ {Ag}^{+}]$$[ {CN}^{-}]$[ {HCN}]}{[ {H}^{+}]$$[ {CN}^{-}]$}{[ {S}]=\sqrt{;{K_{ {sp}}$[ {H}^{+}]$}/{K_{ {a}}}} \Rightarrow ;\frac{2.2 × 10^{-16}}{6.2 × 10^{-10}}=;\frac{[S][S]}{10^{-3}}}{[ {S}]^{2}=;\frac{2.2 × 10^{-16}}{6.2 × 10^{-10}} × 10^{-3}}S=1.9 × 10^{-5}$

Q34JEE Main 2021MCQ4MStates of Matter

In Freundlich adsorption isotherm at moderate pressure, the extent of adsorption $(\;\frac{x}{m})$ is directly proportional to $p^{x}$ . The value of $x$ is

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📖 Explanation

According to Freundlich isotherm, at moderate pressure, extent of adsorption $(\;\frac{x}{m}) \propto (p)^{\;\frac{1}{n}}$ At moderate pressure, $\;\frac{x}{m}=k(p)^{1 / n}$ $\;\frac{x}{m} \propto (p)^{1 / n}$ . . . (i) $\;\frac{x}{m} \propto p^{x} \;\text{ (Given \in question) }\;$ . . . (ii) Compare Eqs. (i) and (ii), $(p)^{1 / n} \propto p^{x}$ $x=\;\frac{1}{n}$

Q35JEE Main 2021MCQ4MChemical Thermodynamics

Ellingham diagram is a graphical representation of

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📖 Explanation

Ellingham diagram is the graphical representation of $\Delta G$ vs $T$ . For the formation of oxides of elements similar diagrams can also be constructed for sulphides and halides. Such diagrams can help us in pridicting the feasibility of thermal reduction of an ore.

Q36JEE Main 2021MCQ4MRedox Reactions

Which of the following equation depicts the oxidising nature of ${H}_{2} {O}_{2}$ ?

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📖 Explanation

Oxidation involves increase in oxidation number and reduction involves decrease in oxidation number. $2 {I}^{-}+2 {H}^{+}+ {H}_{2} { {O}^{1}}_{2} \longrightarrow { {I}^{1}}_{2}+2 { {H}^{+1}}_{2} { {O}^{-2}_{2}}$ In this reaction, ${H}_{2} {O}_{2}$ oxidises ${I}^{-}$ to ${I}_{2}$ and itself gets reduced to ${H}_{2} {O}$ , so the reaction depicts oxidising nature of ${H}_{2} {O}_{2}$ . While in other reactions ${H}_{2} {O}_{2}$ does not oxidise ${KlO}_{4}, {I}_{2}$ and ${Cl}_{2}$ .

Q37JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

The correct statement about ${B}_{2} {H}_{6}$ is

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📖 Explanation

Statement (c) is correct, whereas all other statements are incorrect. Correct statements are as follows - Both B-H-B bridge bond having same bond length. - B-H-B bond angle is $90^{\circ} .$ - ${BH}_{3}$ is electron deficient species and therefore act as Lewis acid.

Q38JEE Main 2021MCQ4Md and f Block Elements

Given below are two statements: Statement I ${CeO}_{2}$ can be used for oxidation of aldehydes and ketones. Statement II Aqueous solution of EuSO $_{4}$ is a strong reducing agent. In the light of the above statements, choose the correct answer from the options given below.

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📖 Explanation

Both statement I and statement II are true. The $+3$ oxidation state of lanthanide is most stable and therefore lanthanide in $+4$ oxidation state has strong tendency to gain electrons and converted into $+3$ and therefore act as strong oxidising agent, e.g. ${Ce}^{4+}$ . $\therefore {CeO}_{2}$ is used to oxidised alcohol, aldehyde and ketones. Lanthanides in $+2$ oxidation state has strong tendency to loss electron and converted into $+3$ oxidation state e.g. ${Eu}^{+2}$ . $\therefore$ EuSO $_{4}$ acts as strong reducing agent.

Q39JEE Main 2021MCQ4MStructure of Atom

In which of the following pairs, the outer most electronic configuration will be the same?

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📖 Explanation

(a) ${V}^{2+}-[ {Ar}] 3 d^{3} ; {Cr}^{+}-[ {Ar}] 3 d^{5}$ (b) ${Cr}^{+}-[ {Ar}] 3 {d}^{5} ; \;\; {Mn}^{2+}-[ {Ar}] 3 {d}^{5}$ (c) ${Ni}^{2+}-[ {Ar}] 3 d^{8} ; \;\; {Cu}^{+}-[ {Ar}] 3 d^{10}$ (d) ${Fe}^{2+}-[ {Ar}] 3 d^{6} ; \;\; {Co}^{+}-[ {Ar}] 3 {d}^{7} 4 s^{1}$ Thus, in option (b), both ions have same outer most electronic configuration.

Q40JEE Main 2021MCQ4MCoordination Compounds

The hybridisation and magnetic nature of $[ {Mn}( {CN})_{6}]^{4-}$ and $[ {Fe}( {CN})_{6}]^{3-}$ , respectively are

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📖 Explanation

$[ {Mn}( {CN})_{6}]^{4-}$ In this complex, oxidation state of ${Mn}$ is $+2$ Hybridisation- $d^{2} s p^{3}$ Magnetic nature-Paramagnetic $[ {Fe}( {CN})_{6}]^{-3}$ Hybridisation- $d^{2} s p^{3}$ Magnetic nature-Paramagnetic

Q41JEE Main 2021MCQ4Mp Block Elements

Given below are two statements: Statement I An allotrope of oxygen is an important intermediate in the formation of reducing smog. Statement II Gases such as oxides of nitrogen and sulphur present in troposphere contribute to the formation of photochemical smog. In the light of the above statements, choose the correct answer from the options given below.

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📖 Explanation

Reducing smog is a mixture of smoke, fog and sulphur dioxide. Therefore, statement $I$ is false. Tropospheric pollutants such as nitrogen oxide and sulphur oxide contribute to the formation of photochemical smog. So, statement II is true.

Q42JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

Complete combustion of $1.80 {g}$ of an oxygen containing compound $( {C}_{x} {H}_{y} {O}_{z})$ gave $2.64 {g}$ of ${CO}_{2}$ and $1.08 {g}$ of ${H}_{2} {O}$ . The percentage of oxygen in the organic compound is

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📖 Explanation

${{C}_{x} {H}_{y} {O}_{z}}_{1.8 {g}}+ {O}_{2} \longrightarrow { {CO}_{2}}_{2.64 g}+ {{H}_{2} {O}}_{1.08 g}$ $\;n_{ {C}}=\;{ {CO}}_{\;\text{ (Moles) }\;_{2}}=\;\frac{2.64 \;\text{ (Given mass) }\;}{44 \;\text{ (Molecular mass) }\;}=0.06$ $\;n_{ {H}}=2 \times n_{ {H}_{2} {O}}=\;\frac{1.08}{18} \times 2=0.12$ Weight of oxygen in ${C}_{x} {H}_{y} {O}_{z}$ $\;\; =1.80-12 \times \;\frac{2.64}{44}-\;\frac{1.08}{18} \times 2$ $\;\; =1.80-0.72-0.12=0.96 {g}$ $\%$ of oxygen by weight $=\;\frac{0.96}{1.80} \times 100=53.33 \%$

Q43JEE Main 2021MCQ4MHydrocarbons

Identify $A$ in the given chemical reaction.

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📖 Explanation

Oxides of molybdenum, vanadium and chromium, i.e. ${Mo}_{2} {O}_{3}, {V}_{2} {O}_{5}$ and ${Cr}_{2} {O}_{3}$ work as catalyst. When they are operated under certain temperature and pressure with $n$ -heptane then they form toluene which is aromatic compound. ${Mo}_{2} {O}_{3}$ behave as aromatising agent.

Q44JEE Main 2021MCQ4MHaloalkanes and Haloarenes

Identify $A$ and $B$ in the chemical reaction.

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📖 Explanation

Ist reaction is Markownikoff's addition of ${HCl}$ on double bond while 2 nd reaction is halide substitution by Finkelstein reaction in which chlorine get displaced by iodine.

Q45JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

The major product of the following chemical reaction is ${\text{CH}}_{3} {\text{CH}}_{2} {\text{CN}}{ \to }^{(ii) {\text{SOCl}}_{2}}^{(i) {H}_{3} {O}^{+}, \Delta }_{(iii) {Pd} / {\text{BaSO}}_{4}, {H}_{2}}$ ?

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Q46JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Which one of the following reactions will not form acetaldehyde?

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📖 Explanation

Since, ${CrO}_{3} \cdot {H}_{2} {SO}_{4}$ behave as strong oxidising agent and it converts alcohol directly to carboxylic acid. Thus, reaction (b) will not form acetaldehyde.

Q47JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are

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📖 Explanation

The reactions of given compound with sodium bicarbonate solution are as followsAs ${H}_{2} {CO}_{3}$ is weak acid it dissociate to liberate ${CO}_{2}$ gas and ${H}_{2} {O}$ .Equilibrium favours forward direction and ${CO}_{2}$ is liberated. In the above two reactions, ${H}_{2} {CO}_{3}$ is comparatively weak acid.Equilibrium favours backward direction and ${CO}_{2}$ is not liberated. Thus, only $B$ and $C$ will liberate carbon dioxide with sodium bicarbonate solution.

Q48JEE Main 2021MCQ4MAmines

Which of the following reaction(s) will not give $p$ -aminoazobenzene?

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📖 Explanation

Nitrobenzene in presence of ${NaBH}_{4}, {NaOH}$ and aniline will not give $p$ -amino azobenzene.In case of $(A)$ and $(C)$ , coupling reaction takes place as the medium is quite acidic follows

Q49JEE Main 2021MCQ4MBiomolecules

Which statement is correct?

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📖 Explanation

Only statement (d) is correct whereas other statements are incorrect. The correct statements are as follows Buna- ${N}$ is a synthetic polymer. Buna- $S$ is an elastomer. Buna- $S$ is polymerised by addition polymerisation method, which needs radical initiator for chain propagation step. Nascent oxygen can be used as an radical initiator.Neoprene is a synthetic rubber not an addition copolymer.

Q50JEE Main 2021MCQ4MBiomolecules

Which of the glycosidic linkage between galactose and glucose is present in lactose?

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📖 Explanation

Glycosidic linkage is a type of covalent bond that joins carbohydrate molecules to another group. In lactose, glycosidic linkage is formed between C-1 of galactose and $C-4$ , glycosidic of glucose.

Q51JEE Main 2021NAT4MIonic Equilibrium

$0.4 {g}$ mixture of ${NaOH}, {Na}_{2} {CO}_{3}$ and some inert impurities was first titrated with $\;\frac{ {N}}{10} {HCl}$ using phenolphthalein as an indicator, $17.5 {mL}$ of ${HCl}$ was required at the end point. After this methyl orange was added and titrated. $1.5 {mL}$ of same ${HCl}$ was required for the next end point. The weight percentage of ${Na}_{2} {CO}_{3}$ in the mixture is $. . . . . . . . . . . . . .$ (Rounded off to the nearest integer).

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📖 Explanation

As given, ${NaOH}$ and ${Na}_{2} {CO}_{3}$ is titrated with ${N} / 10 {HCl}$ . For ${NaOH}$ , Equivalents of ${NaOH}=$ Equivalents of ${HCl}$ Equivalents of ${HCl}=$ Normality $\times$ Volume $( {L})$ $=0.1 \times \;\frac{17.5}{1000}$ Equivalent of ${HCl}=1.75 \times 10^{-3}$ Equivalents of ${NaOH}=1.75 \times 10^{-3}$ Weight of ${NaOH}=$ Equivalent of ${NaOH} \times$ equivalent weight of $=40 \times 1.75 \times 10^{-3}=0.07 {g}$ ${NaOH}$ Now, weight % of ${NaOH}$ $=\;\frac{0.07}{0.4} \times 100=\;\frac{70}{4}=17.5 \%$ Similarly for ${Na}_{2} {CO}_{3}$ , Equivalent of ${HCl}=$ Equivalent of ${Na}_{2} {CO}_{3}$ Equivalent of ${Na}_{2} {CO}_{3}=0.1 \times \;\frac{1.5}{1000}$ $=0.15 \times 10^{-3}$ Weight of ${Na}_{2} {CO}_{3}=$ Equivalent of ${Na}_{2} {CO}_{3}$ $\times$ equivalent weight of ${Na}_{2} {CO}_{3}$ $=0.15 \times 10^{-3} \times 106=15.9 \times 10^{-3} {g}$ Weight % of ${Na}_{2} {CO}_{3}=\;\frac{15.9 \times 10^{-3}}{0.4} \times 100$ $=0.039 \times 100=3.9 \% ∼eq 4 \%$

Q52JEE Main 2021NAT4MStates of Matter

A car tyre is filled with nitrogen gas at 35 psi at $27^{\circ} {C}$ . It will burst if pressure exceeds 40 psi. The temperature in ${ }^{\circ} {C}$ at which the car tyre will burst is ........ (Rounded-off to the nearest integer).

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📖 Explanation

$p_{1}=35 {\psi}, T_{1}=27^{\circ} {C}=300 {K}$ $p_{2}=40 {\psi}, T_{2}=\;\text{ ? }\;$ According to Charle's law, $p\;\propto T$ $\;\frac{p_{2}}{p_{1}} \;\; =\;\frac{T_{2}}{T_{1}} \Rightarrow \;\frac{40}{35}=\;\frac{T_{2}}{300}$ $T_{2} \;\; =\;\frac{300 \times 40}{35}=342.85 {K}=69.7^{\circ} {C}$ $T_{2}\;∼eq 70^{\circ} {C}$ Hence, answer is 70 .

Q53JEE Main 2021NAT4MChemical Thermodynamics

The reaction of cyanamide, ${NH}_{2} {CN}( {s})$ with oxygen was run in a bomb calorimeter and $\Delta U$ was found to be $-742.24 {kJ} {mol}^{-1}$ . The magnitude of $\Delta {H}_{298}$ for the reaction ${NH}_{2} {CN}( {s})+\;\frac{3}{2} {O}_{2}(g) \longrightarrow {N}_{2}(g)+ {O}_{2}(g)+ {H}_{2} {O}(l)$ is ............. ${kJ}$ (Rounded off to the nearest integer). $[Assume ideal gases and $R=8.314 {J} {mol}^{-1} {K}^{-1}$ ]$

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📖 Explanation

$\Delta U=-742.24 {kJ} {mol}^{-1}$ $\Delta n_{g}=$ [Number of gaseous molecules of products - Number of gaseous molecules of reactants] ${NH}_{2} {CN}( {s})+\;\frac{3}{2} {O}_{2}( {g}) \longrightarrow {N}_{2}( {g})+ {O}_{2}( {g})+ {H}_{2} {O}(l)$ $\Delta n_{ {g}}=2-\;\frac{3}{2}=\;\frac{1}{2}$ $\Delta {H}=\Delta U+\Delta n_{g} R T$ $=-742.24+\;\frac{1}{2} \times \;\frac{8.314}{1000} \times 298$ $=-741 {kJ} / {mol}$ Hence, answer is 741 .

Q54JEE Main 2021NAT4MSolutions

1 molal aqueous solution of an electrolyte $A_{2} B_{3}$ is $60 \%$ ionised. The boiling point of the solution at $1 {atm}$ is ........... K (Rounded off to the nearest integer). $[Given, $K_{b}$ for $( {H}_{2} {O})$ $=0.52 {K} {kg} {mol}^{-1}$ ]$

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📖 Explanation

Given, $K_{b}( {H}_{2} {O})=0.52 {Kkg} {mol}^{-1}$ $A_{2} B_{3} \longrightarrow 2 A^{+3}+3 B^{-2}$ No. of ions, $n=5$ , concentration, $m=1$ molal (Given) $\alpha \;\; =\;\frac{60}{100}=0.6$ $\Delta T_{b} \;\; =i \cdot K_{f} \cdot m$ $\;\; =[(1+(n-1) \alpha] \times K_{f} \times m$ $\Delta T_{b} \;\; =[(1+(5-1) 0.6] \times 0.52 \times 1$ $\;\; =(1+2.4) \times 0.52$ $\Delta T_{b} \;\; =1.768$ $T_{b} \;\; =1.768+373.15$ $\;\; =374.91=375 {K}$

Q55JEE Main 2021NAT4MRedox Reactions

In basic medium ${CrO}_{4}^{2-}$ oxidises ${S}_{2} {O}_{3}^{2-}$ to form ${SO}_{4}^{2-}$ and itself changes into ${Cr}( {OH})_{4}^{-}$ . The volume of $0.154 {M} {CrO}_{4}^{2-}$ required to react with $40 {mL}$ of $0.25 {MS}_{2} {O}_{3}^{2-}$ is ........... ${mL}$ (Rounded off to the nearest integer).

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📖 Explanation

Given, Molarity of ${CrO}_{4}^{2-}(M_{1})=0.154 {M}$ Molarity of ${S}_{2} {O}_{3}^{2-}(M_{2})=0.25 {M}$ Volume of ${S}_{2} {O}_{3}^{2-}( {V}_{2})=40 {mL}$ Volume of ${CrO}_{4}^{2-}(V_{1})=$ ? ${Cr} {O}_{4}+ {S}_{2} {O}_{3} \longrightarrow {SO}_{4}^{2-}+ {Cr}( {OH})_{4}^{+6}$ Gram equivalent of ${CrO}_{4}^{2-}=$ Gram equivalent of ${S}_{2} {O}_{3}^{2-}$ $N_{1} V_{1}=N_{2} V_{2}$ Normality $=$ Molarity $\times n$ factor $n$ for ${Cr}=6-3=3$ $n$ for ${S} \Rightarrow {S}_{2} {O}_{3}^{2-} \longrightarrow 2 {SO}_{4}^{2-}+8 {e}^{-}$ $\;\; =8$ $0.154 \times 3 \times V_{1} \;\; =0.25 \times 40 \times 8$ $V_{1} \;\; =173 {mL}$

Q56JEE Main 2021NAT4MChemical Kinetics

For the reaction, $a A+b B \to c C+d D$ , the plot of $\log k v s \;\frac{1}{T}$ is given belowThe temperature at which the rate constant of the reaction is $10^{-4} {s}^{-1}$ is ......... K (Rounded off to the nearest integer). $[Given : The rate constant of the reaction is $10^{-5} {s}^{-1}$ at $500 {K}$ ]$

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📖 Explanation

According to Arrhenius equation, $\log K \;\; =\log A-\;\frac{E_{a}}{2.303 R T}$ $\;\text{ Given, }\; \;\text{ Slope }\; \;\; =\;\frac{E_{a}}{2.303 R}=10,000$ $K_{1} \;\; =10^{-5}, K_{2}=10^{-4}$ $T_{1} \;\; =500 {K}$ $\log _{10}[\;\frac{K_{1}}{K_{2}}] \;\; =\;\frac{E a}{2.303 R}[\;\frac{1}{T_{1}}-\;\frac{1}{T_{2}}]$ $\log \$ [;\frac{10^{-4}}{10^{-5}}]$ ;; =10000[;\frac{1}{500}-;\frac{1}{T_{2}}]\log 10 ;; =10000[;\frac{1}{500}-;\frac{1}{T_{2}}]T_{2} ;; =526.31 ∼eq 526 {K}$

Q57JEE Main 2021NAT4MChemical Thermodynamics

The ionisation enthalpy of ${Na}^{+}$ formation from ${Na}(g)$ is $495.8 {kJ} {mol}^{-1}$ , while the electron gain enthalpy of ${Br}$ is $-325.0 {kJ} {mol}^{-1}$ . Given, the lattice enthalpy of ${NaBr}$ is $-728.4 {kJ} {mol}^{-1}$ . The energy for the formation of ${NaBr}$ ionic solid is $(-) . . . . . . . . . . . . \times 10^{-1} {kJ} {mol}^{-1}$ .

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📖 Explanation

${Na}(g) {\longrightarrow }^{ {IE}_{1}} {Na}^{+}(g)[ {IE}_{1}=495.8 {kJ} {mol}^{-1}]$ ${Br}( {g}) {\to }_{\text{gain} \text{enthalpy}}^{\;\text{ Electron }\;}(E G) {Br}^{-}( {g})[ {Gain}$ enthalpy $=-325 {kJ} {mol}^{-1}]$ ${Na}^{+}+ {Br}^{-} {\to }_{\text{energy} }^{\;\text{ Lattice }\;}(L E {NaBr}(s)[$ Lattice energy $=-728.4 {kJ} {mol}^{-1}]$ $\Delta H_{\;\text{Formation }\;}= {IE}_{1}+$ Gain enthalpy $+$ Lattice energy $\Delta H \;\; =495.8+(-325)+(-728.4)$ $\;\; =-557.6 {kJ} / {mol}$ $\;\; =-5576 \times 10^{-1} {kJ}$

Q58JEE Main 2021MCQ4Mp Block Elements

Among the following, the number of halide(s) which is/are inert to hydrolysis is

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📖 Explanation

(1) ${\text{BCl}}_3+3 {\text{H}}_2 {\text{O}} \longrightarrow {\text{H}}_3 {\text{BO}}_3+3 {\text{HCl}}$ (2) ${\text{SiCl}}_4+4 {\text{H}}_2 {\text{O}} \longrightarrow {\text{SiO}}_4 {\text{H}}_4+4 {\text{HCl}}$ $(3) {\text{PCl}}_5+ {\text{H}}_2 {\text{O}} \longrightarrow {\text{POCl}}_3+2 {\text{HCl}}$ (4) Due to steric crowding by six F-atoms, ${\text{SF}}_6$ is inert towards hydrolysis.

Q59JEE Main 2021NAT4MAldehydes, Ketones and Carboxylic Acids

Consider the following chemical reaction. ${CH} \equiv {CH} {\to }_{(2) \; {\text{CO}}, {\text{HCl}}, {\text{AlCl}}_{3}}^{\;\text{ (1) Red hot Fe tube, }\; 873 {K}}$ Product The number of $s p^{2}$ hybridised carbon atom(s) present in the product is .............

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📖 Explanation

The reaction take place as follows Every carbon atom found in benzaldehyde (6 carbon in benzene ring $+1$ in aldehyde group), forms three bonds with neighbouring atoms-(one double bond + two single bond). Therefore, in benzaldehyde total number of $s p^{2}{ }^{'} C^{'}$ are 7 .

Q60JEE Main 2021NAT4MPractical Organic Chemistry

Using the provided information in the following paper chromatogram. The calculated $R_{f}$ value of $A$ $\times 10^{-1}$

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📖 Explanation

The retention factor $(R_{f})$ value can be used to identify the components of a mixture (solutes). $R_{f}=\;\frac{\;\text{ Distance travelled by solute }\;}{\;\text{ Distance travelled by solvent front }\;}$ On chromatogram, distance travelled by compound is $2 {cm}$ . Distance travelled by solvent $=5 {cm}$ $\therefore \;\; R_{f}=\;\frac{2}{5}=4 \times 10^{-1}$

Mathematics30 questions

Q61JEE Main 2021MCQ4MProbability

When a missile is fired from a ship, the probability that it is intercepted is $\;\frac{1}{3}$ and the probability that the missile hits the target, given that it is not intercepted, is $\;\frac{3}{4}$ . If three missiles are fired independently from the ship, then the probability that all three hit the target, is

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📖 Explanation

Probability of missile to get intercepted $=\;\frac{1}{3}$ $\therefore$ Probability of missile to not get intercepted $=1-\;\frac{1}{3}=\;\frac{2}{3}$ Probability of missile to hit the target $=\;\frac{3}{4}$ $\therefore$ Probability of three missiles to hit the target $=(\;\frac{2}{3} \times \;\frac{3}{4}) \times (\;\frac{2}{3} \times \;\frac{3}{4}) \times (\;\frac{2}{3} \times \;\frac{3}{4})=\;\frac{1}{8}$

Q62JEE Main 2021MCQ4MSequences and Series

If $0< \theta, \varphi < \;\frac{\pi}{2}, x=sum_{n=0}^{\infty } \cos ^{2 n} \theta, y=sum_{n=0}^{\infty } \sin ^{2 n} \varphi$ and $z=sum_{n=0}^{\infty } \cos ^{2 n} \theta \cdot \sin ^{2 n} \varphi$ , then

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📖 Explanation

Given, $x=sum_{n=0}^{\infty } \cos ^{2 n} \theta$ $\;y \;\; =sum_{n=0}^{\infty } \sin ^{2 n} \varphi$ $\;z=sum_{n=0}^{\infty } \cos ^{2 n} \theta \cdot \sin ^{2 n} \varphi$ $\Rightarrow \;$ $x \;\; =1+\cos ^{2} \theta+\cos ^{4} \theta+. . . \infty$ $\;x=\;\frac{1}{1-\cos ^{2} \theta}$ . . . (i) $\;1-\cos ^{2} \theta=\;\frac{1}{x} \Rightarrow \cos ^{2} \theta=1-\;\frac{1}{x}$ $\Rightarrow \;y=1+\sin ^{2} \varphi +\sin ^{4} \varphi +. . . \infty$ $\therefore \;y=\;\frac{1}{1-\sin ^{2} \varphi }$ . . . (ii) $\;1-\sin ^{2} \varphi =\;\frac{1}{y} \Rightarrow \sin ^{2} \varphi =1-\;\frac{1}{y}$ $\Rightarrow \;z=1+\cos ^{2} \theta \cdot \sin ^{2} \varphi +\cos ^{4} \theta \sin ^{4} \varphi +. . . \infty$ $\therefore \;z=\;\frac{1}{1-\cos ^{2} \theta \sin ^{2} \varphi }$ . . . (iii) From Eqs. (i), (ii) and (iii), we get $\;z=\;\frac{1}{1-(1-\;\frac{1}{x})(1-\;\frac{1}{y})}$ $\begin{bmatrix}\because \cos ^{2} \theta=1-\\ \\ \frac{1}{x} \\ \because \sin ^{2} \varphi =1-\\ \\ \frac{1}{y}\end{bmatrix}$ $\;z=\;\frac{x y}{x y-(x-1)(y-1)}$ $\;z=\;\frac{x y}{x y-x y+x+y-1}$ $\Rightarrow \;\; x z+y z-z=x y$ $\Rightarrow \;\; x y+z=(x+y) z$

Q63JEE Main 2021MCQ4MFunctions

Let $f, g: N \to N$ , such that $f(n+1)=f(n)+f(1) ∀ n \in N$ and $g$ be any arbitrary function. Which of the following statements is not true?

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📖 Explanation

Given, $f(n+1)=f(n)+f(1), ∀ n \in N$ $\Rightarrow \;\; f(n+1)-f(n)=f(1)$ It is an AP with common difference $=f(1)$ Also, general term $=T_{n}=f(1)+(n-1) f(1)=n f(1)$ $\Rightarrow \;\; f(n)=n f(1)$ Clearly, $f(n)$ is one-one. For fog to be one-one, $g$ must be one-one. For $f$ to be onto, $f(n)$ should take all the values of natural numbers. As, $f(x)$ is increasing, $f(1)=1$ $\Rightarrow \;\; f(n)=n$ If $g$ is many-one, then fog is many one. So, if $g$ is onto, then fog is onto.

Q64JEE Main 2021MCQ4MThree Dimensional Geometry

The equation of the line through the point $(0,1,2)$ and perpendicular to the line $\;\frac{x-1}{2}=\;\frac{y+1}{3}=\;\frac{z-1}{-2}$ is

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📖 Explanation

Given, line $\Rightarrow \;\frac{x-1}{2}=\;\frac{y+1}{3}=\;\frac{z-1}{-2}=\lambda$ (let) Any point on this line is $B(2 \lambda+1,3 \lambda-1,-2 \lambda+1)$ and direction ratios of this line $=(2,3,-2)=d_{1}$ Let given point be $A(0,1,2)$ . Then direction ratio of $A B=(2 \lambda+1,3 \lambda-2,-2 \lambda-1)=d_{2}$ $\because$ Both lines are perpendicular to each other. $\therefore \;\; d_{1} \cdot d_{2}=0$ $\;2(2 \lambda+1)+3(3 \lambda-2)-2(-2 \lambda-1)=0$ $\Rightarrow \;\; 4 \lambda+2+9 \lambda-6+4 \lambda+2=0$ $\Rightarrow \;\; 17 \lambda=2$ $\;\lambda \;\; \lambda=2 / 17$ $\therefore \;\text{ Direction ratio of required line }\; d_{2}=(21,-28,-21)$ $\;\; =(3,-4,-3)=(-3,4,3)$ This line passes through $A(0,1,2)$ . $\therefore$ Required equation of line $\Rightarrow \;\frac{x-0}{-3}=\;\frac{y-1}{4}=\;\frac{z-2}{3}$

Q65JEE Main 2021MCQ4MThree Dimensional Geometry

Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations $I+m-n=0$ and $I^{2}+m^{2}-n^{2}=0$ . Then, the value of $\sin ^{4} \alpha+\cos ^{4} \alpha$ is

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📖 Explanation

Given, $l+m-n=0$ . . . (i) and $I^{2}+m^{2}-n^{2}=0 \;\; . . .$ (ii) On squaring Eq. (i), we get $(l+m)^{2}=n^{2}$ $\Rightarrow \;\; l^{2}+m^{2}+2 l m=n^{2}$ . . . (iii) From Eqs. (ii) and (iii), $I^{2}+m^{2}-n^{2}=0$ $I^{2}+m^{2}+2 l m=n^{2}$ $\;\frac{-- \;\; - \;\; -}{-n^{2}-2 l m=-n^{2}}$ $\Rightarrow \;\; 2 l m=0 \Rightarrow I m=0$ $\Rightarrow \;\; I=0 \;\text{ or }\; m=0$ $\;\text{ Case I When I = 0 }\;$ $\Rightarrow \;\; 0+m-n=0$ $\Rightarrow \;\; m=n$ $\;\text{ and }\; I^{2}+m^{2}+n^{2}=1$ $\Rightarrow \;\; m^{2}+m^{2}=1 \;\;$ $[ \because n=m$ and $l=0]$ $\Rightarrow \;\; m^{2}=\;\frac{1}{2}$ $\;\;\; m=\pm \;\frac{1}{\sqrt{2}}=n$ $\therefore \;\; (I, m, n)=(0, \;\frac{1}{\sqrt{2}}, \;\frac{1}{\sqrt{2}}) \;\text{ or }\;(0, \;\frac{-1}{\sqrt{2}}, \;\frac{-1}{\sqrt{2}})$ Case II When $m=0$ then, $l+m-n=0$ $\Rightarrow \;\; I=n$ and $l^{2}+m^{2}+n^{2}=1$ $[ \because n=I$ and $m=0]$ $\Rightarrow \;\; I^{2}+0+I^{2}=1$ $I=\pm \;\frac{1}{\sqrt{2}}\;\;$ $[ \because n=l$ and $m=0]$ $\Rightarrow$ $\therefore \;\; (I, m, n)=(\;\frac{1}{\sqrt{2}}, 0, \;\frac{1}{\sqrt{2}})$ or $(\;\frac{-1}{\sqrt{2}}, 0, \;\frac{-1}{\sqrt{2}})$ $\Rightarrow \;\; a=(0, \;\frac{1}{\sqrt{2}}, \;\frac{1}{\sqrt{2}})$ and $b=(\;\frac{1}{\sqrt{2}}, 0, \;\frac{1}{\sqrt{2}})$ Then $\cos \alpha=\;\frac{a \cdot b}{|a||b|}=\;\frac{1}{2}$ $\therefore \sin \alpha=\pm \;\frac{\sqrt{3}}{2}$ Now, $\cos ^{4} \alpha+\sin ^{4} \alpha=\;\frac{1}{16}+\;\frac{9}{16}=\;\frac{10}{16}=\;\frac{5}{8}$

Q66JEE Main 2021MCQ4MIndefinite Integrals

The value of the integral $\int{\sin\theta \; \sin\2\theta(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}/{ 1-\cos2theta}\text{d}\theta$ is (where, $c$ is a constant of integration)

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📖 Explanation

Let $\int\$ [{\sin theta ⋅ \sin 2 theta(\sin ^{6} theta+\sin ^{4} theta+\sin ^{2} theta) \sqrt{2 \sin ^{4} theta+3 \sin ^{2} theta+6}}/{ 1-\cos2theta}]$\text{d}\theta\because ;; \sin 2 A=2 \sin A \cos A $and$ 1-\cos 2 A=2 \sin ^{2} A\sin \theta \cdot 2 \sin \theta(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta)I=\int ;{\sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}/{2 \sin ^{2} \theta} d \thetaI=\int \cos \theta(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta)\sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6} d \theta=\int(t^{6}+t^{4}+t^{2}) \sqrt{2 t^{4}+3 t^{2}+6} d t=\int(t^{5}+t^{3}+t) \sqrt{2 t^{6}+3 t^{4}+6 t^{2}} d t $Let$ 2 t^{6}+3 t^{4}+6 t^{2}=z\therefore ;; d z=(12 t^{5}+12 t^{3}+12 t) d t\therefore ;; d z=12(t^{5}+t^{3}+t) d t $Now,$ ;\frac{1}{12} \int \sqrt{z} d z=;\frac{1}{12} \times ;\frac{z^{3 / 2}}{3 / 2}+c=;\frac{1}{18} z^{3 / 2}+c=;\frac{1}{18}$[2 t^{6}+3 t^{4}+6 t^{2}]^{3 / 2}+c=;\frac{1}{18}$[2 \sin ^{6} \theta+3 \sin ^{4} \theta+6 \sin ^{2} \theta]^{3 / 2}+c=;\frac{1}{18}$[(1-\cos ^{2} theta){2(1-\cos ^{2} theta)^{3}+3 -3 \cos ^{2} theta+6}]^{3 / 2}+c=;\frac{1}{18}$[(1-\cos ^{2} \theta)(2 \cos ^{4} \theta -7 \cos ^{2} \theta+11)]^{3 / 2}+c=;\frac{1}{18}$[-2 \cos ^{6} theta+9 \cos ^{4} theta-18 \cos ^{2} theta+11]^{3 / 2} +c=;\frac{1}{18}$[11-18 \cos ^{2} \theta+9 \cos ^{4} \theta-2 \cos ^{6} \theta]^{3 / 2}$

Q67JEE Main 2021MCQ4MDefinite Integrals

The value of $int_{-1}^{1} x^{2} e^{[x^{3}]\$ } d x $, where$ [t] $denotes the greatest integer$ ≤ t$, is

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📖 Explanation

Given, $int_{-1}^{1} x^{2} e^{[x^{3}]\$ } d x $, where$ [t] $is greatest integer function.$ \because ;; $[x^{3}]$=0 ∀ x in(0,1) $and$ [x^{3}]=-1 ∀ x in(-1,0) $So,$ int_{-1}^{1} x^{2} e^{[x^{3}]$} d x=int_{-1}^{0} x^{2} e^{-1} d x+int_{0}^{1} x^{2} e^{0} d x;; =;\frac{1}{e} int_{-1}^{0} x^{2} d x+int_{0}^{1} x^{2} d x=;\frac{1}{e} \times $[;\frac{x^{3}}{3}]{-1}^{0}+$[;\frac{x^{3}}{3}]{0}^{1};; =;\frac{1}{e} \times [0+;\frac{1}{3}]+[;\frac{1}{3}]=;\frac{1}{3 e}+;\frac{1}{3}=(;\frac{1+e}{3 e})$

Q68JEE Main 2021MCQ4MTrigonometric Ratios and Identities

A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point $A$ , with uniform speed. At that point, angle of depression of the boat with the man's eye is $30^{\circ}$ (ignore man's height). After sailing for $20 {s}$ , towards the base of the tower (which is at the level of water), the boat has reached a point $B$ , where the angle of depression is $45^{\circ}$ . Then, the time taken (in seconds) by the boat from $B$ to reach the base of the tower is

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📖 Explanation

$\ln \Delta P Q B, \tan 45^{\circ} \;\; =\;\frac{P Q}{B Q}$ $1 \;\; =\;\frac{P Q}{B Q}$ $P Q \;\; =B Q=h(\;\text{ let }\;)$ In $\triangle P A Q, \tan 30^{\circ}=\;\frac{P Q}{A Q}$ $\Rightarrow \;\; \;\frac{1}{\sqrt{3}}=\;\frac{h}{x+h} \Rightarrow x+h=\sqrt{3} h$ $\Rightarrow \;\; x=(\sqrt{3}-1) h$ $\because \;\;$ Speed $=\;\frac{\;\text{ Distance }\;}{\;\text{ Time }\;}$ $\therefore \;\text{ Speed from }\; A \;\text{ to }\; B=\;\frac{A B}{20}=\;\frac{x}{20}\;$ $\;\text{ Also, distance from }\; B \;\text{ to }\; Q=h\;$ $\therefore \;\text{ Time taken to reach }\; Q \;\text{ from }\; B=\;\frac{B Q}{\;\text{ Speed }\;}\;$ $\;=\;\frac{h}{x / 20}=\;\frac{h \times 20}{x}\;{[ \because x=(\sqrt{3}-1) h]}$ $\;=\;\frac{h}{(\sqrt{3}-1) h} \times 20\;$ $\;=\;\frac{(\sqrt{3}+1) \times 20}{(\sqrt{3}-1)(\sqrt{3}+1)}=\;\frac{(\sqrt{3}+1) \times 20}{2}\;$ $\;=10(\sqrt{3}+1)\;$

Q69JEE Main 2021MCQ4MParabola

A tangent is drawn to the parabola $y^{2}=6 x$ , which is perpendicular to the line $2 x+y=1$ . Which of the following points does not lie on it?

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📖 Explanation

Given, parabola $\Rightarrow y^{2}=6 x$ $\Rightarrow \;\; y^{2}=4(\;\frac{3}{2}) x$ and given, line $\Rightarrow 2 x+y=1 .$ $\because$ Equation of any tangent to the parabola having slope $m$ is $y=m x+3 / 2 m$ Slope of line $2 x+y=1$ is $m_{1}=-2$ $\because$ Tangent is perpendicular to this line, $\therefore$ Slope of tangent $=m_{2}=-\;\frac{1}{m}=\;\frac{1}{2}$ $\therefore$ Equation of tangent will be $y=\;\frac{1}{2} x+\;\frac{3}{2} \times 2$ $y=\;\frac{x}{2}+3$ or $\;\; 2 y=x+6$ or $\;\; x-2 y+6=0$ Clearly, on putting the coordinates of point $(5,4)$ , the equation of tangent is not satisfied. $\therefore$ Point $(5,4)$ does not lie on this tangent.

Q70JEE Main 2021MCQ4MTrigonometric Ratios and Identities

All possible values of $\theta \in[0,2 \pi]$ for which $\sin 2 \theta+\tan 2 \theta>0$ lie in

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📖 Explanation

$\sin 2 \theta+\tan 2 \theta > 0$ $\Rightarrow \;\; \sin 2 \theta+\;\frac{\sin 2 \theta}{\cos 2 \theta} > 0$ $\Rightarrow \;\; \sin 2 \theta(1+\;\frac{1}{\cos 2 \theta}) > 0$ $\Rightarrow \;\; \sin 2 \theta(\;\frac{\cos 2 \theta+1}{\cos 2 \theta}) > 0$ $\Rightarrow \;\; \tan 2 \theta(2 \cos ^{2} \theta) > 0$ $\cos 2 \theta \ne 0$ $\Rightarrow \;\; 1-2 \sin ^{2} \theta \ne 0$ $\sin \theta \ne \pm \;\frac{1}{\sqrt{2}}$ Now, $\tan 2 \theta(1+\cos 2 \theta) > 0$ $\Rightarrow \tan 2 \theta > 0$ as $1+\cos 2 \theta > 0$ $\Rightarrow 2 \theta \in(0, \;\frac{\pi}{2}) \cup(\pi, \;\frac{3 \pi}{2}) \cup(2 \pi, \;\frac{5 \pi}{2}) \cup(3 \pi, \;\frac{7 \pi}{2})$ $\therefore \theta \in(0, \;\frac{\pi}{4}) \cup(\;\frac{\pi}{2}, \;\frac{3 \pi}{4}) \cup(\pi, \;\frac{5 \pi}{4}) \cup(\;\frac{3 \pi}{2}, \;\frac{7 \pi}{4})$ Since, $\sin \theta \ne \pm \;\frac{1}{\sqrt{2}}]$

Q71JEE Main 2021MCQ4MComplex Numbers

Let the lines $(2-i) z=(2+i) \overline{z}$ and $(2+i) z+(i-2) \overline{z}-4 i=0$ , (here $i^{2}=-1$ ) be normal to a circle $C$ . If the line $i z+\overline{z}+1+i=0$ is tangent to this circle $C$ , then its radius is

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📖 Explanation

Given, $(2-i) z=(2+i) \overline{z}$ Let $z=x+i y$ , then $\overline{z}=x-i y$ $\Rightarrow \;\; (2-i)(x+i y)=(2+i)(x-i y)$ $\Rightarrow 2 x-i x+2 i y+y=2 x+i x-2 i y+y$ $\Rightarrow \;\; 2 i x-4 i y=0$ $\therefore$ Equation of line $L_{1} \Rightarrow x-2 y=0$ . . . (i) Also, $(2+i) z+(i-2) \overline{z}-4 i=0$ $\Rightarrow \;\; (2+i)(x+i y)+(i-2)(x-i y)-4 i=0$ $\Rightarrow \;\; 2 x+i x+2 i y-y+i x-2 x+y+2 i y-4 i=0$ $\Rightarrow \;\; 2 i x+4 i y-4 i=0$ $\therefore$ Equation of line $L_{2} \Rightarrow x+2 y-2=0$ . . . (ii) From Eqs. (i) and (ii), $4 y=2 \;\text{ or }\; y=1 / 2 \;\text{ and }\; x=1$ Hence, centre $=(1,1 / 2)$ Equation of third line $L_{3} \Rightarrow i z+\overline{z}+1+i=0$ $\Rightarrow \;\; i(x+i y)+(x-i y)+1+i=0$ $\Rightarrow \;\; i x-y+x-i y+1+i=0$ $\Rightarrow \;\; (x-y+1)+i(x-y+1)=0$ $\therefore$ Radius $=$ Distance of point $(1,1 / 2)$ to the line $x-y+1=0$ $\therefore \;\; r=\;{|1-\frac{1}{2}+1|}/{\sqrt{1^{2}+1^{2}}}=\;\frac{3}{2 \sqrt{2}}$

Q72JEE Main 2021MCQ4MStraight Lines and Pair of Straight Lines

The image of the point $(3,5)$ in the line $x-y+1=0$ , lies on

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📖 Explanation

Image of $P(3,5)$ on the line $x-y+1=0$ is $\;\frac{x-3}{1} \;\; =\;\frac{y-5}{-1}=\;\frac{-2(3-5+1)}{2}$ $\Rightarrow \;\; \;\frac{x-3}{1} \;\; =\;\frac{y-5}{-1}=1$ $\Rightarrow \;\; \;\frac{x-3}{1} \;\; =1 \;\text{ and }\; \;\frac{y-5}{-1}=1$ $x \;\; =4, y=4$ $\therefore$ Required image is at $(4,4)$ . Clearly, this point lies on $(x-2)^{2}+(y-4)^{2}=4$ as $(4,4)$ satisfies this equation.

Q73JEE Main 2021MCQ4MEllipse

If the curves, $\;\frac{x^{2}}{a}+\;\frac{y^{2}}{b}=1$ and $\;\frac{x^{2}}{c}+\;\frac{y^{2}}{d}=1$ , intersect each other at an angle of $90^{\circ}$ , then which of the following relations is true?

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📖 Explanation

Given, curves $\;\frac{x^{2}}{a}+\;\frac{y^{2}}{b}=1$ and $\;\frac{x^{2}}{c}+\;\frac{y^{2}}{d}=1$ $\because \;\; \;\frac{x^{2}}{a}+\;\frac{y^{2}}{b}=1$ On differentiating both sides w.r.t. $x$ , we get $\;\frac{2 x}{a}+\;\frac{2 y}{b} \cdot \;\frac{d y}{d x} \;\; =0$ $\;\frac{d y}{d x} \;\; =\;\frac{-b x}{a y}$ . . . (i) Also, $\;\frac{x^{2}}{c}+\;\frac{y^{2}}{d}=1$ On differentiating both sides w.r.t. $x$ , we get $\;\frac{2 x}{c}+\;\frac{2 y}{d} \cdot \;\frac{d y}{d x} \;\; =0$ $\;\frac{d y}{d x} \;\; =\;\frac{-d x}{c y}$ $\because$ Both the curves intersect each other at $90^{\circ}$ . $\therefore$ Tangents at point of intersection must be perpendicular to each other. $\therefore$ Product of slope of tangents $=-1$ $\;\frac{-b x}{a y} \times \;\frac{-d x}{c y}=-1 \;\;$ [from Eqs. (i) and (ii)] $\Rightarrow \;\; b d x^{2}=-a c y^{2}$ ... (iii) Also, on subtracting the equation of given curves, we get $(\;\frac{x^{2}}{a}+\;\frac{y^{2}}{b}-1)-(\;\frac{x^{2}}{c}+\;\frac{y^{2}}{d}-1)=0$ $\Rightarrow \;\; x^{2}(\;\frac{1}{a}-\;\frac{1}{c})+y^{2}(\;\frac{1}{b}-\;\frac{1}{d})=0$ or $x^{2}(\;\frac{1}{a}-\;\frac{1}{c})=-y^{2}(\;\frac{1}{b}-\;\frac{1}{d})$ ... (iv) Dividing Eq. (iii) by Eq. (iv), $\;\;\frac{b d}{(\;\frac{1}{a}-\;\frac{1}{c})} \;\; =\;\frac{a c}{(\;\frac{1}{b}-\;\frac{1}{d})}$ $\Rightarrow \;\;\frac{b d \times a c}{c-a} \;\; =\;\frac{a c \times b d}{d-b}$ $\Rightarrow \;c-a \;\; =d-b$ $\;\text{ or }\;\;c-d \;\; =a-b$ $\;\text{ or }\;\;a-b \;\; =c-d$

Q74JEE Main 2021MCQ4MLimits, Continuity and Differentiability

$\lim _{n \to \infty } 1+(\;\frac{1+\frac{1}{2}+. . . . . .+\frac{1}{n}}{n^{2}})^{n}$ is equal to

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📖 Explanation

Given, limit form is $1^{\infty }$ . $L=e^{\lim _{n \to }(\;\frac{1+\frac{1}{2}+\frac{1}{3}+. . .+\frac{1}{n}}{n})}$ Let $S=1+(\;\frac{1}{2}+\;\frac{1}{3})+(\;\frac{1}{4}+\;\frac{1}{5}+\;\frac{1}{6}+\;\frac{1}{7})+. . .$ Clearly, $S < 1+(\;\frac{1}{2}+\;\frac{1}{2})+(\;\frac{1}{4}+\;\frac{1}{4}+\;\frac{1}{4}+\;\frac{1}{4})+. . .+$ $⏟{(\;\frac{1}{2^{n}}+. . .+\;\frac{1}{2^{n}})}_{2^{n} \;\text{ \times }\;}$ $S < 1+1+1+1+. . .+1$ $S < n+1$ $\therefore \;\; L=e^{n \to -(\;\frac{n+1}{2^{n+1}-1})}\;\Rightarrow L=e^{0}$ $\therefore \;\; L=1\;L=1$

Q75JEE Main 2021MCQ4MProbability

The coefficients $a, b$ and $c$ of the quadratic equation, $a x^{2}+b x+c=0$ are obtained by throwing a dice three times. The probability that this equation has equal roots is

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📖 Explanation

Given, $a x^{2}+b x+c=0$ According to the question, $a, b, c \in\{1,2,3,4,5,6\}$ $\therefore \;\; n(S)=6 \times 6 \times 6=216$ For equal roots $\;D \;\; =0$ $\Rightarrow \;b^{2}-4 a c \;\; =0$ $\Rightarrow \;b^{2} \;\; =4 a c$ $\Rightarrow \;a c \;\; =\;\frac{b^{2}}{4}$ $b=2, a c=1 \Rightarrow (a=1, c=1)$ $b=4, a c=4 \Rightarrow (a=1, c=4)$ or $(a=4, c=1)$ $(a=2, c=2)$ $b=6, a c=9 \Rightarrow (a=3, c=3)$

Q76JEE Main 2021MCQ4MPermutations and Combinations

The total number of positive integral solutions $(x, y, z)$ , such that $x y z=24$ is

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📖 Explanation

Given, $x y z=24$ $\Rightarrow \;\; x y z \;\; =2^{3} \cdot 3^{1}$ $\;\text{ Let }\; \;\; x \;\; =2^{a_{1}} \cdot 3^{b_{1}}$ $y \;\; =2^{a_{2}} \cdot 3^{b_{2}}$ $z \;\; =2^{a_{3}} \cdot 3^{b_{3}}$ where, $a_{1}, a_{2}, a_{3} \in\{0,1,2,3\}$ $b_{1}, b_{2}, b_{3} \in\{0,1\}$ Case $I a_{1}+a_{2}+a_{3}=3$ $\therefore$ Non-negative solution $={ }^{3+3-1} C_{3-1}={ }^{5} C_{2}=10$ Case II $b_{1}+b_{2}+b_{3}=1$ $\therefore$ Non-negative solution $={ }^{1+3-1} C_{3-1}={ }^{3} C_{2}=3$ $\therefore$ Total solutions $=10 \times 3=30$

Q77JEE Main 2021MCQ4MQuadratic Equation and Inequalities

The integer ' $k$ ', for which the inequality $x^{2}-2(3 k-1) x+8 k^{2}-7 > 0$ is valid for every $x$ in $R$ , is

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📖 Explanation

Given, $x^{2}-2(3 k-1) x+8 k^{2}-7 > 0$ Here, $a > 0$ $\; \therefore D< 0\;$ $\Rightarrow \;{[2(3 k-1)]^{2}-4(8 k^{2}-7) < 0}$ $\;4(9 k^{2}+1-6 k)-4(8 k^{2}-7) < 0$ $\Rightarrow \;k^{2}-6 k+8 < 0$ $\Rightarrow \;(k-4)(k-2) < 0$ $k \in(2,4)$ $\therefore$ Required integer, $k=3$

Q78JEE Main 2021MCQ4MDifferential Equations

If a curve passes through the origin and the slope of the tangent to it at any point $(x, y)$ is $\;\frac{x^{2}-4 x+y+8}{x-2}$ , then this curve also passes through the point

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📖 Explanation

Given, slope $=\;\frac{x^{2}-4 x+y+8}{x-2}$ $\Rightarrow \;\; \;\frac{d y}{d x} \;\; =\;\frac{x^{2}-4 x+y+8}{x-2}=\;\frac{(x-2)^{2}+(y+4)}{(x-2)}$ $\;\; =(x-2)+\;\frac{y+4}{x-2}$ Let $(x-2)=t \Rightarrow d x=d t$ $\;\text{ and }\;(y+4)=u \Rightarrow d y=d u$ $\therefore \;\; \;\frac{d y}{d x}=\;\frac{d u}{d t}$ $\;\text{ Now, }\; \;\frac{d y}{d x}=(x-2)+\;\frac{(y+4)}{(x-2)}$ $\Rightarrow \;\; \;\frac{d u}{d t}=t+\;\frac{u}{t} \Rightarrow \;\frac{d u}{d t}-\;\frac{u}{t}=t$ Here, integrating factor (IF) $=1 / t$ $\Rightarrow \;\; u \cdot (\;\frac{1}{t})=\int t(\;\frac{1}{t}) d t \Rightarrow u / t=t+c$ $\Rightarrow \;\; \;\frac{(y+4)}{(x-2)}=(x-2)+c$ $\because$ It passes through origin [i.e. $(0,0)$ ], then $\therefore \;\; \;\frac{4}{-2}=-2+c$ $\Rightarrow \;\; -2+2=c \Rightarrow c=0$ $\;\text{ Hence, }\; \;\frac{(y+4)}{(x-2)}=(x-2)+0$ $\Rightarrow \;\; y+4=(x-2)^{2}$ Clearly, this curve passes through $(5,5)$ as it satisfies the equation.

Q79JEE Main 2021MCQ4MSets and Relations

The statement $A \to (B \to A)$ is equivalent to

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📖 Explanation

Given, statement $A \to (B \to A)$ $\;\equiv A \to (∼ B ∨ A)$ $\;\equiv ∼ A ∨(∼ B ∨ A)$ $\;\equiv (∼ A ∨ A) ∨ ∼ B$ $\;\equiv T ∨ ∼ B \equiv T$ $\therefore T ∨ B=T\;\equiv (∼ A ∨ A) ∨ B$ $\;\equiv ∼ A ∨(A ∨ B)$ $\;\equiv A \to (A ∨ B)$

Q80JEE Main 2021MCQ4MApplication of Derivatives

If Rolle's theorem holds for the function $f(x)=x^{3}-a x^{2}+b x+4, x \in[1,2]$ with $f^{'}(\;\frac{4}{3})=0$ , then ordered pair $(a, b)$ is equal to

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📖 Explanation

Given, $f(x)=x^{3}-a x^{2}+b x+4, x \in[1,2]$ Here, $f(1)=f(2)$ $\Rightarrow \;\; 1-a+b-4=8-4 a+2 b-4$ $\Rightarrow 3 a-b=7$ . . . (i) Also, $f^{'}(x)=3 x^{2}-2 a x+b$ According to the question, $f^{'}(\;\frac{4}{3})=0$ $\Rightarrow \;\; 3 \times (\;\frac{4}{3})^{2}-2 a(\;\frac{4}{3})+b=0$ $\Rightarrow -8 a+3 b=-16$ . . . (ii) From Eqs. (i) and (ii), $a \;\; =5, b=8$ $(a, b) \;\; =(5,8)$ $\therefore \;\; (a, b)=(5,8)$

Q81JEE Main 2021NAT4MApplication of Derivatives

Let $f(x)$ be a polynomial of degree 6 in $x$ , in which the coefficient of $x^{6}$ is unity and it has extrema at $x=-1$ and $x=1$ . If $\lim _{x \to 0} \;\frac{f(x)}{x^{3}}=1$ , then $5 f(2)$ is equal to ................

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📖 Explanation

$f(x)=x^{6}+a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f$ As, $\lim _{x \to 0} \;\frac{f(x)}{x^{3}}=1$ non-zero finite So, $d= {e}=f=0$ and $f(x)=x^{3}(x^{3}+a x^{2}+b x+c)$ Hence, $\lim _{x \to 0} \;\frac{f(x)}{x^{3}}=c=1$ Now, as $f(x)=x^{6}+a x^{5}+b x^{4}+x^{3}$ and $f^{'}(x)=0$ at $x=1$ and $x=-1$ i.e. $f^{'}(x)=6 x^{5}+5 a x^{4}+4 b x^{3}+3 x^{2}$ Now, $f^{'}(1)=0$ $\Rightarrow \;\; 6+5 a+4 b+3=0$ $\Rightarrow \;\; 5 a+4 b=-9$ . . . (i) and $f^{'}(-1)=0$ $\Rightarrow \;\; -6+5 a-4 b+3=0$ $\Rightarrow \;\; 5 a-4 b=3$ . . . (ii) From Eqs. (i) and (ii), $a=-3 / 5 \;\text{ and }\; b=-3 / 2$ $\therefore \;\; f(x)=x^{6}-\;\frac{3}{5} x^{5}-\;\frac{3}{2} x^{4}+x^{3}$ $\therefore \;\; 5 f(2) \;\; =5[2^{6}-\;\frac{3}{5}(2)^{5}-\;\frac{3}{2}(2)^{4}+(2)^{3}]$ $\;\; =5[64-\;\frac{3 \times 32}{5}-\;\frac{3 \times 16}{2}+8]$ $\;\; =320-96-120+40$ $\;\; =144$

Q82JEE Main 2021NAT4MLimits, Continuity and Differentiability

The number of points at which the function $f(x)=|2 x+1|-3 x+2|+x^{2}+x-2|, x \in R$ is not differentiable, is ........... .

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📖 Explanation

Given, $f(x) \;\; =|2 x+1|-3|x+2|+|x^{2}+x-2|$ $\;\; =|2 x+1|-3|x+2|+|x+2| \times |x-1|$ Here, critical points are $x=\;\frac{-1}{2},-2,1$ $\therefore f(x)=\begin{cases}x^{2}+2 x+3 \\ x < -2 \\ -x^{2}-6 x-5 \\ -2 < x < \\ \\ \frac{-1}{2} \\ -x^{2}-2 x-3 \\ \\ \\ \frac{-1}{2} < x < 1 \\ x^{2}-7 \\ x > 1\end{cases}$ Now, $f^{'}(x)=\begin{cases}2 x+2 \\ x < -2 \\ -2 x-6 \\ -2 < x < \\ \\ \frac{-1}{2} \\ -2 x-2 \\ \\ \\ \frac{-1}{2} < x < 1 \\ 2 x \\ \\ \\ \\ x > 1\end{cases}$ Now, $f^{'}(x)$ at $1,-2$ and $-1 / 2$ are For $x=1$ , $f^{'}(x) \;\; =2 x=2 \times 1=2$ $\;\text{ and }\;-2 x-2 \;\; =-(2 \times 1)-2=-4$ $f^{'}(x)=2 x=$ and $-2 x-2=-(2$ both are not equal. $\therefore$ Non-differentiable at $x=1$ Similarly, for $x=-2, f^{'}(x)=2 x+2=2 \times (-2)+2=-2$ and $-2 x-6=-2 \times (-2)-6=-2$ both are equal. $\therefore$ Differentiable at $x=-2$ and for $x=-1 / 2, f^{'}(x)=-2 x-6$ $=-2 \times (\;\frac{-1}{2})-6=-5 \;\text{ and }\;$ $-2 x-2=-2 \times (\;\frac{-1}{2})-2=-1$ both are not equal. $\therefore$ Non-differentiable at $x=-1 / 2$ $\therefore$ The number of points at which $f(x)$ is non-differentiable is 2 .

Q83JEE Main 2021NAT4MArea Under The Curves

The graph of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area $A$ . Then $A^{4}$ is equal to ......... .

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📖 Explanation

Required area of shaded region $A \;\; =int_{\pi / 4}^{5 \pi / 4}\{\sin x-\cos x) d x$ $\;\; =[-\cos x-\sin x]_{\pi / 4}^{5 \pi / 4}$ $\;\; =-[(\cos \;\frac{5 \pi}{4}+\sin \;\frac{5 \pi}{4})-(\cos \;\frac{\pi}{4}+\sin \;\frac{\pi}{4})]$ $\;\; =-[(-\;\frac{1}{\sqrt{2}}-\;\frac{1}{\sqrt{2}})-(\;\frac{1}{\sqrt{2}}+\;\frac{1}{\sqrt{2}})]$ $\therefore \;\; A \;\; =\;\frac{4}{\sqrt{2}}=2 \sqrt{2}$ $\Rightarrow \;\; A^{4} \;\; =(2 \sqrt{2})^{4}=64$

Q84JEE Main 2021NAT4MSequences and Series

Let $A_{1}, A_{2}, A_{3}, . . . . . . . .$ be squares, such that for each $n \ge 1$ , the length of the side of $A_{n}$ equals the length of diagonal of $A_{n+1}$ . If the length of $A_{1}$ is $12 {cm}$ , then the smallest value of $n$ for which area of $A_{n}$ is less than one, is ......... .

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📖 Explanation

According to the question, length of side of $A_{1}$ square is $12 {cm}$ . $\because$ Side lengths are in GP. (Side of $n$ th square i.e. $A_{n}$ ) $\therefore \;\; T_{n}=\;\frac{12}{(\sqrt{2})^{n-1}}$ $\therefore$ Area $=(\;\text{ Side }\;)^{2}=(\;\frac{12}{(\sqrt{2})^{n-1}})^{2}=\;\frac{144}{2^{n-1}}$ According to the question, the area of $A_{n}$ square $< 1$ $\;\frac{144}{2^{n-1}} < 1$ $\Rightarrow \;\; 2^{n-1} > 144$ Here, the smallest possible value of $n$ is $9 .$

Q85JEE Main 2021NAT4MMatrices and Determinants

Let $A=\begin{bmatrix}x & y & z \\ y & z & x \\ z & x & y\end{bmatrix}$ , where $x, y$ and $z$ are real numbers, such that $x+y+z > 0$ and $x y z=2$ . If $A^{2}=I_{3}$ , then the value of $x^{3}+y^{3}+z^{3}$ is ........... .

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📖 Explanation

Here, $A=\begin{bmatrix}x & y & z \\ y & z & x \\ z & x & y\end{bmatrix}$ $\therefore \;\; |A|=\begin{vmatrix}x & y & z \\ y & z & x \\ z & x & y\end{vmatrix}=x^{3}+y^{3}+z^{3}-3 x y z$ $\;|A^{2}|=|I_{3}|=1\;$ $\therefore \;\; |(x^{3}+y^{3}+z^{3}-3 x y z)^{2}| \;\; =1\;$ $\Rightarrow \;\; x^{3}+y^{3}+z^{3}-3 x y z \;\; =1\;$ $\Rightarrow \;\; x^{3}+y^{3}+z^{3} \;\; =1+3 x y z \;[ \because x+y+z>0]$ $\Rightarrow \;\; \;\; =1+3(2)\;$ $\; \;\; =7 \;[ \because x y z=2]$

Q86JEE Main 2021NAT4MMatrices and Determinants

If $A=\begin{bmatrix}0 & -\tan (\\ \\ \frac{\theta}{2}) \\ \tan (\\ \\ \frac{\theta}{2}) & 0\end{bmatrix}$ and $(I_{2}+A)(I_{2}-A)^{-1}=\begin{bmatrix}a & -b \\ b & a\end{bmatrix}$ , then $13(a^{2}+b^{2})$ is ......... .

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📖 Explanation

$A=\begin{bmatrix}0 & -\tan (\\ \\ \frac{\theta}{2}) \\ \tan (\\ \\ \frac{\theta}{2}) & 0\end{bmatrix}$ and $(I_{2}+A)(I_{2}-A)^{-1}=\begin{bmatrix}a & -b \\ b & a\end{bmatrix}$ $\Rightarrow |(I_{2}+A)(I_{2}-A)^{-1}|=a^{2}+b^{2}$ $\Rightarrow \;\; a^{2}+b^{2}=\;\frac{|I_{2}+A|}{|I_{2}-A|}$ . . . (i) Now, $I_{2}+A=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}+\begin{bmatrix}0 & -\tan (\\ \\ \frac{\theta}{2}) \\ \tan (\\ \\ \frac{\theta}{2}) & 0\end{bmatrix}$ $= \;\; \begin{bmatrix}1 & -\tan (\\ \\ \frac{\theta}{2}) \\ \tan (\\ \\ \frac{\theta}{2}) & 1\end{bmatrix}$ $\;\text{ Similarly, I }\;_{2}-A= \;\; \begin{bmatrix}1 & \tan (\\ \\ \frac{\theta}{2}) \\ -\tan (\\ \\ \frac{\theta}{2}) & 1\end{bmatrix}$ Here, $|I_{2}+A|=|I_{2}-A|=(1+\tan ^{2}(\;\frac{\theta}{2}))$ $\Rightarrow \;\; \;\frac{|I_{2}+A|}{|I_{2}-A|}=1$ . . . (ii) From Eqs. (i) and (ii), $a^{2}+b^{2}=1$ Now, $13(a^{2}+b^{2})=13 \times 1=13$

Q87JEE Main 2021NAT4MPermutations and Combinations

The total number of numbers, lying between 100 and 1000 that can be formed with the digits $1,2,3,4,5$ , if the repetition of digits is not allowed and numbers are divisible by either 3 or 5 , is ........... .

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📖 Explanation

Given, digits $=\{1,2,3,4,5\}$ Numbers divisible by 3 (sum of digits divisible by 3 ).Case I When sum is $12 \to 3,4,5 \to 3 !=6$ Case II When sum is $9 \to 2,3,4 \to 3 !=6$ Case III When sum is $9 \to 1,3,5 \to 3 !=6$ Case IV When sum is $6 \to 1,2,3 \to 3 !=6$ So, total numbers divisible by $3=6 \times 4=24$ Numbers divisible by 5 (ending with 5) So, total numbers divisible by $5=12$ Numbers divisible by 15 , are $145,415,345,435$ i.e. total 4 numbers are divisible by both 3 and $5 .$ i.e. divisible by 15 . Hence, the required numbers which are divisible by 3 or 5 $=24+12-4=32$

Q88JEE Main 2021NAT4MVector Algebra

Let $a=\hat{i}+2 \hat{j}-\hat{k}, b=\hat{i}-\hat{j}$ and $c=\hat{i}-\hat{j}-\hat{k}$ be three given vectors. If $r$ is a vector such that $r \times a=c \times a$ and $r \cdot b=0$ , then $r \cdot a$ is equal to ......... .

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📖 Explanation

$b=\hat{i}-\hat{j}$ $c=\hat{i}-\hat{j}-\hat{k}$ $r \times a=c \times a$ $\Rightarrow \;\; r \times a-c \times a=0$ $\Rightarrow \;\; (r-c) \times a=0$ $\therefore \;\; r-c=\lambda a$ $\Rightarrow \;\; r=\lambda a+c$ $\Rightarrow \;\; r \cdot b=\lambda a \cdot b+c \cdot b \;\;$ (taking dot ${w}$ ) $\Rightarrow \;\; 0=\lambda a \cdot b+c \cdot b$ $\Rightarrow \lambda(\hat{i}+2 \hat{j}-\hat{k}) \cdot (\hat{i}-\hat{j})+(\hat{i}-\hat{j}-\hat{k}) \cdot (\hat{i}-\hat{j})=0$ $\Rightarrow \;\; \lambda(1-2)+2=0$ $\Rightarrow \;\; \lambda=2$ $\therefore \;\; r=2 a+c$ $\Rightarrow \;\; r \cdot a=2 a \cdot a+c \cdot a \;\;$ [taking dot with $a$ ] $=2|a|^{2}+a \cdot c$ $=2(1+4+1)+(1-2+1)$ $r \cdot a=12$

Q89JEE Main 2021NAT4MMatrices and Determinants

If the system of equations $\;k x+y+2 z=1$ $\;\; -2 x-2 y-4 z=3$ $\;\; 3 x-y-2 z=2$ has infinitely many solutions, then $k$ is equal to ..........

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📖 Explanation

Given equations, $k x+y+2 z=1$ $3 x-y-2 z=2$ $-2 x-2 y-4 z=3$ For infinitely many solutions, $\Delta =\Delta x=\Delta y=\Delta z=0$ Here, $\Delta y=\begin{vmatrix}k & 1 & 2 \\ 3 & 2 & -2 \\ -2 & 3 & -4\end{vmatrix}=0$ $\Rightarrow \;k(-8+6)-1(-12-4)+2(9+4)=0$ $\Rightarrow \;\; -2 k+16+26=0$ $\Rightarrow \;2 k=42$ $\therefore \;k=21$

Q90JEE Main 2021NAT4MHyperbola

The locus of the point of intersection of the lines $(\sqrt{3}) k x+k y-4 \sqrt{3}=0$ and $\sqrt{3} x-y-4(\sqrt{3}) k=0$ is a conic, whose eccentricity is ............ .

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📖 Explanation

Given, lines are $\sqrt{3} k x+k y-4 \sqrt{3}=0$ . . . (i) $\sqrt{3} x-y-4 \sqrt{3} k=0 \;\; . . . \;\text{ (ii) }\;$ Multiply Eq. (ii) $\times k$ and then adding Eqs. (i) and (ii), $\sqrt{3} k x+k y-4 \sqrt{3}=0$ $\;\frac{\sqrt{3} k x-k y-4 \sqrt{3} k^{2}=0}{(2 \sqrt{3} x) k=4 \sqrt{3}+4 \sqrt{3} k^{2}}$ $\therefore \;\; x=\;\frac{4 \sqrt{3}(1+k^{2})}{2 \sqrt{3} k}=2(k+\;\frac{1}{k})$ Subtracting Eq. (i) from Eq. (ii), $\sqrt{3} k x+k y-4 \sqrt{3}=0$ $\sqrt{3} k x-k y-4 \sqrt{3} k^{2}=0$ $\;\frac{- \;\; + \;\; +}{2 k y=4 \sqrt{3}-4 \sqrt{3} k^{2}}$ $y=\;\frac{4 \sqrt{3}(1+k^{2})}{2 k}=2 \sqrt{3}(k+\;\frac{1}{k})$ We have, $x=2(k+\;\frac{1}{k})$ and $y=2 \sqrt{3}(\;\frac{1}{k}-k)$ $\;\frac{x}{2}=(k+\;\frac{1}{k})$ . . . (iii) $\;\frac{y}{2 \sqrt{3}}=(\;\frac{1}{k}-k)$ . . . (iv) Squaring and subtracting Eq. (iii) from Eq. (iv), $\;\frac{x^{2}}{4}-\;\frac{y^{2}}{12} \;\; =(k^{2}+\;\frac{1}{k^{2}}+2)-(\;\frac{1}{k^{2}}+k^{2}-2)$ $\;\frac{x^{2}}{4}-\;\frac{y^{2}}{12} \;\; =4$ $\;\text{ or }\; \;\; \;\frac{x^{2}}{16}-\;\frac{y^{2}}{48} \;\; =1$ Clearly, this is a hyperbola. $\therefore \;\; {e}^{2} \;\; =1+\;\frac{b^{2}}{ {a}^{2}}$ $\;\; =1+\;\frac{48}{16}=1+3$ $\; {e}^{2}=4$ $\; {e}=\sqrt{4}=2$ ( $\because$ e is positive)

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